Geoffrey R. Grimmett - One Thousand Exercises in Probability Solution
May 1, 2017 | Author: meko1986 | Category: N/A
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1 Events and their probabilities
1.2 Solutions. Events as sets 1. (a) Let a E (U Ai )c. Then a ¢ U Ai, so that a E A~ for all i. Hence (U Ai )c ~ n A~. Conversely, if a E A~, then a ¢ Ai for every i. Hence a ¢ U Ai, and so A~ ~ (U Ai )c. The first De Morgan law follows.
n
n
(b) Applying part (a) to the family {Af : i E /},we obtain that Taking the complement of each side yields the second law.
(Ui
Aft =
ni(Af)c
=
ni
Ai.
2. Clearly (i) An B = (Ac U Bc)c, (ii) A\ B =An Be= (Ac U B)c, (iii) A D. B =(A\ B) U (B \A)= (Ac U B)c U (AU Bc)c. Now :F is closed under the operations of countable unions and complements, and therefore each of these sets lies in :F.
3. Let us number the players 1, 2, ... , 2n in the order in which they appear in the initial table of draws. The set of victors in the first round is a point in the space Vn = { 1, 2} x {3, 4} x .. · x {2n - 1, 2n}. Renumbering these victors in the same way as done for the initial draw, the set of second-round victors can be thought of as a point in the space Vn-I, and so on. The sample space of all possible outcomes of the tournament may therefore be taken to be Vn x Vn -I x · · · x VI, a set containing . 22n-122n-2 . . . 2I = 22n-I pomts. Should we be interested in the ultimate winner only, we may take as sample space the set {1, 2, ... , 2n} of all possible winners. 4. We must check that g, satisfies the definition of a a-field: (a) 0 E :F., and therefore 0 = 0 n B E g,, (b) if AI, A2, ... E :F., then Ui(Ai n B)= (Ui Ai) n BEg,, (c) if A E :F., then Ac E :Fso that B \(An B)= Ac n B E g,.
Note that g, is a a-field of subsets of B but not a a-field of subsets of Q, since C E that cc = Q \ C E g,. 5.
(a), (b), and (d) are identically true; (c) is true if and only if A~ C.
1.3 Solutions. Probability 1.
(i) We have (using the fact that lP' is a non-decreasing set function) that lP'(A
n B) = lP'(A)
+ lP'(B) -
lP'(A U B) :::: lP'(A)
135
+ lP'(B) -
1=
fz.
g, does not imply
[1.3.2]-[1.3.4]
Solutions
Events and their probabilities
Also, since An B ~A and An B ~ B, li"(A n B) :::: min{lP'(A), lP'(B)} = ~These bounds are attained in the following example. Pick a number at random from {1, 2, ... , 12}. Taking A = {1, 2, ... , 9} and B = {9, 10, 11, 12}, we find that An B = {9}, and so li"(A) =
i.
li"(B) = ~. li"(A n B) B = {1, 2, 3, 4}.
= /2 .
To attain the upper bound for lP'(A n B), take A
= {1, 2, ... , 9} and
(ii) Likewise we have in this case lP'(A U B) :::: min{lP'(A) + li"(B), 1} = 1, and lP'(A U B) 2:: max{lP'(A), lP'(B)} = These bounds are attained in the examples above.
i·
2.
(i) We have (using the continuity property oflP') that lP'(no head ever)
= n--+oo lim lP'(no head in first n tosses) = lim 2-n = 0, n--+oo
so that lP'(some head turns up) = 1 - lP'(no head ever) = 1. (ii) Given a fixed sequences of heads and tails of length k, we consider the sequence of tosses arranged in disjoint groups of consecutive outcomes, each group being of length k. There is probability 2-k that any given one of these is s, independently of the others. The event {one of the first n such groups iss} is a subset of the event {s occurs in the first nk tosses}. Hence (using the general properties of probability measures) we have that ll"(s turns up eventually) =
lim lP'(s occurs in the first nk tosses)
n--+oo
2:: lim lP'(s occurs as one of the first n groups) n--+oo
= 1-
=
1-
lim lP'(none of the first n groups is s)
n--+oo
lim (1 -
n--+oo
rk( =
1.
3. Lay out the saucers in order, say as RRWWSS. The cups may be arranged in 6! ways, but since each pair of a given colour may be switched without changing the appearance, there are 6!-:- (2 !) 3 = 90 distinct arrangements. By assumption these are equally likely. In how many such arrangements is no cup on a saucer of the same colour? The only acceptable arrangements in which cups of the same colour are paired off are WWSSRR and SSRRWW; by inspection, there are a further eight arrangements in which the first pair of cups is either SW or WS, the second pair is either RS or SR, and the third either RW or WR. Hence the required probability is 10/90 = ~.
4. Weprovethisbyinductiononn,consideringfirstthecasen = 2. Certainly B = (AnB)U(B\A) is a union of disjoint sets, so that lP'(B) = li"(A n B)+ li"(B \A). Similarly AU B = AU (B \A), and so li"(A U B)= li"(A) + li"(B \A)= li"(A) + {li"(B) -li"(A n B)}. Hence the result is true for n = 2. Let m 2:: 2 and suppose that the result is true for n :::: m. Then it is true for pairs of events, so that 1 lP'Cu Ai) 1
= lP'(UAi) + li"(Am+d -11"{ (UAi) 1
n Am+1}
1
= lP'(u Ai) + li"(Am+d -11"{ U 0, x E variable. If a < 0,
~.then
{w: aX(w)
{w: aX(w)
~ x} =
~
x}
= {w:
{w: X(w) 2: xja} = {
X(w) ~ xja} E .T'since X is a random
U1 { w: X(w) ~ ~- ~}
n2'::
r
which lies in r since it is the complement of a countable union of members of :F. If a
{w: aX(w)
~ x} =
{
~
= 0,
if X< 0, ifx 2: 0;
in either case, the event lies in :F. (ii) For wE n, X(w)- X(w) = 0, so that X- X is the zero random variable (that this is a random variable follows from part (i) with a = 0). Similarly X (w) +X (w) = 2X (w ). 2.
Set Y = aX +b. We have that
lP'(X ~ (y- b)/a)= F((y- b)/a) lP'(Y < y) = { JP>(X 2: (y- b)/a) = 1 -1imxt(y-b)fa F(x) Finally, if a= 0, then Y = b, so that JP>(Y
~
y) equals 0 if b > y and 1 if b
if a> 0, if a < 0. ~
y.
3. Assume that any specified sequence of heads and tails with length n has probability z-n. There are exactly (k) such sequences with k heads. If heads occurs with probability p then, assuming the independence of outcomes, the probability of any givensequenceofk headsandn-k tails is pk (1- p)n-k. The answer is therefore (k) pk(l- p)n-k.
4. Write H = "AF + (1 - "A)G. Then limx-+-oo H(x) = 0, limx-+oo H(x) non-decreasing and right-continuous. Therefore His a distribution function.
= 1, and clearly His
5. The function g(F(x)) is a distribution function whenever g is continuous and non-decreasing on [0, 1], with g(O) = 0, g(l) = 1. This is easy to check in each special case.
151
[2.2.1]-[2.4.1]
Solutions
Random variables and their distributions
2.2 Solutions. The law of averages 1. Let p be the potentially embarrassed fraction of the population, and suppose that each sampled individual would truthfully answer "yes" with probability p independently of all other individuals. In the modified procedure, the chance that someone says yes is p + ~(1 - p) = ~(1 + p). If the proportion of yes's is now ifJ, then 2¢J- 1 is a decent estimate of p. The advantage of the given procedure is that it allows individuals to answer "yes" without their being identified with certainty as having the embarrassing property. 2.
Clearly Hn
+ Tn
= n, so that (Hn - Tn)/n = (2Hn/n)- 1. Therefore
lP' ( 2p - 1 -
as n
--+
E~ ~ (Hn -
Tn)
~ 2p - 1 + E) = lP' (I~ Hn -
pI ~ n
--+
1
oo, by the law of large numbers (2.2.1 ).
3. Let/n (x) be the indicator function of the event {Xn ~ x}. By the law of averages, n -I I:~= 1 I r (x) converges in the sense of (2.2.1) and (2.2.6) to lP'(Xn ~ x) = F(x).
2.3 Solutions. Discrete and continuous variables 1.
With 8 = supm lam- am-JI, we have that IF(x)- G(x)l
~
IF(am)- F(am-dl
~
IF(x
+ 8)- F(x- 8)1
for x E [am-I, am). Hence G(x) approaches F(x) for any x at which F is continuous. 2.
For y lying in the range of g, {Y ~ y} ={X~ g- 1 (y)}
3.
Certainly Y is a random variable, using the result of the previous exercise (2). Also
E
:F.
lP'(Y ~ y) = IP'(F- 1 (X) ~ y) = IP'(X ~ F(y)) = F(y)
as required. IfF is discontinuous then p- 1 (x) is not defined for all x, so that Y is not well defined. If F is non-decreasing and continuous, but not strictly increasing, then p-I (x) is not always defined uniquely. Such difficulties may be circumvented by defining p-I (x) = inf {y : F (y) ~ x}.
4. The function A.f +(1- A. )g is non-negative and integrable over llHo 1. Finally, f g is not necessarily a density, though it may be: e.g., iff= g = 1, 0 ~ x ~ 1 then f(x)g(x) = 1, 0 ~ x ~ 1. 5. (a) If d > 1, then J100 cx-d dx = c/(d- 1). Therefore f is a density function if c = d- 1, and F(x) = 1- x-(d-I) when this holds. If d ~ 1, then f has infinite integral and cannot therefore be a density function. (b) By differentiating F(x) =ex j(l +ex), we see that F is the distribution function, and c = 1.
2.4 Solutions. Worked examples 1.
(a) If y ~ 0,
lP'(X 2 ~ y) = lP'(X ~
(b) We must assume that
X~
0. If y
,JY) -lP'(X ~
<
0,
152
-,JY) =
F(,J'Y)- F(-,J'Y).
Solutions [2.4.2]-[2.5.6]
Random vectors (c)If-1
~
y
~
1, 00
lP'(sinX ~ y) =
L
IP'((2n
+ l)rr- sin- 1 y ~X~ (2n + 2)rr + sin- 1 y)
n=-oo 00
=
L {F((2n + 2)rr + sin- 1 y)- F((2n + l)rr- sin- 1 y) }· n=-oo
(d) lP'(G- 1 (X) ~ y) = lP'(X ~ G(y)) = F(G(y)). (e) IfO ~ y ~ 1, then lP'(F(X) ~ y) = lP'(X ~ p- 1 (y)) = F(F- 1 (y)) = y. There is a small difficulty ifF is not strictly increasing, but this is overcome by defining p- 1 (y) = sup{x : F (x) = y}. (f) lP'(G- 1 (F(X)) ~ y) = lP'(F(X) ~ G(y)) = G(y). 2.
~.
It is the case that, for x E
Fy(x) and Fz(x) approach F(x) as
a~
-oo, b
~
oo.
2.5 Solutions. Random vectors 1. Write fxw = lP'(X = x, W = w). Then foo = h1 = x,w. 2.
(a) We have that /x,y(x,y)
~
{
r-
:!, /10 = i. and fxw = 0 for other pairs
if (x, y) = (1, 0), if (x, y) = (0, 1), otherwise.
p
(b) Secondly, 1- p
!x,z(x, z) =
{
if (x, z) = (0, 0), if (x, z) = (1, 0),
~
otherwise.
3.
Differentiating gives fx,y(x, y) =e-x /{rr(l
4.
Let A = {X
~
+ y 2 )}, x
::: 0, y E R.
b, c < Y ~ d}, B = {a < X ~ b, Y ~ d}. Clearly
lP'(A) = F(b, d) - F(b, c),
lP'(B) = F(b, d) - F(a, d),
lP'(A U B) = F(b, d) - F(a, c);
now lP'(A n B)= lP'(A) + lP'(B) -lP'(A U B), which gives the answer. Draw a map of~2 and plot the regions of values of (X, Y) involved.
5.
The given expression equals lP'(X = x, Y
Secondly, for 1
~
~
y) -lP'(X = x, Y
~
y- 1) = lP'(X = x, Y = y).
x ~ y ~ 6,
if X < y,
ifx = y. 6.
No, because F is twice differentiable with a2 F jaxay < 0.
153
[2.7.1]-[2.7.6]
Solutions
Random variables and their distributions
2. 7 Solutions to problems 1.
By the independence of the tosses, JP>(X > m) = JP>(first m tosses are tails)= (1- p)m.
Hence JP>(X < x) = {
-
1-(1-p)lxJ
if X::: 0,
0
if X< 0.
Remember that Lx J denotes the integer part of x. (a) If X takes values {xi : i ::: 1} then X = 2:::~ 1 Xi IAi where Ai = {X= Xi}. (b) Partition the real line into intervals of the form [k2-m, (k + 1)2-m), -oo < k < oo, and define Xm = L~-oo k2-m h,m where h,m is the indicator function of the event {k2-m :::: X < (k + 1)2-m}. Clearly Xm is a random variable, and Xm(w) t X(w) as m--+ oo for all w. 2.
(c) Suppose {Xm} is a sequence of random variables such that Xm(w) t X(w) for all w. Then {X :::: X} = nm {Xm :::: X}, which is a countable intersection of events and therefore lies in :F.
3.
(a) We have that 00
{X+Y::::x}=n n=1
U
({X::::r}n{Y::::x-r+n- 1 })
rEI()>+
where IQI+ is the set of positive rationals. In the second case, if X Y is a positive function, then X Y = exp{log X +log Y}; now use Exercise (2.3.2) and the above. For the general case, note first that IZI is a random variable whenever Z is a random variable, since {IZI :::: a} = {Z :::: a} \ {Z < -a} for a ::: 0. Now, if a ::: 0, then {XY ::::a}= {XY < 0} U {IXYI ::::a} and {XY < 0}
=
({X < 0}
n {Y
> 0}) U ({X > 0}
n {Y
< 0}).
Similar relations are valid if a < 0. Finally {min{X, Y} > x} = {X > x} n {Y > y }, the intersection of events. (b) It is enough to check that aX+ {JY is a random variable whenever a, f3 E lR and X, Yare random variables. This follows from the argument above. If Q is finite, we may take as basis the set {IA : A E
4.
(a) F(~)- F(~)
(b) F(2)- F(1) =
(c)
JP>(X 2 ::::
1·
=
J1 of all indicator functions of events.
i·
X)= JP>(X:::: 1) =
i·
1) = l· i) = JP>(X:::: i) = !-
(d) li"(X:::: 2X 2 ) = li"(X::: (e) JP>(X +
x
2 ::::
(f) JP>(vfx:::: z) = JP>(X:::: z2 ) =
5.
iz2 ifO:::: z:::: .J2.
JP>(X = -1) = 1- p, JP>(X = 0) = 0, li"(X::: 1) =
-!P·
6. There are 6 intervals of 5 minutes, preceding the arrival times of buses. Each such interval has 5 = 1 , so the answer IS . 6 · 12 1 = 1. probab1"l"1ty 60 12 2
154
Solutions [2.7.7]-[2.7.12]
Problems
7. Let T and B be the numbers of people on given typical flights of TWA and BA. From Exercise (2.1.3), ~T-~-
(10) k
(~)k (_.!._) 10-k , 10
JP'(B =k) = ( 20) ( -9 k 10
10
)k (-1 )20-k 10
Now lP'(TWA overbooked)= JP'(T = 10) = ( [0 ) 10 , lP'(BAoverbooked) = JP'(B:::: 19) = 20(-fu) 19(fo) + (-fu) 20, of which the latter is the larger.
8.
Assuming the coins are fair, the chance of getting at least five heads is (~) 6 + 6(~) 6 =
9.
(a) We have that lP'(x+ < x) = { 0 -
F(x)
-l4.
if X< 0, ifx :=:: 0.
(b) Secondly, lP'(X- :::0 x) = { O . 1 - hmyt-x F(y)
if X < 0, if X:::: 0.
(c) lP'(IXI:::: x) = JP'(-x:::: X:::: x) if x :=:: 0. Therefore if X < 0,
lP'(IXI < x) = { 0 -
F(x) -limyt-x F(y)
ifx:::: 0.
(d) lP'(-X:::: x) = 1 -limyt-x F(y). 10. By the continuity of probability measures (1.3.5), lP'(X = xo) = lim lP'(y m, by the right-continuity ofF, a contradiction). Hence m is a median, and is smallest with this property. A similar argument may be used to show that M = sup{x : F (x) :::: ~} is a median, and is largest with this property. The set of medians is then the closed interval [m, M].
12. Let the dice show X andY. WriteS= X+ Y and fi = lP'(X = i), gi = JP'(Y = i). Assume that lP'(S = 2) = lP'(S = 7) = lP'(S = 12) = 1\. Now lP'(S = 2) = lP'(X = l)lP'(Y = 1) = flg1, lP'(S = 12) = lP'(X = 6)lP'(Y = 6) = f6g6, lP'(S = 7) :::: lP'(X = 1)lP'(Y = 6) + lP'(X = 6)lP'(Y = 1) = flg6 + f6g1. It follows that
f1 g1 =
f6g6, and also
- 1 = lP'(S = 7) :::: flg1 11
(g6-g1 + -fl!6) = -111 ( 155
X
+ -1) X
[2.7.13]-[2.7.14]
Solutions
Random variables and their distributions
where x = g6/ gJ. However x +x-i > 1 for all x > 0, a contradiction.
13. (a) Clearly dL satisfies (i). As for (ii), suppose that dL(F, G)
= 0. Then
F(x) ::0 lim{G(x +E)+ E} = G(x) E,l-0
and F(y) 0:: lim{G(y- E)- E} E,l-0
= G(y-).
Now G(y-) :::: G(x) if y > x; taking the limit as y ,j, x we obtain F(x):::: limG(y-):::: G(x), y,l-x
implying that F (x) = G (x) for all x. Finally, if F(x) ::0 G(x +E)+ E and G(x) ::0 H(x + 8) + 8 for all x and some E, 8 > 0, then F(x) ::0 H(x + 8 +E) + E + 8 for all x. A similar lower bound for F(x) is valid, implying that dL(F, H) ::: dL(F, G)
+ dL(G, H).
(b) Clearly drv satisfies (i), and drv(X, Y) inequality, llP'(X
= k) -lP'(Z = k)l
:S llP'(X
= 0 if and only if lP'(X = Y) = 1. By the usual triangle = k) -lP'(Y = k)l + llP'(Y = k) -lP'(Z = k)l,
and (iii) follows by summing over k. We have that 2llP'(X E A) - lP'(Y E A) I = I (lP'(X E A) -lP'(Y E A)) - (lP'(X E A c) - lP'(Y E A c)) I
= ~~(lP'(X = k) -lP'(Y = k))lA(k)i where J A (k) equals 1 if k E A and equals -1 if k E Ac. Therefore, 2llP'(X E A) -lP'(Y E A)l :S LllP'(X = k) -lP'(Y = k)l·llA(k)l :S drv(X, Y). k
Equality holds if A= {k: lP'(X = k) > lP'(Y = k)}.
14. (a) Note that a2F - = -e -x- Y axay
0,
ifO ::0 x :S y, ifO==oy==ox,
a2 F --dxdy = 1. in0 oo inoo 0 axay
Hence F is a joint distribution function, and easy substitutions reveal the marginals: Fx(x) Fy(y)
= y-+oo lim F(x, y) =
1 -e-x,
x :::: 0,
= X-+00 lim F(x, y) = 1- e-Y- ye-Y, 156
y:::: 0.
Solutions [2.7.15]-[2.7.20]
Problems
15. Suppose that, for some i i= j, we have Pi < Pj and Bi is to the left of Bj. Write m for the position of Bi and r for the position of Bj, and consider the effect of interchanging Bi and Bj. For k :::;: m and k > r, lP'(T :::: k) is unchanged by the move. Form < k :::;: r, lP'(T :::: k) is decreased by an amount p j - Pi, since this is the increased probability that the search is successful at the m th position. Therefore the interchange of Bi and Bj is desirable. It follows that the only ordering in which lP'(T :::: k) can be reduced for no k is that ordering in which the books appear in decreasing order of probability. In the event of ties, it is of no importance how the tied books are placed.
16. Intuitively, it may seem better to go first since the first person has greater choice. This conclusion is in fact false. Denote the coins by C1, C2, C3 in order, and suppose you go second. If your opponent chooses C1 then you choose C3, because 1P'(C3 beats C1) = ~ + ~ · ~ = i~ > Likewise 1P'(C1 beats C2) = 1P'(C2 beats C3) = ~ > Whichever coin your opponent picks, you can arrange to have a better than evens chance of winning.
1·
1·
17. Various difficulties arise in sequential decision theory, even in simple problems such as this one. The following simple argument yields the optimal policy. Suppose that you have made a unsuccessful searches "ahead" and b unsuccessful searches "behind" (if any of these searches were successful, then there is no further problem). Let A be the event that the correct direction is ahead. Then lP'(A
I current know1e dge)
lP'(current knowledge I A)lP'(A) d lP'(current knowle ge)
=
(1- p)aa
1
which exceeds if and only if (1 - p)aa > (1 - p)b(l -a). The optimal policy is to compare (1 - p)aa with (1 - p)b(1 -a). You search ahead if the former is larger and behind otherwise; in the event of a tie, do either.
i)
i).
18. (a) There are (6 possible layouts, of which 8+8+2 are linear. The answer is 18/(6 (b) Each row and column must contain exactly one pawn. There are 8 possible positions in the first row. Having chosen which of these is occupied, there are 7 positions in the second row which are admissible, 6 in the third, and so one. The answer is 8!/( 0. (c) The density function is f(x) = F'(x) = 2(ex + e-x)- 2 , x E .R (d) This is not a distribution function because F 1 (1) < 0.
20. We have that lP'(U
=
V)
= {
fu,v(u, v) du dv
= 0.
J{(u,v):u=v)
The random variables X, Y are continuous but not jointly continuous: there exists no integrable function f : [0, 1]2 ---+ lR such that lP'(X :5 x, Y :5 y) =
1x 1Y
f(u, v)dudv,
u=O v=O
157
0:5 x, y :51.
3 Discrete random variables
3.1 Solutions. Probability mass functions
c- 1 = I::i"' 2-k = 1. (b) c- = I::i"' rk fk = Iog2. (c) c- 1 = I::i"' k- 2 = :rr 2 /6. (d) c- 1 = I::i"' 2k /k! = e 2 - 1. 1.
(a)
1
2. (i) ~; 1- (2log2)- 1 ; 1- 6:rr- 2; (e 2 - 3)j(e 2 - 1). (ii) 1; 1; 1; 1 and 2. (iii) It is the case that JP>(X even) = I:;~ 1 JP>(X = 2k), and the answers are therefore (d) We have that (a) ~,(b) 1 -(log 3)j(log4), (c)
!·
so the answer is
-! (1 -
e- 2 ).
3. The number X of heads on the second round is the same as if we toss all the coins twice and count the number which show heads on both occasions. Each coin shows heads twice with probability p 2 , so JP>(X = k) = (~)p2k(1 _ p2)n-k.
4.
Let Dk be the number of digits (to base 10) in the integer k. Then
5. (a) The assertion follows for the binomial distribution because k(n - k) :::: (n - k The Poisson case is trivial. (b) This follows from the fact that k8 2: (k2 - 1) 4 . (c) Thegeometricmassfunctionf(k) =qpk, k 2:0.
3.2 Solutions. Independence 1.
We have that JP>(X
= 1, Z = 1) = JP>(X = 1, Y = 1) = ! = JP>(X = 1)1P'(Z = 1). 158
+ l)(k + 1).
Solutions
Independence
[3.2.2]-[3.2.3]
This, together with three similar equations, shows that X and Z are independent. Likewise, Y and Z are independent. However lP'(X = 1, Y = 1, Z = -1) = 0 ~ ~ = lP'(X = 1)1P'(Y = 1)1P'(Z = -1), so that X, Y, and Z are not independent.
2.
(a) If x :::: 1, JP>( min{X, Y} ~ x) = 1 -JP'(X > x, Y > x) = 1 -lP'(X > x)lP'(Y > x) = 1- 2-x . 2-x = 1- 4-x.
(b) lP'(Y >X)= lP'(Y < X) by symmetry. Also lP'(Y > X)+ lP'(Y < X)+ lP'(Y =X)= 1.
Since
lP'(Y =X)= LlP'(Y =X =X)= Lz-x .z-x = ~, X
X
we have that lP'(Y > X)= ~ (c) ~ by part (b). 00
(d)
lP'(X :::: kY) = L JP>( X :::: kY, y = y) y=1 00
00
= LlP'(X:::: ky, y = y) = 2::JP>(X:::: ky)lP'(Y = y) y=1 y=1 00 00 2 - " ' " ' 2-ky-xz-y - ..,...,.-;-.......-- L..., L..., - 2k+ 1 - 1 · y=1 x=O
00
(e)
00
00
lP'(X divides Y) = L lP'(Y = kX) = L L lP'(Y = kx, X= x) k=1 k=1x=1 =
~~z-kxz-x
L..., L...,
k=1x=1
=
~
1
L..., 2k+ 1 -
k=1
1
.
(f) Let r = mfn where m and n are coprime. Then 00 00 1 lP'(X = rY) = L lP'(X = km, y = kn) = L 2-km2-kn = m+n- . 2 1 k=1 k=1
3.
(a) We have that 1P'(X1 < Xz < X3) =
L
(1- P1)(1- pz)(l- P3)P~- 1 p~- 1 p~- 1
i(Y = y, Z = z
x)
y,z
y,z
=aEyJP>(Y=y
I X=x)+bEzJP>(Z=z I X=x).
y
Parts (b)-(e) are verified by similar trivial calculations. Turning to (f),
E{E(Y
I
X,Z)IX~x} ~ ~ { ~yP(Y~y I X ~x,Z ~z)P(X~x.z~z I X ~x)} = L LY JP>(Y = y, X = x, Z = z) . JP>(X = x, Z = z) z Y JP>(X=x,Z=z) JP>(X=x)
IX
= LY JP>(Y = y
= X) = E(Y
IX
= X)
y
= m:{ E(Y I X) I X= x, Z = z }.
by part (e).
2. If¢ and 1{1 are two such functions then E( (c/J(X) -l{I(X))g(X)) = 0 for any suitable g. Setting g(X) = l(X=x! for any x E lR such that JP>(X = x) > 0, we obtain ¢(x) = l{l(x). Therefore JP>(cp(X) = l{I(X)) = 1. 3. We do not seriously expect you to want to do this one. However, if you insist, the method is to check in each case that both sides satisfy the appropriate definition, and then to appeal to uniqueness, deducing that the sides are almost surely equal (see Williams 1991, p. 88).
4.
The natural definition is given by
Now, var(Y)
= E({Y- EY}2) = E = E(var(Y
I X))
[m:( {Y- E(Y I X)+ E(Y I X)- EY} 2
+ var (E(Y
1
X)]
I X))
since the mean ofE(Y I X) is EY, and the cross product is, by Exercise (le),
2m:[m:( {Y- E(Y 1 X) }{E(Y 1 X)- EY} x) J 1
=2m:[ {E(Y 167
1
X)- EY}E{Y- E(Y
1
X)
1
x}] = o
[3.7.5]-[3.7.10]
Solutions
Discrete random variables
since lE{Y -JE(Y I X)/ X}= lE(Y I X) -JE(Y I X)= 0.
5.
We have that
L IP'(T > t + r I T > t) = r=O L IP'(TIP'(T> >t +t) r) . r=O 00
JE(T - t I T > t) =
00
N-t N- t - r
(a)
JE(T- t I T > t) =
L
N- t
= ~(N- t
+ 1).
r=O oo 2-(t+r)
(b)
lE(T - t I T > t) =
L
2=t
= 2.
r=O
6.
Clearly lE(SIN=n)=lE(txi) =Jm, 1=1
and hence JE(S I N) = J-tN. It follows that JE(S) = lE{JE(S I N)} =
lE(~-tN).
7. A robot passed is in fact faulty with probability n = {¢(1 - 8)}/(1- ¢8). Thus the number of faulty passed robots, given Y, is bin(n- Y, n), with mean (n- Y){¢(1 - 8)}/(1 - ¢8). Hence lE(X I Y) = Y
+
(n - Y).4.(1 - 8) '+' . 1-¢8
8. (a) Let m be the family size, 4>r the indicator that the rth child is female, and J-tr the indicator of a male. The numbers G, B of girls and boys satisfy m
B =
L
J-tr,
lE(G) = ~m = JE(B).
r=1
(It will be shown later that the result remains true for random m under reasonable conditions.) We have not used the property of independence. (b) With M the event that the selected child is male,
lE(G I M) = lE (
m-1
L
)
4>r
= ~(m- 1) = JE(B).
r=1
The independence is necessary for this argument.
9. Conditional expectation is defined in terms of the conditional distribution, so the first step is not justified. Even if this step were accepted, the second equality is generally false. 10. By conditioning on Xn-1• lEXn = lE[lE(Xn I Xn-1)] = lE[p(Xn-1
+ 1) + (1- p)(Xn-1 + 1 + Xn)]
where Xn has the same distribution as Xn. Hence lEXn = (1 lEX1=p- 1.
168
+ lEXn-1)/p.
Solve this subject to
Solutions
Sums of random variables
[3.8.1)-[3.8.5)
3.8 Solutions. Sums of random variables 1.
By the convolution theorem, lP'(X + Y
= z) = LlP'(X = k)lP'(Y = z- k) k
k+1
if 0 ::=: k ::=: m 1\ n,
(m + 1)(n + 1)
+1 + 1)(n + 1)
(m 1\ n) (m
if m 1\ n < k < m v n,
m+n+1-k
if m v n ::=: k ::=: m + n,
(m + 1)(n + 1)
where m 1\ n
2.
= min{m, n} and m v n = max{m, n}.
If z:::: 2, lP'(X
L lP'(X = k, Y = z- k) = z(z c+ 1) . 00
+ Y = z) =
k=1
Also, if z
:::: 0, 00
lP'(X- Y
= z) = L
= k + z, Y = k)
lP'(X
k=1
1
00
- c 2.::: -::-:---,.,--,:-:---,--:-::-:-----:-:k= 1 (2k
-
+ z- 1)(2k + z)(2k + z + 1)
00
= ic {; ,;C
By symmetry, if z ::=: 0, lP'(X- Y 3.
(2k + z- 1)(2k + z) -
oo
1
=
1
{
1
(2k + z)(2k + z + 1)
}
(-l)r+1
~ (r + z)(r + z +
1) ·
= z) = lP'(X- Y = -z) = lP'(X- Y =
z-1
La(l-al-1{3(1-{J)z-r-1= ~1
aR{(1 P
(1
R)Z-1 -p
-
lzl).
a)z-1} -
.
a-{J
4. Repeatedly flip a coin that shows heads with probability p. Let Xr be the number of flips after the r- 1th head up to, and including, the rth. Then Xr is geometric with parameter p. The number of flips Z to obtain n heads is negative binomial, and Z = I:~= 1 X r by construction. 5. Sam. Let Xn be the number of sixes shown by 6n dice, so that Xn+1 same distribution as X 1 and is independent of X n. Then,
= Xn +
Y where Y has the
6
lP'(Xn+1
::=:::
n
+ 1) = LlP'(Xn
::=:::
n
+ 1- r)lP'(Y = r)
r=O 6
= lP'(Xn
::=:::
n)
+ L[lP'(Xn
::=:::
n
+ 1- r) -lP'(Xn
::=:::
n)]lP'(Y
= r).
r=O We set g(k) = lP'(Xn = k) and use the fact, easily proved, that g(n) :::: g(n- 1) :::: · · · :::: g(n- 5) to find that the last sum is no bigger than 6 g(n) L ( r - 1)1P'(Y = r) = g(n)(lE(Y)- 1).
r=O
169
[3.8.6]-[3.9.2]
Solutions
Discrete random variables
The claim follows since lE(Y) = 1. oo
6.
oo
( ) -J..
(i)LHS='2:ng(n)e-J..A.nfn!=A.L~ne
1 A.n-
1 =RHS.
n=1 n - 1).
n=O
(ii) Conditioning on Nand XN, LHS = lE(lE(Sg(S) =
I N)) = lE{ NlE( XN g(S) IN)}
L (:-J..A.n _1)! 1xlE (g (n-1 L Xr +x )) dF(x) n
=A.
r=1
J
xlE(g(S +x)) dF(x) = RHS.
3.9 Solutions. Simple random walk 1. (a) Consider an infinite sequence of tosses of a coin, any one of which turns up heads with probability p. With probability one there will appear a run of N heads sooner or later. If the coin tosses are 'driving' the random walk, then absorption occurs no later than this run, so that ultimate absorption is (almost surely) certain. LetS be the number of tosses before the first run of N heads. Certainly JP>(S > Nr) :::: (1- pNy, since Nr tosses may be divided into r blocks of N tosses, each of which is such a run with probability pN. Hence JP>(S = s) :::: (1 - pN) LsI NJ, and in particular lE(Sk) < oo for all k :::: 1. By the above argument, lE(Tk) < oo also.
2.
If So = k then the first step X 1 satisfies JP>(X 1 = 1 1 W) = JP>(X1 = 1)JP>(W 1 x1 = 1) = PPk+1. lP'(W) Pk
Let T be the duration of the walk. Then
h
= lE(T I So = k, W) = lE(T I So= k, W, X 1 = 1)1P'(X 1 = 1 I So= k, W)
=
+ lE(T I So = k, W, X 1 = -1)1P'(X 1 = -1 I So = (1 + h+ 1) Pk+1P + (1 + lk-I) ( 1 - Pk+1P) Pk
= 1
+ PPk+1Jk+1 + Pk
k, W)
Pk
(Pk- PPk+!) h-1' Pk
as required. Certainly Jo = 0. If p =
1then Pk = 1- (k/ N), so the difference equation becomes
(N- k- 1)h+1 - 2(N- k)h
+ (N- k + 1)h-1
= 2(k- N)
for 1 :::: k :::: N- 1. Setting Uk = (N- k)h, we obtain uk+1 - 2uk
+ Uk-1
= 2(k- N),
with general solution Uk =A+ Bk -1(N- k) 3 for constants A and B. Now uo =UN = 0, and therefore A= 1N3 , B
= -1N2 , implying that Jk
= 1{N 2 - (N- k) 2 }, 0:::: k < N.
170
Solutions [3.9.3]-[3.10.3]
Random walk: counting sample paths
3. The recurrence relation may be established as in Exercise (2). Set uk = (pk- pN)h and use the fact that Pk = (pk- pN)/(1- pN) where p = q jp, to obtain PUk+1 - (1- r)uk
+ quk-1 = PN- pk.
The solution is Uk
=A+ Bpk + k(pk + pN), p-q
for constants A and B. The boundary conditions, uo = u N = 0, yield the answer.
4.
Conditioning in the obvious way on the result of the first toss, we obtain Pmn
= PPm-1,n + (1- P)Pm,n-1•
ifm,n:::: 1.
The boundary conditions are PmO = 0, POn = 1, if m, n :::: 1. 5.
Let Y be the number of negative steps of the walk up to absorption. Then lE(X + Y) = Dk and N - k if the walk is absorbed at N, X-Y= {
-k
if the walk is absorbed at 0.
Hence lE(X - Y) = (N - k)(1 - Pk) - kpk. and solving for lEX gives the result.
3.10 Solutions. Random walk: counting sample paths 1.
Conditioning on the first step X 1, lP'(T = 2n) = -!lP'(T = 2n
I x1
= -!f-1(2n -1)
= 1)
+ -!lP'(T =
2n
I x1
= -1)
+ -!f1(2n -1)
where fb(m) is the probability that the first passage to b of a symmetric walk, starting from 0, takes place at time m. From the hitting time theorem (3.10.14),
f1 (2n- 1) = f-1 (2n- 1) = _1_lP'(S2n-1 = 1) = _1_ (2n- 1) T(2n-1), 2n-1
2n-1
n
which therefore is the value oflP'(T = 2n). For the last part, note first that lP'(T = 2n) = 1, which is to say that lP'(T < oo) = 1; either appeal to your favourite method in order to see this, or observe that lP'(T = 2n) is the coefficient of s 2n in the expansion ofF (s) = 1 - ~- The required result is easily obtained by expanding the binomial coefficient using Stirling's formula.
I:f
2.
By equation (3.10.13) ofPRP, for
r:::: 0,
+ 1) + 1) + lP'(Sn = r)- 2lP'(Sn :::: r + 2) -lP'(Sn = r + 1) = lP'(Sn = r) + lP'(Sn = r + 1) = max{lP'(Sn = r), lP'(Sn = r + 1)}
lP'(Mn = r) = lP'(Mn:::: r) -lP'(Mn:::: r
= 2lP'(Sn :::: r
since only one of these two terms is non-zero. By considering the random walk reversed, we see that the probability of a first visit to S2n at time 2k is the same as the probability of a last visit to So at time 2n - 2k. The result is then immediate
3.
from the arc sine law (3.10.19) for the last visit to the origin.
171
[3.11.1]-[3.11.4]
Solutions
Discrete random variables
3.11 Solutions to problems (a) Clearly, for all a, b E JR,
1.
lP'(g(X) =a, h(Y) =b) =
lP'(X=x,Y=y) x,y: g(x )=a,h(y )=b
L
lP'(X
= x)lP'(Y = y)
x,y: g(x )=a,h(y )=b
lP'(X = x)
L x:g(x)=a
lP'(Y = y)
L y:h(y)=b
= lP'(g(X) = a)lP'(h(Y) =b). (b) See the definition (3 .2.1) of independence. (c) The only remaining part which requires proof is that X andY are independent if fx.r(x, y) = g(x)h(y) for all x, y E JR. Suppose then that this holds. Then fx(x) = L
fx.r(x, y) = g(x) Lh(y),
y
fy(y)
=L
y
fx.r(x, y)
= h(y) Lg(x).
X
X
Now
1 = Lfx(x) = Lg(x) Lh(y), X
y
X
so that fx(x)fy(y)
= g(x)h(y) L
g(x) L
h(y)
2. If E(X 2 ) = I:x x 21P'(X = x) = 0 then lP'(X Therefore, if var(X) = 0, it follows that lP'(X -EX
3.
= g(x)h(y) = fx.r(x, y).
y
X
= x) = 0 for x = 0) = 1.
=/= 0. Hence lP'(X
= 0) =
1.
(a)
E(g(X))
= LYlP'(g(X) = y) = L Y
L
ylP'(X
Y x:g(x)=y
= x) = Lg(x)lP'(X = x) x
as required. (b)
E(g(X)h(Y))
=L
g(x)h(y)fx.r(x, y)
by Lemrna(3.6.6)
x,y
= Lg(x)h(y)fx(x)fy(y)
by independence
x,y
= Lg(x)fx(x) Lh(y)fy(y) = E(g(X))E(h(Y)). X y
= fy(i) = ~fori = 1, 2, 3. (b) (X+ Y)(w,) = 3, (X+ Y)(wz) = 5, (X+ Y)(w3) = 4, and therefore fx+Y(i) = ~fori = 3, 4, 5. (c) (XY)(w!) = 2, (XY)(wz) = 6, (XY)(w3) = 3, and therefore fxy(i) = ~fori = 2, 3, 6.
4.
(a) Clearly fx(i)
172
Problems
Solutions
(d) Similarly fx;r(i) =~fori=
and similarly frlz(3 I 2) =
(a)
lP'(Y = 2, Z = 2) lP'(wJ) lP'(Z = 2) = lP'(wJ u W2) =
I 3) = fz1rO I 1) = 1.
1} = k, and therefore k = L k = k n=l L {1-n - -n=i n(n + 1) n+1 00
2'
i. fr1z(l 11) = 1, and other values are 0.
(f) Likewise fzlr(2 I 2) = fzlr(2
5.
i. ~. 3.
frlz(2 I 2) =
(e)
[3.11.5]-[3.11.8]
00
1.
(b) 2::~ 1 kna = kl;(-a) where 1; is the Riemann zeta function, and we require a< -1 for the sum to converge. In this case k = 1; (-a) -I . 6.
(a) We have that n
lP'(X + Y = n) =
L lP'(X = n -
n
k)lP'(Y = k) =
k=O
L
-A.;..n-k
e-JLJ-tk
_e-..,.-- . - k=O (n- k)! k!
e-A.-JL ~ (n) n-k k e-A.-JL().. + J-t)n =--LJ A. 1-t = - - - - - n! k=O k n!
lP'(X = k, X+ Y = n) lP'(X = k I X+ Y = n) = --lP'-(X_+_Y_=_n_)_
(b)
= lP'(X = k)lP'(Y = n- k) = (n) ;..k/-tn-k . lP'(X + Y = n) k (A.+ J-t)n Hence the conditional distribution is bin(n, A./(A. + J-t)). 7.
(i) We have that
lP'(X=n+k,X >n) lP'(X=n+k I X >n)= - - - - - - lP'(X > n) )n+k-1 (1 P - P . = p(1 - p)k-i = lP'(X = k). L:~n+i p(1- p)J-1
(ii) Many random variables of interest are 'waiting times', i.e., the time one must wait before the occurrence of some event A of interest. If such a time is geometric, the lack-of-memory property states that, given that A has not occurred by time n, the time to wait for A starting from n has the same distribution as it did to start with. With sufficient imagination this can be interpreted as a failure of memory by the process giving rise to A. (iii) No. This is because, by the above, any such process satisfies G(k + n) = G(k)G(n) where G(n) = lP'(X > n). Hence G(k + 1) = G(li+ 1 and X is geometric. 8.
Clearly, k
lP'(X + y = k) = LlP'(X = k- j, y = j) j=O =
t (: j=O
=
k
.)Pk-jqm-k+j J
~
m)
(~)pjqn-j J
k m+n-k LJ ( k pq . (n) . j=O - J J
173
=
k m+n-k pq
(m k+ n)
[3.11.9]-[3.11.13]
Discrete random variables
+ n, p).
which is bin(m
9.
Solutions
Turning immediately to the second request, by the binomial theorem,
as required. Now, IP(N even) =
L (~) pk (1 _ p )n-k k even
= i{ (p + 1 -
P)n
+ (1
- P - P )n}
= ~ {1 + (1 -
2p )n}
in agreement with Problem (1.8.20). 10. There are
(%)
ways of choosing k blue balls, and (~~f) ways of choosing n- k red balls. The
total number of ways of choosing n balls is (~),and the claim follows. Finally,
IP(B
= k) =
(
n) b! (N- b)! (N- n)! k (b- k)! . (N- b- n + k)! . N!
=(~){~·b~1···b-~+1} x { N-; b
~ ( ~) pk(1 -
... N- b-;
as N
p)n-k
+ k + 1} { ~ ...
N-;
+ 1} -I
~ oo.
11. Using the result of Problem (3.11.8),
12. (a) E(X) = c + d, E(Y) = b + d, and E(XY) = d, so cov(X, Y) = d- (c + d)(b +d), and X and Y are uncorrelated if and only if this equals 0. (b) For independence, we require f (i, j) = JP( X = i )JP( Y = j) for all i, j, which is to say that
a=(a+b)(a+c),
b=(a+b)(b+d),
c=(c+d)(a+c),
d=(b+d)(c+d).
Now a+ b + c + d = 1, and with a little work one sees that any one of these relations implies the other three. Therefore X and Yare independent if and only if d = (b + d)(c +d), the same condition as for uncorrelatedness.
13. (a) We have that oo m-!
oo
E(X)
=
L
m=O
m!P(X
= m) =
L L
oo
JP(X
= m) =
m=O n=O
oo
L L
n=Om=n+!
174
oo
JP(X
= m) = LIP(X n=O
> n).
Solutions [3.11.14]-[3.11.14]
Problems (b) First method. Let N be the number of balls drawn. Then, by (a), r
E(N) =
r
L W'(N > n) =LIT>(first n balls are red)
n=O n=O r r r- 1 r- n + 1 r r! (b + r- n)! =.L:b+rb+r-l···b+r-n+l =.L:(b+r)! (r-n)! n=O n=O =
~
t (n
(b+r)! n=O
+b) = b + r + 1 b b+ 1 '
where we have used the combinatorial identity I:~=O the simple identity (r.:I) summation, we find that
+ G)
f>rt
r=O
n=O
=
(nbb)
=
rt!ti).
To see this, either use
(x~I) repeatedly, or argue as follows. Changing the order of
(n+b) b
=
_ 1 f=xn(n+b) l - X n=O b
= (1- x)-(b+2) =
f=xr r=O
(b + ++ 1) r
b
1
by the (negative) binomial theorem. Equating coefficients of xr, we obtain the required identity. Second method. Writing m(b, r) for the mean in question, and conditioning on the colour of the first ball, we find that b r m(b,r) = - - + {1 +m(b,r -1)}--. b+r b+r With appropriate boundary conditions and a little effort, one may obtain the result. Third method. Withdraw all the balls, and let Ni be the number of red balls drawn between the i th and (i + l)th blue ball (No = N, and Nb is defined analogously). Think of a possible 'colour sequence' as comprising r reds, split by b blues into b + 1 red sequences. There is a one-one correspondence between the set of such sequences with No = i, Nm = j (for given i, j, m) and the set of such sequences with No = j, Nm = i; just interchange the 'Oth' red run with the mth red run. In particular E(No) = E(Nm) for all m. Now No+ NI + · · · + Nb = r, so that E(Nm) = r /(b + 1), whence the claim is immediate. (c) We use the notation just introduced. In addition, let Br be the number of blue balls remaining after the removal of the last red ball. The length of the last 'colour run' is Nb + Br, only one of which is non-zero. The answer is therefore rf(b + 1) + bf(r + 1), by the argument of the third solution to part (b).
14. (a) We have that E(Xk) = Pk and var(Xk) = PkO- Pk), and the claims follow in the usual way, the first by the linearity of][ and the second by the independence of the Xi; see Theorems (3.3.8) and (3.3.11). (b) Lets = L:k Pb and let Z be a random variable taking each of the values PI, P2· ... , Pn with equal probability n-I. Now E(Z 2 ) - E(Z) 2 = var(Z) 2:: 0, so that
Lk
1
;;Pt 2::
(
Lk
1 ;;Pk
)2 =
s2 n2
with equality if and only if Z is (almost surely) constant, which is to say that PI = P2 = · · · = Pn· Hence
175
[3.11.15]-[3.11.18]
Solutions
Discrete random variables
with equality if and only if Pl = P2 = · · · = Pn. Essentially the same route may be followed using a Lagrange multiplier. (c) This conclusion is not contrary to informed intuition, but experience shows it to be contrary to much uninformed intuition. 15. A matrix V has zero determinant if and only if it is singular, that is to say if and only if there is a non-zero vector x such that xVx' = 0. However,
Hence, by the result of Problem (3.11.2), L:k Xk(Xk- EXk) is constant with probability one, and the result follows. 16. The random variables X+ Y and IX- Yl are uncorrelated since cov(X + Y, IX- Yl) = E{ (X+ Y)IX- Yl}- E(X + Y)E(IX- Yl) =
! + ! - 1. ~ =
0.
However,
! = lP'(X + Y = 0, IX- Yl = 0) =/= lP'(X + Y = O)lP'(IX- Yl = 0) = ! · ~ = ~. so that X + Y and IX - Y I are dependent. 17. Let h be the indicator function of the event that there is a match in the kth place. Then lP'(h = 1) = n -l , and for k =/= j, lP'(h Now X=
1
= 1, Ij = 1) = lP'(Ij = 1 I h = 1)1P'(h = 1) = -n(n- 1)
L:k=l h, so that E(X) = L:k=l n- 1 = 1 and var(X)
= E(X 2)- (EX) 2 =
m:(t
h)
2
- 1
1
= tE(h) 2 + 1
LE(Ijh) -1
j#
=1
+2(n)
1 -1 2 n(n- 1)
= 1.
We have by the usual (mis)matching argument of Example (3.4.3) that
1 n-r (-1i lP'(X = r) = - ' """"' L.J - z.., ' r. i=O
0 _::: r _::: n- 2,
which tends to e- 1/r! as n--+ oo. 18. (a) Let Y1, Y2, ... , Yn be Bernoulli with parameter P2· and Z1, Z2, ... , Zn Bernoulli with parameter Pl / P2· and suppose the usual independence. Define Ai = Yi Zi, a Bernoulli random variable that has parameter lP'(Ai = 1) = lP'(Yi = 1)1P'(Zi = 1) = Pl· Now (AI, A2, ... , An) .::0 (Y1, Y2, ... , Yn) so that /(A) _::: f(Y). Hence e(pl) = E(f(A)) _::: E(f(Y)) = e(p2)· (b) Suppose first that n = 1, and let X and X' be independent Bernoulli variables with parameter p. We claim that {J(X)- f(X')}{g(X)- g(X')} ~ 0;
176
Solutions [3.11.19]-[3.11.19]
Problems
to see this consider the three cases X = X', X < X', X > X' separately, using the fact that are increasing. Taking expectations, we obtain
f
and g
E({f(X)- f(X')}{g(X)- g(X')}) 2:: 0,
which may be expanded to find that 0 :::; E(f(X)g(X)) - E(f(X')g(X)) - E(f(X)g(X')) =
2{ E(f(X)g(X)) -
+ E(f(X )g(X 1
1 ))
E(f(X))E(g(X))}
by the properties of X and X'. Suppose that the result is valid for all n satisfying n < k where k ::: 2. Now m:(f(X)g(X)) = m: {E(f(X)g(X)
Ix 1, Xz, .. ., xk-1)} ;
here, the conditional expectation given X 1, Xz, ... , Xk-1 is defined in very much the same way as in Definition (3.7.3), with broadly similar properties, in particular Theorem (3.7.4); see also Exercises (3.7.1, 3). If X 1, X 2, ... , Xk-1 are given, then f(X) and g(X) may be thought of as increasing functions of the single remaining variable Xko and therefore
I
I
I
E(f(X)g(X) X1, Xz, ... , Xk-1) ::: m:(f(X) X1, Xz, ... , Xk-1)m:(g(X) X1, Xz, ... , Xk-1)
by the induction hypothesis. Furthermore
I
!'(X)= E(f(X) X1, Xz, ... , Xk-1).
I
g'(X) = m:(g(X) X1, Xz, ... , xk-1),
are increasing functions of the k -1 variables X 1 , X 2, ... , X k-1 , implying by the induction hypothesis that E(f' (X)g' (X)) ::: E(f' (X) )E(g' (X)). We substitute this into (*) to obtain E(f(X)g(X)) 2:: E(f' (X))E(g' (X))= E(f(X))E(g(X))
by the definition of f' and g'.
19. Certainly R(p) = E(IA) = Differentiating, we obtain
LUJ IA(w)IP'(w) and IP'(w)
= pN(UJ)qm-N(UJ) where p + q = 1.
Applying the Cauchy-Schwarz inequality (3.6.9) to the latter covariance, we find that R' (p) :::; (pq)- 1Jvar(IA) var(N). However lA is Bernoulli with parameter R(p), so that var(IA) = R(p)(lR(p)), and finally N is bin(m, p) so that var(N) = mp(l- p), whence the upper bound for R'(p) follows. As for the lower bound, use the general fact that cov( X+ Y, Z) = cov( X, Z) +cov( Y, Z) to deduce that cov(/A, N) = cov(/A, lA) + cov(IA, N- lA)· Now lA and N- lA are increasing functions of w, in the sense of Problem (3.11.18); you should check this. Hence cov(/A, N) 2:: var(/A) + 0 by the result of that problem. The lower bound for R' (p) follows.
177
[3.11.20]-[3.11.21]
Solutions
Discrete random variables
20. (a) Let each edge be blue with probability P1 and yellow with probability P2; assume these two events are independent of each other and of the colourings of all other edges. Call an edge green if it is both blue and yellow, so that each edge is green with probability P1 P2· If there is a working green connection from source to sink, then there is also a blue connection and a yellow connection. Thus IP'(green connection) :::0 IP'(blue connection, and yellow connection) = IP'(blue connection)IP'(yellow connection) so that R(P1 P2) :::0 R(pt)R(p2). (b) This is somewhat harder, and may be proved by induction on the number n of edges of G. If n = 1 then a consideration of the two possible cases yields that either R(p) = 1 for all p, or R(p) = p for all p. In either case the required inequality holds. Suppose then that the inequality is valid whenever n < k where k :::: 2, and consider the case when G has kedges. Let e be an edge of G and write w(e) for the state of e; w(e) = 1 if e is working, and w(e) = 0 otherwise. Writing A for the event that there is a working connection from source to sink, we have that R(pY) = IP'pr (A
I w(e) = 1)pY + IP'pr (A I w(e) = 0)(1- pY)
:::0 IP'p(A 1 w(e) = l)Y pY
+ IP'p(A
1 w(e)
= O)Y (1 - pY)
where IP'a is the appropriate probability measure when each edge is working with probability a. The inequality here is valid since, if w(e) is given, then the network G is effectively reduced in size by one edge; the induction hypothesis is then utilized for the case n = k - 1. It is a minor chore to check that xYpY
+ yY(l- p)Y
:::0 {xp
+ y(l- p)JY
if X 0::: y 0::: 0;
to see this, check that equality holds when x = y :::: 0 and that the derivative of the left-hand side with respect to x is at most the corresponding derivative of the right-hand side when x, y :::: 0. Apply the latter inequality with x = IP'p(A I w(e) = 1) andy= IP'p(A I w(e) = 0) to obtain R(pY) :::0 {IP'p(A
I w(e) = 1)p + IP'p(A I w(e) = 0)(1 - p) V =
R(p)Y.
21. (a) The number X of such extraordinary individuals has the bin(107 , 10- 7) distribution. Hence lEX= 1 and
IP'(X > 1 I X > 1) = IP'(X > 1) = _1_-_IP'_c.(X_=_O_)_-_IP'_(X_=_1_) IP'(X > 0) 1 - IP'(X = 0) 1- (1-10-7)107 -107 ·10-7(1-10-7)107-1 1- (1- 10-7)1o7
1- 2e- 1 1- e-1
:::::::
0.4.
(Shades of (3.5.4) here: X is approximately Poisson distributed with parameter 1.) (b) Likewise 1 -1 1 - 2e -1 - ze :::::::0.3. IP'(X > 2 I X :::: 2) :::::::
1- 2e- 1
(c) Provided m
«N
= 107,
IP'(X = m) =
N! ( 1) m ( 1 ) N -m e -1 1- ::::::: - , m!(N -m)! N N m!
178
Solutions [3.11.22]-[3.11.23]
Problems
the Poisson distribution. Assume that "reasonably confident that n is all" means that lP'(X > n I X :::: n) :::0 r for some suitable small number r. Assuming the Poisson approximation, lP'(X > n) :::0 r lP'(X :::: n) if and only if
1
00
00
1
e- 1 ~ - < re- 1 ~ - . ~ k!~k! k=n+1
k=n
For any given r, the smallest acceptable value of n may be determined numerically. If r is small, then very roughly n::::::: 1/r will do (e.g., if r = 0.05 then n::::::: 20). (d) No level p of improbability is sufficiently small for one to be sure that the person is specified uniquely. If p = w- 7 a, then X is bin(107 , w-7 a), which is approximately Poisson with parameter a. Therefore, in this case, lP'(X > 1 I X:::: 1):::::::
1 - e- 01 - ae- 01 = p, 1- e- 01
say.
An acceptable value of p for a very petty offence might be p ::::::: 0.05, in which case a : : : : 0.1 and so p = w- 8 might be an acceptable level of improbability. For a capital offence, one would normally require a much smaller value of p. We note that the rules of evidence do not allow an overt discussion along these lines in a court of law in the United Kingdom.
22. The number G of girls has the binomial distribution bin(2n, p). Hence
L
lP'(G:::: 2n- G)= lP'(G:::: n) = 2n (2n) k pkq2n-k k=n
:::0 ( 2n) n
= 0.485 and n =
(2n) p n q n - q- , n
q-p
er) :::0 enn) for all k.
where we have used the fact that With p
~ ~p k q 2n-k = k=n
104 , we have using Stirling's formula (Exercise (3.10.1)) that
( 2n) Pn qn _ q - :::0
n
q- p
~ { (1 -
v (nrr)
0.03)(1
+ 0.03)} n 00.50135 .
104 0.515 = - ( 1- -9 ) < 1.23 x
3,Jir
lo4
-
w- 5 .
It follows that the probability that boys outnumber girls for 82 successive years is at least (1 - 1.23 x w- 5 ) 82 :::: o.99899.
23. Let M be the number of such visits. If k =f. 0, then M :::: 1 if and only if the particle hits 0 before it hits N, an event with probability 1- kN- 1 by equation (1.7.7). Having hit 0, the chance of another visit to 0 before hitting N is 1 - N- 1, since the particle at 0 moves immediately to 1 whence there is probability 1 - N- 1 of another visit to 0 before visiting N. Hence lP'(M :::: r I So
= k) = (1 -
k) (1 - N1 )r-1 ,
N
r :::: 1,
so that lP'(M = j
I So =
k) = lP'(M :::: j
I So =
k) - lP'(M :::: j
=(1-:)(1-~y-1~, 179
+ 1 I So = j::::l.
0)
[3.11.24]-[3.11.27]
Solutions
Discrete random variables
24. Either read the solution to Exercise (3.9 .4), or the following two related solutions neither of which uses difference equations. First method. Let Tk be the event that A wins and exactly k tails appear. Then k < n so that lP'(A wins) = L:/::JlP'(Tk). However lP'(Tk) is the probability that m + k tosses yield m heads, k tails, and the last toss is heads. Hence
whence the result follows. Second method. Suppose the coin is tossed m + n - 1 times. If the number of heads is m or more, then A must have won; conversely if the number of heads is m - 1 or less, then the number of tails is nor more, so that B has won. The number of heads is bin(m + n- 1, p) so that lP'(A wins)=
L
m+n-1 (
m
+;
l)
-
pkqm+n-1-k.
k=m
25. The chance of winning, having started from k, is
1- (qjp)k 1- (qjp)N
I
I
1 + (q/p)zk.
1- (q/p)zk
which may be written as
I
1- (qjp)2N
.
I
1 + (qjp)'ZN
'
see Example (3.9.6). If k and N are even, doubling the stake is equivalent to playing the original game with initial fortune k and the price of the Jaguar set at The probability of winning is now
i
iN.
lk
1- (q/p)'Z I
,
1- (qjp)'ZN which is larger than before, since the final term in the above display is greater than 1 (when p <
i,
If p = doubling the stake makes no difference to the chance of winning. If p > to decrease the stake.
i ).
i, it is better
26. This is equivalent to taking the limit as N ~ oo in the previous Problem (3.11.25). In the limit when p =f. the probability of ultimate bankruptcy is
i,
lim (qjp)k _ (qjp)N = { (qjp)k N-+oo 1- (qjp)N 1 where p
+q =
l. If p =
"f p > 2> 1 "f 1 1 p < 2· 1
i, the corresponding limit is limN-+oo(l- k/N) =
l.
27. Using the technique of reversal, we have that lP'(Rn
=
Rn-1
+ 1) = lP'(Sn-1 =f. Sn, Sn-2 =f. Sn, ... , So =f. Sn) = lP'(Xn =f. 0, Xn-1 + Xn =f. 0, ... , X1 + · · · + Xn =f. 0) = 1P'(X1 =f. 0, Xz + X1 =f. 0, ... , Xn + · · · + X1 =f. 0) = lP'(S1 =f. 0, Sz =f. 0, ... , Sn =f. 0) = 1P'(S1 Sz · · · Sn =f. 0).
It follows that E(Rn) = E(Rn-1)
+ lP'(S1S2 · · · Sn =f. 0) for n:::
180
1, whence
Solutions [3.11.28]-[3.11.29]
Problems
since IP'(S1S2 · · · Sm =f. 0) ,[_ IP'(Sk =f. 0 for all k 2:: 1) as m ~ oo. There are various ways of showing that the last probability equals Ip - q I, and here is one. Suppose p > q. If x 1 = 1, the probability of never subsequently hitting the origin equals 1- (q/p), by the calculation in the solution to Problem (3 .11.26) above. If X 1 = -1, the probability of staying away from the origin subsequently is 0. Hence the answer is p(1- (qfp)) + q · 0 = p- q. If q > p, the same argument yields q- p, and if p
=q=
~the answer is 0.
28. Consider first the event that M2n is first attained at time 2k. This event occurs if and only if: (i) the walk makes a first passage to S2k (> 0) at time 2k, and (ii) the walk thereafter does not exceed S2k. These two events are independent. The chance of (i) is, by reversal and symmetry, IP'(S2k-1 < s2k· s2k-2 < S2b ... , So < S2k) = IP'(X2k > o, x 2k_ 1 + x 2k > o, ... , x 1 + ... = IP'(X 1 > = IP'(Si >
+ x 2k > O) o, x 1 + x 2 > o, 00., x 1 + 00 . + x 2k > O) 0 for 1 :::0 i :::0 2k) = ~IP'(Si =f. 0 for 1 :::0 i :::0 2k)
= ~IP'(S2k = 0)
by equation (3.10.23).
As for the second event, we may translate S2k to the origin to obtain the probability of (ii): IP'(S2k+1 :::0 S2b ... , S2n :::0 S2k) = IP'(M2n-2k = 0) = IP'(S2n-2k = 0),
where we have used the result of Exercise (3.10.2). The answer is therefore as given. The probabilities of (i) and (ii) are unchanged in the case i = 2k + 1; the basic reason for this is that S2r is even, and S2r+ 1 odd, for all r. 29. Let Uk = IP'(Sk = 0), fk = IP'(Sk = 0, Si recall from equation (3.10.25)) to obtain
=f. 0 for 1 :::0 i
< k), and use conditional probability (or
n
U2n =
L U2n-2khk· k=1
Now N1 = 2, and therefore it suffices to prove that E(Nn) = E(Nn-1) for n:::: 2. Let N~_ 1 be the number of points visited by the walk S1, S2, ... , Sn exactly once (we have removed So). Then N~_ 1 + 1 if Sk =f. So for 1 :::0 k :::0 n, { Nn = N~_ 1 - 1 if Sk =So for exactly one kin {1, 2, 00., n}, N~_ 1
Hence, writing an = IP'(Sk
otherwise.
=f. 0 for 1 :::0 k :::0 n),
E(Nn) = E(N~_ 1 ) +an -IP'(Sk =So exactly once) = E(Nn-1) +an- {han-2
+ f4an-4 + · · · + hLn/2J}
where LxJ is the integer part of x. Now a2m = a2m+1 = u2m by equation (3.10.23). If n = 2k is even, then E(N2k)- E(N2k-1) = u2k- {f2u2k-2
If n
+ · · · + hkl =
0
+ · · · + hk} =
0
= 2k + 1 is odd, then E(N2k+d- E(N2k) = u2k- {hu2k-2
181
[3.11.30]-[3.11.34]
Solutions
Discrete random variables
In either case the claim is proved.
30. (a) Not much. (b) The rhyme may be interpreted in any of several ways. Interpreting it as meaning that families stop at their first son, we may represent the sample space of a typical family as {B, GB, G2B, ... }, with lP'(GnB) = 2-(n+ 1). The mean number of girls is I:~ 1 nlP'(GnB) = I:~ 1 n2-(n+ 1) = 1; there is exactly one boy. The empirical sex ratio for large populations will be near to 1: 1, by the law of large numbers. However the variance of the number of girls in a typical family is var(#girls) = 2, whilst var(#boys) = 0; #A denotes the cardinality of A. Considerable variation from 1:1 is therefore possible in smaller populations, but in either direction. In a large number of small populations, the number of large predominantly female families would be balanced by a large number of male singletons. 31. Any positive integer m has a unique factorization in the form m = integers m(1), m(2), .... Hence,
lP'(M=m)=
( . . )= IT (1----p1 ) IT lP'N(z)=m(z) i
where
i
c = IJ;(l- i;tJ).
Pi
1 {Jm(i)=C Pi
Now L:m lP'(M = m) = 1, so that
c- 1 =
II; pr(i)
for non-negative
(IT Pi-m(i)){J =mfJc i
L:m m-fJ.
32. Number the plates 0, 1, 2, ... , N where 0 is the starting plate, fix k satisfying 0 < k ::; N, and let Ak be the event that plate number k is the last to be visited. In order to calculate lP'(Ak), we cut the table open at k, and bend its outside edge into a line segment, along which the plate numbers read k, k + 1, ... , N, 0, 1, ... , kin order. It is convenient to relabel the plates as -(N + 1- k), -(Nk), ... , -1, 0, 1, ... , k. Now Ak occurs if and only if a symmetric random walk, starting from 0, visits both -(N- k) and k- 1 before it visits either -(N + 1 - k) or k. Suppose it visits -(N- k) before it visits k - 1. The (conditional) probability that it subsequently visits k - 1 before visiting -(N + 1 - k) is the same as the probability that a symmetric random walk, starting from 1, hits N before it hits 0, a probability of N- 1 by (1.7.7). The same argument applies if the cake visits k- 1 before it visits -(N- k). Therefore lP'(Ak) = N- 1. 33. With j denoting the jth best vertex, the walk has transition probabilities Pjk = ( j - 1)- 1 for 1 ::; k < j. By conditional expectation,
1
rj
j-1
= 1 + -.-- .l.>b
TI = 0.
1- 1 k=1
Induction now supplies the result. Since rj log(~).
~
log j for large j, the worst-case expectation is about
34. Let Pn denote the required probability. If (mr, mr+1) is first pair to make a dimer, then m1 is ultimately uncombined with probability Pr-1· By conditioning on the first pair, we find that Pn = (P1 + P2 + · · · + Pn-z)/(n - 1), giving n(Pn+1 - Pn) = -(Pn - Pn-1). Therefore, n! (Pn+ 1 - Pn) = (-l)n- 1 (pz - P1) = (-l)n, and the claim follows by summing. Finally, n
EUn = LlP'(mr is uncombined)= Pn
+ P1Pn-1 + · · · + Pn-1P1 + Pn.
r=1
since the rth molecule may be thought of as an end molecule of two sequences oflength r and n- r + 1. Now Pn --* e- 1 as n--* oo, and it is an easy exercise of analysis to obtain that n- 1EUn --* e- 2.
182
Solutions [3.11.35]-[3.11.36]
Problems
35. First,
where the last summation is over all subsets {r1, ... , rk} of k distinct elements of {1, 2, ... , n}. Secondly,
Ak :S k!
:S k!
L PqPr2 · · · Prk {q, ... ,rkl L PqPr2 · · · Prk r,, ... ,rk
+ (~)
+ (~)
LPf i
mfXPi
L Pr1Pr2 · · · Prk-2 r,, ... ,rk-2
(~ Pj) k- 1 J
Hence
By Taylor's theorem applied to the function log ( 1-x), there exist t:lr satisfying 0 < t:lr < {2( 1-c)2 )} - 1 such that n
II (1- Pr) =II exp{-Pr- t:lrp;} = exp{ -A.- A.O(mfXPi) }· r=1
r
Finally, lP'(X = k) =
(IIr (1 -
Pr ))
L
Pq ... Prk
{r ,, ... ,rk } (1 - Pr1) · .. (1 - Prk)
The claim follows from(*) and(**).
36. It is elementary that -
1 N
1 N
n
E(Y) =- LE(Xr) =- LXr
n r=1 We write Y
·-=f.-(.
n r=1
N
- E(Y) as the mixture of indicator variables thus: N
-Y - E(Y) -
= ~Xr ~ r=1 n
n) .
( Ir - -
N
It follows from the fact i =I= j,
183
.
[3.11.37]-[3.11.38]
Solutions
Discrete random variables
HnC
c HnC
H
nC
· · · · · · · ·· · · ·· · · · · · · · ·· · · ·· · ·· · ·· · (1- y)a
1-y
HnC Figure 3.2. The tree of possibility and probability in Problem (3.11.37). The presence of the disease is denoted by C, and hospitalization by H; their negations are denoted by C and H. that
_
var(Y)
N =L
x?: n n 2 E { ( Ir- N
i#j
r=l
=
~ x?: !!_ ( 1 L
r=l
n2 N
~ 2 N-n = Lxr N2n r=l
=
)2} + L --;:;rE XiXj { ( /i-N n) ( Ij- N n)}
!!_) N
+""' XiXj L n2 i#j
{!!_N Nn -1 _ !!!:.__} - 1 N2
""' N-n - ~XiXj n(N -1)N2 z#J
Nn~;~ 1) {tx?:- ~ tx?:- ~ ~XiXj} r=l
r=l
~
z#J
~
N - n 1 { 2 -2} N - n 1 _2 = n(N-1)N :=txr -Nx = n(N-l)N :=t(xr-x).
37. The tree in Figure 3.2 illustrates the possibilities and probabilities. If G contains n individuals, X isbin(n, yp+ (1- y)a) and Yis bin(n, yp). Itisnotdifficultto seethatcov(X, Y) = nyp(l-v) where v = yp + (1- y)a. Also, var(Y) = nyp(1- yp) and var(X) = nk(1- v). The result follows from the definition of correlation.
38. (a) This is an extension of Exercise (3.5.2). With lP'n denoting the probability measure conditional on N = n, we have that
184
Solutions [3.11.39]-[3.11.39]
Problems where s = n - I:~=l r;. Therefore, 00
lP'(X; = r; for 1 :::0 i :::0 k) = L
lP'n(X; = r; for 1 :::0 i :::0 k)lP'(N = n)
n=O
=IT {
vri J(iY~,e-vf(i)}
i=l
r1 •
E
vs(l-
~(kW e-v(I-F(k)).
s·
s=O
The final sum is a Poisson sum, and equals 1. (b) We use an argument relevant to Wald' s equation. The event {T :::0 n -1} depends only on the random variables X1, X2, ... , Xn-t. and these are independent of Xn. It follows that Xn is independent of the event {T ::: n} = {T :::0 n- l}c. Hence, 00
00
00
lE(S) = LlE(X;/{T;::i}) = LlE(X;)lE(/{T;::i)) = LlE(X;)lP'(T::: i) i=l i=l i=l 00
00
00
t
= L vf(i) LlP'(T = t) = v LlP'(T = t) L f(i) t=i i=l 1=1 i=l 00
= v LlP'(T = t)F(t) = lE(F(T)). t=l
39. (a) Place an absorbing barrier at a+ 1, and let Pa be the probability that the particle is absorbed at 0. By conditioning on the first step, we obtain that
1
1 :::0 n :::0 a.
Pn = - (Po +PI+ P2 + · · · + Pn+I), n+ 2
The boundary conditions are Po = 1, Pa+l = 0. It follows that Pn+l - Pn = (n + 1)(pn - Pn-d for 2 :::0 n :::0 a. We have also that P2 - PI = PI - 1, and Pn+l- Pn =
i(n + 1)! (P2- PI)= i(n + 1)! (PI- Po).
Setting n =a we obtain that -Pa = iCa + 1)! (PI- 1). By summing over 2 :::0 n
= 0, and 1,
whence J-tr+1- J-tr = (r + l)(J-tr- J-tr-1)- 1 for r ~ 2. Therefore, Vr+1 = (J-tr+1- J-tr)/(r satisfies Vr+ 1 - Vr = -1 / (r + 1)! for r ~ 2, and some further algebra yields the value of J-t 1·
+ 1)!
40. We label the vertices 1, 2, ... , n, and we let rr be a random permutation of this set. Let K be the set of vertices v with the property that rr(w) > rr(v) for all neighbours w of v. It is not difficult to see that K is an independent set, whence a( G) ~ IKI. Therefore, a( G) ~ ElK I = 'L:v lP'(v E K). For any vertex v, a random permutation rr is equally likely to assign any given ordering to the set comprising v and its neighbours. Also, v E K if and only if v is the earliest element in this ordering, whence lP'(v E K) = lj(dv + 1). The result follows.
186
4 Continuous random variables
4.1 Solutions. Probability density functions 1.
I
(a) {x(l - x)} -~ is the derivative of sin- 1(2x - 1), and therefore C = n- 1.
(b) C = 1, since
(c) Substitute v = (1
+ x 2 )- 1 to obtain
1
dx
00
-oo -(1-+----,x2=-)-m =
{1 m 3 I 1 1 v -~ (1- v)-~ dv = B(2, m- 2)
lo
where B(·, ·)is a beta function; see paragraph (4.4.8) and Exercise (4.4.2). Hence, if m >
i,
r( 21 )r(m - 21 ) C -1 -_ B(l2'm _ 21) -_ r(m)
2.
(i) The distribution function Fy of Y is Fy(y)
So, differentiating, fy(y)
= lP'(Y :S y) = lP'(aX :S y) = lP'(X :S
yfa)
=
Fx(yfa).
= a- 1 fx(y/a).
(ii) Certainly
= lP'(-X :S x) = lP'(X::: -x) = 1 -lP'(X :S -x) since lP'(X = -x) = 0. Hence f-x(x) = fx( -x). If X and -X have the same distribution function then f-x(x) = fx(x), whence the claim follows. Conversely, if fx( -x) = fx(x) for all x, then, by substituting u = - x, F_x(x)
lP'(-X :S y)
= lP'(X::: -y) =
1
00
-y
fx(x)dx
=
1Y
fx(-u)du
-00
=
1Y
fx(u)du
= lP'(X :S y),
-00
whence X and -X have the same distribution function. 3.
Since a::: 0, f::: 0, and g::: 0, it follows that af
+ (1 -
a)g ::: 0. Also
l{af+(l-a)g}dx=a lfdx+(l-a) lgdx=a+1-a=l.
187
[4.1.4]-[4.2.4]
Solutions
Continuous random variables
If X is a random variable with density f, and Y a random variable with density g, then a f +(1-a) g is the density of a random variable Z which takes the value X with probability a and Y otherwise. Some minor technicalities are necessary in order to find an appropriate probability space for such a Z. If X andY are defined on the probability space (Q, :F, lP'), it is necessary to define the product space (Q, :F, lP') x (:E, g., Q) where :E = {0, 1}, g. is the set of all subsets of :E, and Q(O) = a, Q(l) = 1 -a. For w x a E Q x :E, we define Z(w x a)= {
X(w)
if a= 0,
Y(w)
if a = 1.
.. . 1 F(x +h)- F(x) f(x) (a) By defimtion, r(x) = hm= --h-1-0 h 1- F(x) 1 - F(x) (b) We have that
4.
H(x) x
= .!!_ { .!_ fx r(y) dy} = dx
x
Jo
r(x) - 2_ x x2
r
Jo
r(y) dy
= 2_ x2
r
Jo
[r(x)- r(y)] dy,
which is non-negative if r is non-increasing. (c) H(x)fx is non-decreasing if and only if, for 0 _::::a_:::: 1,
1 1 -H(ax) _:::: -H(x) ax x
~
for all x
0,
which is to say that -a- 1 log[l - F(ax)] _:::: -log[l- F(x)]. We exponentiate to obtain the claim. (d) Likewise, if H(x)/x is non-decreasing then H(at) _:::: aH(t) for 0 _::::a _:::: 1 and t ~ 0, whence H(at) + H(t- at) _:::: H(t) as required.
4.2 Solutions. Independence 1. LetN be the required number. ThenlP'(N distribution with mean [1 - F(K)r 1 .
2.
= n) = F(K)n- 1[1- F(K)] forn
~ 1, the geometric
(i) Max{X, Y} _:::: v if and only if X :S v and Y :S v. Hence, by independence, lP'(max{X, Y} :S
v) = lP'(X :S v, Y :S
v)
= lP'(X
:S v)lP'(Y :S v)
= F(v) 2 .
Differentiate to obtain the density function of V = max{X, Y}. (ii) Similarly min{X, Y} > u if and only if X> u andY> u. Hence lP'(U :S u) giving fu(u)
= 1 -lP'(U
= 2f(u)[l
> u)
=
1 -lP'(X > u)lP'(Y > u)
= 1- [1- F(u)] 2 ,
- F(u)].
3. The 24 permutations of the order statistics are equally likely by symmetry, and thus have equal probability. Hence lP'(Xt < Xz < X3 < X4) = and lP'(Xt > Xz < X3 < X4) = by enumerating the possibilities.
f4,
4. lP'(Y(y) > k) Therefore,
=
F(y)k fork ~ 1. Hence JE:Y(y)
lP'(Y(y) > JE:Y(y))
:A,
=
F(y)/[1 - F(y)]
-+ oo as -+ oo.
= {1- [1- F(y)J}LF(y)/[l-F(y)]J
~ exp {1- [1- F(y)]
188
l
F(y)
1 - F(y)
J}-+ e-
1
as y-+ oo.
y
Solutions [4.3.1]-[4.3.5]
Expectation
4.3 Solutions. Expectation
J
(a)JE:(Xa) = 000 xae-x dx < oo if and only if a> -1. (b) In this case
1.
if and only if -1 < a < 2m - 1. 2.
We have that
X;)
( -L.,n1 -
1 = JE:
n
= LE(X;/Sn).
Sn i=1 By symmetry, JE:(X;/Sn) = JE:(XlfSn) for all i, and hence 1 = nJE:(X1/Sn). Therefore m
E(Sm/Sn) = LE(Xi/Sn) = mE(XlfSn) = mjn. i=1
3.
Either integrate by parts or use Fubini's theorem:
r
roo xr- 11P'(X > x)dx = r roo xr- 1 { roo f(y)dy} dx k k h~ = roo f(y) {1y rxr-1 dx} dy = roo yr f(y)dy. h=o x=O k
An alternative proof is as follows. Let lx be the indicator of the event that X > x, so that J~ lx dx =X. Taking expectations, and taking a minor liberty with the integral which may be made
rigorous, we obtain EX = J000 E(Ix) dx. A similar argument may be used for the more general case.
4. We may suppose without loss of generality that t-t = 0 and a = 1. Assume further that m > 1. In this case, at least half the probability mass lies to the right of 1, whence JE:(X/{x:::mJ) ::: ~- Now O=E(X) =E{X[/{x:::m} +l{X. Using this identity and integrating by parts repeatedly,
1- (x) =
1oo
tj>(u)du = -
= tj>(x) _ tj>(x) x
1oo
-1
X
x3
X
4>'(u) tj>(x) --du = U
00 3¢'(u) du = u5
x
+
1oo -
X
X
tj>(x) _ tj>(x) x x3
q/(u) 3-du U
+ 3¢(x) x5
-100 x
15¢(u) du. u6
4.5 Solutions. Dependence 1.
(i) As the product of non-negative continuous functions, g(x) = ~e-lxl
1-oo00
1
f
is non- negative and continuous. Also
I 2 2
e-zx Y dy = ~e-lxl
J2nx-2
if x =f. 0, since the integrand is the N(O, x- 2 ) density function. It is easily seen that g(O) = 0, so that g is discontinuous, while
1-00oo
g(x)dx =
1oo ~e-lxl -00
dx = 1.
(ii) Clearly f Q ::: 0 and
00 100 1-oo -oo !Q(x,y)dxdy = 2::00 n=l
Ur
.1
=
1.
Also f Q is the uniform limit of continuous functions on any subset of ~2 of the form [- M, M] x ~; hence f Q is continuous. Hence f Q is a continuous density function. On the other hand
1
00
-oo
!Q(x, y)dy =
00 2:: O·r g(x- qn).
n=l
where g is discontinuous at 0. (iii) Take Q to be the set of the rationals, in some order. 2. We may assume that the centre of the rod is uniformly positioned in a square of size a x b, while the acute angle between the rod and a line of the first grid is uniform on [0, ~ n]. If the latter angle is e then, with the aid of a diagram, one finds that there is no intersection if and only if the centre of the rod lies within a certain inner rectangle of size (a- r cos&) x (b- r sin&). Hence the probability of an intersection is
-
2
7r~
lorr/2 {ab- (a- rcos&)(b- rsin&)} de= -(a+ 2r b- ~r). 7r~
0
3. (i) Let I be the indicator of the event that the first needle intersects a line, and let J be the indicator that the second needle intersects a line. By the result of Exercise (4.5.2), E(/) = E(J) = 2/n; hence Z =I+ J satisfies E(~Z) = 2/n.
191
[4.5.4]-[ 4.5.6]
Solutions
Continuous random variables
(ii) We have that var(iZ) = i{lE(l 2) + JE(J 2) + 2JE(l J)} -JE(iZ)2 1 4 1 4 1 = 4{lE(l) +lE(J) + 2JE(l J)}- 2 = - - 2 + -JE(/ J). 7r JT:Jr: 2
in,
z i
then two intersections occur if < min{sine, cos&} In the notation of (4.5.8), if 0 < e < or 1 - z < min {sin e, cos e}. With a similar component when 1r ::S e < 1r, we find that
i
lE(/ J) = lP'(two intersections) =
i
~ jj
dz de
(z,ll):
O1 :::: ~rr,
o:::: e <
2rr.
The marginals are then fe(B) = (2rr)- 1 , fq,() = ~I cos I, and the conditional density functions are for appropriate e and. Thus E> and are independent. The fact that the conditional density functions are different from each other is sometimes referred to as 'Borel's paradox'. 2.
We have that 1/f(x) =
1oo Y !x,y(x, y) dy -oo fx(x)
and therefom lE(l/f(X)g(X)) = =
y(x y) ' ' g(x)fx(x)dxdy 1-oooo 1oo y fxfx(x)
L: L: -~
{y g(x)}fx,y(x, y)dx dy = JE(Yg(X)).
3. Take Y to be a random variable with mean oo, say fy (y) = y- 2 for 1 :::: y < oo, and let X = Y. Then JE(Y 1 X) = X which is (almost surely) finite.
193
[4.6.4]-[4.6.6] 4.
Solutions
Continuous random variables
(a) We have that
= A.e'- n) =fox Gn-1 (x-u) du =fox Gn-1 (v) dv.
Now Go(v) = 1 for all v
E
(0, 1], and the result follows by induction. Now, 00
EN=
L
lP'(N > n) =ex.
n=O
More generally, GN(s)
oo oo = L:snlP'(N = n) = L:sn n=1
whence var(N)
n=1
(
n) =
n-1 x - ::..._ (n-l)! n!
= G:Z,(l) + G/,r(l)- G/,r(1) 2 = 2xex +ex- e2x. 194
(s- l)esx
+ 1,
dx
Solutions [4.6.7]-[4.7.1]
Functions of random variables 7. We may assume without loss of generality that lEX inequality,
=
lEY
=
0. By the Cauchy-Schwarz
Hence, lE(var(Y
8.
I X)) = lE(Y 2 )
-lE(lE(Y
I X) 2 )
lE(XY) 2 lE(X 2 )
< JEY 2 -
=
(1 - p 2 ) var(Y).
One way is to evaluate
Another way is to observe that min{Y, Z} is exponentially distributed with parameter f-t + v, whence lP'(X < min{Y, Z}) = /...f(/... + f-t + v). Similarly, lP'(Y < Z) = ~-t/(1-t + v), and the product of these two terms is the required answer.
9.
By integration, for x, y > 0,
fy(y) =loy f(x, y) dx whence c
=
=
gcy 3 e-Y,
fx(x)
=
1
1. It is simple to check the values of fXIY(x I y)
and then deduce by integration that JE(X I Y
00
f(x, y) dy =ex e-x,
=
f(x, y)/ fy(y) and fYIX(Y
= y) = 1Y and JE(Y I X = x) = x + 2.
I x),
10. We have that N > n if and only if Xo is largest of {Xo, X 1, ... , Xn}, an event having probability
1/(n + 1). Therefore, lP'(N largest, whence
= n) = 1/{n(n + 1)} for n oo
lP'(XN :::; x)
=
L
F(x)n+l n(n + 1)
n=1
oo
=
L
:=:: 1. Next, on the event {N
F(x)n+l n -
n=1
F(x)n+l n+1
oo
L
= n}, Xn is the
+ F(x),
n=1
as required. Finally,
lP'(M
= m) = lP'(Xo :=:: X1
1
:=:: · · · :=:: Xm-1) -lP'(Xo :=:: X1 :=:: · · · :=:: Xm)
= m!
1
- (m
+ l)! ·
4.7 Solutions. Functions of random variables 1.
We observe that, if 0 :::; u :::; 1,
lP'(XY :::; u)
= lP'(XY
:::; u, Y :::; u)
1
+ lP'(XY :::; u, Y
> u)
= lP'(Y
:::; u)
+ lP'(X:::; ufY,
Y > u)
1u
=u+
-dy=u(1-1ogu).
u y
By the independence of XY and Z,
lP'(XY:::; u,
z 2 :::; v) = lP'(XY
:::; u)lP'(Z :::; .y'v)
195
= u.,fV(l
-log u),
0 < u, v < 1.
[4.7.2]-[4.7.5]
Solutions
Continuous random variables
Differentiate to obtain the joint density function ) _ log(1/u) .fo ,
g(u, v -
0:::: u, v::::
2
Hence
JJ
lP'(XY:::: Z 2 ) =
1.
~-
lo;%u) dudv =
o::::u::::v::::1 Arguing more directly,
Iff
lP'(XY:::: Z 2 ) =
dxdydz =
~
o::::x,y,z::::1 xy::::z 2
2.
The transformation x = uv, y = u- uv has Jacobian
J
=I
vv 1-
u -u
I=
-u ·
Hence Ill= lui, and therefore fu. y(u, v) = ue-u, for 0:::: u < oo, 0:::: v:::: 1. Hence U and V are independent, and fv(v) = 1 on [0, 1] as required. 3.
Arguing directly, 2
lP'(sin X :::: y) = lP'(X :::: sin- 1 y) = - sin- 1 y,
0::::
n
so that fy (y) = variables.
4.
2/ (n ~), for 0 :::: y :::: 1.
Alternatively, make a one-dimensional change of
(a) lP'(sin- 1 X :::: y) = lP'(X :::: siny) = siny, for 0 :::: y ::::
o::::y::::in. (b) Similarly, lP'(sin- 1 X:::: y) = _ln < Y < ln. 2 - 2
y:::: 1,
in.
Hence fy(y) =cosy, for
iO + siny), for -in : : y:::: in. so that fy(y) = i cosy, for
5. Consider the mapping w = x,z = (y-px)/~ withinversex = w,y = pw+zJI=Pf and Jacobian J=
I~ Vl~p 2 1 = R·
The mapping is one-one, and therefore W (= X) and Z satisfy
implying that W and Z are independent N (0, 1) variables. Now {X> 0, y > 0}
= {w >
0,
z
> -Wp
I R},
and therefore, moving to polar coordinates, lP'(X > 0, Y > 0) =
f
~JT
1
oo
1 I 2 -e-zr rdrdB =
Je=a r=O 2n 196
1
~JT 1
a
-dB
2n
Solutions [4.7.6]-[4.7.9]
Functions of random variables where a= -tan- 1 (p/JI=P2") = -sin- 1 p.
6.
We confine ourselves to the more interesting case when p =I= 1. Writing X
=
U, Y
=
pU +
JI=P2 V, we have that U and V are independent N(O, 1) variables. It is easy to check that Y > X if and only if (1 - p) U < JI=P2 V. Turning to polar coordinates, E(max{X, Y}) =
1 ~ [~1{1+rr 00
{pr cosO+ r P sinO} dB+
~~rr r cos edB] dr
where tan 1{1' = ~(1- p)/(1 + p). Some algebra yields the result. For the second part,
by the symmetry of the marginals of X and Y. Adding, we obtain 2E(max{X, Y} 2 ) E(Y 2 ) = 2. 7.
=
E(X 2 ) +
We have that lP'(X < Y, Z > z)
(a) lP'(X
= Z) = lP'(X <
Y)
A = lP'(z z).
A
= --. A+JL
(b) By conditioning on Y, lP'( (X- Y)+
A
= 0) = lP'(X ::;: Y) = A+ tL,
lP'((X- Y)+ >
w)
= _1-L_e-J.w
A+JL
for w > 0.
By conditioning on X, lP'(V > v)
= lP'(IX- Yl
> v)
=
1
00
lP'(Y > v + x)fx(x) dx +
1
00
lP'(Y < x- v)fx(x) dx
v > 0. (c) By conditioning on X, the required probability is found to be
8. that
Either make a change of variables and find the Jacobian, or argue directly. With the convention r2 - u2 = 0 when r 2 - u 2 < 0, we have that
J
F(r,x) = lP'(R::;: r, X::;: x) =
~ 7r
a2 F 2r f(r,x) = - = ~· arax rry r2- x2 9.
jx
Jr2- u2du,
-r
lxl < r < l.
As in the previous exercise, lP'(R::::: r, Z::::: z) = 3- ~z n(r 2 - w 2 )dw.
4n -r
197
[4.7.10]-[4.7.14]
Solutions
Continuous random variables
= ~r for lzl < r < 1. This question may be solved in spherical polars also. The transformations = x + y, r = xf(x + y), has inverse x = rs, y = (1 - r)s and Jacobian
Hence f(r, z)
10. J = s. Therefore, fR(r)
=
roo fR s(r, s) ds = roo fx y(rs, (1 -
lo
·
lo
·
r)s )s ds O::=or::=ol.
11. We have that
whence fy(y) = 2fx(-J(afy)
-1)
=
1 rrJy(a- y)
0 :::0 y :::0 a.
,
12. Using the result of Example (4.6.7), and integrating by parts, we obtain JP(X >a, Y >b)=
1
00
= [1 -
(x) { 1 - (
~)} dx
(a)][1 - (c)]
+
1
00
a
[1 - (x)] (
h
~) 1- p2
1- p2
dx.
Since [1 - (x)]/(x) is decreasing, the last term on the right is no greater than 1 - (a) (a)
1
00
(x)
b- px )
(
p
JT=P2 JT=P2
a
dx,
which yields the upper bound after an integration. 13. The random variable Y is symmetric and, for a > 0,
P(Y > a)
by the transformation v
= JP(O <
-1
X < a
)
ra-1 rr(l + du
= Jo
u2)
=
1aoo rr(1 +
-v-2 dv v-2),
= 1/u. For another example, consider the density function lx- 2 if x > 1
f(x) = { 2
'
1
14. The transformation w J = w, whence f(w, z) = w.
ifO:Sx:Sl.
= x + y, z =xI (x + y) has inverse x = wz, y = (1 -
z)w, and Jacobian
A.(A.wz)a-1 e-A.wz A.(A.(1 _ z)w).B-1 e-A.(1-z)w r(a) . r(.B)
A.(A.w)a+.B-le-A.w
r(a
+ {3)
za-1(1- z).B-l B(a, {3)
w > 0, 0 < z < 1.
Hence Wand Z are independent, and Z is beta distributed with parameters a and {3.
198
Solutions [4.8.1]-[4.8.4]
Sums of random variables
4.8 Solutions. Sums of random variables 1.
= X + Y has density function
By the convolution formula (4.8.2), Z fz(z)
=
rz AJ.-Le-Axe-JL(z-x) dx = ~ (e-AZ- e-JLZ), 11- A
z ~ 0,
lo
if A ::j:. f-L. What happens if A = J1? (Z has a gamma distribution in this case.) 2.
Using the convolution formula (4.8.2), W =aX+ f3Y has density function fw(w)
=
1
00
1
-oo
na(1 + (x/a) 2)
·
1 nf3(1 + {(w- x)/f3}2)
dx,
which equals the limit of a complex integral: lim R-+oo
r
1- . 1 af3 . - dz z2 + a2 (z- w)2 + p2
JD n2
where D is the semicircle in the upper complex plane with diameter [- R, R] on the real axis. Evaluating the residues at z = ia and z = w + if3 yields af32ni { 1 1 1 1 } fw(w) = _n_2_ -2i-a · (ia- w)2 + p2 + -2i-f3 · -(w_+_i_f3--,)2,---+-a""" 2
1
=
1
n(a + f3) · 1 + {wf(a + f3)}2'
< w <
-00
00
after some manipulation. Hence W has a Cauchy distribution also. 3.
Using the convolution formula (4.8.2), fz(z) = loz
4.
~ze-z dy = ~z 2 e-z,
z ~ 0.
Let fn be the density function of Sn. By convolution,
_ , -J.. 1x f 2 (x ) -II.Je
A 2 · -A2 - At
A 2 n A + "-2e , -J..2x · - 1 _ '"""" , -J..,x II - - L...J"-re - -s- . At - A2
r=l
s=l As - Ar
sf.r
This leads to the guess that n
fn(X) = LAre-J..,x r=l
IIn __As_, s=l As - Ar
sf.r
which may be proved by induction as follows. Assume that (*) holds for n ::::: N. Then
199
[4.8.5]-[4.8.8] Solutions
Continuous random variables
for some constant A. We integrate over x to find that
1=L IT _ s _ +A- , N N+1
A
r=1 s=1 As - Ar sf.r
and(*) follows with n
5.
=N
AN+1
+ 1 on solving for A.
The density function of X + Y is, by convolution,
fz(x) = {
ifO:::::x:::::1,
x
2-
X
if 1 ::::::
X ::::::
2.
Therefore, for 1 :::::: x :::::: 2,
f3(x)=
lo1fz(x-y)dy= 11 0
(x-y)dy+
lox-1 (2-x+y)dy=~-(x-~) . 2
x-1
0
Likewise,
A simple induction yields the last part.
6. The covariance satisfies cov(U, V) = E(X 2 - Y 2 ) random variables taking values ± 1, then
= 0, as required.
If X andY are independent N(O, 1) variables, fu,v(u, v) a function of u multiplied by a function of v.
If X andY are symmetric
= (4rr)- 1e-! 0, Y > 0) = 4lP'( (X, Y) E
8.)
= lP'(IUI :::::: zj../i, lVI :::::: z!..fi)
= 2{2(z!..fi)- 1}2 , 200
by symmetry
Solutions [4.9.1]-[4.9.5]
Multivariate normal distribution whence the conditional density function is
f(z)
=
2J2{2(z!J2) -l}tP(z;J2).
Finally, lE(Z
IX
> 0, Y > 0) = 21E(X = 21E(X
IX
> 0, Y > 0)
I X>
0) = 41E(X/[X>O}) = 4
loooo
I 2
X
r;c:e-2x dx. v2rr
4.9 Solutions. Multivariate normal distribution 1. Since Vis symmetric, there exists a non-singular matrix M such that M' = M- 1 and V = MAM- 1, where A is the diagonal matrix with diagonal entries the eigenvalues A1, Az, ... , An of V. I
Let A 2 be the diagonal matrix with diagonal entries
I
.Jil, ../Ai,, ... , $n; A 2 is well defined since
I
V is non-negative definite. Writing W = MA 2M', we have that W = W' and also
I
as required. Clearly W is non-singular if and only if A 2 is non-singular. This happens if and only if Ai > 0 for all i, which is to say that V is positive definite. By Theorem (4.9.6), Y has the multivariate normal distribution with mean vector 0 and covariance matrix
2.
Clearly Y = (X - J.L)a' + J.La' where a = (a1, az, ... , an). Using Theorem (4.9.6) as in the previous solution, (X - J.L )a' is univariate normal with mean 0 and variance aVa'. Hence Y is normal with mean J.La' and variance aVa'.
3.
4. Make the transformation u = x + y, v = x- y, with inverse x = ! a) # lP'(X > a)lP'(Y > a).
6.
Recall from Exercise (4.8.7) that for any pair of centred normal random variables E(X I Y) = cov(X, Y) Y, varY
var(X
I Y)
= {1- p(X, Y) 2} var X.
The first claim follows immediately. Likewise,
7. As in the above exercise, we calculate a = E(X I I ~7 Xr) and b = var(X I I ~7 Xr) using the facts that var XI = vu, var(~7 Xi) = ~ij ViJ• and cov(XI, ~7 Xr) = ~r VIr· 8.
Let p = lP'(X > 0, Y > 0, Z > 0) = lP'(X < 0, Y < 0, Z < 0). Then 1- p = lP'({X > 0} U {Y > 0} U {Z > 0}) = lP'(X > 0)
+ lP'(Y
> 0)
+ lP'(Z
> 0)
+p
- lP'(X > 0, Y > 0) - lP'(Y > 0, Z > 0) - lP'(X > 0, Z > 0)
=
9.
3
[3
. -I . -I . -I }] 2 + p - 4 + 2rr1 {sm PI + sm P2 + sm P3 .
Let U, V, W be independent N(O, 1) variables, and represent X, Y, Z as X = U, Y = PI U
J1- p[V,
z=
p 3U
2 P32 + 2PIP2P3 1 -PI2 - P2-
+ P2- PIP3 V + J1-p[
2
(1 -PI)
We have that U =X, V = (Y- PIX)/V1- Pf and E(Z conditional variance.
w.
I X, Y) follows immediately, as does the
4.10 Solutions. Distributions arising from the normal distribution 1.
x2(m) density function is
First method. We have from (4.4.6) that the
X
The density function of Z = X I
2:0.
+ X 2 is, by the convolution formula,
() lo x2
z lm-I e -!x( ._ z- x )ln-I ._ e -l(z-x)d 2 x
g z = c
0
I
)
I
= cz2(m+n -Ie-2z
I I Jo{I u'Zm-I(lu)Zn-I du
202
+
Solutions [4.10.2]-[4.10.6]
Distributions arising from the normal distribution
by the substitution u = xjz, where cis a constant. Hence g(z) = c'z~(m+n)- 1 e-~z for z.:,:: 0, for an appropriate constant c', as required. Second method. If m and n are integral, the following argument is neat. Let Z1, Z2, ... , Zm+n be independent N(O, 1) variables. Then X1 has the same distribution as Zf + Zi + · · · + Z~, and X2 the same distribution as Z~+ 1 + Z~+ 2 + · · · + Z~+n (see Problem (4.14.12)). Hence X1 + X2 has the same distribution as Zf + · · · + Z~+n• i.e., the x2(m + n) distribution.
2. (i) The t(r) distribution is symmetric with finite mean, and hence this mean is 0. (ii) Here is one way. Let U and V be independent x2 (r) and x2 (s) variables (respectively). Then
by independence. Now U is f(i,
ir) and Vis f(i, is), so that E(U) =rand I
E(V
-1
lo0012-s/2 's-1-'v r(is-1)lo002-:z(s-2) 's-2-'v 1 )= ---v:Z e :Z dv= v:Z e 2 dv=-0 v r(is) 2f(is) 0 r(is- 1) s- 2
if s > 2, since the integrand is a density function. Hence
E
s ( Ujr) Vjs =s-2
if s > 2.
(iii) If s ::; 2 then E(v- 1) = oo.
3.
Substitute r = 1 into the t (r) density function.
First method. Find the density function of X I Y, using a change of variables. The answer is F(2, 2).
4.
Second method. X and Y are independent hence X/Y is F(2, 2).
x2 (2)
variables (just check the density functions), and
5. The vector (X, X1- X, X2- X, ... , Xn- X) has, by Theorem (4.9.6), a multivariate normal distribution. We have as in Exercise (4.5.7) that cov(X, Xr -X) = 0 for all r, which implies that X is independent of each X r. Using the form of the multivariate normal density function, it follows that X is independent of the family {Xr -X : 1 ::; r ::; n}, and hence of any function of these variables. Now s2 = (n- 1)- 1 L:r(Xr- X) 2 is such a function. 6. The choice of fixed vector is immaterial, since the joint distribution of the Xj is spherically symmetric, and we therefore take this vector to be (0, 0, ... , 0, 1). We make the change of variables U2 = Q2 +X~, tan W = Q/ Xn, where Q2 = 2::~;;;;;} and Q .:,:: 0. Since Q has the x2(n - 1) distribution, and is independent of Xn, the pair Q, Xn has joint density function
x;
-~x 2 l(lq)!(n-3)e-~x /( x - _e__ . 2 2 q' ) - ../27i ~r'--(--;i-(n---1-)) -
X E
JR, q > 0.
The theory is now slightly easier than the practice. We solve for U, W, find the Jacobian, and deduce the joint density function fu,'-l!(u, 1/1) of U, W. We now integrate over u, and choose the constant so that the total integral is 1.
203
[4.11.1]-[4.11. 7]
Continuous random variables
Solutions
4.11 Solutions. Sampling from a distribution 1.
Uniform on the set {1, 2, ... , n}.
2.
The result holds trivially when n = 2, and more generally by induction on n.
3. We may assume without loss of generality thatA = 1 (since ZIA. is f(A., t) if Z is f(l, t)). Let U, V be independent random variables which are uniformly distributed on [0, 1]. We set X= -t log V and note that X has the exponential distribution with parameter 1It. It is easy to check that for x > 0, where c = tt e-t+ 1I f(t). Also, conditional on the event A that U < -
xt-1 -t
e f(t)
te-Xft
,
X has the required gamma distribution. This observation may be used as a basis for sampling using the rejection method. We note that A= {log U :=:: (n- l)(log(XIn)- (X In)+ 1) }. We have that IF'( A) = 11c, and therefore there is a mean number c of attempts before a sample of size 1 is obtained. 4. Use your answer to Exercise (4.11.3) to sample X from f(l, a) andY from f(l, {3). By Exercise (4.7.14), Z = XI(X + Y) has the required distribution.
S. (a) This is the beta distribution with parameters 2, 2. Use the result of Exercise (4). (b) The required r (1, 2) variables may be more easily obtained and used by forming X = - log( U1U2) and Y -log(U3 U4) where {Ui : 1 _::: i :=:: 4} are independent and uniform on [0, 1]. (c) Let U1, U2, U3 be as in (b) above, and let Z be the second order statistic U(2)· That is, Z is the middle of the three values taken by the Ui; see Problem (4.14.21). The random variable Z has the required distribution. (d) As a slight variant, take Z = max{U1, U2} conditional on the event {Z :=:: U3}. (e) Finally, let X = .../UlI (.../Ul + ..JU2), Y = .../Ul + ..jUi. The distribution of X, conditional on the event {Y :=:: 1}, is as required. 6. We use induction. The result is obvious when n = 2. Let n 2: 3 and let p = (P1, P2, ... , Pn) be a probability vector. Since p sums to 1, its minimum entry P(1) and maximum entry P(n) must satisfy
1 1 P(1) :S - < - - , n n -1
P(1) + P(n) 2: P(1) +
We relabel the entries of the vector p such that P1 (n -l)p1, 0, ... , 0). Then
1 n -2 P = - -1 v1 + - -1 Pn-1 nn-
where
1-P(1) n- 1
=
1+(n-2)P(1) 1 2: --1 n- 1 n- .
= P(l) and P2 = P(n)• and set v1 = ((n -1) P1, 1-
n-1 ( 1 ) Pn-1 = - 0, P1 + P2- - - , P3, · · · , Pn , n-2 n-1
is a probability vector with at most n - 1 non-zero entries. The induction step is complete. It is a consequence that sampling from a discrete distribution may be achieved by sampling from a collection of Bernoulli random variables.
!
7. It is an elementary exercise to show that lP'( R2 :=:: 1) = TC, and that, conditional on this event, the vector (T1, T2) is uniformly distributed on the unit disk. Assume henceforth that R 2 :=:: 1, and write (R, 8) for the point (T1, T2) expressed in polar coordinates. We have that R and e are independent with joint density function fR,e(r, 8) = r In, 0 :=:: r :=:: 1, 0 :=:: () < 2n. Let (Q, W) be the polar
204
Solutions
Coupling and Poisson approximation
coordinates of (X, Y), and note that 111 =
e and e-2I Q2
[4.11.8]-[4.12.1]
= R2 . The random variables
Q and 111 are
I 2
independent, and, by a change of variables, Q has density function fQ(q) = qe-1q , q > 0. We recognize the distribution of (Q, 111) as that of the polar coordinates of (X, Y) where X and Y are independent N(O, 1) variables. [Alternatively, the last step may be achieved by a two-dimensional change of variables.] 8.
We have that
9.
The polar coordinates (R, 8) of (X, Y) have joint density function fR e(r, e)= '
2r
-, T(
Make a change of variables to find that y I X
= tan e has the Cauchy distribution.
10. By the definition of Z, m-1
JP(Z
=
IT
(1- h(r)) r=O JP(X > O)JP(X > 1 I X> 0) .. ·li"(X
= m) = h(m)
= m I X>
m- 1)
= JP(X = m).
11. Suppose g is increasing, so that h(x) = - g(1- x) is increasing also. By the FKG inequality of Problem (3.11.18b), K = cov(g(U),-g(1 - U)) 2::: 0, yielding the result.
Estimating I by the average (2n) - 1 I:~~ 1 g (Ur) of 2n random vectors U r requires a sample of size 2n and yields an estimate having some variance 2na 2 . If we estimate I by the average (2n)- 1 {2:~= 1 g(Ur) + g(1 - Ur) }, we require a sample of size only n, and we obtain an estimate with the smaller variance 2n(a 2 - K). 12. (a) By the law of the unconscious statistician, lE [g(Y)fx(Y)J fy(Y)
=
j
g(y)fx(Y)f ( )d =I. fy(y) y y y
(b) This is immediate from the fact that the variance of a sum of independent variables is the sum of their variances; see Theorem (3.3.11b). (c) This is an application of the strong law of large numbers, Theorem (7.5.1).
13. (a) If U is uniform on [0, 1], then X = sin(~rr U) has the required distribution. This is an example of the inverse transform method. (b) If U is uniform on [0, 1], then 1- U 2 has density function g(x) = {2~} - 1 , 0 ::=: x ::=: 1. Now g(x) 2::: (rr/4)/(x), which fact may be used as a basis for the rejection method.
4.12 Solutions. Coupling and Poisson approximation 1.
Suppose that lE(u(X)) 2::: lE(u(Y)) for all increasing functions u. Let c E ffi. and set u lc(x) = {
1 if X > . 0 If X :S
205
C, C,
= Ic where
[4.12.2]-[4.13.1]
Solutions
Continuous random variables
to find that lP'(X > c) = lE(lc(X)) 2: lE(lc(Y)) = lP'(Y > c). Conversely, suppose that X 2:st Y. We may assume by Theorem (4.12.3) that X and Y are defined on the same sample space, and that lP'(X 2: Y) = 1. Let u be an increasing function. Then lP'(u(X) 2: u(Y)) 2: lP'(X 2: Y) = 1, whence lE(u(X) - u(Y)) 2: 0 whenever this expectation exists. 2. Let a= fl-f/..., and let Ur : r 2: 1} be independent Bernoulli random variables with parameter a. Then Z =I:;= I lr has the Poisson distribution with parameter J...a = fl-, and Z _:::X. 3.
Use the argument in the solution to Problem (2.7.13).
4.
For any A
~
R,
lP'(X
#-
Y) 2: lP'(X E A, Y E Ac) = IP'(X E A) -lP'(X E A, Y E A)
2: IP'(X E A) - IP'(Y E A), and similarly with X and Y interchanged. Hence, IP'(X
#-
Y) 2: sup jiP'(X E A) -IP'(Y E A)j = ~dTV(X, Y). A9R
5. For any positive x andy, we have that (y- x)+ + x follows that
1\
y = y, where x
2)fx(k)- Jy(k)}+ = 2:)Jy(k)- fx(k)}+ = 1k
L
k
fx(k)
1\
y = min{x, y}. It
A
fy(k),
k
and by the definition of dTv(X, Y) that the common value in this display equals ~dTv(X, Y) = 8. Let U be a Bernoulli variable with parameter 1-8, and let V, W, Z be independent integer-valued variables with IP'(V = k) = Ux(k)- fy(k)}+ /8, lP'(W
= k) = {fy(k)
IP'(Z = k) = fx(k)
- fx(k)}+ /8, A
fy(k)/(1 - 8).
Then X'= UZ + (1- U)V andY'= UZ + (1- U)W have the required marginals, and IP'(X' = Y') = lP'(U = 1) = 1 - 8. See also Problem (7.11.16d). 6. Evidently dTv(X, Y) = Jp- qJ, and we may assume without loss of generality that p 2: q. We have from Exercise (4.12.4) that lP'(X = Y) _::: 1 - (p- q). Let U and Z be independent Bernoulli variables with respective parameters 1- p+q andq/(1-p+q). The pair X'= U(Z-1)+ 1, Y' = UZ has the same marginal distributions as the pair X, Y, and IP'(X' = Y') = IP'(U = 1) = 1 - p + q as required. To achieve the minimum, we set X" = 1 - X' and Y" = Y', so that IP'(X" = Y") = 1 -IP'(X' = Y') = P- q.
4.13 Solutions. Geometrical probability 1. The angular coordinates \II and I; of A and B have joint density f(1/J, u) = (2n)- 2 . We make the change of variables from (p,B) r+ (1/J,u) by p = cos{~(u -1/1)}, e = ~(n +u +1/J), with inverse ,,, r -r p, U = 8COS-I p, 'I' = 8 - 27i -cos
in+
and Jacobian Ill= 2/~.
206
Solutions [4.13.2]-[4.13.6]
Geometrical probability
2. Let A be the left shaded region and B the right shaded region in the figure. Writing A for the random line, by Example (4.13.2), IP'(A meets both Stand S2)
= IP'(A meets both A and B) = IP'(A meets A) + IP'(A meets B) (Ai), we find that S::::: lE(X) ::::: 1 + S where 00
s=
00
i=2
4.
i-1
00
00
2:> -1)1P'(Ai) = LL 1·1P'(Ai) = L L i=2j=1
(a) (i) Let p- 1 (y)
00
J!D(Ai) = LJP>(X:::: j).
j=1 i=j+1
= sup{x: F(x) = y}, so that
lP'(F(X)::::: y) = J!D(X::::: p- 1 (y)) = F(F- 1(y)) = y, (ii) JP>( -log F(X) :::S y)
j=1
= JP>(F(X) = PR,
:::: e-Y)
= 1-
0::::: y:::::
1.
e-Y if y :::: 0.
(b) Draw a picture. With D
JP>(D :::=::d)= lP'(tanPQR :::=::d)= J!D(PQR : :=: tan- 1 d)=~
G·
+tan- 1 d).
Differentiate to obtain the result. 5.
Clearly JP>(X > s + x
IX
> s) =
J!D(X > s + x) e-}.(s+x) JP>(X > s) = e-}.s = e-Ax
if x, s :::: 0, where A is the parameter of the distribution. Suppose that the non-negative random variable X has the lack-of-memory property. Then G (x) = JP>(X > x) is monotone and satisfies G(O) = 1 and G(s + x) = G(s)G(x) for s,x :::: 0. Hence G(s) = e-}.s for some A; certainly A> 0 since G(s) : :=: 1 for all s. Let us prove the hint. Suppose that g is monotone with g(O) = 1 and g(s + t) = g(s)g(t) for s, t :::: 0. For an integer m, g(m) = g(l)g(m - 1) = · · · = g(l)m. For rational x = mfn, g(x)n = g(m) = g(l)m so that g(x) = g(l)x; all such powers are interpreted as exp{x logg(l)}. Finally, if x is irrational, and g is monotone non-increasing (say), then g(u) : :=: g(x) : :=: g(v) for all
210
Solutions [4.14.6]-[4.14.9]
Problems
rationals u and v satisfying v ::0 x ::0 u. Hence g(l)" ::0 g(x) ::0 g(l)v. Take the limits as u {. x and v t x through the rationals to obtain g (x) = eJLx where 1-i = log g ( 1).
6. If X and Y are independent, we may take g f(x, y) = g(x)h(y). Then fx(x)
= g(x)
and 1= Hence fx(x)fy(y)
r:
r:
= g(x)h(y) =
h(y) dy,
fy(y)dy
=
=
fx and h
fy(y)
= h(y)
J: f: g(x)dx
=
fy. Suppose conversely that
r:
g(x) dx
h(y)dy.
f(x, y), so that X andY are independent.
7. TheyarenotindependentsincelP'(Y < 1, X> 1) = OwhereaslP'(Y < 1) > OandlP'(X > 1) > 0. As for the marginals,
fx(x)
=1 2e-x-y dy =2e- x, 00
2
fy(y) =loy 2e-x-y dx
=2e-Y(l- e-Y),
for x, y ::::: 0. Finally,
roo roo xy2e-x-y dx dy = 1 and E(X) =1. E(Y) =~.implying that cov(X, Y) =!· E(XY)
=
Jx=O }y=x
8.
As in Example (4.13.1), the desired property holds if and only if the length X of the chord satisfies X ::0 ,J3. Writing R for the distance from P to the centre 0, and E> for the acute angle between the chord sin E>, and therefore J!D(X ::0 ,J3) = JP>(R sinE> ::::: ~ ). and the line OP, we have that X = 2 1 The answer is therefore
V
lP'
R2 2
(R >- _1_) = ~ r!rr lP' (R > _1_) d(} 2sinE> rr Jo - 2sin8 '
which equals ~ - ,J3j(2rr) in case (a) and ~
9.
+ rr- 1 log tan(rr /12) in case (b).
Evidently, E(U)
=JP>(Y ::0 g(X)) =!1O:o;x,y:::;t dx dy =lot0 g(x) dx, y:::;g(x)
=E(g(X)) =lot g(x)dx, E(W) =1lot {g(x) +g(l-x)}dx =lot g(x)dx. E(V)
Secondly,
2)=E(U) =J, E(V2)=lol g(x)2dx ::0 J since g ::0 1, )=! { 2 lot g(x)2dx + 2 lot g(x)g(1- x) dx} E(W 2 E(U
=E(V 2)-1 lot g(x){g(x)- g(l - x)} dx
rt {g(x)- g(l- x)}
= E(V 2 ) - ! Jo
211
2
dx ::0 E(V 2 ).
[4.14.10]-[4.14.11] Hence var(W)
:::=::
Solutions
var(V)
:::=::
Continuous random variables
var(U).
10. Clearly the claim is true for n = 1, since the r(A., 1) distribution is the exponential distribution. Suppose it is true for n :::=:: k where k :=::: 1, and consider the case n = k + 1. Writing fn for the density function of Sn, we have by the convolution formula (4.8.2) that
which is easily seen to be the r(A., k
+ 1) density function.
11. (a) Let Z1, Z2, ... , Zm+n be independent exponential variables with parameter A.. Then, by Problem (4.14.10), X' = Z1 + · · · + Zm is f(A, m), Y' = Zm+1 + · · · + Zm+n is r(A., n), and X'+ Y' is f(A., m + n). The pair (X, Y) has the same joint distribution as the pair (X', Y 1), and therefore X+ Y has the same distribution as X'+ Y', i.e., f(A., m + n). (b) Using the transformation u = x + y, v = xf(x + y), with inverse x = uv, y = u(l - v), and Jacobian
-I
J we find that U
fu
v v 1-
1-
u - -u ' -u
= X+ Y, V = Xf(X + Y) have joint density function Am+n
'
v(u, v)
= fx ' y(uv, u(l- v))lul = r (m)r (n) (uv)m- 1{u(l- v)}n-le-}."u = {
v)n-1}
J..m+n um+n-le-}.u } { vm-1(1r(m+n) B(m,n)
for u :=::: 0, 0 :::=:: v :::=:: 1. Hence U and V are independent, U being f(A., m + n), and V having the beta distribution with parameters m and n. (c) Integrating by parts,
where X' is f(A., m- 1). Hence, by induction, JP>(X > t) =
m-1
L
(A.tY
e-M _ _ = J!D(Z < m).
k!
k=O
(d) This may be achieved by the usual change of variables technique. Alternatively, reflect that, using the notation and result of part (b), the invertible mapping u = x + y, v = xf(x + y) maps a pair X, Y of independent (r(A., m) and f(A., n)) variables to a pair U, V of independent (f(A., m +n) and B(m, n)) variables. Now UV =X, so that (figuratively) 'T(A., m
+ n) x
B(m, n) = r(A., m)".
Replace n by n - m to obtain the required conclusion.
212
Solutions [4.14.12]-[4.14.15]
Problems
12. (a) Z
= xf satisfies z 2:0,
the f(~, ~)or x2 (1) density function. (b) If z 2: 0, Z
= xf +X~ satisfies
x
the 2 (2) distribution function. (c) One way is to work inn-dimensional polar coordinates! Alternatively use induction. It suffices to show that if X and Y are independent, X being x2 (n) and Y being x2 (2) where n 2: 1 , then Z =X+ Y is x2 (n + 2). However, by the convolution formula (4.8.2),
z 2:0, for some constant c. This is the x2 (n
+ 2) density function as required.
13. Concentrate on where x occurs in fxly(x
f XIY (x I Y ) =
I y); any multiplicative constant can be sorted out later:
2 Jx,y(x,y) = c1 () { 1 (x 2xp,l y exp - - - - 2px(y-M))} fy(y) 2(1 - p 2 ) a[ a[ awz
by Example (4.5.9), where q (y) depends on y only. Hence
X E
IR,
for some cz(y). This is the normal density function with mean 11-1 +pal (y- p,z)/az and variance a[O - p 2 ). See also Exercise (4.8.7).
14. Set u = yjx, v = x, with inverse x = v, y = uv, and Ill= lvl. Hence the pair U = Y/X, V = X has joint density fu. v(u, v) = fx.r(v, uv)lvl for -oo < u, v < oo. Therefore fu(u) = f~oo f(v, uv)lvl dv. 15. By the result of Problem (4.14.14), U = Y I X has density function fu(u)
=
L:
f(y)f(uy)IYI dy,
and therefore it suffices to show that U has the Cauchy distribution if and only if Z = tan -l U is uniform on (-~n, ~n). Clearly IP'(Z::; 0) = IP'(U :StanO),
213
-~n <
e<
~n,
[4.14.16]-[4.14.17]
Solutions
Continuous random variables
= fu(tanO)sec 2 e.
whence fz(O)
Therefore fz(O)
= n- 1 (for 101
1
fu(u) = n(l When
+ u2)'
-00
< j;n)ifandonly if
< u < 00.
f is the N(O, 1) density,
1-oooo
f(x)f(xy)lxldx = 2
looo -e-2x 1 (l+y )xdx, 1 2
2
o 2n
which is easily integrated directly to obtain the Cauchy density. In the second case, we have the following integral:
1-oo
a2 1xl
00
In this case, make the substitution
16. The transformation X
(1 +x4)(1 +x4y4) dx.
z = x 2 and expand as partial fractions.
= r COS 0, y = r sin 0 has Jacobian J = r, SO that r > 0,
0:::::
e < 2n.
Therefore Rand 8 are independent, 8 being uniform on [0, 2n), and R 2 having distribution function
this is the exponential distribution with parameter j; (otherwise known as f(i, 1) or density function of R is fR(r) Now, by symmetry,
I 2
= re-2r
x2 (2)).
The
for r > 0.
In the first octant, i.e., in {(x, y) : 0 :::=:: y :::=:: x}, we have min{x, y} = y, max{x, y} = x. The joint density Jx,Y is invariant under rotations, and hence the expectation in question is
8
!o
y -Jx,y(x,y)dxdy=8
O.:::;y.:::;x X
lerr:f41oo 1:1=0
r=O
tane 1 2 2 - 2-re-'Zr drd0=-log2. 7i
7i
17. (i) Using independence, lP'(U
:::=::
u) = 1-lP'(X > u, Y > u) = 1- (1- Fx(u))(l- Fy(u)).
Similarly lP'(V
(ii) (a) By (i), lP'(U (b) Also, Z = X
:::=::
u)
:::=::
v)
= lP'(X :::=:: v, Y :::=:: v) = Fx(v)Fy(v).
= 1- e- 2u for u :=::: 0.
+ j; Y satisfies
lP'(Z > v) =
looo lP'(Y > 2(v- x) )fx(x) dx = lov e- 2(v-x)e-x dx +
= e-2v(ev-
1)
+ e-v = 1- (1- e-v)2 = lP'(V > 214
v).
1
00
e-x dx
Solutions [4.14.18]-[4.14.19]
Problems Therefore lE(V) = lE(X)
~,and var(V) = var(X)
+ ilE(Y) =
+ ! var(Y) =
i by independence.
18. (a) We have that
(b) Clearly, for w > 0, JP>(U ::: u, W > w) = JP>(U ::: u, W > w, X ::: Y)
+ JP>(U ::: u, W >
w, X > Y).
Now JP>(U :S u, W > w, X :S Y) = J!D(X::: u, Y > X+ w) =lou .l..e-}..xe-JL(x+w) dx
= _J.._e-JLW(l- e-(A+JL)U) .l..+tt
and similarly
Hence, for 0 ::: u ::: u
+w <
oo,
an expression which factorizes into the product of a function of u with a function of w. Hence U and W are independent.
19. U =X+ Y, V =X have joint density function fy(u- v)fx(v), 0::: v::: u. Hence fv u(v I u) = !u, v(u, v) = 1 fu(u) (a) We are given that fvJU(V
I u)
u fy(u- v)fx(v) . fy(u- y)fx(y)dy
fo
= u- 1 for 0::: v::: u; then
1 lou fy(u- y)fx(y)dy fy(u- v)fx(v) = u 0 is a function of u alone, implying that fy(u- v)fx(v) = fy(u)fx(O)
=
fy(O)fx(u)
by setting v = 0 by setting v = u.
In particular fy(u) and fx(u) differ only by a multiplicative constant; they are both density functions, implying that this constant is 1, and fx = fy. Substituting this throughout the above display, we find that g(x) = fx(x)ffx(O) satisfies g(O) = 1, g is continuous, and g(u- v)g(v) = g(u) for 0::: v ::: u. From the hint, g(x) = e-}..x for x 2:: 0 for some./.. > 0 (remember that g is integrable). (b) Arguing similarly, we find that
ru
c fy(u- v)fx(v) = ua+.B- 1 va- 1(u- v).B- 1 Jo fy(u- y)fx(y)dy
215
[4.14.20]-[4.14.22]
Solutions
Continuous random variables
for 0::: v::: u and some constant c. Setting fx(v) = x(v)va- 1 , fy(y) = 1J(y)yf3- 1, we obtain IJ(U- v)x (v) = h(u) for 0::: v::: u, and for some function h. Arguing as before, we find that 17 and x are proportional to negative exponential functions, so that X and Y have gamma distributions.
20. We are given that U is uniform on [0, 1], so that 0 ::: X, Y :::
1almost surely. For 0 <
E
<
i.
E = J!D(X + Y
implying that J!D(X (!- E
i- E)lP'(X (X + Y < g)= J!D(X, Y < g1) -J!D(X, Y < g• J!D(X < 81 , Y < g)= s·
21. Evidently
LlP' (Xrr
lP'(X(l) :S Y1· ... , X(n) :S Yn) =
1
:S Yb ... , Xrrn :S Yn, Xrr1 < · · · < Xrrn)
Jf
where the sum is over all permutations n = (n1, nz, ... , l!n) of (1, 2, ... , n). By symmetry, each term in the sum is equal, whence the sum equals n!J!D(X1 :S Y1· · ··, Xn :S Yn. X1 < Xz < · · · < Xn).
The integral form is then immediate. The joint density function is, by its definition, the integrand. 22. (a) In the notation of Problem (4.14.21), the joint density function of X(2), ... , X(n) is gz(yz, · · ·, Yn)
=
1
Y2
g(y1, · · ·, Yn) dy1
-00
= n! L(yz, ... , Yn)F(yz)f(yz)f(Y3) · · · f(Yn) where F is the common distribution function of the Xi. Similarly X (3), ... , X (n) have joint density
and by iteration, X(k)• ... , X(n) have joint density
nl
gk(Yk. · · ·, Yn) = (k _ 1)! L(Yb · · ·, Yn)F(yk)
k-1
f(Yk) · · · f(Yn).
We now integrate over Yn. Yn-b ... , Yk+l in turn, arriving at f x(k) (Yk ) = (k _ 1)!n!(n _ k)! F(yk) k-1{ 1 - F(yk) }n-kf (Yk ) ·
216
Solutions [4.14.23]-[4.14.25]
Problems
(b) It is neater to argue directly. Fix x, and let I r be the indicator function of the event {Xr ::::; x}, and letS= It + h +···+ln. Then Sis distributed as bin(n, F(x)), and lP'(X(k) ::::; x) = lP'(S;:: k) =
t (;)
F(x) 1(1 - F(x))n-l_
l=k
Differentiate to obtain, with F = F(x), fx(k)(x)
=
t (;)
f(x) {tF 1- 1(1- F)n-l- (n -l)F1(1- F)n-l- 1 }
l=k
= k (:) f(x)Fk-1 (1- F)n-k by successive cancellation of the terms in the series. 23. Using the result of Problem (4.14.21), the joint density function is g(y) = n! L(y)T-n for 0::::; Yi ::::; T, 1 ::::; i ::::; n, where y = (Y1, Y2, ... , Yn). 24. (a) We make use of Problems (4.14.22)-(4.14.23). The density function of X(k) is fk(x) k(~)xk- 1 (1 - x)n-k for 0::::; x ::::; 1, so that the density function of nX(k) is
1 kn(n-l)···(n-k+l) k- 1 ( x)n-k - fk(x jn) = x 1- n k! nk n as n
~
~
1 k- -x - - - x 1e (k - 1)!
cxp. The limit is the f(l, k) density function.
(b) For an increasing sequencex(l), X(2)• ... , X(n) in [0, 1], we define the sequence Un = -n 1ogx(n)• Uk = -klog(x(k)/X(k+l)) for 1 ::::; k < n. This mapping has inverse
X(k) = X(k+1)e
-u
kl
k
n = exp { - ~ 1
·-1
u;
}
,
l=k
with Jacobian J = ( -l)ne-ul-u2-···-un jn!. Applying the mapping to the sequence X(l)• X(2)• ... , X(n) we obtain a family U1, U2, ... , Un of random variables with joint density g(u1, u2, ... , un) = e-ul-u2-···-un for u; ;:: 0, 1 ::::; i ::::; n, yielding that the U; are independent and exponentially distributed, with parameter 1. Finally log X (k) = - 'L/t=k i - 1U;. (c) In the notation of part (b), Zk = exp( -Uk) for 1 ::::; k ::::; n, a collection of independent variables. Finally, Uk is exponential with parameter 1, and therefore lP'(Zk ::::; z) = lP'(Uk :=:: -log z) = elogz = z,
O 0. Substitute t = x 1fqy- 11P where X
=
_....:.1X__:._l......{EIXPI}1/p'
(we may as well suppose that lP'(XY = 0) :1= l) to find that IX!P pEIXPI
IYiq
+ qEIYql
IXYI 2: {EIXPI}l/P{EIYql}l/q.
Holder's inequality follows by taking expectations.
218
Solutions [4.14.28]-[4.14.31]
Problems (b) We have, with Z =IX+
Yl.
E(ZP) = E(Z · zP- 1 )::::; E(IXIZp-I) + E(IYizp-I)
::::; {EIXP1} 1/P{JE:(ZP)} 1/q + {EIYPI} 1/P{JE:(ZP)}l/q by Holder's inequality, where p- 1 + q- 1 = 1. Divide by {E(ZP)} 1/q to get the result. 28. Apply the Cauchy-Schwarz inequality to IZI!(b-a) and IZI!(b+a), where 0::::; a::::; b, to obtain {EIZbl} 2 ::::; Elzb-al Elzb+al. Now take logarithms: 2g(b)::::; g(b- a)+ g(b +a) for 0::::; a::::; b. Also g(p) --+ g(O) = 1 asp .J, 0 (by dominated convergence). These two properties of g imply that g is convex on intervals of the form [0, M) if it is finite on this interval. The reference to dominated convergence may be avoided by using HOlder instead of Cauchy-Schwarz. By convexity, g(x)jx is non-decreasing inx, and therefore g(r)/r 2': g(s)js ifO < s::::; r. 29. Assume that X, Y, Z are jointly continuously distributed with joint density function E(X I Y = y, Z = z) =
J
xfxiY z(x I y, z) d x = '
J
x
f.
Then
f(x,y,z)d x. !r,z(y, z)
Hence
I
E{E(X I Y, Z) Y =
y} = =
=
j E(X I Y = y, Z = z)fzly(z I y)dz
rr
X
f(x, y, z) fy,z(y, z) dxdz
11 fr.z(y, z) fy(y) f(x,y,z) ~r x fy(y) dxdz=E(XIY=y).
Alternatively, use the general properties of conditional expectation as laid down in Section 4.6. 30. The first car to arrive in a car park of length x + 1 effectively divides it into two disjoint parks of length Y and x - Y, where Y is the position of the car's leftmost point. Now Y is uniform on [0, x], and the formula follows by conditioning on Y. Laplace transforms are the key to exploring the asymptotic behaviour of m (x) / x as x --+ oo.
31. (a) If the needle were oflength d, the answer would be 2/:rr as before. Think about the new needle as being obtained from a needle of length d by dividing it into two parts, an 'active' part of length L, and a 'passive' part of length d- L, and then counting only intersections involving the active part. The chance of an 'active intersection' is now (2/:rr)(L/d) = 2Lj(:rrd). (b) As in part (a), the angle between the line and the needle is independent of the distance between the line and the needle's centre, each having the same distribution as before. The answer is therefore unchanged. (c) The following argument lacks a little rigour, which may be supplied as a consequence of the statement that S has finite length. ForE > 0, let XI, x2, ... , Xn be points on S, taken in order along S, such that xo and Xn are the endpoints of S, and Ixi+ 1 - Xi I < E for 0 ::::; i < n; lx - y I denotes the Euclidean distance from x to y. Let Ji be the straight line segment joining Xi to xi+ 1, and let /i be the indicator function of {Ji n A :1= 0}. If E is sufficiently small, the total number of intersections between Jo U lt U · · · U ln-1 and S has mean n-1
'L:EUi) i=O
2 n-1 =lxi+l -xil
L
:rrd i=O
219
[4.14.32]-[4.14.34]
Solutions
Continuous random variables
by part (b). In the limit as E .j.. 0, we have that L::i JE(/i) approaches the required mean, while
2
2L(S)
n-I
-:rrd L lxi+l- xil--+ - - . i=O :rrd 32. (i) Fix Cartesian axes within the gut in question. Taking one end of the needle as the origin, the other end is uniformly distributed on the unit sphere of R 3 . With the X-ray plate parallel to the (x, y )-plane, the projected length V of the needle satisfies V ~ v if and only if IZ I ::::; ~. where Z is the (random) z-coordinate of the 'loose' end of the needle. Hence, for 0::::; v ::::; 1, ;-::--;:) ;-::--;:) 4:rr ~ ;-::--;:) lP'(V~v)=lP'(-Yl-v~::;z::;vl-v2)= =vl-v 2 , 4:rr since 4:rr ~is the surface area of that part of the unit sphere satisfying lzl ::::; ~(use Archimedes's theorem of the circumscribing cylinder, or calculus). Therefore V has density function
fv(v) = vj~ forO::::; v::::; 1. (ii) Draw a picture, if you can. The lengths of the projections are determined by the angle 8 between the plane of the cross and the X-ray plate, together with the angle w of rotation of the cross about an axis normal to its arms. Assume that 8 and Ware independent and uniform on [0, ~:rr]. If the axis system has been chosen with enough care, we find that the lengths A, B of the projections of the arms are given by B = Vsin 2 w + cos2 w cos2
A= Vcos2 w+ sin 2 wcos2 e, with inverse
e,
w =tan -~M-A2 --2. 1- B
Some slog is now required to calculate the Jacobian J of this mapping, and the answer will be !A,B(a, b)= 41ll:rr- 2 for 0 1.
33. The order statistics of the Xi have joint density function
on the set I of increasing sequences of positive reals. Define the one-one mapping from I onto (0, oo)n by Yr = (n + 1- r)(xr -Xr-1) for 1 < r ::5 n, YI = nxt, with inverse Xr = L:k=l Yk/(n - k + 1) for r ~ 1. The Jacobian is (n !)- 1 , whence the joint density function of Yt, Y2, ... , Yn is
~Ann! exp(- txi(Y)) =Anexp(- tYk)·
n.
i=l
k=l
34. Recall Problem (4.14.4). First, Zi = F(Xi), 1 ::::; i::::; n, is a sequence of independent variables with the uniform distribution on [0, 1]. Secondly, a variable U has the latter distribution if and only if - log U has the exponential distribution with parameter 1.
220
Solutions [4.14.35]-[4.14.37]
Problems
ltfollowsthatLi = -log F(Xi), 1 ::S i ::S n, isasequenceofindependentvariables with the exponential distribution. The order statistics L(l)• ... , L(n) are in order -log F(X(n)), ... , -log F(X(l)), since the function -log FO is non-increasing. Applying the result of Problem (4.14.33), Et -n log F(X(n)) and Er
= -(n + 1- r){log F(X(n+l-r)
1 < r ::S n,
-log F(X(n+2-r)) },
are independent with the exponential distribution. Therefore exp(- Er ), 1 ::S r ::S n, are independent with the uniform distribution. 35. One may be said to be in state j if the first j - 1 prizes have been rejected and the jth prize has just been viewed. There are two possible decisions at this stage: accept the jth prize if it is the best so far (there is no point in accepting it if it is not), or reject it and continue. The mean return of the first decision equals the probability j In that the j th prize is the best so far, and the mean return of the second is the maximal probability f (j) that one may obtain the best prize having rejected the first j. Thus the maximal mean return V (j) in state j satisfies V(j)
= max{jln, f(j)}.
Now j In increases with j, and f(j) decreases with j (since a possible stategy is to reject the (j + l)th prize also). Therefore there exists J such that j In ::=:: f (j) if and only if j ::S J. This confirms the optimal stategy as having the following form: reject the first J prizes out of hand, and accept the subsequent prize which is the best of those viewed so far. If there is no such prize, we pick the last prize presented. Let n, be the probability of achieving the best prize by following the above strategy. Let Ak be the event that you pick the kth prize, and B the event that the prize picked is the best. Then,
Ln
rr, =
lP'(B
1
Ak)lP'(Ak) =
k=J+l
(k)( -
Ln k=J+l
n
1)
1 -k- ·-1 k
Ln
J =-
n
k=J+l
1
-k- , -1
and we choose the integer J which maximizes this expresion. When n is large, we have the asymptotic relation rr, ~ (J In) log(nl J). The maximum of the function hn(x) = (xln) log(nlx) occurs atx = nle, and we deduce that J ~ nle. [A version of this problem was posed by Cayley in 1875. Our solution is due to Lindley (1961).] 36. The joint density function of (X, Y, Z) is f(x, y, z)
=
1 exp{ -!(r 2 - 2A.x- 2p,y- 2vz 2 (2Jra ) 312
+ ;.2 + J-L 2 + v2 )}
where r 2 = x 2 + y 2 + z 2 . The conditional density of X, Y, Z given R = r is therefore proportional to exp{A.x +fLY+ vz}. Now choosing spherical polar coordinates with axis in the direction (A., J-L, v), we obtain a density function proportional to exp(a cosO) sine, where a = r)A.2 + J-L2 + v2. The constant is chosen in such a way that the total probability is unity. 37. (a) ¢ 1 (x) = -x¢(x), so H1 (x) = x. Differentiate the equation for Hn to obtain Hn+l (x) = xHn(x)- H~(x), and use induction to deduce that Hn is a polynomial of degree n as required. Integrating by parts gives, when m ::S n,
L:
Hm(x)Hn(x)¢(x) dx = (-l)n
=
L: L:
(-l)n-1
Hm(x)¢(n)(x) dx H:n(x)¢(n-1)(x)dx
= ... = (-l)n-m ~oo H~m\x)¢(n-m)(x) dx, -00
221
[4.14.38]-[4.14.39]
Solutions
Continuous random variables
and the claim follows by the fact that H~m) (x) = m!. (b) ~y Taylor's theorem and the first part, tn t)n L -Hn(x) = L --=--cp(n)(x) = ¢(x- t), oo
l/J(x)
oo (
n=O n!
n!
n=O
whence
38. The polynomials of Problem (4.14.37) are orthogonal, and there are unique expansions (subject to mild conditions) of the form u(x) = 2::;~ 0 arHr(x) and v(x) = 2::;~ 0 brHr(x). Without loss of generality, we may assume that E(U) = E(V) = 0, whence, by Problem (4.14.37a), ao = bo = 0. By (4.14.37a) again, 00
00
var(U) = E(u(X) 2 ) = L:a?r!, r=1
L b?r!.
var(V) =
r=1
By considering the coefficient of sm tn, pnnl E(Hm(X)Hn(Y)) = { 0 ·
ifm=n, ifm =/=n,
and so
where we have used the Cauchy-Schwarz inequality at the last stage.
= X(r) - X(r-1) with the convention that X(Q) = 0 and X(n+1) = 1. By Problem (4.14.21) and a change of variables, we may see that Y1, Y2, ... , Yn+1 have the distribution of a point chosen uniformly at random in the simplex of non-negative vectors y = (Y1, Y2, ... , Yn+ 1) with sum 1. [This may also be derived using a Poisson process representation and Theorem (6.12.7).] Consequently, the Yj are identically distributed, and their joint distribution is invariant under permutations
39. (a) Let Yr
of the indices of the Yj. Now I;~~f Yr = 1 and, by taking expectations, (n E(X(r)) = rE(Y1) = r/(n + 1). (b) We have that E(Y 2 ) 1
=
loo [(I: 1
x 2n(1 - x)n- 1 dx
1= E
2
Yr) ] = (n
=
2 (n
+ 1)(n + 2)
+ 1)E(Yl) =
,
+ 1)E(Y[) + n(n + 1)E(Y1 Y2),
r=1
222
1, whence
Solutions [4.14.40]-[4.14.42]
Problems implying that I JE(Y1 Y2) = (n + I)(n + 2), and also
2 r(s + I) JE(X(r)X(s)) = rlE(Y1 ) + r(s- I)JE(Y1 Y2) = -:-(n-+--'-:-I):-:(-n-'-+-:2-:-)
The required covariance follows.
40. (a) By paragraph (4.4.6), X 2 is f(!, !) and Y2 + z 2 is f(!, I). Now use theresultsofExercise (4.7.I4). (b) Since the distribution of x 2I R 2 is independent of the value of R 2 = x 2 + Y2 + z 2 ' it is valid also if the three points are picked independently and uniformly within the sphere.
41. (a) Immediate, because the N(O, I) distribution is symmetric. (b) We have that
L:
2¢(x)(A.x) dx = =
L: L:
{¢(x)(A.x) + ¢(x)[l- (-A.x)J} dx
L:
¢(x)dx +
¢(x)[(A.x)- (-J...x)]dx =I,
because the second integrand is odd. Using the symmetry of¢, the density function of IYI is ¢(x) + ¢(x){ (A.x)- (-J...x)} + ¢(-x) + ¢(-x){ (-A.x)- (A.x)} = 2¢(x). (c) Finally, make the change of variables W = IYI. Z =(X+ A.IYI)/)I + A. 2, with inverse IYI = w, x = zJI + J...2- A.w, and Jacobian )I+ A. 2. Then
Jw,z(w,
z) =)I+ A. 2fx.IYI (zJI
+ A. 2 - A.w, w)
=)I+ A. 2 · ¢(zJI + A. 2 - A.w) · 2¢(w),
W
> 0,
X
E
R.
The result follows by integrating over w and using the fact that
42. The required probability equals
where U1, U2 are N(O, therefore p =
i), V1, V2 are N(O, ~),and U1, U2, V1, V2 are independent. The answer is
JP>(i(Nf + Nj') ::S iCN} + N}))
= 1P'(K1 ::::: ~ K2) K1 = lP' ( K1 + K2 ::::;
I)
4
where the
Ni
are independent N(O, 1)
where the Ki are independent chi-squared 1
1
= lP'(B ::::; 4) = 4
223
x2(2)
[4.14.43]-[4.14.48]
Solutions
Continuous random variables
where we have used the result of Exercise (4.7.14), and B is a beta-distributed random variable with parameters 1, 1. 43. The argument of Problem (4.14.42) leads to the expression
IP'(Uf + u:} + uj :::
Vf +
vf + vf)
= lP'(K 1 :::
1K2)
where the K; are x2 (3)
.J3 3- 4n'
1 1 = JP'(B ::; 4 ) =
where B is beta-distributed with parameters ~. ~· 44. (a) Simply expand thus: JE[(X - p,) 3 ] = JE[X 3 - 3X2p, + 3Xp,2 - p, 3 ] where p, = lE(X).
(b) var(Sn) = na 2 and lE[(Sn - np,) 3 ] = nlE[(X 1 - p,) 3 ] plus terms which equal zero because lE(X1- p,) = 0. (c) If Y is Bernoulli with parameter p, then skw(Y) = (1 - 2p)/ .fii(j, and the claim follows by (b). (d) m1 =A, m2 =A+ A2, m3 = A3 + 3A 2 +A, and the claim follows by (a). (e)SinceAXisf(l, t), wemayaswellassumethatA = 1. ItisimmediatethatlE(Xn) = f(t+n)/f(t), whence t(t + l)(t + 2)- 3t. t(t + 1) + 2t 3 2 skw(X) = t 312 = 0 . 45. We find as above that kur(X) = (m4- 4m3m1 + 6m2mi- 3mj)/a 4 where mk = lE(Xk).
(a) m4 = 3a 4 for the N(O, a 2 ) distribution, whence kur(X) = 3a 4 ja 4 . (b) mr = r !jAr, and the result follows. (c) In this case, m4 = A4 + 6A 3 + n 2 +A, m3 = A3 + 3A2 +A, m2 = A2 +A, and m1 =A. (d) (varSn) 2 = n 2a 4 andlE[(Sn- nm1) 4 ] = nlE[(X1 -m1) 4 ] + 3n(n -l)a 4 . 46. We have as n
--+
oo that -x)n
JP'(X(n) ::; x + logn) = {1- e-(x+Iogn)}n = ( 1- ~
--+ e-e-x.
By the lack-of-memory property,
1 1 lE(X(2)) =- + - -1 , n nwhence, by Lemma (4.3.4),
!oo
oo
{1- e-e -x }dx = lim lE(X(n) -logn) = lim n-+oo
n-+oo
(
-1 + -1- + · · · + 1-logn ) = y. n n- 1
47. By the argument presented in Section 4.11, conditional on acceptance, X has density function fs. You might use this method when fs is itself particularly tiresome or expensive to calculate. If a(x) and b(x) are easy to calculate and are close to fs, much computational effort may be saved. 48. M = max{U1, U2, ... , Uy} satisfies
e1 - 1 JP'(M::; t) = lE(ty) = - - . e -I
224
Solutions [4.14.49]-[4.14.51]
Problems Thus,
lP'(Z ~
z) = lP'(X =
~
Y ~
LzJ + 2) + IP'(X = LzJ + 1,
(e- l)e-LzJ-2 1- e-1
+ (e-
LzJ + 1- z)
LJ 1 elzJ+1-z- 1 . e- 1
= e-z
l)e- z -
49. (a) Y has density function e-Y for y > 0, and X has density function fx(x) = ae-ax for x > 0. Now Y ~ (X - a) 2 if and only if
!
I 2
which is to say that aVfx(X) ~ f(X), where a = a- 1e2" 01 .J2[ii. Recalling the argument of Example (4.11.5), we conclude that, conditional on this event occurring, X has density function f. (b) The number of rejections is geometrically distributed with mean a- 1, so the optimal value of a is I 2
that which minimizes ae- 2" 01 (c) Setting Z = { +X -X
.fii72, that is, a
= 1.
1} !
with probability with probability
conditional on Y >
1(X -
a )2 ,
we obtain a random variable Z with the N (0, 1) distribution.
50. (a) lE(X)
=
fo
2
1
~du = n/4.
In
(b) lE(Y) = -
{2" sin B dB = 2/rr.
n lo
51. You are asked to calculate the mean distance of a randomly chosen pebble from the nearest collection point. Running through the cases, where we suppose the circle has radius a and we write P for the position of the pebble, (i)
lEIOPI =
(ii)
lEIAPI
__!__2 na
lE(IAPI
(iii)
A
IBPI)
=-
fa r 2dr dB
2
o
= 2a.
3
lo!n lo2acos8
2
na
{ 2n
Jo Jo
o
r 2 drdB
32a
= -. 9n
= __.;. [ [!n fasece r 2drdB + [!n f 2acose r 2drde] na
=
lo lo
4a { 16 3rr 3
l!n lo
~ 1 ~ } - 617 v2+ 2 1og(l +v2)
:::::::
2a
3
x 1.13.
(iv) lE(IAPI
A
IBPI 1\ ICPI) = ---; { na
r r 2dr dB+ [!n f 2acose r 2dr dB} lo J!n lo
[!n
lo
where x =a sin(jrr) cosec(~rr- B)
1 = -2a {loj-n -3.J3 cosec 3 (n - +e) 7r
0
= 2a { 16 7r
3
8
3
dB+
.!...!.,J3 + 3.J3 log~} ::::::: 2a 4 16 2 3 225
~!n 8 cos 3 B dB }
!n
x 0.67.
[4.14.52]-[4.14.55]
Solutions
Continuous random variables
52. By Problem (4.14.4 ), the displacement of R relative to P is the sum of two independent Cauchy random variables. By Exercise (4.8.2), this sum has also a Cauchy distribution, and inverting the transformation shows that e is uniformly distributed. ~ occurs if and only if the sum of any two parts exceeds the length of the third part. (a) If the breaks are at X, Y, where 0 < X < Y < 1, then ~occurs if and only if 2Y > 1, and 2(Y - X) < 1 and 2X < 1. These inequalities are satisfied with probability
53. We may assume without loss of generality that R has length 1. Note that
i.
(b) The length X of the shorter piece has density function fx (x) = 2 for 0 ~ x ~ ~. The other pieces are oflength (1- X)Y and (1- X)(l- Y), where Y is uniform on (0, 1). The event~ occurs if and only if2Y < 2XY + 1 and X+ Y- XY > and this has probability
i.
2
(c) The three lengths are X,
lo0 ~ { 2(1 1- X) -
1- 2x } dx = log(4/e). 2(1 - X)
io -X), iO- X), where X is uniform on (0, 1). The event~ occurs
if and only if X < ~(d) This triangle is obtuse if and only if
1
lx 2
1
2 (1- X)
which is to say that X >
.fi -
>
;;:;• v2
1. Hence,
IP'(obtuse
IM =
IP'( .fi- 1 < X < ! ) 2 = 3- 2./i. 1 IP'(X < 2)
54. The shorter piece has density function fx(x) = 2 for 0 ~ x ~ IP'(R
~
r) = IP' ( -X1-X
~
i· Hence,
r ) = -2r -, l+r
with density function fR(r) = 2/(1 - r) 2 for 0 ~ r ~ 1. Therefore, 1 lo1 1 r JE(R) = lo IP'(R > r) dr = - - dr = 2log2- 1, 0 o 1 +r 1 lo1 2r(l - r) JE(R 2 )= lo 2r!P'(R>r)dr= dr=3-4log2, 0 0 1 +r and var(R) = 2 - (2log 2) 2. 55. With an obvious notation,
By a natural re-scaling, we may assume that a = 1. Now, X1 - X2 and Y1 - Y2 have the same triangular density symmetric on (-1, 1), whence (X 1 - X2) 2 and (Y1 - Y2) 2 have distribution
226
Solutions [4.14.56]-[4.14.59]
Problems I
function F(z) = 2../i- z and density function fz(z) = z-2" - 1, for 0:::::: z:::::: 1. Therefore R 2 has the density f given by
f(r) =
1 r (-1 _ 1) (~ _ 1) dz { lo 1 ../i
1 (-../i
1 - 1) ( 1 - 1) dz ~
r-1
if0::S:r::S:1, if 1 ::S: r ::S: 2.
The claim follows since
1
b1-1 - dz = 2 (. sm a ../i~
-1 ~ .-1l) - - sm r
r
56. We use an argument similar to that used for Buffon's needle. Dropping the paper at random amounts to dropping the lattice at random on the paper. The mean number of points of the lattice in a small element of area dAis dA. By the additivity of expectations, the mean number of points on the paper is A. There must therefore exist a position for the paper in which it covers at least rA l points. 57. Consider a small element of surface dS. Positioning the rock at random amounts to shining light at this element from a randomly chosen direction. On averaging over all possible directions, we see that the mean area of the shadow cast by d S is proportional to the area of d S. We now integrate over the surface of the rock, and use the additivity of expectation, to deduce that the area A of the random shadow satisfies E(A) = cS for some constant c which is independent of the shape of the rock. By considering the special case of the sphere, we find c = It follows that at least one orientation of the rock gives a shadow of area at least S.
;! .
;!
58. (a) We have from Problem (4.14.1lb) that Yr = Xr/(X1 +···+X,) is independent of X1 + ···+X,, and therefore of the variables Xr+1· Xr+2· · · ·, Xk+1• X1 + · · · + Xk+1· Therefore Yr is independent of {Yr+s : s ~ 1}, and the claim follows. (b) Lets= x1 + ... + xk+1· The inverse transformation X1 = Z1S, xz = zzs, ... ' Xk = ZkS, xk+1 = s - z1s- zzs- · · · - ZkS has Jacobian s
0
0
s
0 0
Z1 zz =sk.
1=
0
0
0
Zk
-s
-s
-s
1- Z1- · · ·- Zk
The joint density function of X 1 , X 2, ... , X k, S is therefore (with a = I:~!} fJr),
{IT
r=1
)._fJr (zrs)f3r-1e-'Azrs } . )..fJk+! {s(l _ z 1 _ ... _ Zk)}fJk+! -1e-'As(l-zJ-···-zk)
f'(fJr)
f'(fJk+J) k
=
f(A.,
f3, s)
(IT
zf'- 1) (1 - Z1 - · · ·- Zk)fJk+J- 1,
r=1
where
f
is a function of the given variables. The result follows by integrating overs.
59. Let C = (crs) be an orthogonal n x n matrix with Cni = 1/ ,fii for 1 ::S: i ::S: n. Let Yir = 2::~= 1 XisCrs. and note that the vectors Y, = (Y1, Yz, ... , Ynr), 1 ::S: r ::S: n, are multivariate normal. Clearly EYir = 0, and E(YirYjs) = LCrtCsuE(XitXju) = LCrtCsu8tuVij = LCrtCstVij = 8rsVij• t,u t,u t
227
[4.14.60]-[4.14.60]
Solutions
Continuous random variables
where Dtu is the Kronecker delta, since C is orthogonal. It follows that the set of vectors Y r has the same joint distribution as the set of Xr. Since C is orthogonal, Xir = 2::~= 1 Csr Yis, and therefore
n-1 = LYisYjs- YinYjn = LYisYjs· s=1
s
This has the same distribution as Tij because the Yr and the X7 are identically distributed.
60. We sketch this. Let EIPQRI = m(a), and use Crofton's method. A point randomly dropped in S(a + da) lies in S(a) with probability 2 2da ( -a- ) = 1 - +o(da). a+da a
Hence
dm 6m 6mb -=--+--, da a a where mb(a) is the conditional mean of IPQRI given that Pis constrained to lie on the boundary of S(a). Let b(x) be the conditional mean of IPQRI given that Plies a distance x down one vertical edge.
X
T1
R1
T2
p
R2 By conditioning on whether Q and R lie above or beneath P we find, in an obvious notation, that
x)2 (a-x) 2 b(x) = ( ~ mR 1 + -amR2
+ 2x(a-x) a2 mR 1,Rz·
By Exercise (4.13.6) (see also Exercise (4.13.7)), mR 1 ,R2 = icia)(ia) = ~a 2 . In order to find m R 1, we condition on whether Q and R lie in the triangles T1 or T2, and use an obvious notation.
i, ·
iax. Next, arguing as we did in Recalling Example (4.13.6), we have that mT1 = mT2 = that example, 1 }. mT1 ,T2 = 21 · 94{ ax- 41 ax- 41 ax- gax Hence, by conditional expectation,
m R!
=
141
4 · 'I7 · 2 ax
1 143 13 + 41· 4 'I7 · zax + 2 · 9 · gax = TOSax.
We replace x by a - x to find m Rz, whence in total
x _ (::_)213ax b( ) - a 108
+
(a -x) 2 13a(a -x) a 108
+
228
2x(a -x). a 2 _ ~a 2 _ 12ax a2 8 - 108 108
+
12x 2
108 ·
Solutions [4.14.61]-[4.14.63]
Problems
Since the height of P is uniformly distributed on [0, a], we have that
11oa b(x) dx =
mb(a) = -
a
o
11a 2 --. 108
We substitute this into the differential equation to obtain the solution m(a) = -{;ha 2 . Turning to the last part, by making an affine transformation, we may without loss of generality take the parallelogram to be a square. The points form a convex quadrilateral when no point lies inside the triangle formed by the other three, and the required probability is therefore 1 - 4m (a) I a 2 =
1- ~ = ~· 61. Choose four points P, Q, R, S uniformly at random inside C, and let T be the event that their convex hull is a triangle. By considering which of the four points lies in the interior of the convex hull of the other three, we see that IP'(T) = 41P'(S E PQR) = 4EIPQRIIICI. Having chosen P, Q, R, the four points form a triangle if and only if S lies in either the triangle PQR or the shaded region A. Thus, IP'(T) ={I AI+ EIPQRI}IICI, and the claim follows on solving for IP'(T). 62. Since X has zero means and covariance matrix I, we have that E(Z) = fL + E(X)L = fL, and the covariance matrix of Z is E(L'X'XL) = L'IL = V. 63. Let D = (dij) = AB-C. The claim is trivial if D = 0, and so we assume the converse. Choose i, k such that dik i= 0, and write Yi = L,}=t dijXj = S + dikXk· Now IP'(Yi = 0) = E(IP'(xk = - SI dik I S)). For any given S, there is probability at least ~ that Xk i= - S I dik. and the second claim follows. Let Xt, x2, ... , Xm be independent random vectors with the given distribution. If D i= 0, the probability that Dxs = 0 for 1 :::; s :::; m is at most ( ~ )m, which may be made as small as required by choosing m sufficiently large.
229
5
Generating functions and their applications
5.1 Solutions. Generating functions 1.
(a)lflsln
1
~· 1
lxl
< 1,
X
by the calculation in the solution to Exercise (5.3.2). Using the fact that F(x) = 1 - ~.we deduce that H(s, t) = 1/VO- t 2)(1- s2t2). The coefficient of s 2kt 2n is IP'(Lzn=2k)= (n-_!k)(-l)n-k.
(~~)(-1)k 2)Zn =
2k) (2n= ( k n _ k2k) ( 1
IP'(Szk = O)IP'(Szn-2k = 0).
6. We show that all three terms have the same generating function, using various results established for simple symmetric random walk. First, in the usual notation, oo 2m ' 2s2 L 4mlP'(Szm = O)s = 2sP0 (s) = 2 312 . m=O (1- s ) Secondly, by Exercise (5.3.2a, b), m
lE(T
1\
m
2m)= 2mlP'(T > 2m)+ L 2kfo(2k) = 2m1P'(Szm = 0) + L IP'(Szk-2 = 0). k=l k=l
236
Solutions [5.3.7]-[5.3.8]
Random walk Hence, oo 2m s2Po(s) ' L s E(T A 2m)= - -2- +sP0 (s) = m=O 1- s
2s2 2 312 .
(1 - s )
Finally, using the hitting time theorem (3.10.14), (5.3.7), and some algebra at the last stage, oo
L
m=O
oo
s 2m2EIS2m I = 4
L
oo
m
s 2m
m=O
m
L 2k1P'(S2m = 2k) = 4 L
s 2m
m=O
k=1
L 2mf2k(2m) k=1
d ~ 2m~ d F1(s) 2 2s 2 = 4s- L s L hk(2m) = 4s2 = 2 312 . ds m=O k= 1 ds 1 - F 1(s) (1 - s )
7. Let In be the indicator of the event {Sn = 0}, so that Sn+1 = Sn + Xn+1 +ln. In equilibrium, E(So) = E(So) + E(X 1) + E(/o), which implies that IP'(So = 0) = E(/o) = -E(X 1) and entails E(X t) ::::: 0. Furthermore, it is impossible that IP'(So = 0) = 0 since this entails IP'(So =a) = 0 for all a < oo. Hence E(X 1) < 0 if Sis in equilibrium. Next, in equilibrium,
Now, E{ zSn+Xn+l +In (1 - In)} = E(zSn
I Sn
> O)E(zxl )IP'(Sn > 0)
E(zSn+Xn+I+ln In)= zE(zx 1 )IP'(Sn = 0).
Hence E(zsO) = E(zxi) [{E(zso) - IP'(So = 0)} +ziP'( So = 0)] which yields the appropriate choice for E(zsO). 8.
The hitting time theorem (3.10.14), (5.3.7), states that IP'(Tob = n) = (lbl/n)IP'(Sn =b), whence E(Tob
I Tob
< oo)
The walk is transient if and only if p =f:. p =f:.
1· Suppose henceforth that p =f:. 1·
=
b
L IP'(Sn =b).
IP'(Tob < oo) n
1, and therefore E(Tob
I Tob
< oo) < oo if and only if
The required conditional mean may be found by conditioning on the first step, or alternatively
as follows. Assume first that p < q, so that IP'(Tob < oo) = (p/q)b by Corollary (5.3.6). Then L:n IP'(Sn =b) is the mean of the number N of visits of the walk to b. Now
IP'(N=r)=(~rpr- 1 (1-p),
r0::1,
where p = IP'(Sn = 0 for some n 0:: 1) = 1 - IP- ql. Therefore E(N) = (pfq)b liP- ql and E(Tob I Tob < oo)
=
b (p/q)b (p/q)b · IP _ ql.
We have when p > q that IP'(Tob < oo) = 1, and E(TOI) = (p- q)- 1. The result follows from the fact that E(Tob) = bE(To1).
237
[5.4.1]-[5.4.4]
Solutions
Generating functions and their applications
5.4 Solutions. Branching processes 1. Clearly E(Zn I Zm) = Zm!Ln-m since, given Zm, Zn is the sum of the numbers of (n - m)th generation descendants of Zm progenitors. Hence E(ZmZn I Zm) = Z~{Ln-m and E(ZmZn) = E{E(Zm Zn I Zm)} = E(Z~)!Ln-m. Hence cov(Zm, Zn)
= fLn-mE(Z~)- E(Zm)E(Zn) =
f.ln-m var(Zm),
and, by Lemma (5.4.2),
2. Suppose 0 :::0 r :::0 n, and that everything is known about the process up to timer. Conditional on this information, and using a symmetry argument, a randomly chosen individual in the nth generation has probability 1/ Zr of having as rth generation ancestor any given member of the rth generation. The chance that two individuals from the nth generation, chosen randomly and independently of each other, have the same rth generation ancestor is therefore 1/Zr. Therefore lP'(L < r)
= E{lP'(L
< r
I Zr)} = E(1- z; 1)
and so lP'(L
= r) = lP'(L
< r
+ 1) -lP'(L
< r)
= E(Z; 1)- E(z;J 1),
0:::0 r < n.
If 0 < lP'(ZI = 0) < 1, then almost the same argument proves that lP'(L Y/r- Ylr+l forO :::0 r < n, where Y/r = E(Z; 1 I Zn > 0).
3.
=
r
I
Zn > 0)
=
The number Zn of nth generation decendants satisfies
lP'(Zn
= 0) = Gn(O) =
{n:
if p = q,
1
q (p n -q n)
pn+l _ qn+l
if p # q,
whence, for n :::: 1, 1 lP'(T
= n) = lP'(Zn = 0) -lP'(Zn-1 = 0) =
{ n(n
+ 1) p
n-1 n 2 q (p- q)
if p
= q,
if p # q.
It follows that E(T) < oo if and only if p < q.
4.
(a) As usual,
This suggests that Gn(s) = 1 - al+.B+·+.Bn-I (1 - s).Bn for n :::: 1; this formula may be proved easily by induction, using the fact that Gn(s) = G(Gn-1 (s)). (b) As in the above part (a),
238
Solutions
Age-dependent branching processes
[5.4.5]-[5.5.1]
whereP2(s) = P(P(s)). SimilarlyGn(s) = /- 1 (Pn(/(s)))forn:::: l,wherePn(s) = P(Pn-I(s)). (c) With P(s) =as /{1- (1 - a)s} where a = y- 1 , it is an easy exercise to prove, by induction, that Pn(s) = ans/{1- (1- an)s} for n:::: 1, implying that
5.
Let Zn be the number of members of the nth generation. The (n + 1)th generation has size
Cn+l + ln+l where Cn+I is the number of natural offspring of the previous generation, and ln+l is
the number of immigrants. Therefore by the independence,
whence Gn+I(s) 6.
= lE(sZn+l) = lE{G(s)Zn}H(s) = Gn(G(s))H(s).
By Example (5.4.3),
z
lE(s n) = Differentiate and set s
n - (n - 1)s n- 1 1 = -- + , n + 1- ns n n2(1 + n-1 - s)
n:::: 0.
= 0 to find that
Similarly, oo
n
oo
00
n2
00
1
oo
1
1 2
oo
1
lE(V2)
=
2:: (n + 1)3 = n=O 2:: (n + 1)2 - n=O 2:: (n + 1)3 = 0 JT - n=O 2:: (n + 1)3 ' n=O
lE(V3)
=
2:: (n + 1)4 = n=O 2:: n=O
(n + 1) 2 - 2(n + 1) + 1 (n + 1)4
=
1 2 1 4 ()JT + 90JT - 2
00
The conclusion is obtained by eliminating ~n (n + 1) - 3.
5.5 Solutions. Age-dependent branching processes 1.
(i) The usual renewal argument shows as in Theorem (5.5.1) that Gt(s)
=
l
G(Gt-u(s))fr(u) du +
1
00
sfr(u) du.
Differentiate with respect to t, to obtain
a -Gt(s) = G(Go(s))fr(t) +
at
Now Go(s) = s, and
a
lot -a {G(Gt-u(s))} fr(u) du- sfr(t). o
- {G(Gt-u(s))}
at
at
a =--a {G(Gr-u(s))}, u 239
1
2:: -(n-+-1)-.,-3. n=O
[5.5.2]-[5.5.2]
Solutions
Generating functions and their applications
so that, using the fact that fr (u) = Ae-A.u if u 2: 0, {' ~ {G(Gt-u(s))} fr(u)du =- [G(Gt-u(s))Jr(u)] 1 -A {' G(Gt-u(s))fr(u)du, lo at o lo
having integrated by parts. Hence :t G 1 (s) = G(s)Ae-At
+ { -G(s)Ae-A.t + AG(G 1 (s))}- A { G 1 (s)
-1
00
sfr(u) du}- sAe-At
=A {G(Gt(s))- G 1 (s)}.
(ii) Substitute G(s)
= s 2 into the last equation to obtain aGt
ift = A(G 12 -
Gt)
with boundary condition Go(s) = s. Integrate to obtain At+ c(s) = log{l - G( 1} for some function c(s). Using the boundary condition att = 0, we find thatc(s) = log{l- G01} = log{1-s- 1}, and hence G 1 (s) =se-At /{1- s(1- e-At)}. Expand in powers of s to find that Z(t) has the geometric distribution JID(Z(t) = k) = (1- e-At)k- 1e-A.t fork 2: 1.
2.
The equation becomes aGt
ift with boundary condition Go(s)
1 = 2(1
+ G21 ) - Gt
= s. This differential equation is easily solved with the result
4/t G t (s ) -_ 2s + t(l - s) -_ -,-------,-,--2+t(1-s) 2+t(l-s)
2-t
We pick out the coefficient of sn to obtain 4 - ( -tJP>Zt -n - - ( () - ) - t(2 + t) 2 + t
)n ,
n 2: 1,
and therefore
( ()- )-?:;
JP>Zt >k-
00
4 - ( -t-t(2 + t) 2 + t
)n -- -2t ( -2 +t-t )k ,
It follows that, for x > 0 and in the limit as t -+ oo,
JID(Z(t) 2: xt
I Z(t)
> 0)
=
JID(Z(t) > xt)
-
JID(Z(t) 2: 1)
(
t ) LxtJ-1 2+t
= --
240
(
2)
= 1 +t
1-LxtJ
-+ e- 2x.
Solutions
Characteristic functions
[5.6.1]-[5.7.2]
5.6 Solutions. Expectation revisited 1. Set a = lE(X) to find that u(X) ::=: u(lEX) + A.(X -lEX) for some fixed A. Take expectations to obtain the result.
2. Certainly Zn = L:?=l Xi and Z dominated convergence.
3.
Apply Fatou's lemma to the sequence {- Xn : n ::=: 1} to find that lE(limsupXn) n->oo
4.
= L:~ 1 IXi I are such that IZn I :::: Z, and the result follows by
=
-lE(liminf -Xn) ::=: -liminflE(-Xn) n->oo
n->oo
= limsuplE(Xn). n->oo
Suppose that lEIXrl < oo where r > 0. We have that, if x > 0, xrlP'(IXI :::: x) ::::
r
ur dF(u)--+ 0
as X--+
00,
Jrx,oo)
where F is the distribution function of IX 1. Conversely suppose that xrlP'(IXI ::=: x) --+ 0 where r ::=: 0, and let 0 ::S s < r. Now lEIXs I = limM->oo J0M us dF(u) and, by integration by parts,
The first term on the right-hand side is negative. The integrand in the second term satisfies sus - 11P'(I X I > u) :::: sus-l · u-r for all large u. Therefore the integral is bounded uniformly in M, as required.
5. Suppose first that, for all E > 0, there exists 8 = 8(E) > 0, such that lE(IXI/A) < E for all A satisfying IP'(A) < 8. FixE > 0, and find x (> 0) such that IP'(IXI > x) < 8(E). Then, for y > x,
Hence
J!.y lui dFx(u) converges as y--+
oo, whence lEI XI < oo.
ConverselysupposethatlEIXI < oo. ItfollowsthatlE (IXI/{IXI>yJ)--+ Oasy--+ oo. LetE > 0, and find y such thatlE (IXI/{IXI>yJ) < iE. For any event A, lA :::: /Anne+ Is where B = {lXI > y}. Hence lE(IXI/A):::: lE(IXI/Ansc) +lE(IXI/s):::: ylP'(A) + iE. Writing 8 = E/(2y), we have thatlE(IXI/A) < E iflP'(A) < 8.
5.7 Solutions. Characteristic functions 1. Let X have the Cauchy distribution, with characteristic function c/>(s) = e-lsl. Setting Y = X, we have that ¢x+r(t) = ¢(2t) = e- 2 ltl = c/>x(t)c/>y (t). However, X and Yare certainly dependent.
2.
(i) It is the case that Re{¢(t)} = lE(cos t X), so that, in the obvious notation, Re{1 - ¢(2t)}
=
L: L:
:::: 4
{1- cos(2tx)} dF(x) {1- cos(tx)} dF(x)
=2
L:
= 4Re{1
241
{1 - cos(tx)}{1 + cos(tx)} dF(x) - ¢(t)}.
[5.7.3]-[5.7.6]
Solutions
Generating functions and their applications
(ii) Note first that, if X and Y are independent with common characteristic function ¢, then X - Y has characteristic function
1/f(t) = lE(eitX)lE(e-itY) = c/>(t)c/>(-t) = c/>(t)c/>(t) = lc/>(t)l 2 . Apply the result of part (i) to the function 1/1 to obtain that 1 -lc/>(2t)1 2 :;:; 4(1 -lc/>(t)l 2 ). However lc/>(t)l :;:; 1, so that 1 - lc/>(2t)l :;:; 1 - lc/>(2t)l 2 :;:; 4(1 - lc/>(t)l 2 ) :;:; 8(1 - lc/>(t)l).
3.
(a) With mk = lE(Xk), we have that
say, and therefore, for sufficiently small values of(), Kx(e) =
Loo x(T)I = 1 for some T E (0, 2nlb). Then c/>x(T) = eic for some c E JR. Now JE(cos(TX _c))= ~lE(eiTX-ic
+ e-iTX+ic)
= 1,
using the fact that 1E(e-iTX) = cJ>x(T) = e-ic. However cosx :;:; 1 for all x, with equality if and only if x is a multiple of 2rr. It follows that T X -cis a multiple of 2rr, with probability 1, and hence that X takes values in the set L(ciT, 2rr IT). However 2rr IT > b, which contradicts the maximality of the span b. We deduce that no such T exists. (b) This follows by the argument above.
6.
This is a form of the 'Riemann-Lebesgue lemma'. It is a standard result of analysis that, forE > 0, there exists a step function gE such that f~oo If (x) - gE (x) I dx < E. Let ¢E (t) = f~oo eitx gE (x) dx. Then
If we can prove that, for each E, lc/>E(t)l --+ 0 as t--+ ±oo, then it will follow that lc/>x(t)l < 2E for all large t, and the claim then follows.
242
Solutions [5.7.7]-[5.7.9]
Characteristic functions
Now gE(x) is a finite linear combination of functions of the form ciA (x) for reals c and intervals A, that is gE(x) = 'LJ=l q!Ak (x); elementary integration yields
where ak and bk are the endpoints of Ak. Therefore
2 K II/IE (t) I < -
2:: Ck -+ 0,
as t -+ ±oo.
- t k=l
7.
If X is N(f1, 1), then the moment generating function of X 2 is
M 2(s) X
= lE(esX 2 ) =
100
_ 00
1 I (x -JL )2 dx esx 2 --e-2
..(iii
=
1
.J[=2S
112s ) , exp ( 1 - 2s
if s < ~, by completing the square in the exponent. It follows that
My(s) =
ll
n {
1
.Jf=2S
11~s
1 exp ( - 1- ) } = exp ( -se- ) . 1 - 2s (1 - 2s)nf2 1 - 2s
It is tempting to substitute s = it to obtain the answer. This procedure may be justified in this case using the theory of analytic continuation.
8. (a) T 2 = X 2 /(Y jn), where X 2 is x2 (1; 112 ) by Exercise (5.7.7), andY is F(1, n; 112 ). (b) F has the same distribution function as
x2 (n).
Hence T 2 is
(A 2 + B)jm Z= --------'-'--Vjn
where A, B, V are independent, A being N(.fO, 1), B being x2 (m-1), and V being x2 (n). Therefore lE(Z) = =
~ { JE(A2)JE ( ~) + (m- 1)lE ( Bj~~ 1))} 2_ {(1 + 0)___!!__2 + (mm n-
1)___!!__2} = n~m + ~~, nmn-
where we have used the fact (see Exercise (4.10.2)) that the F(r, s) distribution has mean sj(s- 2) if s > 2. 9. ~t X be independent of X with the same distribution. Then 11/11 2 is the characteristic function of X - X and, by the inversion theorem,
-1
100 lt/l(t)l 2 e-ltx. dt = fx_x_(x) = 100 f(y)f(x + y)dy.
2n -oo
-oo
Now set x = 0. We require that the density function of X 243
X be differentiable at 0.
[5.7.10]-[5.8.2]
Solutions
Generating functions and their applications
10. By definition, e-ityt/Jx(y)
=
j_:
eiy(x-t) fx(x)dx.
Now multiply by fy (y ), integrate over y E JR, and change the order of integration with an appeal to Fubini's theorem.
J:
11. (a) We adopt the usual convention that integrals of the form g(y) dF(y) include any atom of the distribution function Fat the upper endpoint v but not at the lower endpoint u. It is a consequence that Fr: is right-continuous, and it is immediate that Fr: increases from 0 to 1. Therefore Fr: is a distribution function. The corresponding moment generating function is
=
Mr:(t)
1
00
etx dFr:(x)
-oo
= -1M(t)
1
00
etx+r:x dF(x)
-oo
=
M(t + r) . M(t)
(b) The required moment generating function is Mx+r(t
+ r)
Mx(t
Mx+y(t)
+ r)My(t + r)
Mx(t)My(t)
the product of the moment generating functions of the individual tilted distributions.
5.8 Solutions. Examples of characteristic functions 1. (i) We have that {j)(t) = E(eitX) = E(e-itX) = tP-x(t). (ii) If X 1 and X 2 are independent random variables with common characteristic function ¢, then t/Jx 1+Xz (t) = ¢x 1 (t)¢x2 (t) = ¢(t) 2 . (iii) Similarly, ¢x 1-x2 (t)
= ¢x1 (t)¢-x2 (t) = ¢(t)¢(t) =
l¢(t)l 2 .
(iv) Let X have characteristic function ¢, and let Z be equal to X with probability otherwise. The characteristic function of Z is given by
1and to -X
where we have used the argument of part (i) above.
1,
1
(v) If X is Bernoulli with parameter then its characteristic function is ¢ (t) = ~ + eit. Suppose Y is a random variable with characteristic function 1/f(t) = l¢(t)l. Then 1/f(t) 2 = ¢(t)¢(-t). Written in terms of random variables this asserts that Y1 + Y2 has the same distribution as X 1 - Xz, where the Y; are independent with characteristic function 1/f, and the X; are independent with characteristic function¢. Now Xj E {0, 1}, so that Xr- Xz E {-1, 0, 1}, and therefore Yj E Write
a
= lP'(Yj =
{-1, 1l-
J). Then
+ Yz = l) = a 2 = lP'(Xr- Xz = 1) = ~. lP'(Yr + Yz = -1) = (1- a) 2 = lP'(Xr- Xz = -1) = lP'(Yr
implying that a 2 = (1 - a) 2 so that a such variable Y exists. 2.
Fort~
=
i, contradicting the fact that a 2 = ~- We deduce that no
0,
Now minimize overt
~
~.
0.
244
Solutions [5.8.3]-[5.8.5]
Examples of characteristic functions 3.
The moment generating function of Z is
Mz(t)
= lE { lE(etXY I Y)} = lE{Mx(tY)} = lE {C.~ tY) m} -lo1 (-A.-)m yn-1(1 _ y)m-n-1 ~· o A.- ty B(n, m - n)
Substitute v = 1/y and integrate by parts to obtain that
1
oo (v _ l)m-n-1
Imn =
(A.v - t)m
1
dv
satisfies 1 (v-l)m-n- 1 ] 00 Imn =[- A.(m- 1) (A.v- t)m-1 1
+
m-n-1 A.(m- 1) Im-1,n =cmnA.I ( , , ) m-1,n
for some c(m, n, A.). We iterate this to obtain
Imn = c'In+1 n = c' {oo dv 1 ' J1 (A.v - t)n+
c'
for somec' depending onm, n, A.. Therefore Mz(t) = c" (A.- t)-n for some c" depending on m, n, A.. However Mz(O) = 1, and hence c" = A.n, giving that Z is r(A., n). Throughout these calculations we have assumed that tis sufficiently small and positive. Alternatively, we could have set t =is and used characteristic functions. See also Problem (4.14.12).
4.
We have that
1E(e 11. x2 ) =
1
00
-oo
. 2 1 - exp ( - (x - ~-t) 2 ) dx ettx -2 J2na2 2a
1
= oo
1 ( ---exp -oo J2na2
=
1
y'1- 2a2it
exp (
t) (
[x- ~-tO- 2a 2it)- 1 . 2a 2 (l - 2a 2 zt)- 1
itf-/,2 ) . dx 1 - 2a 2zt
exp
itf-/,2 ) . 1- 2a 2it
The integral is evaluated by using Cauchy's theorem when integrating around a sector in the complex plane. It is highly suggestive to observe that the integrand differs only by a multiplicative constant from a hypothetical normal density function with (complex) mean ~-t(1 - 2a 2 it)- 1 and (complex) variance a 2 (1- 2a 2 it)- 1. 5.
(a) Use the result of Exercise (5.8.4) with f-t
characteristic function of the
= 0 and a 2
I
1: 4>xz(t) = (1- 2it)-2, the I
x2 ( 1) distribution.
(b) From (a), the sumS has characteristic function t/Js(t) of the x2 (n) distribution. (c) We have that
245
I
= (1- 2it)-z.n, the characteristic function
[5.8.6]-[5.8.6]
Solutions
Now
Generating functions and their applications
E(exp{-it2/X~}) =1oo -00
_l_exp (- t22- x2) dx. ./2ii 2x 2
There are various ways of evaluating this integral. Using the result of Problem (5.12.18c), we find that the answer is e-ltl, whence X 1/ Xz has the Cauchy distribution. (d) We have that E(eitX1X2) = E{E(eitX1X2 I Xz)} = E(if>x 1(tXz)) = =
1
E(e-~t 2 X~)
~exp{-ix 2 (1 +t 2 )}dx = ~, 1 + t2
00
-oo v2rr
1
on observing that the integrand differs from the N (0, (1 + t 2 )- 2) density function only by a multiplicative constant. Now, examination of a standard work of reference, such as Abramowitz and Stegun (1965, Section 9.6.21), reveals that
!o
oo cos(xt) d
0
~ y 1 -r r-
( ) t=Kox,
where Ko(x) is the second kind of modified Bessel function. Hence the required density, by the inversion theorem, is f(x) = Ko(lxl)jn. Note that, for small x, Ko(x) ~ -logx, and for large positive x, Ko(x) ~e-x .Jnx/2. As a matter of interest, note that we may also invert the more general characteristic function if>(t) = (1- it)-a(l + it)-f3. Setting 1- it= -zjx in the integral gives
1 1oo
f(x) = 2rr
e-itx
e-xxa-I 1-x+ixoo e-z dz dt = ----;;;---oo (1- it)a(l + it)f3 2f32ni -x-ixoo (-z)a (1 + z/(2x))f3 1
ex (2x) z;(f3-a)
----'---'---wz;(a-f3), 1 1 (2x) f(a) 2 (I-a-f3) where W is a confluent hypergeometric function. When a = f3 this becomes I
f(x)
=
(xj2)a-z; f(a).J]r Ka-~ (x)
where K is a Bessel function of the second kind. (e) Using (d), we find that the required characteristic function is ¢>x 1x 2 (t)¢>x 3 x 4 (t) = (1 + t 2)- 1. In order to invert this, either use the inversion theorem for the Cauchy distribution to find the required density to be f(x) = ie-lxl for -oo < x < oo, or alternatively express (1 + t 2)- 1 as partial it)- 1 + (1 + it)- 1}, and recall that (1- it)- 1 is the characteristic fractions, (1 + t 2)- 1 = function of an exponential distribution.
iW-
6. The joint characteristic function of X= (X 1, Xz, ... , Xn) satisfies if>x(t) = E(eitX') = E(eiY) where t = (tJ, tz, ... , tn) E JRn andY= tX' = t1X 1 + · · · + tnXn. Now Y is normal with mean and variance n
E(Y) = _2)jE(Xj) = tfL',
n
var(Y) =
L
tjtkcov(Xj, Xk) = tVt',
j,k=I
j=l
where fL is the mean vector of X, and Vis the covariance matrix of X. Therefore if>x(t) = if>y(l) = exp(itp/- itVt') by (5.8.5).
246
Solutions [5.8.7]-[5.9.2]
Inversion and continuity theorems
Let Z = X- JL. It is easy to check that the vector Z has joint characteristic function ¢Jz(t) =
e- !tvt', which we recognize by (5.8.6) as being that of the N (0, V) distribution. I 2
I 2
7. We have that E(Z) = 0, E(Z 2 ) = 1, and E(e 12 ) = E{E(e12 I U, V)} = E(e2;t ) = e2 1 • If X and Y have the bivariate normal distribution with correlation p, then the random variable Z = (UX + VY)jy'u2 + 2pUV + y2 is N(O, 1). 8.
By definition, E(eitX)
loo
oo
= E(cos(tX)) + iE(sin(tX)).
A2 cos(tx )Ae -Ax dx = - 2, 2 A +t
loo
By integrating by parts,
oo .
sm(tx)Ae
-Ax
dx
At = -2 2, A +t
and A2 +iA.t A A2 + t 2 = A - it .
9.
(a) We have that e-lxl =e-x l{x:c:O) +ex l{x 0) =
~
("' ("' {
op lo lo
a 1 {{ = op 4n 2 llJRz = - 12 4n
1~ JR2
~ 4n
{ { exp( -itx'- itVt') dt} dx
1}JRz
exp(-itVt')
dt
(it 1 )(it2 )
1 2Jl' .JjV=Tf = 1 exp(- 2J tVt)dt= ~· 2 4n 2n y 1 - p2
We integrate with respect top to find that, in agreement with Exercise (4.7.5), IP'(X1 > 0, X2 > 0) =
1
1
4 + 2n
1
sin- p.
5.10 Solutions. Two limit theorems 1. (a) Let{ Xi : i 2': 1} be a collection of independent Bernoulli random variables with parameter Then Sn =~I Xi is binomially distributed as bin(n, Hence, by the central limit theorem,
i).
where is the N(O, 1) distribution function.
249
i.
[5.10.2]-[5.10.4]
Solutions
Generating functions and their applications
(b) Let {Xi : i :=:: 1} be a collection of independent Poisson random variables, each with parameter 1. Then Sn = Xi is Poisson with parameter n, and by the central limit theorem
'J:'i
nk (ISn-nl k! = IP' Jn ~ x ) ~ (x)- (-x),
as above.
k:
lk-nl::=:xy'n 2. A superficially plausible argument asserts that, if all babies look the same, then the number X of correct answers inn trials is a random variable with the bin(n, distribution. Then, for large n,
i)
X- ln ) IP' ( -~-2 - > 3 ::::::1- (3)::::::
zJn
rJrm
by the central limit theorem. For the given values of n and X,
X- ~n
--= ~Jn
910- 750
~s
5v'15 - ·
Now we might say that the event {X- ~n > iJn} is sufficiently unlikely that its occurrence casts doubt on the original supposition that babies look the same. A statistician would level a good many objections at drawing such a clear cut decision from such murky data, but this is beyond our scope to elaborate. 3.
Clearly l/Jy(t) = E{E(eitY
I X)}= E{exp(X(eit -1))} = (
~t
1 - (e 1
-
1)
)s
1 -. = (1 2- e1
)s
It follows that
1 I E(Y) = --:-l/Jy(O) = s, l
whence var(Y) = 2s. Therefore the characteristic function of the normalized variable Z = (Y EY)/Jvar(Y) is Now, 1og{ifJy(t/5s)} = -s log(2- eitf,ffs) = s(eitj,ffs- 1) =
+ ~s(eitj,ffs- 1) 2 + o(l)
it/f!- !-t 2 - !-t 2 + o(l),
-it
2 ass~ oo, and the result follows where the o(l) terms are ass~ oo. Hence log{l/Jz(t)} ~ by the continuity theorem (5.9.5). Let P1, P2, ... be an infinite sequence of independent Poisson variables with parameter 1. Then Sn = P1 + P2 + · · · + Pn is Poisson with parameter n. Now Y has the Poisson distribution with parameter X, and so Y is distributed as Sx. Also, X has the same distribution as the sum of s independent exponential variables, implying that X ~ oo as s ~ oo, with probability 1. This suggests by the central limit theorem that Sx (and hence Y also) is approximately normal in the limit as s ~ oo. We have neglected the facts that s and X are not generally integer valued.
4. Since X1 is non-arithmetic, there exist integers nJ, n2, ... , nk with greatest common divisor 1 and such that IP'(X 1 = ni) > 0 for 1 ~ i ~ k. There exists N such that, for all n :=:: N, there exist nonnegative integers a1, a2, ... , ak such that n = a1n1 + · · ·+aknk. If xis a non-negative integer, write
250
Solutions
Two limit theorems
[5.10.5]-[5.10.5]
N = f31 n 1 + · · ·+ f3knk> N + x = Y1 n 1 + · · ·+ Yknk for non-negative integers f31, ... , f3b Y1 ... , Yk· Now Sn = X1 + · · · + Xn is such that k
IP'(Sn = N) 2: IP'(Xj = n; for B;-1 < j:::: B;, 1 ::::
i:::: k)
=
IJ IP'(X1 = n;)f3i > 0 i=1
where Bo = 0, B; = f31 + f32 + G = Y1 + Y2 + · · · + Yk· Therefore
·· · + {3;,
B = Bk. Similarly IP'(Sa = N
IP'(Sa- SG,B+G = x) 2: IP'(Sa = N
+ x)IP'(Sn =
+ x)
> 0 where
N) > 0
where SG,B+G = ~f=+J'+ 1 X;. Also, IP'(Sn- Sn,B+G = -x) > 0 as required.
5.
Let X 1, X 2, ... be independent integer-valued random variables with mean 0, variance 1, span 1, I 2
and common characteristic function cf>. We are required to prove that JnlP'(Un = x)--+ e -z-x as n --+ oo where 1 X1 + X2 + · · · + Xn Un = JnSn = Jn
;...fiJi
and xis any number of the form kf Jn for integral k. The case of generalt-t and a 2 is easily derived from this. By the result of Exercise (5.9.4), for any such x,
IP'(Un
= x) =
1
r.;;
2n v n
Jrr -fii e-ztxcf>Un . (t) dt, -rr -fii
since Un is arithmetic. Arguing as in the proof of the local limit theorem (6),
2n IJnlP'(Un = x)- f(x)l :S In+ In where
f
is the N(O, 1) density function, and
Now In = 2...fiii ( 1 - (n Jn)) --+ 0 as n --+ oo, where is the N (0, 1) distribution function. As for In, pick 8 E (0, n). Then
I 2
The final term involving e -z-t is dealt with as was ln. By Exercise (5.7.5a), there exists). such that lc/>(t)l o{g(t)} where g(t) =e-at M(t). Now g has a minimum when e 1 = .J(1 + a)/(1 -a), where it takes the value 1/V(l + a)l+a(l- a)l-a as required. If a 2: 1, then IP'(Sn >an)= 0 for all n.
2. (i) Let Yn have the binomial distribution with parameters n and i. Then 2Yn - n has the same distribution as the random variable Sn in Exercise (5.11.1). Therefore, if 0 < a < 1,
and similarly for IP'(Yn - in < -ian), by symmetry. Hence
(ii) This time let Sn = X 1+ · · · + Xn, the sum of independent Poisson variables with parameter 1. Then Tn = en!P'(Sn > n(l +a)). The moment generating function of X1 - l is M(t) = exp(e 1 - 1 - t), and the large deviation theorem gives that r,f!n ~ einf1 >o{g(t)} where g(t) =e-at M(t). Now g' (t) = (e 1 -a - l) exp(e 1 -at - t - 1) whence g has a minimum at t = log(a + l). Therefore r,!ln ~ eg(log(1 +a)) = {ej(a + 1)}a+l. 3. SupposethatM(t) = E(e 1x) isfiniteontheinterval [-8, 8]. Now, fora> 0, M(8) 2: e 8a!P'(X > a), sothat!P'(X >a)_:::: M(8)e- 8a. Similarly, IP'(X 3.
t=l
Applying the first argumentto the mean, we find that f1, = E(X) satisfies f1, = q(1 + f-1,) + pq(2+ ~-t) + p 2q(3 + ~-t) + 3p3 and hence 1-t = (1 + p + p2)jp3. As for HTH, consider the event that HTH does not occur in n tosses, and in addition the next three tosses give HTH. The number Y until the first occurrence of HTH satisfies lP'(Y > n)p 2q = lP'(Y = n + 1)pq + lP'(Y = n + 3),
n
:=:::
2.
Sum over n to obtain E(Y) = (pq + l)j(p2q). (b) G N(s) = (q + ps)n, in the obvious notation. (i) 1!"(2 divides N) = ~{GN(1) + GN(-1)}, since only the coefficients of the even powers of s contribute to this probability. (ii) Let w be a complex cube root of unity. Then the coefficient of lP'(X = k) in ~ {G N (1) + G N(w) + GN(w 2 )} is
~{1
+ w 3 + w6 } = ~{1 +w +w 2 } =
~{1+w 2
1,
if k = 3r,
0,
ifk = 3r + 1,
+w4 }=0,
ifk=3r+2,
254
Solutions
Problems
~{GN(l)
for integers r. Hence + GN(w) + GN(w2 )} = is a multiple of 3. Generalize this conclusion.
2::;!~J JP(N =
[5.12.3]-[5.12.6]
3r), the probability that N
3. We have that T = k if no run of n heads appears in the first k - n - 1 throws, then there is a tail, and then a run of n heads. Therefore IP(T = k) = JP(T > k- n- 1)qpn fork :=::: n + 1 where p + q = 1. Finally JP(T = n) = pn. Multiply by sk and sum to obtain a formula for the probability generating function G of T: 00
00
~
G(s)-pnsn=qpn L sk L JP(T=j)=qpnLIP(T=j) L sk k=n+1 j>k-n-1 j=1 k=n+1
=
qpnsn+1 oo . qpnsn+1 1-s LIP(T=j)(l-sl)= 1-s (1-G(s)). j=1
Therefore
4.
The required generating function is G(s) =
~ sk L...
(k- l)
k=r
r- 1
pr (1- p)k-r =
(___!!!_)r l - qs
where p+q = 1. The mean is G'(l) = rjp and the variance is G"(l) +G'(l)- {G'(l)} 2 = rqjp 2 . 5.
It is standard (5.3.3) that Po(2n) =
Po(2n) ~
e::) (pq)n. Using Stirling's formula,
( 2n)2n+~ e-2n .J27f 1
{nn+2e-n.J2]r}2
(pq)n
=
(4 pq)n
;:;;-;; .
v7rn
The generating function Fo(s) for the first return time is given by Fo(s) = 1 - Po(s)- 1 where Po(s) = l:n s 2n Po(2n). Therefore the probability of ultimate return is Fo(l) = 1 -A - 1 where, by Abel's theorem, .f 1 1 p=q=2> A= LPo(2n){ = 00 if p =J:. q. n 1, and
ldl
-p- Jp2 +4pq 2p
=
pe3o
+ qeoo = pef0 + q.
The
d= -p+Jp2+4pq. 2p
:=:: 1 if and only if p 2 + 4pq :=:: 9p 2 which is to say that p _::: ~-It follows that
e10 = 1 if p _::: ~.so that eao = 1 if p _::: ~-
When p > ~.we have that d < 1, and it is actually the case that e10
eao
=
(
-p + Jp2 +4pq)a 2p
= d, and hence
'f p > 3• 1
1
In order to prove this, it suffices to prove that eao < 1 for all large a; this is a minor but necessary chore. Write Tn = Sn - So = I:i= 1 Xi, where Xi is the value of the ith step. Then eao
= lP'(Tn
_::: -a for some n :=:: 1)
= lP'(nJL-
Tn :=:: nJL +a for some n :=:: 1)
00
.::0
L lP'(nJL- Tn 2':: nJL +a) n=1
where Jl = E(X1) = 2p- q > 0. As in the theory of large deviations, fort> 0,
+
where X is a typical step. Now E(e 1 (~t-X)) = 1 + o(t) as t 0, and therefore we may pick t > 0 such that e(t) = e- 11LE(e 1(tL-X)) < 1. It follows that eao .::0 I;~ 1 e-tae(t)n which is less than 1 for all large a, as required. 8.
We have that
where p + q = 1. Hence Gx,y(s, t) = G(ps + qt) where G is the probability generating function of X + Y. Now X and Y are independent, so that G(ps
+ qt) = Gx(s)Gy(t) = Gx,y(s, 1)Gx,Y(1, t) = G(ps + q)G(p + qt). 256
Solutions [5.12.9]-[5.12.11]
Problems
Write f(u) = G(l + u), x = s- 1, y = t - 1, to obtain f(px + qy) = f(px)f(qy), a functional equation valid at least when -2 < x, y ::; 0. Now f is continuous within its disc of convergence, and also f(O) = 1; the usual argument (see Problem (4.14.5)) implies that f(x) = eA.x for some A, and therefore G(s) = f(s- 1) = eA(s- 1). Therefore X+ Y has the Poisson distribution with parameter A. Furthermore, Gx(s) = G(ps + q) = e)cp(s-1), whence X has the Poisson distribution with parameter Ap. Similarly Y has the Poisson distribution with parameter Aq. In the usual notation, Gn+1(s) = Gn(G(s)). It follows that G~+ 1 (1) = G~(l)G 1 (1) 2 G~(l)G"(l) so that, after some work, var(Zn+1) = ~-t 2 var(Zn) + ~-tno- 2 . Iterate to obtain
9.
var(Zn+1) = o-2(~-tn
+ fl,n+1 + ... + f1,2n)
for the case f1, =F 1. If f1, = 1, then var(Zn+d = o- 2 (n
(52fl,n(l _
=
1
+
fl,n+1)
-fl,
,
n ::=::
0,
+ 1).
10. (a) Since the coin is unbiased, we may assume that each player, having won a round, continues to back the same face (heads or tails) until losing. The duration D of the game equals kif and only if k is the first time at which there has been either a run of r - 1 heads or a run of r - 1 tails; the probability of this may be evaluated in a routine way. Alternatively, argue as follows. We record S (for 'same') each time a coin shows the same face as its predecessor, and we record C (for 'change') otherwise; start with a C. It is easy to see that each symbol in the resulting sequence is independent of earlier symbols and is equally likely to be S or C. Now D = k if and only if the first run of r - 2 S's is completed at time k. It is immediate from the result of Problem (5.12.3) that
( .!st-2 (1 - .!s)
GD(s)= 2 1- s
2
+ (~sy- 1
.
(b) The probability that Ak wins is 00
1rk = ~)P'(D = n(r- 1)
+ k- 1).
n=1
Let w be a complex (r - 1)th root of unity, and set
1 { G D(s) r-1
Wk(s) = - -
1 1 2 ) G D(w s) + kTG D(ws) + ~(k ww -1 1 r-2 } + ... + w(r-2)(k-1) G D (w s)
.
It may be seen (as for Problem (5.12.2)) that the coefficient of si in Wk(s) is lP'(D = i) if i is of the form n(r- 1) + (k- 1) for some n, and is 0 otherwise. Therefore lP'(Ak wins)= Wk(1). (c) The pool contains £ D when it is won. The required mean is therefore JE(D I Ak wins)=
E(DI
·
{Ak wms}
)
lP'(Ak wins)
W 1 (1)
= _k_. Wk(1)
(d) Using the result of Exercise (5.1.2), the generating function of the sequence lP'(D > k), k T(s) = (1- G D(s))/(1 - s). The required probability is the coefficient of sn in T(s).
::=::
0, is
11. Let Tn be the total number of people in the first n generations. By considering the size Z 1 of the first generation, we see that ZJ
Tn = 1 + LTn-1(i) i=1
257
[5.12.12]-[5.12.14]
Solutions
Generating functions and their applications
where Tn-1 (1), Tn-1 (2), ... are independent random variables, each being distributed as Tn-1· Using the compounding formula (5.1.25), Hn(s) = sG(Hn-1 (s)).
12. We have that lP'(Z > N n
IZ m
= O) = lP'(Zn > N, Zm = 0) lP'(Zm = 0) =
f
lP'(Zm = 0
I Zn
=
f
+ r)lP'(Zn
lP'(Zm-n = O)N+rlP'(Zn = N
= N
+ r)
+ r)
lP'(Zm = 0)
r= 1
:S
= N
lP'(Zm = 0)
r= 1
lP'(Zm = O)N+ 1 lP'(Zm = O)
00
L lP'(Zn = N + r) :S lP'(Zm = 0) N = Gm(O) N . r=1
13. (a) We have that Gw(s) = GN(G(s)) = eJ.(G(s)- 1). Also, Gw(s) 11n = eJ.((G(s)- 1)/n, the same probability generating function as Gw but with A replaced by Ajn. (b) We can suppose that H (0) < 1, since if H (0) = 1 then H (s) = 1 for all s, and we may take A = 0 and G(s) = l. We may suppose also that H(O) > 0. To see this, suppose instead that H(O) = 0 so that H(s) = sr "L.'f':=o sj hj+r for some sequence (hk) and some r 2: 1 such that hr > 0. Find a positive integer n such that r fn is non-integral; then H(s) 1/n is not a power series, which contradicts the assumption that H is infinitely divisible. Thus we take 0 < H(O) < 1, and so 0 < 1 - H(s) < 1 for 0::: s < l. Therefore log H(s) =log( 1 - {1 - H(s)l) =A ( -1
+ A(s))
where A = -log H(O) and A(s) is a power series with A(O) = 0, A(1) 00 . "'£j= 1 ajsl, we have that
l. Writing A(s)
as n --+ oo. Now H(s) 11n is a probability generating function, so that each such expression is nonnegative. Therefore aj 2: 0 for all j, implying that A(s) is a probability generating function, as required.
14. It is clear from the definition of infinite divisibility that a distribution has this property if and only if, for each n, there exists a characteristic function 1/Jn such that cf>(t) = 1/Jn (t)n for all t. (a) The characteristic functions in question are "t
cf> (t) = e' Poisson (A) : r(A, ~-t):
4> (t) =
JL- 'I1 a 2t2 "t
eJ.(e' -1)
cf>(t) =(-A. A -It
)JL.
In these respective cases, the 'nth root' 1/Jn of 4> is the characteristic function of the N(~-t/n, a 2 jn), Poisson (A/n), and r(A, ~-t/n) distributions.
258
Solutions [5.12.15]-[5.12.16]
Problems
(b) Suppose that ifJ is the characteristic function of an infinitely divisible distribution, and let 1/Jn be a characteristic function such thatifJ(t) = 1/Jn(t)n. Now lifJ(t)l ::0 l for all t, so that 11/ln(t)l = lifJ(t)l1/n--+ { l
~ lifJ(t)l
0
lf lifJ(t)l
=I= 0,
= 0.
For any value oft such that ifJ(t) =I= 0, it is the case that 1/Jn (t) --+ 1 as n --+ oo. To see this, suppose < 2n such that 1/Jn (t) --+ eie along some subsequence. instead that there exists satisfying 0 < Then 1/Jn (t)n does not converge along this subsequence, a contradiction. It follows that
e
e
1/J(t)
=
lim
1/ln (t) = { 1 ~f ifJ (t) 0
n---+oo
=I= 0,
tf ifJ(t) = 0.
Now ¢J is a characteristic function, so that ifJ(t) =I= 0 on some neighbourhood of the origin. Hence 1/f(t) = 1 on some neighbourhood of the origin, so that 1/J is continuous at the origin. Applying the continuity theorem (5.9.5), we deduce that 1/J is itself a characteristic function. In particular, 1/J is continuous, and hence 1/f(t) = 1 for all t, by ( *). We deduce that ¢J (t) =I= 0 for all t.
15. We have that lP'(N = n I S = N) =
ll"(S = n IN= n)lP'(N = n)
l:k lP'(S = k I N = k)lP'(N = k)
pnlP'(N = n)
=
2::~1 pkJP'(N = k)
.
Hence lE(xN I S = N) = G(px)jG(p). If N is Poisson with parameter A, then e)c(px-1)
lE(xN I S = N) = eA(p-1) = e)cp(x-1) = G(x)P. Conversely, suppose that JE(xN I S = N) = G(x)P. Then G(px) = G(p)G(x)P, valid for lxl ::0 1, 0 < p < 1. Therefore f(x) = logG(x) satisfies f(px) = f(p) + pf(x), and in addition f has a power series expansion which is convergent at least for 0 < x ::0 1. Substituting this expansion into the above functional equation for f, and equating coefficients of pi xj, we obtain that f(x) = -A(1- x) for some A :=:: 0. It follows that N has a Poisson distribution.
16. Certainly 1 - (Pl + P2)) n Gx(s) = Gx,Y(s, 1) = ( 1 , - P2- PIS
Gy(t) = Gx,Y(l, t) =
1- (Pl + P2) Gx+Y(s) = Gx,Y(s, s) = ( 1 _ (Pl + p 2 )s
)n
+ P2)) n , 1 - P1 - P2t
( 1 - (P1
,
giving that X, Y, and X+ Y have distributions similar to the negative binomial distribution. More specifically,
ll"(X = k) =
C+~- )ci(l1
ll"(X
+y
fork 2':: 0, where a= pl/(1- P2), ,8
a)n,
= k) = ( n
= P2!(1
lP'(Y = k) = (n
+~ -
- Pl), Y
259
+~ -
1) yk (1 - y )n,
= Pl + P2·
1),ak(1 _ ,B)n,
[5.12.17]-[5.12.19]
Solutions
Generating functions and their applications
Now JE(sx I y = y) =
JE(sx I
{Y=yJ
)
JP'(Y = y)
A
B
where A is the coefficient of tYinG x,Y(s, t) and B is the coefficient of tY in Gy(t)o Therefore JE(sx
Iy
P2)n (____!!l:_)y/ {( 1- Pl- P2)n (____!!l:_)y}
= y) = ( 1- Pl1-pls
=(
1-pls
1 - Pl ) n+y
1-pl
0
1- PIS 17. As in the previous solution,
18. (a) Substitute u = y I a to obtain
(b) Differentiating through the integral sign,
-ai = !ooo {-2b - exp( -a2 u 2 ab
0
=-
by the substitution u
= b/ y
u2
b 2 u -2 ) } du
!ooo 2exp(-a 2b2y- 2- y 2)dy =
-2I(l,ab),
0
(c) Hence a I jab= -2ai, whence I= c(a)e- 2ab where c(a)=I(a,O)=
loo
oo -a2u2
(d) We have that
by the substitution x = y 2 (e) Similarly
0
by substituting x = y-20
19. (a) We have that
260
e
..jii
du=-0 2a
1-pl
Solutions [5.12.20]-[5.12.22]
Problems
in the notation of Problem (5.12.18). Hence U has the Cauchy distribution. (b) Similarly
fort> 0. Using the result of Problem (5.12.18e), V has density function f(x) = _1_e-lf(2x),
x > 0.
J2nx3 (c) We have that
w- 2 =
x- 2 + y- 2 + z- 2 . Therefore, using (b),
fort > 0. It follows that w- 2 has the same distribution as 9V = 9X- 2 , and so W 2 has the same distribution as ~ X 2 . Therefore, using the fact that both X and W are symmetric random variables, W has the same distribution as
1X, that is N (0, ~).
20. It follows from the inversion theorem that F(x +h)- F(x)
1
h
2n
------=-
1N
1"
1m
-N
N-+oo
1 - e-ith -itx,~.( ) d e '~" t t.
it
Since 14>1 is integrable, we may use the dominated convergence theorem to take the limit as h ..j, 0 within the integral: f(x)
=
lim
___!__
2n
N-+oo
1N
e-itxrf>(t)dt.
-N
The condition that 4> be absolutely integrable is stronger than necessary; note that the characteristic function of the exponential distribution fails this condition, in reflection of the fact that its density function has a discontinuity at the origin.
21. Let Gn denote the probability generating function of Zn. The (conditional) characteristic function of Zn/fl.-n is lE(eitZn/JLn
IZ
>
0) =
Gn(eit/JLn)- Gn(O).
1 - Gn(O)
n
It is a standard exercise (or see Example (5.4.3)) that
whence by an elementary calculation as n -+ oo, the characteristic function of the exponential distribution with parameter 1 - p., -l . 22. The imaginary part of rf>x(t) satisfies
261
[5.12.23]-[5.12.24]
Solutions
Generating functions and their applications
for all t, if and only if X and -X have the same characteristic function, or equivalently the same distribution.
23. (a) U = X + Y and V = X - Y are independent, so that r/Ju + v = r/Jur/Jv, which is to say that ¢2X = ¢X+YrfJX-Y• or
Write 1/f(t) = rjJ(t)!¢(-t). Then 1fr(2t) =
¢(2t) = ¢(t)3¢(-t) = 1/f(t)2 ¢(-2t) ¢(-t)3¢(t)
Therefore 1/f(t) = 1fr(1t)2 = 1fr(!t)4 = ... = 1/f(t/2n)2n
for n:::: 0.
However, as h -+ 0,
so that 1/f(t) = { 1 + o(t 2 j22n)} zn -+ 1 as n -+ oo, whence 1/f(t) ¢(-t) = rjJ(t). It follows that rp(t) = ¢(1t)3¢(-1t) = rfJ(it)4 = rp(t/2nln =
{1-! ..£_ 2
22n
+o(t2/22n)}z2n-+ e-1t2
1 for all t, giving that
for n :::: 1 as n-+ oo,
so that X andY are N(O, 1). (b) With U =X+ Y and V =X- Y, we have that 1/f(s, t) = E(eisU+itV) satisfies 1/f(s, t) = E(ei(s+t)X+i(s-t)Y) = rp(s + t)rjJ(s- t). Using what is given,
However, by (*),
a2t/ at
=2{¢"(s)¢(s)-¢'(s)2},
t=o
yielding the required differential equation, which may be written as d ' /¢) -(¢ ds
= -1. 1 2
Hence log¢(s) =a+ bs -1s 2 for constants a, b, whence ¢(s) = e-:zs . 24. (a) Using characteristic functions, rpz(t) = r/Jx(t/n)n = e-ltl. (b) EIX;I = oo.
262
Solutions [5.12.25]-[5.12.27]
Problems
25. (a) See the solution to Problem (5.12.24). (b) This is much longer. Having established the hint, the rest follows thus: fX+Y(Y)
= =
L:
f(x)f(y -x)dx
1
2
100 {J(x) +
JT(4 + y ) -oo
f(y- x)} dx + J g(y)
=
2
2
JT(4 + y )
+ J g(y)
where 1= L:{xf(x)+(y-x)f(y-x)}dx
=
lim
M,N---+oo
[__!_{log(l+x 2)-log(1+(y-x) 2)}]N 2JT
-M
=0.
Finally, fz(z) = 2fx+Y(2z) =
1 JT(1
+ z2 )
.
26. (a)X1 +X2+···+Xn. (b) X 1 - X!, where X 1 and X! are independent and identically distributed. (c) XN, where N is a random variable with W'(N = j) = P) for 1 :::: j :::: n, independent of X1, X2, ... , Xn. (d) ~~ 1 ZJ where Z1, Z2, ... are independent and distributed as X1, and M is independent of the Zj with W'(M = m) = n) = - - - L.J L.J 1-s
n=l
n=O
270
Solutions
Problems
[5.12.48]-[5.12.52]
where E(s Wr) = Gr(s) is given in Problem (5.12.46).
48. We have that Xn+l =
{ iXn I
zXn
with probability
i,
with probability
i.
+ Yn
Hence the characteristic functions satisfy '
¢n+l (t) = E(e
itX I
I I I I A n+l) = z¢n(zt) + z¢n(zt)--. A -zt ,
A-
I . zlt
I
,
A-
I ·
4!(
= ¢n(zt)-,-.- = c/Jn-l(4t)~t = · · · = ¢1(t2 A-lf
-n
A-!
,
A-
)
A
·
lf
2-n .t
-!
,
A
-+ -,-------;-t A-!
as n -+ oo. The limiting distribution is exponential with parameter A. 49. We have that
(a) (1-e-A.)/A, (b) -(pjq 2 )(q+1og p), (c) (1-qn+l )/[(n+ 1) p ], (d) -[l+(p/q) log p]jlog p. (e) Not if lP'(X + 1 > 0) = 1, by Jensen's inequality (see Exercise (5.6.1)) and the strict concavity of the function f(x) = 1jx. If X+ 1 is permitted to be negative, consider the case when lP'(X + 1 = -1) =lP'(X+ 1 = 1) =
i·
50. By compounding, as in Theorem (5.1.25), the sum has characteristic function GN(¢x(t)) =
P¢x(t) - _AP_ 1- q¢x(t) Ap- it'
whence the sum is exponentially distributed with parameter Ap.
51. Consider the function G(x) = {E(X 2 )}- 1 J~oo y 2 dF(y). This function is right-continuous and increases from 0 to 1, and is therefore a distribution function. Its characteristic function is
52. By integration, fx(x) = fy(y) = are not independent. Now, fx+Y(Z)=
1 1
i, lxl <
1,
lyl < 1. Since
f(x,z-x)dx=
{
,\(z + 2) 1
4 (2-z)
-1
f(x,
y) i=
if - 2 <
fx(x)fy(y), X andY
z<
0,
if0 0, if Br = 0,
Yr
where Yr is a geometrically distributed random variable with parameter Bo, B2, · · · , Br. Hence B is Markovian with Pij=
1
{
(-65)j-1_61
(i) If Xn = i, then Xn+! E {i- 1, i
3.
(*) lP'(Xn+! = i
+ 1 I Xn
if j = i - 1 ~ 0, .f. 0 . 1 ll=,j~.
+ 1}. Now, fori
~
1,
= i, B) = lP'(Xn+! = i + 1 I Sn = i, B)lP'(Sn = i I Xn = i, B) + lP'(Xn+! = i + 1 I Sn = -i, B)lP'(Sn = -i I Xn = i, B)
where B = {Xr = ir for 0 lP'(Xn+! = i
! ,independent of the sequence
~
r < n} and io, i1, ... , in-! are integers. Clearly
+ 1 I Sn =
i, B)= p,
lP'(Xn+! = i
+ 1 I Sn
= -i, B)= q,
where p (= 1 - q) is the chance of a rightward step. Let l be the time of the last visit to 0 prior to the time n, l = max{r : ir = 0}. During the time-interval (1, n], the path lies entirely in either the positive integers or the negative integers. If the former, it is required to follow the route prescribed by the event B n {Sn = i }, and if the latter by the event B n {Sn = -i}. The absolute probabilities of these two routes are
whence lP'(Sn = i
n, +
I Xn = i, B) =
Jl"i
7r2
=
L = 1 - lP'(Sn = -i I Xn = i, B). pz + qz
Substitute into (*) to obtain
. . pi+!+ qi+! lP'(Xn+! = l + 1 I Xn = l, B)= . . = 1 -lP'(Xn+l = i - 1 I Xn = i, B). pl +ql Finally lP'(Xn+l = 1 I Xn = 0, B) = 1. (ii) If Yn > 0, then Yn - Yn+! equals the (n + l)th step, a random variable which is independent of the past history of the process. If Yn = 0 then Sn = Mn, so that Yn+! takes the values 0 and 1 with respective probabilities p and q, independently of the past history. Therefore Y is a Markov chain with transition probabilities fori > 0,
Pij =
{
ifj=i-1 'f . .+1 1 J = l '
p
q
p
POj = { q
ifj=O ifj=l.
The sequence Y is a random walk with a retaining barrier at 0.
4.
For any sequence io, i" ... of states, lP'(Yk+! = ik+!
I Yr = ir for 0
~ r ~
k) =
= =
lP'(Xns =is for 0 < s < k + 1) . - lP'(Xns = ls forO~ s ~ k)
TI~=O Pis .is+! (ns+ I
-
ns)
k-!
Tis=O Pis,is+l (ns+l- ns) Pikoik+l
273
(nk+! - nk) = lP'(Yk+! = ik+!
I Yk = ik),
[6.1.5]-[6.1.9]
Markov chains
Solutions
where Pij (n) denotes the appropriate n-step transition probability of X. (a) With the usual notation, the transition matrix of Y is p2 lrjj
=
if j = i
{ 2pq q2
if j
+ 2,
= i,
if j = i- 2.
(b) With the usual notation, the transition probability lrij is the coefficient of sj in G(G(s))i. 5.
Writing X= (X1, X2, ... , Xn), we have that
.)
( I
lP' F /(X)= 1, Xn = z =
lP'(F,/(X)=l.Xn=i) ( .) lP' /(X)= 1,Xn = I
where F is any event defined in terms of Xn, Xn+1· .... Let A be the set of all sequences x = (x1, x2, ... , Xn-1, i) of states such that I (x) = 1. Then lP'(F, /(X)= 1, Xn =
i)
=
L lP'(F, X= x) = lP'(F I Xn = i) L lP'(X = x) XEA
XEA
I
by the Markov property. Divide through by the final summation to obtain lP'(F /(X) = 1, Xn = i) = lP'(F I Xn = i). 6.
Let Hn = {Xk = Xk for 0:::: k < n, Xn = i}. The required probability may be written as lP'(XT+m = j, HT)
l:n lP'(XT+m = j, HT, T = n)
lP'(HT)
lP'(HT)
Now lP'(XT+m = j I HT, T = n) = lP'(Xn+m = j I Hn, T = n). Let I be the indicator function of the event Hn n {T = n }, an event which depends only upon the values of X 1, X2, ... , Xn. Using the result of Exercise (6.1.5), lP'(Xn+m = j
I Hn,
T = n) = lP'(Xn+m = j
I Xn
= i) = Pij(m).
Hence lP'(X
7.
-
. · (m) "' lP'(H T - n) . I H ) - PlJ LJn n, -
T +m - 1
T
-
lP'(HT)
-
··(
)
- P•J m ·
Clearly lP'(Yn+1 = j
I Yr
= ir for 0 ::S r ::S n) = lP'(Xn+1 = b
I Xr
= ar for 0 ::S r ::S n)
where b = h- 1(j), ar = h- 1Cir ); the claim follows by the Markov property of X.
It is easy to find an example in which h is not one-one, for which X is a Markov chain but Y is not. The first part of Exercise (6.1.3) describes such a case if So =f::. 0. 8.
Not necessarily! Take as example the chains SandY of Exercise (6.1.3). The sum is Sn
+ Yn
Mn, which is not a Markov chain. 9.
All of them. (a) Using the Markov property of X, lP'(Xm+r = k
I Xm
= im, · · ·, Xm+r-1 = im+r-1) = lP'(Xm+r = k
274
I Xm+r-1
= im+r-1).
=
Solutions [6.1.10]-[6.2.1]
Classification of states
(b) Let {even}= {X2r = i2r forO:::: r:::: m} and {odd}= {X2r+1 = i2r+1 forO :S r :S m- 1}. Then,
""''lP'(X2m+2 = k, X2m+1 = i2m+1• even, odd) lP'(X2m+2 = k I even) = L..J lP'(even) = L,lP'(X2m+2 = k, X2m+1 = i2m+1 I X2m = i2m)IP'(even, odd) lP'(even) = lP'(X2m+2 = k I X 2m= i2m), where the sum is taken over all possible values of is for odd s. (c) With Yn = (Xn, Xn+1),
I
I (k, l) I Yn =
IP'(Yn+1 = (k, l) Yo = (io, i1), ... , Yn = Cin, k)) = IP'(Yn+1 = (k, l) Xn+1 = k) = IP'(Yn+1 =
Cin, k)),
by the Markov property of X.
10. We have by Lemma (6.1.8) that, with 11-Ji) = lP'(Xi = j),
11. (a) Since Sn+ 1 = Sn
+ Xn+ 1, a sum of independent random variables, S is a Markov chain.
(b) We have that
lP'(Yn+1 = k I Yi =Xi
+ Xi-1 for 1 :S i
:S n) = lP'(Yn+1 = k I Xn = Xn)
by the Markov property of X. However, conditioning on Xn is not generally equivalent to conditioning on Yn = Xn + Xn-1, so Y does not generally constitute a Markov chain. (c) Zn = nX1 + (n -l)X2 + · · · + Xn, so Zn+1 is the sum of Xn+1 andacertainlinearcombination of Z1, Z2, ... , Zn, and so cannot be Markovian. (d) Since Sn+1 = Sn + Xn+1• Zn+1 = Zn + Sn + Xn+1• and Xn+1 is independent of X1, ... , Xn, this is a Markov chain.
12. With 1 a row vector of 1's, a matrix Pis stochastic (respectively, doubly stochastic, sub-stochastic) ifP1' = 1 (respectively, 1P = 1, P1' :::: 1, with inequalities interpreted coordinatewise). By recursion, P satisfies any of these equations if and only if pn satisfies the same equation.
6.2 Solutions. Classification of states 1. Let Ak be the event that the last visit to i, prior to n, took place at time k. Suppose that Xo = i, so that Ao, A 1, ... , An-1 form a partition of the sample space. It follows, by conditioning on the Ai, that n-1 Pi} (n) = L Pii (k)lij (n - k) k=O for i of. j. Multiply by sn and sum over n (0:: 1) to obtain Pij (s) = Pii (s )Lij (s) for i of. j. Now Pij (s) = Fij (s)Pjj (s) if i of. j, so that Fij (s) = Lij (s) whenever Pii (s) = Pjj (s). As examples of chains for which Pii (s) does not depend on i, consider a simple random walk on the integers, or a symmetric random walk on a complete graph. 275
[6.2.2]-[6.3.1]
Solutions
Markov chains
2. Let i (# s) be a state of the chain, and define ni = min{n: Pis(n) > 0}. If Xo = i and Xn; = s then, with probability one, X makes no visit to i during the intervening period [1, n;- 1]; this follows from the minimality of n;. Now s is absorbing, and hence lP'(no return to i
I Xo = i)
;::: lP'(Xn;
= s I Xo = i)
> 0.
3. Let h be the indicator function of the event {Xk = i}, so that N = I:~o visits to i. Then 00
h is the number of
00
JE:(N) = "LE(h) = LPii(k) k=O
k=O
which diverges if and only if i is persistent. There is another argument which we shall encounter in some detail when solving Problem (6.15.5).
4. We write lP'iO = lP'(· I Xo = i). One way is as follows, another is via the calculation of Problem (6.15.5). Note that lP';(Yj ;::: 1) = lP';(1j < oo). (a) We have that
by the strong Markov property (Exercise (6.1.6)) applied at the stopping time Ti. By iteration, lP'i (Vi ;::: n) = lP'i ( V; ;::: l)n, and allowing n -+ oo gives the result. (b) Suppose i ¥= j. Form ;::: 1,
by the strong Markov property. Now let m -+ oo, and use the result of (a). 5. Let e = lP'(1j < Ti before visiting i. Then
1
Xo
= i) = lP'(Ti
lP'(N:::: 1 I Xo Likewise, lP'(N:::: k I Xo
< 1j
1
Xo
= i) = lP'(1j
= i) =eo - e)k-! fork::::
= j), and let N
< T;
I Xo = i)
be the number of visits to j
=e.
1, whence
00
E(N 1 Xo
= i) = "Leo - e)k-l = 1. k=l
6.3 Solutions. Classification of chains If r = 1, then state i is absorbing for i ;::: 1; also, 0 is transient unless ao = l. Assume r < 1 and let J = sup{j : aj > 0}. The states 0, 1, ... , J form an irreducible persistent class; they are aperiodic if r > 0. All other states are transient. For 0 _:::: i _:::: J, the recurrence time Ti of i satisfies lP'(Ti = 1) = r. If T; > 1 then Ti may be expressed as the sum of 1.
r/') := time to reach 0, given that the first step is leftwards, r?) := time spent in excursions from 0 not reaching i, T-(3) := l
time taken to reach i in final excursion.
276
Solutions
Classification of chains
[6.3.2]-[6.3.3]
It is easy to see that JE:(Tp)) = 1 + (i - 1)/(1 - r) if i 0:: 1, since the waiting time at each intermediate point has mean (1- r)- 1. The number N of such 'small' excursions has mass function f'(N = n) = ai(l- ai)n, n :=: 0, where ai = L~i aj; hence JE:(N) = (1- ai)/ai. Each such small excursion has mean duration i-1 (
Eo
.
)
- 1-
1-r
+1
__!!1_- 1 +
i-1
.
1-ai-
Eo
1{ = (1 - ai) ai
i-1 +L
Jaj
(1-ai)(l-r)
and therefore
E(T?))
.
}
Jaj
. 1- r ]=0
.
By a similar argument,
E(T?))
1 =-""' 00
a·~
(
j=i
l
1 + J• - t' ) aj. 1-r
Combining this information, we obtain that
and a similar argument yields JE:(To) = 1 + Lj jaj j(l- r). The apparent simplicity of these formulae suggests the possibility of an easier derivation; see Exercise (6.4.2). Clearly JE:(Ti) < oo for i ::::: J whenever Lj jaj < oo, a condition which certainly holds if J < oo. 2. Assume that 0 < p < 1. The mean jump-size is 3p - 1, whence the chain is persistent if and only if p = see Theorem (5.10.17).
1;
3. (a) All states are absorbing if p = 0. Assume henceforth that p =f. 0. Diagonalize P to obtain P = BAB- 1 where
B~ (:
~}
1 0 -1
P"
~ BA'n-
1
B-1=
0 ~ 0 -gp)' 0 l-2p 0
A=
Therefore
(! -\). I
B
(I
whence Pij (n) is easily found. In particular,
and P33(n) = Pll (n) by symmetry. 277
2 0
I
I
2
4
0
) B- 1
0 ) . l-4p
(1- 4p)n
[6.3.4]-[6.3.5]
Markov chains
Solutions
Now Fu(s) = 1- Pu(s)- 1 , where
Pn (s)
=
p 33 (s)
=
Pzz(s) =
1 4(1 - s)
+ 2{1 -
1 s(l- 2p)}
1 2(1 - s)
+ 2{1 -
1 . s(l- 4p)}
+ 4{1 -
1 s(l- 4p)}'
Mter a little work one obtains the mean recurrence times J-ti = F{i (1): t-tl = t-t3 = 4, t-t2 = 2. (b) The chain has period 2 (if p =f. 0), and all states are non-null and persistent. By symmetry, the mean recurrence times J-ti are equal. One way of calculating their common value (we shall encounter an easier way in Section 6.4) is to observe that the sequence of visits to any given state j is a renewal process (see Example (5.2.15)). Suppose for simplicity that p =f. 0. The times between successive visits to j must be even, and therefore we work on a new time-scale in which one new unit equals two old units. Using the renewal theorem (5.2.24), we obtain 2
Pij (2n) -+ - i f Jj- i I is even, 1-tj '
Pij(2n
+ 1)-+
2
- i f Jj1-tj
il is odd;
note that the mean recurrence time of j in the new time-scale is it-tj. Now "E,j Pij (m) = 1 for all m, and so, letting m = 2n -+ oo, we find that 4/ t-t = 1 where t-t is a typical mean recurrence time. There is insufficient space here to calculate Pij (n). One way is to diagonalize the transition matrix. Another is to write down a family of difference equations of the form PI2 (n) = p · P22 (n 1) + (1 - p) · P4z(n- 1), and solve them. 4. (a) By symmetry, all states have the same mean-recurrence time. Using the renewal-process argument of the last solution, the common value equals 8, being the number of vertices of the cube. Hence t-tv = 8. Alternatively, lets be a neighbour of v, and lett be a neighbour of s other than v. In the obvious notation, by symmetry,
- 1 + 4/-tsv. 3 l-tvI
1-ttv = 1 + 21-tsv
I I + 4/-ttv + 41-twv,
1-tsv = 1 + 1-twv = 1 +
I
I + 'J.I-ttv. I 3 41-twv + 4/-ttv,
4/-tsv
a system of equations which may be solved to obtain t-tv = 8. (b) Using the above equations, t-twv =
'j0 , whence t-tvw = 'j0 by symmetry.
(c) The required number X satisfies lP'(X = n) = en-! (1 - e) 2 for n ?: 1, where e is the probability that the first return of the walk to its starting point precedes its first visit to the diametrically opposed vertex. Therefore 00
JE(X)
=
I>en-1(1- e)2 = 1. n=I
5.
(a) Let lP'i(·) = lP'(·
I Xo =
i). Since i is persistent,
1 = 1P'i(Vi = oo) = 1P'i(Vj = 0, Vi= oo) _:::: lP'j(Vj = 0)
+ 1P'i(7} < oo, vi= oo) + 1P'i(7} < oo)lP'j(Vi
_:::: 1 -lP'i(7} < oo)
+ lP'i(V;
> 0, Vi= oo)
= oo),
by the strong Markov property. Since i -+ j, we have that lP'j (Vi = oo) ?: 1, which implies IJji = 1. Also, lP'i (1j < oo) = 1, and hence j -+ i and j is persistent. This implies IJij = 1. (b) This is an immediate consequence of Exercise (6.2.4b).
278
Classification of chains
Solutions
6. Let lP'i 0 = lP'(· I Xo = i). It is trivial that T/j = 1 for j step and use the Markov property to obtain TJi =
2.:: PiklP'(TA < oo
1
x1
= k) =
rf A, condition on the first
For j
E A.
[6.3.6]-[ 6.3.9]
2.:: PjkTJk· k
kES
If x = (Xj : j E S) is any non-negative solution of these equations, then Xj = 1 ::: T/j for j E A. For j rf A, Xj
=L kES
PjkXk = L Pjk kEA
= IP'j(TA = 1)
+
+L
1)
+L
PjkXk
krjA
L Pjk{ L Pki krjA iEA
We obtain by iteration that, for j
=
PjkXk = IP'j (TA
krjA
+L
PkiXi} = IP'j(TA ::::: 2)
irjA
+L krjA
Pjk L PkiXi. irjA
rf. A,
where the sum is over all k1, k2, ... , kn
rf.
A. We let n --+ oo to find that Xj ::: IP'j(TA < oo) = T/j.
7. The first part follows as in Exercise (6.3.6). Suppose x = (xj : j E S) is a non-negative solution to the equations. As above, for j rf A, Xj = 1 + LPjkXk = IP'j(TA::: 1)
+L
= IP'j(TA ::: 1)
Pjk(1
+ LPkiXi)
~A
k
+ IP'j(TA
::: 2)
+ · · · + IP'j(TA
i~
::: n)
+L
Pjk 1 Pk1k2 · • · Pkk-!knXkn
n
::: L lP'(TA::: m), m=i
where the penultimate sum is over all paths of length n that do not visit A. We let n --+ oo to obtain thatxj :::Ej(TA)=Pj·
8. Yes, because the Sr and Tr are stopping times whenever they are finite. Whether or not the exit times are stopping times depends on their exact definition. The times Ur = min{k > Ur-i : Xu, E A, Xu,+! rf. A} are not stopping times, but the times Ur + 1 are stopping times. 9. (a) Using the aperiodicity of j, there exist integers TJ, r2, ... , rs having highest common factor 1 and such that Pjj (rk) > 0 for 1 ::::: k ::::: s. There exists a positive integer M such that, if r ::: M, then r = Z::~=l akrk for some sequence aJ, a2, ... , as of non-negative integers. Now, by the ChapmanKolmogorov equations, s
Pjj (r) :::
IT Pjj (rk)ak > 0, k=i
sothatpjj(r) > Oforallr::: M. Finally, find m such that Pij (m) > 0. Then Pij(r
+ m)::: Pij(m)Pjj(r)
> 0
ifr::: M.
(b) Since there are only finitely many pairs i, j, the maximum R(P) = max{N(i, j) : i, j E S} is finite. Now R(P) depends only on the positions of the non-negative entries in the transition matrix P.
279
[6.3.10]-[6.3.10]
Solutions
Markov chains
There are only finitely many subsets of entries of P, and so there exists f(n) such that R(P) ::": f(n) for all relevant n x n transition matrices P. (c) Consider the two chains with diagrams in the figure beneath. In the case on the left, we have that Pll (5) = 0, and in the case on the right, we may apply the postage stamp lemma with a = n and b=n-1. 3 2
4
2
n 10. Let Xn be the number of green balls after n steps. Let ej be the probability that Xn is ever zero when Xo = j. By conditioning on the first removal,
j+2 j ej = 2 (j + 1) ej+1 + 2 (j + 1) ej-1·
j::: 1,
with eo = 1. Solving recursively gives q1 q1q2 .. •qj-1 } ej = 1- (1 - e1) { 1 + - + · · · + , P1 P1P2···Pj-1
(*)
where
j+2 Pj = 2(j + 1)'
It is easy to see that j-1
"'
q1q2 ... qj-1
f::QP1P2···Pj-1
2
= 2 - -- ~ 2
j+1
as J. ~ oo.
By the result of Exercise (6.3.6), we seek the minimal non-negative solution (ej) to(*), which is attained when 2(1 - q) = 1, that is, e1 = Hence
i·
j-1
ej =
1-! L
q1q2 .. ·qj-1 2 r=O P1P2 · · · Pj-1
j
+1
For the second part, let dj be the expected time until j - 1 green balls remain, having started with j green balls and j + 2 red. We condition as above to obtain j dj = 1 + 2(j + 1) {dj+1 +dj}.
We set ej = dj - (2j + 1) to find that (j + 2)ej = jej+b whence ej = ij (j + 1)e1. The expected time to remove all the green balls is n n n dj = ej + 2(j - 1)} = n(n + 2) + e1 j (j + 1).
L
L{
L-!
j=1
j=1
j=1
The minimal non-negative solution is found by setting q = 0, and the conclusion follows by Exercise (6.3.7).
280
Solutions
Stationary distributions and the limit theorem
[6.4.1]-[6.4.2]
6.4 Solutions. Stationary distributions and the limit theorem 1. Let Yn be the number of new errors introduced at the nth stage, and let G be the common probability generating function of the Yn. Now Xn+1 = Sn + Yn+1 where Sn has the binomial distribution with parameters Xn and q (= 1- p). Thus the probability generating function Gn of Xn satisfies Gn+1(s) = G(s)lE(sSn) = G(s)lE{lE(sSn I Xn)} = G(s)JE{(p + qs)Xn} = G(s)Gn(P + qs) = G(s)Gn(1- q(1- s)).
Therefore, for s < 1, Gn(s)
= G(s)G(l- q(l- s))Gn-2(1- q 2 (l- s)) = · · · n-1 = Go(1- qn(l- s))
oo
II G(1- qr(l- s))---+ II G(1- qr(l- s)) r=O
as n ---+ oo, assuming q < 1. This infinite product is therefore the probability generating function of the stationary distribution whenever this exists. If G(s) = el.(s-1), then
IT
G(1 - qr (1- s)) = exp{ A.(s- 1)
r=O
f
qr} = el.(s- 1)/P,
r=O
so that the stationary distribution is Poisson with parameter A./ p. 2. (6.3.1): Assume for simplicity that sup{j : aj > 0} = oo. The chain is irreducible if r < 1. Look for a stationary distribution :rr with probability generating function G. We have that
:rro = ao:rro + (1 - r):rr1,
:rri = aino + rni + (1 - r)ni+1 fori ::: 1.
Hence sG(s) = nosA(s) + rs(G(s)- no)+ (1- r)(G(s)- no) where A (s) =
2:,]:_0 aj s j, and therefore G(s) =no (
Taking the limit ass
t
sA(s)- (1- r + sr)) . (1 - r)(s- 1)
1, we obtain by L'H6pital's rule that
A'(1)+1-r) G(1) =no ( . 1-r There exists a stationary distribution if and only if r < 1 and A' (1) < oo, in which case sA(s)- (1 - r + sr) G(s)- - - - - - - - (s- 1)(A'(l) + 1- r) · Hence the chain is non-null persistent if and only if r < 1 and A' (1) < oo. The mean recurrence time j;.,i is found by expanding G and setting f-Li = 1/ni. (6.3.2): Assume that 0 < p < 1, and suppose first that p =f. ~- Look for a solution {Yj : j =f. 0} of the equations Yi =
L PijYj. j-10
281
i =f. 0,
[6.4.3]-[6.4.3]
Solutions
Markov chains
as in (6.4.10). Away from the origin, this equation is Yi = qyi-1 + PYi+2 where p + q = 1, with auxiliary equation pe 3 - e + q = 0. Now pe 3 - e + q = p(e - 1)(8 - a)(8 - /3) where a=
-p- y'p2 +4pq < -1 , 2p
Note that 0 < f3 < 1 if p > ~,while f3 > 1 if p < ~ For p > ~,set . = { A+ Bf3i Yz C + Dai
if i :::: 1, if i :::; -1,
the constants A, B, C, D being chosen in such a manner as to 'patch over' the omission of 0 in the equations (*): Y-2 = qy-3,
Y-1 = qy_2 + PY1·
Y1 = PY3·
The result is a bounded non-zero solution {Yj} to (*), and it follows that the chain is transient. For p < ~, follow the same route with if i :::: 1, ifi:::;-1, the constants being chosen such that Y-2 = qy_3, Y-1 = qY-2· Finally suppose that p = ~,so that a= -2 and f3 = 1. The general solution to(*) is Yi = { A+ Bi + Cai. D+ Ei +Fa 1
if i :::: 1, ifi::::: -1,
subject to (** ). Any bounded solution has B = E = C = 0, and (**) implies that A = D = F = 0. Therefore the only bounded solution to(*) is the zero solution, whence the chain is persistent. The equation x = xP is satisfied by the vector x of 1's; by an appeal to (6.4.6), the walk is null.
(!, 1, !)
(6.3.3): (a) Solve the equation 1r = 1rP to find a stationary distribution 1r = when p =I= 0. Hence the chain is non-null and persistent, with f.L1 = rr1 1 = 4, and similarly f.L2 = 2, f.L3 = 4. . a stationary . d"tstn"but10n, . . "1 ar1y, 1r = ( 41:, 41:. 41 , 41) ts (b) Snru an d J.Li = ni-1 = 4 . (6.3.4): (a) The stationary distribution may be found to be Jri = ~ for all i, so that J.Lv = 8. 3.
The quantities X 1, X2, ... , Xn depend only on the initial contents of the reservoir and the rainfalls Yo, Y1, ... , Yn-1· The contents on day n + 1 depend only on the value Xn of the previous contents and the rainfall Yn. Since Yn is independent of all earlier rainfalls, the process X is a Markov chain. Its state space isS= {0, 1, 2, ... , K- 1} and it has transition matrix
[go+ lP'=
go 0
0
Kl
g2 g1 go
g3 g2 g1
gK-1 gK-2 gK-3
GK-1 Gx GK-2
0
0
go
G1
l
where gi = lP'(Y1 = i) and Gi = 2:-'f:=i gj. The equation 1r = 1rP is as follows: no= no(go + g1) + rr1go, 1rr = nogr+1 + n1gr + · · · + 1rr+1go,
1rK-1 = noGK +rr1GK-1 + · · · +rrK-1G1.
282
0 < r < K- 1,
Solutions
Stationary distributions and the limit theorem
[6.4.4]-[6.4.5]
The final equation is a consequence of the previous ones, since L;~(/ Jri = 1. Suppose then that v = ( v1, v2, ... ) is an infinite vector satisfying
Multiply through the equation for Vr by sr+ 1, and sum over r to find (after a little work) that 00
N(s) =
L
00
G(s) = Lgisi
VjSi,
i=O
satisfy sN(s) = N(s)G(s)
i=O
+ vogo(s- 1), and hence 1 go(s - 1) -N(s) = . vo s- G(s)
The probability generating function of the Jri is therefore a constant multiplied by the coefficients of s0 , s 1, ... , sK - 1 in go(s -1)/(s- G(s)), the constant being chosen in such a way that L;~0 1 rri = 1.
When G(s) = p(1 - qs)- 1, then go= p and go(s- 1)
p(l - qs)
s-G(s)
p-qs
-=-=--'------- =
= p
q + _ ______;:...__ 1-(qsjp)
The coefficient of s0 is 1, and of si is qi+ 1j pi if i ::: 1. The stationary distribution is therefore given byrri =qno(qfpi fori::: 1,where
+ 1) if p =
if p
#
4.
The transition matrices
q, and no= 2/(K
q =
i·
P,~ (~
1 2 1 2
0 0
0 0 1 2 1 2
i i p, i (1 -
have respective stationary distributions n 1 = (p, 1 - p) and n 2 = ( p, for any 0 ~ p ~ 1.
p),
i(1 -
P))
5. (a) Set i = 1, and find an increasing sequence n1 (1), n1 (2), ... along which x1 (n) converges. Now set i = 2, and find a subsequence of (n1(j) : j ::: 1) along which x2(n) converges; denote this subsequence by n2(1), n2(2), .... Continue inductively to obtain, for each i, a sequence Di = (ni (j) : j ::: 1), noting that: (i) Di+1 is a subsequence ofni, and (ii) limr-+ooXi(ni(r)) exists for all i. Finally, define m k = n k ( k). For each i ::: 1, the sequence m i, m i+1, . . . is a subsequence of Di , and therefore limr-+oo Xi (mr) exists. (b) Let S be the state space of the irreducible Markov chain X. There are countably many pairs i, j of states, and part (a) may be applied to show that there exists a sequence (nr : r ::: 1) and a family (aij : i, j E S), not all zero, such that Pij (nr) --+ Ciij as r --+ oo.
283
[6.4.6]-[6.4.10]
Solutions
Markov chains
Now X is persistent, since otherwise Pi} (n) ---+ 0 for all i, j. The coupling argument in the proof of the ergodic theorem (6.4.17) is valid, so that PaJ (n) - Pbj (n) ---+ 0 as n ---+ oo, implying that aaj = abj for all a, b, j.
6.
Just check that 1r satisfies 1r = 1rP and L:v Irv = 1.
7.
Let Xn be the Markov chain which takes the valuer if the walk is at any of theY nodes at level r. Then Xn executes a simple random walk with retaining barrier having p = 1 - q = ~,and it is thus transient by Example (6.4.15). 8.
Assume that Xn includes particles present just after the entry of the fresh batch Yn. We may write Xn
Xn+l =
L Bi,n + Yn i=l
where the Bi,n are independent Bernoulli variables with parameter 1 - p. Therefore X is a Markov chain. It follows also that
In equilibrium, Gn+l = Gn = G, where G(s) = G(p + qs)el.(s-l). There is a unique stationary distribution, and it is easy to see that G(s) = el.(s-l)fp must therefore be the solution. The answer is the Poisson distribution with parameter 'A/ p. 9. The Markov chain X has a uniform transition distribution P}k = 1/(j + 2), 0 ::::: k ::::: j + 1. Therefore, E(Xn) =lE(E(Xn
I Xn-d) = ~(1 +E(Xn-1)) = ···
= 1- (~)n + (~)nXo. The equilibrium probability generating function satisfies
whence
d ds {0-s)G(s)} = -sG(s),
subject to G(1) = 1. The solution is G(s) = es-l, which is the probability generating function of the Poisson distribution with parameter 1.
10. This is the claim of Theorem (6.4.13). Without loss of generality we may takes = 0 and the Yj to be non-negative (since if the Yj solve the equations, then so do Yj + c for any constant c). LetT be the matrix obtained from P by deleting the row and column labelled 0, and write Tn = (tij (n) : i, j # 0). Then Tn includes all the n-step probabilities of paths that never visit zero. We claim first that, for all i, j it is the case that tij (n) ---+ 0 as n ---+ oo. The quantity tij (n) may be thought of as the n-step transition probability from i to j in an altered chain in which s has been made absorbing. Since the original chain is assumed irreducible, all states communicate with s, and therefore all states other than s are transient in the altered chain, implying by the summability of fij (n) (Corollary (6.2.4)) that tij (n) ---+ 0 as required. Iterating the inequality y :=: Ty yields y :=: Tny, which is to say that 00
00
Yi 0:: Lfij(n)yj 0:: min{Yr+s} L tij(n), . I s> I . I J= J=r+
284
i ::: 1.
Solutions [6.4.11]-[6.4.12]
Stationary distributions and the limit theorem Let An= {Xk
#
0 fork::::: n). Fori
~
1, 00
lP'(Aoo
I Xo = i) = n-+oo lim lP'(An I Xo =
i)
= '"'tij (n) L j=1
+ .
r ::::: lim { Ltij(n)
n-+oo
} . Yi mms:::dYr+s}
j= 1
Let E > 0, and pick R such that
Yi ---'--'--< E. mins:::dYR+sl Taker = Rand let n --+ oo, implying that lP'(A 00 by irreducibility that all states are persistent.
I Xo =
11. By Exercise (6.4.6), the stationary distribution is 1TA
i)
= 0.
It follows that 0 is persistent, and
= 1TB = rrn = 1TE =
j,, nc = 1·
(a) By Theorem (6.4.3), the answer is f-LA = 1/rrA = 6. (b) By the argument around Lemma (6.4.5), the answer is Pn(A)
= nnt-tA = nnfnA =
1.
(c) Using the same argument, the answer is pc(A) = ncfnA = 2. (d)LetlP'iO = lP'(· I Xo = i), let 1j be the time of the first passage to state j, and let Vi By conditioning on the first step,
"th
Wl
1 .
= IP'i(TA
< TE).
5 3 1 1 8, VB = 4> VC = 2, VD = 4 · A typical conditional transition probability Tij = IP'i (X 1 = j I TA < TE) is calculated as follows: SO UtlOn VA
=
and similarly,
We now compute the conditional expectations Iii = lEi (TA I TA < TE) by conditioning on the first step of the conditioned process. This yields equations of the form iiA = 1 + ~JiB + ~ Jic, whose solution gives ji A =
1£:.
(e) Either use the stationary distribution of the conditional transition matrix T, or condition on the first step as follows. With N the number of visits to D, and Tli = lEi (N I TA < TE), we obtain TIA =
~TIB +~Tic,
whence in particular Tl A =
TIB =
0+
~TIC,
TIC=
0+
~TIB + ~(1 +TID),
TID =TIC,
1~.
12. By Exercise (6.4.6), the stationary distribution has n A around Lemma (6.4.5), the answer is PB(A) = 1TBf-LA = 1TB/1TA
285
-h, 1TB = 2.
.t.
Using the argument
[6.5.1]-[6.5.6]
Solutions
Markov chains
6.5 Solutions. Reversibility 1. Look for a solution to the equations JriPij = lrj Pji. The only non-trivial cases of interest are those with j = i + 1, and therefore Ailri = fLi+llri+l for 0::::: i < b, with solution
0::::: i::::: b,
/Ll/L2 ... /Li
an empty product being interpreted as 1. The constant no is chosen in order that the ni sum to 1, and the chain is therefore time-reversible.
2. Letn be the stationary distribution of X, and suppose X is reversible. We have thatniPij = Pjilrj for all i, j, and furthermore n i > 0 for all i. Hence
as required when n = 3. A similar calculation is valid when n > 3. Suppose conversely that the given display holds for all finite sequences of states. Sum over all values of the subsequence h, ... , jn-1 to deduce that Pij (n- 1)Pji = Pij Pji(n- 1), where i = h, j = jn. Take the limit as n ~ oo to obtain lrj Pji = Pij Jri as required for time-reversibility.
3.
With n the stationary distribution of X, look for a stationary distribution v of Y of the form ififjC,
ifi
E
C.
There are four cases. (a) i E C, j fj C: Viqjj = cnif3Pij = cf3njPji = Vjqji,
(b) i, j
E
C: Vjqjj = ClriPij = ClrjPji = VNji,
(c) i fj C, j E C: Viqjj = cf3JriPij = cf3nj Pji = Vjqji, (d) i, j fj C: viqij = cf3niPij = cf3JrjPji = Vjqji· Hence the modified chain is reversible in equilibrium with stationary distribution v, when
In the limit as f3
+0, the chain Y never leaves the set C once it has arrived in it.
4.
Only if the period is 2, because of the detailed balance equations.
5.
With Yn = Xn - im,
Now iterate.
6. (a) The distribution n1 = f3/(a + /3), n2 = aj(a + /3) satisfies the detailed balance equations, so this chain is reversible. (b) By symmetry, the stationary distribution is 1r = (~, ~,~),which satisfies the detailed balance equations if and only if p =
i.
(c) This chain is reversible if and only if p =
i. 286
Solutions
Chains with .finitely many states
[6.5.7]-[6.6.2]
7. A simple random walk which moves rightwards with probability p has a stationary measure TCn = A(p/q)n, in the sense that :n: is a vector satisfying :n: = :n:P. It is not necessarily the case that this :n: has finite sum. It may then be checked that the recipe given in the solution to Exercise (6.5.3)
I
yields n(i, j) = pfp4 L:(r,s)EC PIp~ as stationary distribution for the given process, where Cis the relevant region of the plane, and Pi = pi/ qi and Pi (= 1 - qi) is the chance that the i th walk moves rightwards on any given step.
8.
Since the chain is irreducible with a finite state space, we have that TCi > 0 for all i. Assume the chain is reversible. The balance equations TCi Pij = TCj Pji give Pij = TCj Pji/TCi. Let D be the matrix with entries 1/ni on the diagonal, and S the matrix (nj Pji ), and check that P = DS.
I
Conversely, ifP = DS, then di- 1 Pij = dj- 1 Pji• whence TCi = di- 1 L:k di: 1 satisfies the detailed balance equations. Note that Pij =
~ [!§; Pij y'Jfj.
If the chain is reversible in equilibrium, the matrix M = ( y'ni/nj Pij) is symmetric, and therefore M, and, by the above, P, has real eigenvalues. An example of the failure of the converse is the transition matrix
P=(i ~ 1
0
!). 0
which has real eigenvalues 1, and-! (twice), and stationary distribution :n: = (~,~,b). However, n1 Pl3 = 0
9.
=f::.
b= n3 P31, so that such a chain is not reversible.
Simply check the detailed balance equations TCi Pij = TCj Pji.
6.6 Solutions. Chains with finitely many states 1. Let P = (Pij : 1 ::::= i, j ::::= n) be a stochastic matrix and let C be the subset of JR.n containing all vectors x = (x1, x2, ... , Xn) satisfying Xi 0:: 0 for all i and 2::?= 1 Xi = 1; for x E C, let llxll = maxj{Xj }. Define the linear mapping T : C ~ JR.n by T(x) = xP. Let us check that Tis a continuous function from C into C. First, IIT(x)ll
= mr{ "';XiPij}
::::: allxll
l
where
hence IIT(x)- T(y)ll
::::=
allx- Yll. Secondly, T(x)j 0:: Oforall j, and L T(x)j = LLXiPij = LXi LPij = 1. j j i j
Applying the given theorem, there exists a point :n: in C such that T (n) = n, which is to say that
n = nP. 2.
Let P be a stochastic m x m matrix and letT be them x (m Pi'- Oi'J t··{ 1J lj -
287
if j
::::=
+ 1) matrix with (i, j)th entry
m,
if j = m + 1,
[6.6.3]-[6.6.4]
Solutions
Markov chains
where Dij is the Kronecker delta. Let v = (0, 0, ... , 0, 1) is valid, there exists y = (YI, Y2· ... , Ym+l) such that
E
JRm+l. If statement (ii) of the question
m
"' 0 for 1 _
Ym+l < 0,
j=l this implies that m
for all i,
L Pij Yj :0:: Yi - Ym+l > Yi j=l
and hence the impossibility that 'L/}=I Pij Yj > maxi {Yi}. It follows that statement (i) holds, which is to say that there exists a non-negative vector x = (xl, x2, ... , Xm) such that x(P - I) = 0 and I:t=l Xi = 1; such an xis the required eigenvector.
3. Thinking of Xn+ 1 as the amount you may be sure of winning, you seek a betting scheme x such that Xn+ 1 is maximized subject to the inequalities n
for 1 :S j :Sm.
Xn+l :S LXitij i=l
Writing aij = -tij for 1 :S i :S nand an+l,j = 1, we obtain the linear program: maximize
Xn+l
subject to
n+l L Xiaij :S 0 i=l
for 1 :S j :Sm.
The dua11inear program is: m
minimize
0
subject to
L aij Yj = 0 j=l
for 1 :S i :S n,
m
Lan+l,jYj = 1, j=l
Yj :0::0
for 1 :S j :Sm.
Re-expressing the aij in terms of the tij as above, the dual program takes the form: m
minimize
0
subject to
L tij Pj j=l
=0
for 1 :S i :S n,
m
L Pj = 1, j=l
Pj :::: 0
for 1 :S j :Sm.
The vector x = 0 is a feasible solution of the primal program. The dual program has a feasible solution if and only if statement (a) holds. Therefore, if (a) holds, the dual program has minimal value 0, whence by the duality theorem of linear programming, the maximal value of the primal program is 0, in contradiction of statement (b). On the other hand, if (a) does not hold, the dual has no feasible solution, and therefore the primal program has no optimal solution. That is, the objective function of the primal is unbounded, and therefore (b) holds. [This was proved by De Finetti in 1937.] 4. Use induction, the claim being evidently true when n = 1. Suppose it is true for n = m. Certainly pm+l is of the correct form, and the equation pm+ 1x' = P(Pmx') with x = (1, w, w2 ) yields in its first row
288
Solutions
Branching processes revisited
[6.6.5]-[6.7.2]
as required. 5. The first part follows from the fact that Jr1 1 = 1 if and only if 1rU = 1. The second part follows from the fact that ni > 0 for all i if P is finite and irreducible, since this implies the invertibility of 1-P+ U.
6. The chessboard corresponds to a graph with 8 x 8 = 64 vertices, pairs of which are connected by edges when the corresponding move is legitimate for the piece in question. By Exercises (6.4.6), (6.5.9), we need only check that the graph is connected, and to calculate the degree of a comer vertex. (a) For the king there are 4 vertices of degree 3, 24 of degree 5, 36 of degree 8. Hence, the number of edges is 210 and the degree of a comer is 3. Therefore tt(king) = 420/3 = 140. (b) tt(queen) = (28 x 21 + 20 x 23 + 12 x 25 + 4 x 27)/21 = 208/3. (c) We restrict ourselves to the set of 32 vertices accessible from a given comer. Then tt(bishop) = (14x7+10x9+6x 11+2x 13)/7=40. (d) tt(knight) = (4 X 2 + 8 X 3 + 20 X 4 + 16 X 6 + 16 X 8)/2 = 168. (e) tt(rook) = 64 x 14/14 = 64. 7. They are walking on a product space of 8 x 16 vertices. Of these, 6 x 16 have degree 6 x 3 and 16 x 2 have degree 6 x 5. Hence tt(C) = (6
X
16
X
6
X
3 + 16
X
2
X
6
X
5)/18 = 448/3.
IP- All =(A.- l)(A. + !)(A.+ ~). Tedious computation yields the eigenvectors, and thus
8.
0 0)
-4 -1
3
-3
5
-1 1
.
6. 7 Solutions. Branching processes revisited 1. We have (using Theorem (5.4.3), or the fact that Gn+1(s) = G(Gn(s))) that the probability generating function of Zn is n- (n- l)s Gn(s) = , n + 1- ns so that n )k-1 nk-1 ( n )k+1 IP(Zn = k) = n + 1 n+ 1 n+ 1 - (n + l)k+ 1
(n -1) (
fork :=::: 1. Therefore, for y > 0, as n --+ oo,
1 llD(Z 0)j[
2.
n-
l2ynJ
- 1 - Gn(O)
n
"""' t:J
nk-1
(
(n + l)k+1
=1-
1 ) - L2ynJ 1+-n -+l-e- 2Y.
Using the independence of different lines of descent, JE(s
z
n
.
.
~ skiP(Zn = k, extinction)
I extmct10n) = L
k=O
IP( extinction)
where Gn is the probability generating function of Zn.
289
~ skiP(Zn = k)r/ 1 = L = -Gn(sry), k=O
TJ
TJ
[6.7.3]-[6.8.1] 3.
Markov chains
Solutions
We have that 17 = G(17). In this case G(s) = q(l - ps)- 1, and therefore 17 = q I p. Hence
P q{pn _ qn _ p(sqlp)(pn-1 _ qn-1)} 1 - G n (s 17) = - · ..::.-.::---,-,..-::---,-7--=..:._::_-"----=---'17 q pn+1 _ qn+1 _ p(sqlp)(pn _ qn) p{qn _ Pn _ qs(qn-1 _ Pn-1)} qn+1 _ pn+1 _ qs(qn _ pn) ' which is Gn (s) with p and q interchanged.
4.
(a) Using the fact that var(X I X > 0) ::=: 0, JE(X 2 ) = JE(X 2
IX
> O)JP(X > 0) :::: lE(X
IX
> 0) 2 JP(X > 0) = lE(X)lE(X
IX
> 0).
(b) Hence
lE(Z~)
n
2
lE(Znl t-t I Zn > 0) < = lE(Wn) - t-tnlE(Zn) where Wn = ZnllE(Zn). By an easy calculation (see Lemma (5.4.2)),
where 0" 2 = var(Z 1 ) = plq 2 . (c) Doing the calculation exactly, lE Z n ( nlt-t
Z > 0 - lE(Znlt-tn) 1 ) - IP'(Zn > 0)- 1- Gn(O)--+ 1-11
I n
where 17 =II"(ultimate extinction) = q I p.
6.8 Solutions. Birth processes and the Poisson process 1. Let F and W be the incoming Poisson processes, and let N(t) = F(t)+ W(t). Certainly N(O) = 0 and N is non-decreasing. Arrivals of flies during [0, s] are independent of arrivals during (s, t ], if s < t; similarly for wasps. Therefore the aggregated arrival process during [0, s] is independent of the aggregated process during (s, t]. Now IP'(N(t +h)=
n +II N(t)
=
n)
= JP(A Ll B)
where
A = {one fly arrives during (t, t + h l},
B = {one wasp arrives during (t, t + h l}.
We have that JP(A Ll B) = JP(A) + JP(B) - II"( A
n B)
= A.h + t-th- (A.h)(t-th) + o(h) = (A.+ t-t)h + o(h). Finally
IP'(N(t +h)> n +II N(t) =
n)::::: IP'(A n B)+ JP(C u D),
290
Solutions, [6.8.2]-[6.8.4]
Birth processes and the Poisson process
where C ={two or more flies arrive in (t, t + h]} and D ={two or more wasps arrive in (t, t + h]}. This probability is no greater than (A.h)(p,h) + o(h) = o(h). 2. Let I be the incoming Poisson process, and let G be the process of arrivals of green insects. Matters of independence are dealt with as above. Finally, IP'(G(t +h)= n +II G(t) =
n)
n)
= p!P'(I(t +h)= n +II I(t) =
+ o(h) = pA.h + o(h),
11"( G(t +h) > n +II G(t) = n) ::=:II" (I (t +h) > n +II I (t) = n) = o(h). 3.
Conditioning on T1 and using the time-homogeneity of the process,
1P'(E(t)>xiT1=u)=
{
IP'(E(t-u)>x)
if u :::: t,
~
iftt+x,
(draw a diagram to help you see this). Therefore IP'(E(t) > x) = fooo IP'(E(t) >xI T1 = u)A.e-Audu = t!P'(E(t-u) >x)A.e-Audu+1
Jo
00
A.e-Audu.
t+x
You may solve the integral equation using Laplace transforms. Alternately you may guess the answer and then check that it works. The answer is IP'(E(t) ::=: x) = I - e-h, the exponential distribution. Actually this answer is obvious since E(t) > x if and only if there is no arrival in [t, t + x ], an event having probability e -Ax. 4.
The forward equation is i :::: j,
with boundary conditions Pi} (0) = Oij, the Kronecker delta. We write Gi (s, t) = "E-1 si Pi} (t), the probability generating function of B(t) conditional on B(O) = i. Multiply through the differential equation by si and sum over j:
a partial differential equation with boundary condition Gi (s, 0) = si. This may be solved in the usual way to obtain Gi(s, t) = g(eAt(l- s- 1)) for some function g. Using the boundary condition, we find that g(I- s- 1) = si and so g(u) = (1 - u)-i, yielding I (se-At)i Gi(s, t) = {I- eAI(I- s-1 )}i = {I- s(l- e-At)}i.
The coefficient of si is, by the binomial series, j
as required.
291
?:_
i,
[6.8.5]-[ 6.8.6]
Solutions
Markov chains
Alternatively use induction. Set j =ito obtain pji(t) = -Aipu(t) (remember Pi,i-1 (t) = 0), and therefore Pi i (t) = e- Ai t. Rewrite the differential equation as d ( Pij(t)e A1't) = A(j -l)Pi,j-t(t)e A1't . dt
Set j = i + 1 and solve to obtain Pi,i+l (t) = ie-Ait ( 1- e-At). Hence(*) holds, by induction. The mean is E(B(t)) = !_GJ(S,
as
t)l
=/eM, s=l
by an easy calculation. Similarly var(B(t)) = A + E(B(t)) - E(B(t)) 2 where A= -a22 G1(s,t)
as
I
.
s=l
Alternatively, note that B(t) has the negative binomial distribution with parameters e-At and I.
5.
The forward equations are P~(t) = An-lPn-t(t)- AnPn(t),
n 2: 0,
where Ai = i A+ v. The process is honest, and therefore m(t) = L::n npn(t) satisfies 00
00
m'(t) = Ln[(n -l)A
+ v]Pn-t(f)-
n=l
Ln(nA + v)pn(t) n=O
00
= L
{ A[(n + l)n- n 2 ] + v[(n + 1)- nl} Pn(t)
n=O 00
=
L(An n=O
+ v)pn(t)
= Am(t)
+ v.
Solve subject to m(O) = 0 to obtain m(t) = v(eJ..t- 1)/A.
6. Using the fact that the time to the nth arrival is the sum of exponential interarrival times (or using equation (6.8.15)), we have that
is given by
which may be expressed, using partial fractions, as
where n
a·z-
A.
IT-1A·-A·
j=O 1
Hi
292
z
Solutions
Continuous-time Markov chains
[6.8.7]-[6.9.1]
so long as Ai =!= A.J whenever i =!= j. The Laplace transform Pn may now be inverted as
See also Exercise (4.8.4).
7. Let Tn be the time of the nth arrival, and letT= Exercise (6.8.6), AnPn(())
limn~oo
Tn = sup{t: N(t) < oo}. Now, as in
. =ITn __). z= lE(e-eTn)
+ () Xo +X 1 + · · ·+ Xn where Xk is the (k + l)th interarrival time, a random variable which is i=O Ai
since Tn = exponentially distributed with parameter Ak· Using the continuity theorem, lE(e-eTn) -+ 1E(e-eT) as n-+ oo, whence AnPn(()) -+ 1E(e-eT) as n-+ oo, which may be inverted to obtain AnPn(t) -+ f(t) as n-+ oo where f is the density function ofT. Now
I
lE(N(t) N(t) < oo) = L:=;onPn(t) L:=n=O Pn (t) which converges or diverges according to whether or not L:=n npn (t) converges. However Pn (t) ~ A.; 1 f (t) as n -+ oo, so that L:=n npn (t) < oo if and only if L:=n nA.; 1 < oo. When A.n = (n
+ ~) 2 , we have that lE(e-eT) =
IT 00
{
n=O
1+
() 1 2 (n + 2)
}-1
= sech
(nv'e).
Inverting the Laplace transform (or consulting a table of such transforms) we find that
where 81 is the first Jacobi theta function.
6.9 Solutions. Continuous-time Markov chains 1.
(a) We have that
where P12 = 1 - Plio P21 = 1 - P22· Solve these subject to Pi} (t) = 8ij, the Kronecker delta, to obtain that the matrix P1 = (Pi} (t)) is given by
(b) There are many ways of calculating Gn; let us use generating functions. Note first that G0 =I, the identity matrix. Write
n ::=: 0,
293
[6.9.2]-[6.9.4]
Solutions
Markov chnins
and use the equation Gn+ 1 = G · Gn to find that
Hence an+1 = -(pfA.)cn+1 for n 2: 0, and the first difference equation becomes an+1 = -(A.+ p,)an, n 2: 1, which, subject to a1 = -p,, has solution an = (-l)np,(A. + p,)n- 1, n 2: 1. Therefore Cn = (-l)n+ 1A.(A. + p,)n- 1 for n 2: 1, and one may see similarly that bn =-an, dn = -en for n 2: 1. Using the facts that ao =do = 1 and bo = co = 0, we deduce that L:~0 (tn jn!)Gn = Pt where P 1 is given in part (a). (c) With1r = (n1, n2), we have that -p,n1 + A.n2 = 0 and p,n1- A.n2 = 0, whence n1 = (A.jp,)n2. In addition, n1 + n2 = 1 if n1 = A./(A. + p,) = 1- n2. 2.
(a) The required probability is lP'(X(t) = 2, X(3t) = 1 I X(O) = 1P'(X(3t) = 1 I X(O) = 1)
1)
p!2(t)P21 (2t)
Pll (3t)
using the Markov property and the homogeneity of the process. (b) Likewise, the required probability is p!2(t)p21 (2t)pu (t)
Pu (3t)pu (t)
the same as in part (a).
3. The interarrival times and runtimes are independent and exponentially distributed. It is the lackof-memory property which guarantees that X has the Markov property. The state space isS= {0, 1, 2, ... } and the generator is
G~
c ~
A. -(A.+ p,) p,
0 A. -(A.+ p,)
0 0 A.
...
)
Solutions of the equation 1rG = 0 satisfy -A.no + p,n1 = 0,
A.nj-1 -(A.+ p,)nj + P,7rj+1 = 0 for j =:: 1,
with solutionni = no(A.jp,)i. We have in addition that:Li ni = 1 ifA < p,andno = {1- (A./p,)}- 1. 4.
One may use the strong Markov property. Alternatively, by the Markov property, lP'(Yn+1 = j I Yn = i, Tn = t, B)= lP'(Yn+1 = j I Yn = i, Tn = t)
for any event B defined in terms of {X (s) : s :S Tn}. Hence lP'(Yn+1 = j
I Yn =
i, B)= fooo lP'(Yn+1 = j
I Yn =
i, Tn = t)fTn(t)dt
= lP'(Yn+1 = j I Yn = i), so that Y is a Markov chain. Now qij = lP'(Yn+1 = j I Yn = i) is given by
% = { 00 Pij(t)A.e-M dt,
lo
294
Solutions
Continuous-time Markov chains
[6.9.5]-[6.9.8]
by conditioning on the (n + 1)th interarrival time of N; here, as usual, Pij (t) is a transition probability of X. Now
The jump chain Z = {Zn : n ::=: 0} has transition probabilities hij = 8i} / 8i, i -=f. j. The chance that Z ever reaches A from j is also IJj, and 11} = Lk hjkiJk for j as required.
5.
6. Let T1 = inf {t : X (t) -=f. X (0)}, and more generally let Tm be the time of the mth change in value of X. For j ¢:. A, /J-j = 1Ej(T1)
+ Lhjk/J-k> kf-j
where Ej denotes expectation conditional on Xo = j. Now Ej (Tl) = g1- 1 , and the given equations follow. Suppose next that (ak : k E S) is another non-negative solution of these equations. With Ui = Ti+1 - Ti and R = min{n 2: 1 : Zn E A}, we have for j l)) + · · · + Ej(Unl(R>nJ) Url(R>r})
= Ej (min{Tn, HA})
-+ Ej(HA)
r=O
as n -+ oo, by monotone convergence. Therefore, f.L is the minimal non-negative solution.
7. First note that i is persistent if and only if it is also a persistent state in the jump chain Z. The integrand being positive, we can write
where {Tn : n ::=: 1} are the times of the jumps of X. The right side equals
1
00
00
L JE(T1 I X (0) = i)hu (n) = -:- L hu (n) n=O
gz n=O
where H = (hij) is the transition matrix of Z. The sum diverges if and only if i is persistent for Z.
8. Since the imbedded jump walk is persistent, so is X. The probability of visiting m during an excursion is a= (2m)- 1, since such a visit requires an initial step to the right, followed by a visit to m before 0, cf. Example (3.9.6). Having arrived at m, the chance of returning tom before visiting 0 is 1 -a, by the same argument with 0 and m interchanged. In this way one sees that the number N of visits tom during an excursion from 0 has distribution given by IP'(N ::=: k) = a(l - a)k- 1, k ::=: 1. The 'total jump rate' from any state is A., whence T may be expressed as :L~o Vi where the Vi are exponential with parameter A.. Therefore, JE(e eT )=GN ( -A.- ) ).-()
a). =(1-a)+a--. a).-()
295
[6.9.9]-[6.9.11]
Solutions
Markov chains
The distribution ofT is a mixture of an atom at 0 and the exponential distribution with parameter a)..
9. The number N of sojourns in i has a geometric distribution IP'(N = k) = fk-l(l- f), k =:: 1, for some f < 1. The length of each of these sojourns has the exponential distribution with some parameter 8i. By the independence of these lengths, the total time T in state i has moment generating function gi(l-f) 8i0- f ) -
e
The distribution ofT is exponential with parameter 8i (1 - f).
10. The jump chain is the simple random walk with probabilities ).j (). + JJ-) and JJ-/ (). + JJ-), and with POl = 1. By Corollary (5.3.6), the chance of ever hitting 0 having started at 1 is JJ-/A, whence the probability of returning to 0 having started there is f = JJ-/A. By the result of Exercise (6.9.9),
as required. Having started at 0, the walk visits the state r =:: 1 with probability 1. The probability of returning to r having started there is
and each sojourn is exponentially distributed with parameter 8r = ). + JJ-. Now g 7 (1 - fr) = ). - JJ-, whence, as above, JE(eevr) = ). - /JA-JJ--e The probability of ever reaching 0 from X (0) is (JJ-/A)x (O), and the time spent there subsequently is exponential with parameter ). - JJ-. Therefore, the mean total time spent at 0 is
11. (a) The imbedded chain has transition probabilities
where 8k = -gkk· Therefore, for any state j,
where we have used the fact thatJrG = 0. Also irk =:: 0 and L:k irk = 1, and thereforeJi is a stationary distribution of Y. Clearly fik = nk for all k if and only if 8k = l::i ni 8i for all k, which is to say that 8i = 8k for all pairs i, k. This requires that the 'holding times' have the same distribution. (b) Let Tn be the time of the nth change of value of X, with To= 0, and let Un = Tn+l - Tn. Fix a state k, and let H = min{n =:: 1 : Zn = k}. Let Yi (k) be the mean time spent in state i between two consecutive visits to k, and let Yi(k) be the mean number of visits to i by the jump chain in between two
296
Solutions
Birth-death processes and imbedding
[6.9.12]-[6.11.2]
visits to k (so that, in particular, Yk(k) = g/; 1 and n(k) = 1). With Ej and 1P'j denoting expectation and probability conditional on X (0) = j, we have that Yi(k) =lEk(f:unl(Zn=i,H>n}) = f:Ek(Un I f(Zn=iJ)lP'k(Zn = i, H > n) n=O n=O
1 1 -lP'k(Zn = i, H > n) = -yi(k). n=O 8i 8i 00
= L
The vector y(k) = (Yi (k) : i E S) satisfies y(k)H = y(k), by Lemma (6.4.5), where H is the transition matrix of the jump chain Z. That is to say, for j E S, whence I:i Yi (k)gij = 0 for all j E S. If tlk = I:i Yi (k) < oo, the vector (Yi (k)/ t-tk) is a stationary distribution for X, whence :rri = Yi(k)/t-tk for all i. Setting i = k we deduce that Trk = 1/(gktLk). Finally, ifi;i Tri8i < oo, then
~
1
tlk =-:::::--- = Trk
I:i Tri8i "' = tlk L:rri8i· Trk8k i
12. Define the generator G by gu = -vi, 8ij = vihij, so that the imbedded chain has transition matrix H. A root of the equation 1rG = 0 satisfies
0 = L:rri8ij = -TrjVj i
+
L (:rrivi)hij i:i-/-j
whence the vectors = (:rrjVj : j E S) satisfies s = tH. Therefore s = av, which is to say that Trj Vj = avj ,for some constant a. Now Vj > 0 for all j, so that Trj = a, which implies that I:j Trj =!= 1. Therefore the continuous-time chain X with generator G has no stationary distribution.
6.11 Solutions. Birth-death processes and imbedding 1.
The jump chain is a walk {Zn} on the setS= {0, 1, 2, ... } satisfying, fori ::: 1, lP'(Zn+I = j
I Zn = i) = {
if j = i + 1, ifj=i-1,
Pi 1- Pi
where Pi= Ai/(Ai + t-ti)· Also lP'(Zn+l = 1 I Zn = 0) = 1. 2.
The transition matrix H = (hij) of Z is given by i tL h .. - { A+it-t lJA -A+ it-t
'f .
.
1
1 ]=!-'
ifj=i+l.
To find the stationary distribution of Y, either solve the equation 1r = 1rQ, or look for a solution of the detailed balance equations :rrihi,i+l = :rri+Ihi+ 1,i. Following the latter route, we have that
297
[6.11.3]-[6.11.4]
Solutions
Markov chains
whence TCi =reo pi (1 + i j p )/ i! fori 2:: 1. Choosing reo accordingly, we obtain the result. It is a standard calculation that X has stationary distribution v given by vi = pie -p j i! fori 2:: 0. The difference between 1r and v arises from the fact that the holding-times of X have distributions which depend on the current state. 3.
We have, by conditioning on X(h), that
I
l](t +h)= lE{lP'(X(t +h) = 0 X(h))} = JJ-h · 1 + (1- A.h- JJ-h)l](t) + A.h~(t) + o(h) where~ (t) = lP'(X (t) = 0 I X (0) = 2). The process X may be thought of as a collection of particles each of which dies at rate fJ- and divides at rate A., different particles enjoying a certain independence; this is a consequence of the linearity of A.n and JJ-n. Hence ~ (t) = 11 (t ) 2 , since each of the initial pair is required to have no descendants at time t. Therefore
1J 1(t) = M- (A.+ JJ-)IJ(f) + A.17(t) 2
subject to 11 (0) = 0. Rewrite the equation as 11'
--,-,---------,___:_-----,-- = 1
(1 - IJ)(JJ- - A.IJ)
and solve using partial fractions to obtain if).=/)-,
Finally, if 0 < t < u, lP'(X(t) = 0 I X(u) = 0) = lP'(X(u) = 0 I X(t) = 0)
4.
lP'(X(t) = 0) lP'(X(u) = 0)
17(t) = -. 17(u)
The random variable X (t) has generating function JJ-(1 - s)- (M- A.s)e-t(A.-JL) G (s, t) = ;____:___ ___:____:__ ____:_..,.,-;------,).(1 - s)- (M- A.s)e-t(A.-JL)
as usual. The generating function of X(t), conditional on {X(t) > 0}, is therefore
~ sn lP'(X(t) = n) = G(s, t)- G(O, t). ~
lP'(X(t) > 0)
Substitute for G and take the limit as t
~
H(s)
=
1 - G(O, t)
oo to obtain as limit (M- A.)s /)--AS
oo
= Ls
n
Pn
n=l
where, with p = A./ JJ-, we have that Pn = pn-l (1 - p) for n 2:: 1.
298
Solutions
Special processes 5.
[6.11.5]-[6.12.1]
Extinction is certain if A. < JJ-, and in this case, by Theorem (6.11.10),
IfA > fJ- then lP'(T < oo) = JJ-/A., so JE(T
IT
< oo)
=
!oo
oo {
)..
1 - -lE(sX(t))l =O M s
}
dt
=
looo (A. - JJ-)e 0. Then Pu(m
+ r + n):::: apjj(r).
Set r = 0 to find that Pii (m + n) > 0, and so d(i) I (m + n). If d(i) f r then Pii (m + r that Pjj(r) = 0; therefore d(i) I d(j). Similarly d(j) I d(i), giving that d(i) = d(j).
+ n) =
0, so
4. (a) See the solution to Exercise (6.3.9a). (b) Let i, j, r, s E S, and choose N(i, r) and N(j, s) according to part (a). Then
I
lP'(Zn = (r, s) Zo = (i,
n)
= Pir(n)Pjs(n) > 0
if n:::: max{N(i, r), N(j, s)}, so that the chain is irreducible and aperiodic. (c) SupposeS= {1, 2} and
P=C 6)· In this case { {1, 1}, {2, 2}} and { {1, 2}, {2, 1}} are closed sets of states for the bivariate chain. 5. Clearly lP'(N = 0) = 1 - /ij, while, by conditioning on the time of the nth visit to j, we have that lP'(N :::: n + 1 I N :::: n) = fjj for n :::: 1, whence the answer is immediate. Now lP'(N = oo) = 1 - l:~o lP'(N = n) which equals 1 if and only if /ij = fjj = 1. 6. Fix i =/= j and let m = min{n : Pij (n) > 0}. If Xo = i and Xm = j then there can be no intermediate visit to i (with probability one), since such a visit would contradict the minimality of m.
304
Solutions
Problems
[6.15.7]-[6.15.8]
Suppose Xo = i, and note that (1 - f}i) Pij (m) :5 1 - Iii, since if the chain visits j at time m and subsequently does not return to i, then no return to i takes place at all. However fii = 1 if i is persistent, so that fj i = 1. 7.
(a) We may takeS= {0, 1, 2, ... }. Note that qij (n)::: 0, and 00
L%(n) =
%(n
1,
+ 1) = Lqu(l)qu(n),
j
1=0
whence Q = (% (1)) is the transition matrix of a Markov chain, and Qn = (% (n)). This chain is persistent since for all i, n
n
and irreducible since i communicates with j in the new chain whenever j communicates with i in the original chain. That i =I= j, n ::: 1,
is evident when n = 1 since both sides are qij (1). Suppose it is true for n = m where m ::: 1. Now lji(m
L
+ 1) =
i =I= j,
ljk(m)Pki·
k:kfj
so that
x·
_l_ lji(m
L
+ 1) =
Xi
i =I= j,
gkj(m)qik(1),
k:kfj
which equals gij (m + 1) as required. (b) Sum (*) over n to obtain that i =I= j,
where Pi(j) is the mean number of visits to i between two visits to j; we have used the fact that I:n gij (n) = 1, since the chain is persistent (see Problem (6.15.6)). It follows that Xi = XOPi (0) for all i, and therefore x is unique up to a multiplicative constant. (c) The claim is trivial when i = j, and we assume therefore that i =I= j. Let Ni (j) be the number of visits to i before reaching j for the first time, and write lP'k and lEk for probability and expectation conditional on Xo = k. Clearly, lP'j(Ni(j):::: r) = hjiO- hijy-l for r::: 1, whence
The claim follows by(**).
8.
(a) If such a Markov chain exists, then n
Un =
L
fiUn-i•
i=l
305
n:::
1,
[6.15.9]-[6.15.9] where
Solutions
Marlwv chains
fi is the probability that the first return of X to its persistent starting point s takes place at time
i. Certainly uo = 1. Conversely, suppose u is a renewal sequence with respect to the collection Um : m 2: 1). Let X be a Markov chain on the state space S = {0, 1, 2, ... } with transition matrix
.. _ { lP'(T 2: i + 2 I T 2: i + 1) PI] 1 - lP'(T 2: i + 2 I T 2: i + 1)
if j = i + 1, if j = 0,
where Tis a random variable having mass function fm = lP'(T = m). With Xo = 0, the chance that the first return to 0 takes place at time n is
lP'( Xn = 0,
IT
xi
i= 0 I Xo
= 0) = P01P12 ... Pn-2,n-1Pn-1,0
1
IT
= ( 1 _ G(n + 1)) G(i + 1) G(n) i=l G(i)
= G(n) -
G(n
+ 1) =
fn
where G(m) = lP'(T 2: m) = L:~m fn· Now Vn = lP'(Xn = 0 I Xo = 0) satisfies n
vo
= 1,
forn::;::1,
Vn = LfiVn-i i=l
whence Vn = Un for all n. (b) Let X andY be the two Markov chains which are associated (respectively) with u and v in the above sense. We shall assume that X andY are independent. The product (unvn: n 2: 1) is now the renewal sequence associated with the bivariate Markov chain (Xn, Yn).
2n (2n)
9. Of the first steps, let there be i rightwards, j upwards, and k inwards. Now Xzn = 0 if and only if there are also i leftwards, j downwards, and k outwards. The number of such possible !/ {(i! j! k!) 2 }, and each such combination has probability (t) 2U+ j+k) = (t) 2 combinations is The first equality follows, and the second is immediate. Now
n.
1 ) 2n lP'(Xzn = 0) ::S ( 2 where
M=max{ 3 n·~!·lkl: l. J . .
(2n) M n
L
i+J+k=n
n! 3n·1 "lkl
l.J. ·
i,j,k::;::O, i+j+k=n}.
It is not difficult to see that the maximum M is attained when i, j, and k are all closest to :} n, so that
Furthermore the summation in (*) equals 1, since the summand is the probability that, in allocating n balls randomly to three urns, the urns contain respectively i, j, and k balls. It follows tliat (2n)! lP'(X2 = 0) < -----'-----7----::n
- 12nn!
306
(Lj-nJ !)3
Solutions [6.15.10]-[6.15.13]
Problems
3
which, by an application of Stirling's formula, is no bigger than Cn -2 for some constant C. Hence l:n lP'(X2n = 0) < oo, so that the origin is transient.
10. No. The line of ancestors of any cell-state is a random walk in three dimensions. The difference between two such lines of ancestors is also a type of random walk, which in three dimensions is transient.
11. There are one or two absorbing states according as whether one or both of a and f3 equal zero. If af3 =f. 0, the chain is irreducible and persistent. It is periodic if and only if a = f3 = 1, in which case it has period 2.
IfO < af3 < 1 then 1r=
(a!f3'a:f3)
is the stationary distribution. There are various ways of calculating pn; see Exercise (6.3.3) for example. In this case the answer is given by
proof by induction. Hence as n ---+ oo. The chain is reversible in equilibrium if and only if n1 P12
= 7r2P21, which is to say that af3 = f3a
!
12. The transition matrix is given by
Pi}=
(NN-i)2 ifj=i+1, (NN- i) 2 if j = i, 1 - (Ni ) if j = i - 1, (Ni )2 2
_
for 0 :::; i :::; N. This process is a birth-death process in discrete time, and by Exercise (6.5.1) is reversible in equilibrium. Its stationary distribution satisfies the detailed balance equation Jri Pi,i+l = ni+lPi+l,i for 0:::; i < N, whence ni = no(~) 2 for 0:::; i :::; N, where
_!_
no
=
t (~) i=O
z
2
=
(2N). N
13. (a) The chain X is irreducible; all states are therefore of the same type. The state 0 is aperiodic, and so therefore is every other state. Suppose that Xo = 0, and let T be the time of the first return to 0. Then lP'(T > n) = aoa1 · · · an-1 = bn for n :::: 1, so that 0 is persistent if and only if bn ---+ 0 as n ---+ oo. (b) The mean of T is 00
JE(T)
=
00
L lP'(T > n) = L bn. 307
[6.15.14]-[6.15.14]
Solutions
Markov chains
The stationary distribution 1r satisfies 00
no= L 7l'k(l- ak), k=O
Jl'n = 7l'n-1Gn-1 for n 2: 1.
Hence Jl'n = nobn and n 1 = :L~o bn if this sum is finite. (c) Suppose ai has the stated form for i 2: I. Then
0
n-1
II (1 -
bn =hi
Ai-f3),
n 2: /.
i=I
Hence bn --+ 0 if and only if :Li Ai-fJ = oo, which is to say that persistent if and only if f3 ::=: 1. (d) We have that 1 - x :::: e-x for x 2: 0, and therefore
f3 ::=:
1. The chain is therefore
if{J 1, there is a constant c I such that oo oo { n-1 1 } oo oo Lbn::SbiLexp - A L i ::SciLexp{-Alogn}=ciLn-A 0 for some t. Now
implying that (Gn)ij > 0 for some n, which is to say that
for some sequence k1, k2, ... , kn of states. Suppose conversely that (*)holds for some sequence of states, and choose the minimal such value of n. Then i, k1, k2, ... , kn, j are distinct states, since otherwise n is not minimal. It follows that (Gn)ij > 0, while (Gm)ij = 0 for 0::: m < n. Therefore 00
Pij (t) = t
n"'l
k
k
L..J k! t (G )ij
k=n
is strictly positive for all sufficiently small positive values oft. Therefore i communicates with j.
16. (a) Suppose X is reversible, and let i and j be distinct states. Now lP'(X(O) = i, X(t) =
j)
= lP'(X(t) = i, X(O) =
j),
which is to say that Tri Pij (t) = Trj Pji (t). Divide by t and take the limit as t
+0 to obtain that
Trigij = Trjgji·
Suppose now that the chain is uniform, and X (0) has distribution 1r. If t > 0, then lP'(X (t) = j) =
2.:: Tri Pij (t) = Trj, i
so that X(t) has distribution 1r also. Now lett < 0, and suppose that X(t) has distribution J.L. The distribution of X (s) for s ::: 0 is J.LPs-t = 1r, a polynomial identity in the variables - t, valid for all s ::: 0. Such an identity must be valid for all s, and particularly for s = t, implying that J.L = 1r. Suppose in addition that Tri gij = Trj gji for all i, j. For any sequence k1, k2, ... , kn of states, Trigi,ktgkt,kz ···gkn,j = gkJ,iTrktgkt,kz · · ·gkn,j = ··· = gk1,igkz,k1 ·· ·gj,knTrj.
Sum this expression over all sequences k1, k2, ... , kn oflength n, to obtain Tri(Gn+1)ij = Trj(Gn+1)ji•
n::: 0.
It follows, by the fact that Pt = etG, that
foralli,j,t. Fort1 < t2 < ··· < tn, lP'(X(t1) = i1, X(t2) = i2, ... , X(tn) =in)
= Tri! Pit ,iz (t2 - t1) · · · Pin_ 1,in (tn - tn-1) = Piz,i1 (t2- tl)TrizPiz,i3 (t3- t2) · · · Pin-J,in Ctn- tn-1) = · · · = Pi2 ,i 1 (t2- t1) · · · Pin.in- 1 Ctn- tn-1)1rin = lP'(Y(t1) = i1, Y(t2) = i2, ... , Y(tn) =in),
309
[6.15.17]-[6.15.19]
Markov chains
Solutions
giving that the chain is reversible. (b) LetS = {1, 2} and
G
=(-a a) f3
-{3
where af3 > 0. The chain is uniform with stationary distribution
and therefore Jqgl2 = n2g2i· (c) Let X be a birth--death process with birth rates A.; and death rates J.Li. The stationary distribution n satisfies
Therefore nk+iiLk+i = Jl'kAk fork 2: 0, the conditions for reversibility.
17. Consider the continuous-time chain with generator G = (-{3
y
f3 ) .
-y
It is a standard calculation (Exercise (6.9.1)) that the associated semigroup satisfies ({3
+
+ f3h(t) y(1- h(t))
)P = ( y y
t
{3(1- h(t))) f3 + yh(t)
where h(t) = e-t(,B+y). Now P1 = Pifandonlyify + f3h(1) = f3 + yh(1) to say that f3 = y = -1log(2a - 1), a solution which requires that a > 1·
= a(f3 +
y), which is
18. The forward equations for Pn(t) = lP'(X(t) = n) are Pb(t) = JLPi - A.po,
P~ (t) = APn-i - (A.+ nJL) Pn + JL(n + 1) Pn+i, In the usual way,
aa at
=(s- 1)(A.G- 11
no::
I.
aa) as
with boundary condition G(s, 0) =sf. The characteristics are given by
ds
dt= - - JL(S - 1)
dG A.(s - 1)G'
and therefore G = eP(s-i) f ( (s-l)e-JU), for some function f, determined by the boundary condition to satisfy eP(s-l) f(s- 1) =sf. The claim follows. As t-+ oo, G(s, t)-+ eP(s-i), the generating function of the Poisson distribution, parameter p.
19. (a) The forward equations are
a
-p;;(s, t) = -A.(t)p;;(s, t), at
a
atPij(s,t) = -A.(t)pij(S, t) +A.(t)Pi,j-J(t),
310
i < j.
Solutions [6.15.20]-[6.15.20]
Problems Assume N(s) = i and s < t. In the usual way, 00
G(s, t; x)
= 2:xilP'(N(t) = j
I N(s)
= i)
J=i
satisfies iJG
at= A.(t)(x- 1)G.
We integrate subject to the boundary condition to obtain G(s, t; x) =xi exp { (x- 1)
1t
A.(u) du},
whence Pi} (t) is found to be the probability that A = j - i where A has the Poisson distribution with
J:
parameter A.(u) du. The backward equations are
a
osPij(S, t) = A.(s)Pi+l,j(S, t)- A.(s)Pij(S, t);
using the fact that Pi+ I,} (t) = Pi,J-1 (t), we are led to iJG - - = A.(s)(x- !)G.
OS
The solution is the same as above. (b) We have that lP'(T > t)
= Poo(t) = exp {
so that fT(t) = A.(t)exp {
In the case A.(t) = cj(l
-l
-
lot
A.(u) du},
t 2:0.
A.(u)du }•
+ t), E(T)
= !o oo lP'(T 0
> t)dt
= !ooo o
du
(1
+ u)c
which is finite if and only if c > 1. 20. Let s > 0. Each offer has probability 1 - F (s) of exceeding s, and therefore the first offer exceeding sis the Mth offer overall, where lP'(M = m) = F(s)m-l [I - F(s)], m 2: 1. Conditional on {M = m}, the value of XM is independent of the values of X1, X2, ... , XM-1, with lP'(XM > u I M = m) =
1- F(u) , I - F(s)
0<
S ::": U,
and X 1, X 2, ... , X M -I have shared (conditional) distribution function F(u) G(u Is)= F(s),
311
0:::: u:::: s.
[6.15.21]-[6.15.21]
Solutions
Markov chains
For any event B defined in terms of X 1, X 2, ... , X M -! , we have that 00
IP'(XM > u, B)=
L
IP'(XM > u, B
IM =
m)IP'(M = m)
m=i 00
=
L
IM =
IP'(XM > u
m)IP'(B
IM =
m)IP'(M = m)
m=i 00
= IP'(XM > u)
L IP'(B I M = m)IP'(M = m) m=i
0<
= IP'(XM > u)IP'(B),
S _:::: U,
where we have used the fact that IP'(XM > u I M = m) is independent of m. It follows that the first record value exceeding s is independent of all record values not exceeding s. By a similar argument (or an iteration of the above) all record values exceeding s are independent of all record values not exceeding s. The chance of a record value in (s, s + h] is IP'(s < XM < s +h)= -
F(s +h)- F(s)
1- F(s)
=
J(s)h
1- F(s)
+o(h).
A very similar argument works for the runners-up. Let XM 1 , XM2 , .•. be the values, in order, of offers exceeding s. It may be seen that this sequence is independent of the sequence of offers not exceeding s, whence it follows that the sequence of runners-up is a non-homogeneous Poisson process. There is a runner-up in (s, s + h] if (neglecting terms of order o(h)) the first offer exceeding s is larger than s + h, and the second is in (s, s + h]. The probability of this is ( 1-F(s+h)) (F(s+h)-F(s)) +o(h ) = f(s)h +o (h) . 1-F(s) 1-F(s) 1-F(s)
21. Let F 1 (x) = IP'(N*(t) _:::: x), and let A be the event that N has a arrival during (t, t +h). Then Fr+h(x) = A.h!P'(N*(t +h)_::::
xI A)+ (1- A.h)Fr(x) + o(h)
where IP'(N*(t +h)_::::
Hence
~Fr(x) = at
xI A)=
-A.Fr(x)
+
L:
t.../
00
-oo
Fr(x- y)f(y)dy.
Fr(x- y)f(y)dy.
Take Fourier transforms to find that ¢ 1 (8) = IE(eifiN*(t)) satisfies
an equation which may be solved subject to ¢o(8) = 1 to obtain ¢ 1 (8) = eAI((e)-ll. Alternatively, using conditional expectation,
where N (t) is Poisson with parameter A.t.
312
Solutions [6.15.22]-[6.15.25]
Problems
22. We have that E(sN(t)) = E{E(sN(t)
I A)}=
1{eJ.It(s-1)
+ eJ.2t(s-l)},
= ~(A.J + A.z)t and var(N(t)) = 1(A.J + A.z)t + i t), or in the sink with probability p = lP'(U +X < t < U +X+ Y), or departed with probability 1 - p - n. Thus, IE(xA(t)YB(t)) = IE{IE(xA(t)YB(t) I N(t))}
= IE{ (n x + py + 1 -
n - p )N(t)}
= eh(x-1) eJ..p(y-1).
Thus A(t) and B(t) are independent Poisson-distributed random variables. If the ants arrive in pairs and then separate,
319
[6.15.42]-[6.15.45]
Solutions
Markov chains
where y = 1- n- p. Hence,
whence A(t) and B(t) are not independent in this case.
42. The sequence {Xr} generates a Poisson process N(t) = max{n : Sn _:::: t}. The statement that Sn =tis equivalent to saying that there are n- 1 arrivals in (0, t), and in addition an arrival at t. By Theorem (6.12.7) or Theorem (6.13.11), the first n - 1 arrival times have the required distribution. Part (b) follows similarly, on noting that fu(u) depends on u = (UJ, u2, ... , un) only through the constraints on the Ur.
43. Let Y be a Markov chain independent of X, having the same transition matrix and such that Yo has the stationary distribution Jr. LetT= rnin{n ::: 1 : Xn = Yn} and suppose Xo = i. As in the proof of Theorem (6.4.17), IPij (n)- Tl:j I =
IL
nk(Pij (n)- Pkj (n))
k
I_: L
ll:klP'(T > n) = lP'(T > n).
k
Now,
lP'(T > r where
E
+1 IT
for r::: 0,
> r) _:::: 1- E2
= rnin;j {Pi}} > 0. The claim follows with "A = 1 -
E2 •
44. Let/k(n)betheindicatorfunctionofavisittokattimen,sothatE(h(n)) =lP'(Xn =k) =ak(n), say. By Problem (6.15.43), lak(n)- nkl _:::: ;..n. Now,
Lets = rnin{m, r} and t = lm- rl. The last summation equals
= n12
L L {(a;(s)- n;)(Pii (t)- n;) + n;(p;; (t)- n;) r
m
+ n; (a; (s) -
n;) - n; (a; (r) - n;) - n; (a; (m) - n;)}
1 _:: : 2 LL(;..s+t +"At +"As +"Ar +"Am) n r m as n--+ oo, where 0 < A < oo. For the last part, use the fact that I:~;:J f(Xr) = l:::iES f(i)V; (n). The result is obtained by Minkowski's inequality (Problem (4.14.27b)) and the first part.
45. We have by the Markov property that f(Xn+i I Xn, Xn-i, ... , Xo) = f(Xn+i I Xn), whence E(log f(Xn+i I Xn, Xn-J, ... , Xo) I Xn, ... , Xo) = E(log f(Xn+i I Xn) I Xn).
320
Problems
Solutions
[6.15.46]-[ 6.15.48]
Taking the expectation of each side gives the result. Furthermore, H(Xn+! I Xn) =- 2)Pij log Pij )lP'(Xn = i). i,j
Now X has a unique stationary distribution is finite, and the claim follows.
JC,
so that lP'(Xn = i) ---+ Jri as n ---+ oo. The state space
46. Let T = inf{t : X 1 = Y1 }. Since X and Y are persistent, and since each process moves by distance 1 at continuously distributed times, it is the case that lP'(T < oo) = 1. We define
Zt= {
Xt
if t < T,
Yt
ift 2: T,
noting that the processes X and Z have the same distributions. (a) By the above remarks, IIP'(Xt = k) -lP'(Yt = k)l = llP'(Zt = k) -lP'(Yt = k)l :::silP'CZt=k, T:::St)+lP'(Zt=k, T>t)-lP'(Yr=k, T:::St)-lP'(Yt=k, T>t)l :::S lP'(Xt = k, T > t)
+ lP'(Yt
= k, T > t).
We sum over k E A, and let t ---+ oo. (b) We have in this case that Z 1 :::S Y1 for all t. The claim follows from the fact that X and Z are processes with the same distributions. 47. We reformulate the problem in the following way. Suppose there are two containers, Wand N, containing n particles in all. During the time interval (t, t + dt), any particle in W moves toN with probability [tdt +o(dt), and any particle inN moves toW with probability A.dt +o(dt). The particles move independently of one another. The number Z(t) of particles in W has the same rules of evolution as the process X in the original problem. Now, Z (t) may be expressed as the sum of two independent random variables U and V, where U is bin(r, e1 ), Vis bin(n- r, 1/11), and e1 is the probability that a particle starting in W is in W at timet, 1/11 is the probability that a particle starting inN at 0 is in W at t. By considering the two-state Markov chain of Exercise (6.9.1),
and therefore E(sX(t)) = E(sU )E(s V) = (set
+1-
sr (s1flt
+1-
s)n-r.
Also, E(X(t)) = re 1 + (n- r)1/1 1 and var(X(t)) = re 1 (1- e1 ) + (n- r)1/f1 (1 -1/11 ). In the limit as n ---+ oo, the distribution of X (t) approaches the bin(n, A./(A. + ft)) distribution. 48. Solving the equations
gives the first claim. We have that y = l:i (Pi - qi )Jri, and the formula for y follows. Considering the three walks in order, we have that: A. Jri =
1for each i, and YA = -2a < 0.
B. Substitution in the formula for YB gives the numerator as 3{- ~a +o(a) }, which is negative for small a whereas the denominator is positive.
321
[6.15.49]-[6.15.51]
Solutions
Markov chains
i (Jb -
C. The transition probabilities are the averages of those for A and B, namely, Po = a)+ a) = a, and so on. The numerator in the formula for YC equals +o(l), which is positive for small a.
i E}. Hence EIX' I 2: Eli; I = lP'(IXI > E), implying that lP'(IXI > E) = 0 for all E > 0. The converse is trivial.
2.
(a) E({aX
+ bY}Z) = + E({X -
(b) E({X + Y} 2 ) (c) Clearly
aE(XZ)
+ bE(YZ). + 2E(Y 2 ).
Y} 2 ) = 2E(X 2 )
3. Let f(u) = ~E, g(u) = 0, h(u) = -~E, for all u. Then df(f, g)+ dE(g, h) = 0 whereas df(f, h)= 1. ~
Either argue directly, or as follows. With any distribution function F, we may associate a graph F obtained by adding to the graph of F vertical line segments connecting the two endpoints at each discontinuity ofF. By drawing a picture, you may see that .../2 d (F, G) equals the maximum distance between F and G measured along lines of slope -1. It is now clear that d(F, G) = 0 if and only if F = G, and that d(F, G) = d(G, F). Finally, by the triangle inequality for real numbers, we have that d(F, H) ~ d(F, G)+ d(G, H). 5.
Take X to be any random variable satisfying E(X 2 ) = oo, and define Xn = X for all n.
7.2 Solutions. Modes of convergence 1.
(a) By Minkowski's inequality,
let n --+ oo to obtain lim infn---+oo EIX~ I 2: EIX' 1. By another application ofMinkowski's inequality,
whencelimsupn 400 EIX~I ~ EIX'I.
323
[7.2.2]-[7.2.4]
Solutions
Convergence of random variables
(b) We have that IE(Xn)- E(X)I = IE(Xn- X) I ~ EIXn- XI--+ 0 as n --+ oo. The converse is clearly false. If each Xn takes the values ±1, each with probability then E(Xn) = 0, but EIXn - 01 = 1.
i,
(c) By part (a), E(X~) --+ E(X 2 ). Now Xn _..;. X by Theorem (7.2.3), and therefore E(Xn) --+ E(X) by part (b). Therefore var(Xn) = E(X~) - E(Xn) 2 --+ var(X).
2. Assume that Xn ~ X. Since IXnl ~ Z for all n, it is the case that lXI Zn = IXn -XI satisfies Zn ~ 2Z a.s. In addition, if E > 0,
~ Z a.s. Therefore
As n--+ oo, lP'(IZnl >E) --+ 0, and therefore the last term tends to 0; to see this, use the fact that E(Z) < oo, together with the result of Exercise (5.6.5). Now let E ,!.. 0 to obtain that EIZnl --+ 0 as n --+ oo. 3. We have that X- n- 1 ~ Xn ~ X, so that E(Xn) --+ E(X), and similarly E(Yn) --+ E(Y). By the independence of Xn and Yn,
E
{( 1) ( 1) } X - -;;_
Y - -;;_
= E(XY) -
E(X) + E(Y) 1 n + n 2 --+ E(XY)
as n --+ oo, so that E(Xn Yn) --+ E(XY).
4. Let F1, F2, ... be distribution functions. As in Section 5.9, we write Fn --+ F if Fn(x) --+ F(x) for all x at which F is continuous. We are required to prove that Fn --+ F if and only if d (Fn, F) --+ 0. Suppose that d(Fn, F) --+ 0. Then, forE > 0, there exists N such that F(x- E)- E ~ Fn(x)
~
F(x +E)+ E
for all x.
Take the limits as n --+ oo and E --+ 0 in that order, to find that Fn (x) --+ F (x) whenever F is continuous at x. Suppose that Fn --+ F. Let E > 0, and find real numbers a = x1 < x2 < · · · < Xn = b, each being points of continuity of F, such that (i) Fi(a) < E for all i, F(b) > 1- E, (ii) lxi+1 -Xi I < E for 1 ~ i < n. In order to pick a such that Fi (a) < E for all i, first choose a 1 such that F(a 1) < iE and F is continuous at a', then find M such that IFm (a')- F(a 1 ) I < iE form ::::: M, and lastly find a continuity point a ofF such that a~ a 1 and Fm(a) < E for 1 ~ m < M. There are finitely many points Xi, and therefore there exists N such that IFm(Xi)- F(xi)l < E for all i and m ::::: N. Now, if m ::::: N and Xi ~ x < Xi+ 1,
and similarly Fm(x)::::: Fm(Xi) > F(xi)- E :0:: F(x- E)- E.
Similar inequalities hold if x d(Fm, F) --+ 0 as m --+ oo.
~
a or x ::::: b, and it follows that d(Fm, F) < E if m ::::: N. Therefore
324
Modes of convergence 5.
n:::::
Solutions [7.2.5]-[7.2.7]
(a) Suppose c > 0 and pick 8 such that 0 < 8 < c. Find N such that lP'(/Yn - cJ > 8) < 8 for N. Now, for x::::: 0,
lP'(XnYn.::; x) .::; IP'(XnYn .::; X, /Yn- cJ .:'S
8) +
IP'(/Yn- c/ >
8)
.:'S lP' ( Xn .:'S c ~ 8 ) + 8,
and similarly
lP'(XnYn > x).::; IP'(XnYn >X, /Yn- cJ .:'S
8)
+ 8 .:'S lP' ( Xn > c: 8 ) + 8.
Taking the limits as n --+ oo and 8 .,),. 0, we find that lP'(XnYn .::; x) --+ lP'(X .::; xfc) if xfc is a point of continuity of the distribution function of X. A similar argument holds if x < 0, and we conclude that XnYn
~ eX if c > 0. No extra difficulty arises if c < 0, and the case c = 0 is similar.
For the second part, it suffices to prove that Y,;- 1 ~ c- 1 if Yn ~ c (# 0). This is immediate from the fact that IYn- 1 - c- 1 1 < E/{/c/(/c/- E)} if /Yn - c/ < E ( < Jcl). (b) Let E > 0. There exists N such that
lP'(/Xn/ >E)< E,
IP'(/Yn- Y/ >E)< E,
ifn
::0:: N,
and in addition lP'(/YI > N) 0) such that
/g(x 1, y 1) - g(O, y)/ <
if Jx'J .::; 8, Jy'-
E
y/ .::; 8.
If /Xn/.::; 8, /Yn- Y/ .::; 8, and IYI .::; N, then Jg(Xn, Yn)- g(O, Y)J < E, so that
IP'(/g(Xn, Yn)- g(O, Y)J
::0:: E).::;
for n::::: N. Therefore g(Xn, Yn)
6.
lP'(/Xn/ > 8) + IP'(IYn- Y/ > 8) + lP'(/Y/ > N).::; 3E,
p ---+ g(O,
Y) as n--+ oo.
The subset A of the sample space Q may be expressed thus:
nU n 00
A=
00
00
{/Xn+m- Xn/ < k- 1 },
k=1 n=1m=1
a countable sequence of intersections and unions of events. For the last part, define
X(w) = {
limn---+oo Xn(w)
ifw E A
0
ifw
rf. A.
The function X is .1"-measurable since A E :F.
7. (a) If Xn(w)--+ X(w) then cnXn(w)--+ cX(w). (b) We have by Minkowski's inequality that, as n --+ oo,
E(/cnXn- cXJ').::; /cn/'E(/Xn- XI')+ /en- cJ'EJX'J--+ 0. (c) If c = 0, the claim is nearly obvious. Otherwise c # 0, and we may assume that c > 0. For 0 < E < c, there exists N such that /en - cJ < E whenever n ::::: N. By the triangle inequality, /cnXn- eX/ .::; /cn(Xn- X)/+ /(en- c)X/, so that, for n ::0:: N,
IP'(/cnXn- eX/> E).::; IP'(cn/Xn- X/> iE) + IP'(/cn- cJ·JXJ > iE) .:'S lP' (/Xn -X/ > --+ 0
E ) + lP' (/X/ > E ) 2(c +E) 2/cn - cJ as n --+ oo.
325
[7.2.8]-[7.3.1]
Solutions
Convergence of random variables
_!;. X, find random variables ~ cY, so that cnYn _!;. cY,
(d) A neat way is to use the Skorokhod representation (7.2.14). If Xn Yn, Y with the same distributions such that Yn implying the same conclusion for the X's.
8.
~
Y. Then cnYn
If X is not a.s. constant, there exist real numbers c and E such that 0 < E <
lP'(X > c +E) > 2E. Since Xn
i and lP'( X < c) > 2E,
!.. X, there exists N such that
lP'(XnE,
lP'(Xn>c+E)>E,
ifn;:::N.
Also, by the triangle inequality, IXr- Xsl _:::: IXr- XI+ IXs- XI; therefore there exists M such that IP'(IX, - Xs I > E) < E3 for r, s ::::: M. Assume now that the Xn are independent. Then, for r, s ::::: max{M, N}, r =f. s, E3 > IP'(IXr - Xs I > E)
IP'(Xr < c, Xs > c +E) = lP'(Xr < c)lP'(Xs > c +E) > E2 ,
::0::
a contradiction. 9. Either use the fact (Exercise (4.12.3)) that convergence in total variation implies convergence in distribution, together with Theorem (7.2.19), or argue directly thus. Since luOI :::: K < oo, IE(u(Xn))- E(u(X))I = jL:u(k){fn(k)- /(k)}l _:::: KL lfn(k)- /(k)l--+ 0. k
10. The partial sum Sn
= 2::~= 1
k
X, is Poisson-distributed with parameter an
= 2::~= 1 A.,.
For fixed
x, the event {Sn _:::: x} is decreasing inn, whence by Lemma (1.3.5), if an --+ a < oo and xis a non-negative integer, oo
lP' ( LXr r=1
)
:::=x =n~~lP'(Sn ::;x)
=
x
-crj
)=0
} ·
L~·
Hence if a < oo, 2::~ 1 X, converges to a Poisson random variable. On the other hand, if an --+ oo,
L:J=
then e-crn 0 aj I j! --+ 0, giving that IP'(2::~ 1 X, > x) diverges with probability 1, as required.
=
I for all
x,
and therefore the sum
7.3 Solutions. Some ancillary results 1.
(a) If IXn - Xm I > E then either IXn- XI > iE or IXm -XI > iE, so that
as n, m --+ oo, forE > 0. Conversely, suppose that {Xn} is Cauchy convergent in probability. For each positive integer k, there exists nk such that forn,
m::::: nk.
The sequence (nk) may not be increasing, and we work instead with the sequence defined by N1 = n1, Nk+1 = max{Nk + 1, nk+d· We have that LlP'(IXNk+J- XNkl::::: Tk) < k
326
00,
Solutions
Some ancillary results
[7.3.2]-[7.3.3]
whence, by the first Borel-Cantelli lemma, a.s. only finitely many of the events {IXNk+l - XNk I 2:: 2-k} occur. Therefore, the expression 00
X= XN 1 + L(XNk+i - XNk) k=1 converges absolutely on an event C having probability one. Define X(w) accordingly for wE C, and X(w) = 0 for w rf. C. We have, by the definition of X, that XNk ~ X ask-+ oo. Finally, we 'fill in the gaps'. As before, forE > 0,
as n, k-+ oo, where we are using the assumption that {Xn} is Cauchy convergent in probability. (b) Since Xn
!.. X, the sequence {Xn} is Cauchy convergent in probability. Hence IP'(IYn- Yml >E)= IP'(IXn- Xml >E)-+ 0
as n, m-+ oo,
forE > 0. Therefore {Yn} is Cauchy convergent also, and the sequence converges in probability to some limit Y. Finally, Xn 2.
.£.. X and Yn .£.. X, so that X andY have the same distribution.
Since An ~ U;;'=n Am, we have that
lJ
lJ
limsuplP'(An) _:::: lim lP'( Am) = lP'( lim Am) = lP'(An i.o.), n--+oo n--+oo m=n n--+oo m=n where we have used the continuity of lP'. Alternatively, apply Fatou's lemma to the sequence (4 ~ of indicator functions. 3. (a) Suppose Xzn = 1, Xzn+1 = -1, for n 2:: 1. Then {Sn = 0 i.o.} occurs if X 1 = -1, and not if X 1 = 1. The event is therefore not in the tail a-field of the X's. (b) Here is a way. As usual, lP'(Szn = 0) =
e:)
{p(l - p W, so that
LlP'(Szn = 0) < oo
if p
=f.
i.
n
implying by the first Borel-Cantelli lemma that lP'(Sn = 0 i.o.) = 0. (c) Changing the values of any finite collection of the steps has no effect on I = lim inf Tn and J = lim sup Tn, since such changes are extinguished in the limit by the denominator '.[ii'. Hence I and J are tail functions, and are measurable with respectto the tail a -field. In particular, {I _:::: - x} n {J ::::: x} lies in the a -field. Take x = 1, say. Then, lP'(/ _:::: -1) = lP'(J 2:: 1) by symmetry; using Exercise (7.3.2) and the central limit theorem, lP'(J 2:: 1) 2:: lP'(Sn 2:: .[ii i.o.) 2:: lim sup lP'(Sn 2:: n--+oo
,Jfi,) =
1 - (l) > 0,
where is theN (0, 1) distribution function. Since {J 2:: 1} is a tail event of an independent sequence, it has probability either 0 or 1, and therefore lP'(/ _:::: -1) = lP'(J 2:: 1) = 1, and also lP'(/ _:::: -1, J 2:: 1) = 1. That is, on an event having probability one, each visit of the walk to the left of -.[ii is followed by a visit of the walk to the right of .[ii, and vice versa. It follows that the walk visits 0 infinitely often, with probability one.
327
[7.3.4]-[7.3.8]
Solutions
Convergence of random variables
4. Let A be exchangeable. Since A is defined in terms of the Xi, it follows by a standard result of measure theory that, foreachn, thereexistsaneventAn E a(X1, X2, ... , Xn), suchthatlP'(Ab.An)--+ 0 as n --+ oo. We may express An and A in the form An = {Xn E Bn},
where Xn
A = {X E B},
= (X 1, X2, ... , Xn), and Bn and Bare appropriate subsets ofll~n and JR 00 • Let A~ ={X~
E
Bn},
A'= {X 1 E B},
where X~ = (Xn+1, Xn+2, ... , X2n) and X'= (Xn+1, Xn+2, ... , X2n, X1, X2, ... , Xn, X2n+1, X2n+2, · · · ). Now !P'(An n A~) = lP'(An)lP'(A~), by independence. Also, !P'(An) = lP'(A~), and therefore
!P'(A
By the exchangeability of A, we have that lP'(A t. A~) = !P'(A' t. A~), which in turn equals t. An), using the fact that the Xi are independent and identically distributed. Therefore, liP'( An
n A~)
- lP'(A) I ::: lP'(A
t.
An)
+ !P'(A t. A~) --+ 0
as n --+ oo.
Combining this with(*), we obtain that lP'(A) = lP'(A) 2, and hence lP'(A) equals 0 or 1.
5. The value of Sn does not depend on the order of the first n steps, but only on their sum. If Sn = 0 i.o., then S~ = 0 i.o. for all walks {S~} obtained from {Sn} by permutations of finitely many steps. 6. Since f is continuous on a closed interval, it is bounded: If (y) I ::: c for all y E [0, 1]for some c. Furthermore f is uniformly continuous on [0, 1], which is to say that, if E > 0, there exists 8 (> 0), such that lf(y)- f(z)l < E if IY- zl :S 8. With this choice of E, 8, we have that IE(Z/Ac)l < E, and IE(ZIA)I ::: 2clP'(A) ::: 2c ·
x(1 - x)
n 82
by Chebyshov's inequality. Therefore 2c IE(Z)I < E + n 82 , which is less than 2E for values of n exceeding 2c/(E8 2).
7. If {Xn} converges completely to X then, by the first Borel-Cantelli lemma, IXn -XI > E only finitely often with probability one, for all E > 0. This implies that Xn ~ X; see Theorem (7.2.4c). Suppose conversely that {Xn} is a sequence of independent variables which converges almost surely to X. By Exercise (7.2.8), X is almost surely constant, and we may therefore suppose that Xn ~ c where c E JR. It follows that, for E > 0, only finitely many of the (independent) events {IXn - cl > E} occur, with probability one. Using the second Borel-Cantelli lemma, LlP'(IXn- cl >E)<
00.
n
8.
Of the various ways of doing this, here is one. We have that
(n) 2
-1
" ' XiXj L · · 1~l1+E logn nl+E'
for lEI < 1.
By the Borel-Cantelli lemmas, the events An = {Xn/logn -1 < E ::=: 0, and a.s. only finitely often for E > 0.
~
1 + E} occur a.s. infinitely often for
10. (a) Mills's ratio (Exercise (4.4.8) or Problem (4.14.1c)) informs us that 1- 0, and the Borel--Cantelli lemmas imply the claim. (b) This is an easy implication of the Borel--Cantelli lemmas. 11. Let X be uniformly distributed on the interval [-1, 1], and define Xn = /{X:S(-l)n /n)· The distribution of Xn approaches the Bernoulli distribution which takes the values ± 1 with equal probability The median of X n is 1 if n is even and - 1 if n is odd.
i.
12. (i) We have that
for x > 0. The result follows by the second Borel--Cantelli lemma. (ii) (a) The stationary distribution 1l is found in the usual way to satisfy
Hence nk = {k(k + 1)}- 1 fork~ 1, a distribution with mean L;~ 1 (k + 1)- 1 (b) By construction, lP'(Xn ::=: Xo + n) = 1 for all n, whence
. sup -Xn ::=: 1) lP' ( hm n-+oo n It may in fact be shown that lP'(lim supn-+oo Xn/n
=
= 0) = 1.
329
1.
= oo.
[7 .3.13]-[7.4.3]
Solutions
Convergence of random variables
13. We divide the numerator and denominator by Jna. By the central limit theorem, the former converges in distribution to the N(O, 1) distribution. We expand the new denominator, squared, as
-
12:n
na 2
(Xr -
~-t)
2
2-
- -(Xna 2
r=1
~-t)
I:n (Xr
-
~-t)
1-
+-(Xa2
r=1
~-t)
2
.
By the weak law of large numbers (Theorem (5.10.2), combined with Theorem (7.2.3)), the first term converges in probability to 1, and the other terms to 0. Their sum converges to 1, by Theorem (7.3.9), and the result follows by Slutsky's theorem, Exercise (7.2.5).
7.4 Solutions. Laws of large numbers 1.
LetSn=X1+X2+···+Xn.Then n
.
2
'"'L 1 -< nE(S 2) = n i= 2 log i - log n
and therefore Snfn ~ 0. On the other hand, I::i lP'(IXil 2: i) = 1, so that IXil 2: i i.o., with probability one, by the second Borel-Cantelli lemma. For such a value of i, we have that ISi - Si -11 2: i, implying that Sn/ n does not converge, with probability one.
2.
Let the Xn satisfy
1 n
lP'(Xn = -n) = 1 - 2•
lP'(Xn
= n3 -
n)
= 21 , n
whence they have zero mean. However,
implying by the first Borel-Cantelli lemma that lP'(Xnfn-+ -1) = 1. It is an elementary result of real analysis that n - 1 I::~= 1 Xn -+ -1 if Xn -+ -1, and the claim follows. 3. The random variable N(S) has mean and variance AlSI = crd, where cis a constant depending only on d. By Chebyshov's inequality,
lP'
(I
I )
N(S) ISl-A :::
E
A (A) ~
.:5 E21SI =
By the first Borel-Cantelli lemma, I1Ski- 1N(Sk)- AI 2:
E
2
era·
for only finitely many integers k, a.s.,
where Skis the sphere of radius k. It follows that N(Sk)/ISkl ~ A ask-+ oo. The same conclusion holds ask -+ oo through the reals, since N(S) is non-decreasing in the radius of S.
330
Solutions
Martingales
[7.5.1]-[7.7.1]
7.5 Solutions. The strong law 1.
Let Iij be the indicator function of the event that X J lies in the i th interval. Then n
log Rm
=L
n
Zm (i) log Pi
=L
i=l
where, for 1 .::0 j .::0 m, Yj variables with mean
m
L
m
lij log Pi
=L
i=l }=I
l:t=I Iij
Yj
}=I
log Pi is the sum of independent identically distributed n
lE(Yj) =
L Pi log Pi = -h. i=l
By the strong law, m- 1 log Rm ~ -h. 2. The following two observations are clear: (a) N(t) < n if and only if Tn > t, (b) TN(t) .::0 t < TN(t)+I· If lE(X I) < oo, then lE(Tn) < oo, so that lP'(Tn > t) -+ 0 as t -+ oo. Therefore, by (a), lP'(N(t) < n) = lP'(Tn > t) -+ 0
as t-+ oo,
implying that N(t) ~ oo as t-+ oo. Secondly, by (b), TN(t) < _t_ < TN(t)+I . ( 1 + N(t)-I). N(t) - N(t) N(t) + 1 Take the limit as t -+ oo, using the fact that Tn/n ~ lE(XJ) by the strong law, to deduce that t/N(t) ~ lE(XJ).
By the strong law, Sn/n ~ lE(XJ) =I= 0. In particular, with probability 1, Sn = 0 only finitely often.
3.
7.6 Solution. The law of the iterated logarithm 1.
The sum Sn is approximately N(O, n), so that -~a
lP'(Sn > ylanlogn) = 1- (Jalogn) <
~ alogn
for all large n, by the tail estimate of Exercise (4.4.8) or Problem (4.14.1c) for the normal distribution. This is summable if a > 2, and the claim follows by an application of the first Borel-Cantelli lemma.
7.7 Solutions. Martingales 1.
Suppose i < j. Then
_I]} = lE{ Xi [lE(SJ I So, S1, ... , SJ-I)- s1_I]} = 0
lE(XjXi) = lE{ lE[(SJ- Sj-I)Xi
331
I So, S1, ... , s1
[7. 7.2]-[7.8.2]
Solutions
Convergence of random variables
by the martingale property. 2.
Clearly EISnl < oo for all n. Also, for n 2: 0,
1 { ( 1 _ p,n+1)} E(Sn+1 I Zo, z1' ... ' Zn) = p,n+1 E(Zn+1 I Zo, ... ' Zn)- m 1 - JL 1 { ( 1 _ p,n+1)} = p,n+ 1 m+p,Zn-m =Sn. 1 _p, 3.
CertainlyEISnl < ooforalln. Secondly,forn 2:1, E(Sn+1 I Xo, X1, ... , Xn) = aE(Xn+1 I Xo, ... , Xn) = (aa + 1)Xn +abXn-1,
+ Xn
which equals Sn if a= (1 - a)- 1. 4. The gambler stakes Zi = /i-1 (X 1, ... , Xi-1) on the ith play, at a return of Xi per unit. Therefore Si = Si-1 + XiZi fori 2: 2, with S1 = X1Y. Secondly,
where we have used the fact that Zn+1 depends only on X1, X2, ... , Xn.
7.8 Solutions. Martingale convergence theorem 1. It is easily checked that Sn defines a martingale with respect to itself, and the claim follows from the Doob-Kolmogorov inequality, using the fact that n
E(S~) = I:var(Xj). }=1
2. It would be easy but somewhat perverse to use the martingale convergence theorem, and so we give a direct proof based on Kolmogorov 's inequality of Exercise (7 .8 .1 ). Applying this inequality to the sequence Zm, Zm+1' ... where Zi =(Xi- EXi)/i, we obtain that Sn = Z1 + Z2 + · · · + Zn satisfies, for E > 0, lP' ( max ISn- Sml m E ) ::0 2 2 var(Xn). n:=::m E n=m+1 n
.2:::
Now let m -+ oo to obtain (after a small step) lP' ( lim sup ISn - Sm I ::0 m-;.oo n:=::m
E) = 1
for all
E
Any real sequence (xn) satisfying lim sup lxn -
m-;.oo n:=::m
Xm
I ::0
332
E
for all
E
> 0,
> 0.
Solutions
Prediction and conditional expectation
[7.8.3]-[7.9.4]
is Cauchy convergent, and hence convergent. It follows that Sn converges a.s. to some limit Y. The last part is an immediate consequence, using Kronecker's lemma. 3. By the martingale convergence theorem, S = limn-+oo Sn exists a.s., and Sn Exercise (7.2.1c), var(Sn) --+ var(S), and therefore var(S) = 0.
m.s.
~
S. Using
7.9 Solutions. Prediction and conditional expectation 1. (a) Clearly the best predictors are lE(X (b) We have, after expansion, that
I Y) = Y 2 , lE(Y I X) = 0.
since lE(Y) = lE(Y 3 ) = 0. This is a minimum when b = JE(Y 2 ) = ~.and a = 0. The best linear predictor of X given Y is therefore ~ . Note that lE(Y I X) = 0 is a linear function of X; it is therefore the best linear predictor of Y given X. 2.
By the result of Problem (4.14.13), lE(Y I X) =
3.
Write
~-tz
+ paz(X -
~-tdfal,
in the natural notation.
n
g(a) = LaiXi =aX', i=l
and v(a) = lE{ (Y- g(a)) 2 } = lE(Y 2 ) - 2alE(YX1) + aVa'. Let a be a vector satisfying Vi' = lE(YX'). Then v(a) - v(a) = aVa' - 2alE(YX1) + 2ilE(YX1) -iVa' = aVa'- 2aVa' +iVa'= (a- a)V(a- a)' :::: 0, since Vis non-negative definite. Hence v(a) is a minimum when a =a, and the answer is g(i). If V is non-singular, a= JE(YX)V- 1 . 4. Recall that Z = JE(Y I fj,) is the ('almost') unique fJ,-measurable random variable with finite mean and satisfying JE{(Y- Z)/G} = 0 for all G E fJ,.
(i) Q E fJ,, and hence lE{lE(Y I fj,)/n} = lE(Z/n) = lE(Y /g). (ii) U = alE(Y I fj,) + ,BlE(Z I fj,) satisfies JE(U /G) = alE{lE(Y I fj,)/G} + ,BlE{lE(Z I fj,)/G} = alE(Y I G)+ ,BlE(Z/G) = lE{ (aY + ,BZ)IG }, Also, U is fJ,-measurable. (iii) Suppose there exists m (> 0) such that G = {lE(Y I fj,) < -m} has strictly positive probability. Then G E fJ,, and so JE(Y /G) = lE{lE(Y I fj,)/G}. However Y IG :::: 0, whereas lE(Y I fj,)/G < -m. We obtain a contradiction on taking expectations. (iv) Just check the definition of conditional expectation. (v) If Y is independent of fJ,, then lE(Y /G)= lE(Y)lP'(G) for G E fJ,. Hence JE{(Y -JE(Y))/G} = 0 for G E fJ,, as required.
333
[7.9.5]-[7.10.1]
Solutions
Convergence of random variables
(vi) If g is convex then, for all a E JR, there exists A.(a) such that g(y) 2: g(a)
+ (y- a)A.(a);
furthermore A. may be chosen to be a measurable function of a. Set a = E(Y I fJ,) andy = Y, to obtain g(Y) ::: g{E(Y I fJ,)} + {Y - E(Y I fJ,) }A.{E(Y I fJ,) }. Take expectations conditional on g,, and use the fact that E(Y I fJ,) is [J,-measurable. (vii) We have that IECYn I fJ.) -E(Y I fJ.)I :':: E{IYn- Yll fJ.} :':: Vn where Vn = E{supm;c:n IYm- Yll fJ.}. Now {Vn : n::: 1} is non-increasing and bounded below. Hence V = limn--+oo Vn exists and satisfies V ::: 0. Also E(V) : E} satisfies lP'(Bn(E)) < 8 for n > N. Consequently lE(IXn-
xn S
Er
+ E (IXn- Xlr IBn(E)),
n > N,
of which the final term satisfies {lE (IXn - Xlr IBn(E))} lfr S {lE (IX~IIBn(E))} lfr
+ {lE (IXr IIBn(E))} l/r
S 2El/r.
Therefore, Xn ~X. 3.
FixE > 0, and find a real number a such that g(x) > xjE if x >a. If b 2: a, lE (IXnii(IXnl>bJ) < ElE{g(IXnl)} S EsupE{g(IXnl)}, n
whence the left side approaches 0, uniformly inn, as b --+ oo.
4.
Here is a quick way. Extinction is (almost) certain for such a branching process, so that Zn ~ 0,
and hence Zn ~ 0. If {Zn : n 2: 0} were uniformly integrable, it would follow that JE(Zn) --+ 0 as n --+ oo; however lE(Zn) = 1 for all n. 5.
We may suppose that Xn, Yn, and Zn have finite means, for all n. We have that 0 S Yn - Xn S p
Zn- Xn where, by Theorem (7.3.9c), Zn- Xn-+ Z- X. Also
lEIZn- Xn I = lE(Zn- Xn)--+ lE(Z- X)
= lEIZ- XI,
so that {Zn - Xn : n 2: 1} is uniformly integrable, by Theorem (7.10.3). It follows that {Yn - Xn : n 2: 1} is uniformly integrable. Also Yn - Xn ~ Y- X, and therefore by Theorem (7.10.3), lEIYn-Xnl-+ lEIY-XI,whichistosaythatlE(Yn)-lE(Xn)--+ lE(Y)-lE(X);hencelE(Yn)--+ E(Y). It is not necessary to use uniform integrability; try doing it using the 'more primitive' Fatou's lemma.
6. For any event A, lE(IXniiA) S JE(ZIA) where Z = supn IXnl· The uniform integrability follows by the assumption that lE(Z) < oo.
335
[7.11.1]-[7.11.2]
Solutions
Convergence of random variables
7.11 Solutions to problems 1.
lEI X~ I = oo for r 2: 1, so there is no convergence in any mean. However, if E > 0, 2
lP'(IXnl >E)= 1 - - tan- 1 (nE)-+ 0
as n-+ oo,
T(
p
so that Xn -+ 0. You have insufficient information to decide whether or not Xn converges almost surely: (a) Let X be Cauchy, and let Xn = Xjn. Then Xn has the required density function, and Xn ~ 0. (b) Let the Xn be independent with the specified density functions. ForE > 0,
2 sin- 1 ( lP'(IXnl >E)=n
1
,h + n2E2
) "'2, nnE
so that Ln lP'(IXnl >E)= oo. By the second Borel-Cantelli lemma, IXnl > E i.o. with probability one, implying that Xn does not converge a.s. to 0. 2. (i) Assume all the random variables are defined on the same probability space; otherwise it is meaningless to add them together. (a) Clearly Xn(w) + Yn(w) -+ X(w) + Y(w) whenever Xn(w) -+ X(w) and Yn(w) -+ Y(w). Therefore { Xn + Yn...,. X+ Y} ~ {Xn...,. X} U {Yn...,. Y}, a union of events having zero probability. (b) Use Minkowski's inequality to obtain that {lE (IXn + Yn -X- Yn} 1/r .::0 {lE(IXn -
xn} 1/r +
{lE(IYn - Yn} 1/r.
(c) If E > 0, we have that
and the probability of the right side tends to 0 as n -+ oo. (d) If Xn ~ X and the Xn are symmetric, then -Xn ~ X. However Xn + (-Xn) ~ 0, which generally differs from 2X in distribution. (ii) (e) Almost-sure convergence follows as in (a) above. (f) The corresponding statement for convergence in rth mean is false in general. Find a random variable Z such that lEIZrl < oo but lEIZ 2rl = oo, and define Xn = Yn = Z for all n. p
p
(g) Suppose Xn -+ X and Yn -+ Y. Let E > 0. Then
lP'(IXnYn- XYI >E)= lP'(I(Xn- X)(Yn- Y) + (Xn- X)Y + X(Yn-
nl >E)
.::0 lP'(IXn- XI·IYn- Yl >~E) +lP'(IXn- XI·IYI >~E)
+ lP'(IXI ·IYn- Yl > ~E). Now, for 8 > 0, lP'(IXn- XI·IYI >~E) .::0 lP'(IXn- XI> E/(38)) +lP'(IYI >
336
8),
Solutions
Problems
[7 .11.3]-[7.11.5]
which tends to 0 in the limit as n --+ oo and 8 --+ oo in that order. Together with two similar facts, we obtain that XnYn ~ XY. (h) The example of (d) above indicates that the corresponding statement is false for convergence in distribution. 3. Let E > 0. We may pick M such that lP'(IXI 2: M) .::::E. The continuous function g is uniformly continuous on the bounded interval [- M, M]. There exists 8 > 0 such that lg(x)- g(y)l If lg(Xn)- g(X)I >
E,
.:::= E
.:::=
8 and lxl
.:::=
M.
then either IXn- XI > 8 or lXI 2: M. Therefore
lP'(Ig(Xn)- g(X)I
>E) _:: : lP'(IXn- XI > 8)
in the limit as n --+ oo. It follows that g(Xn)
4.
if lx- Yl
~
+ JI»(IXI 2:
M)
--+ lP'(IXI 2:
M) _:: : E,
g(X).
Clearly lE(eitXn)
n . n { 1 1 =II lE(eitYj/101) =II _. - e . . . . 10 1 _ eztf10J
it/10j-l }
]=1
-
]=1
1- eit wn(l-eitjlOn)
1 - eit --+----
it
as n --+ oo. The limit is the characteristic function of the uniform distribution on [0, 1]. Now Xn .::; Xn+1 _:::: 1 for all n, so that Y(w) = limn---+oo Xn(w) exists for all w. Therefore Xn
~
5.
(a) Supposes < t. Then
Y; hence Xn
Er Y, whence Y has the uniform distribution.
lE(N(s)N(t)) = JE(N(s) 2) +lE{N(s)(N(t)- N(s))} = JE(N(s) 2) +lE(N(s))lE(N(t)- N(s)), since N has independent increments. Therefore cov(N(s), N(t)) = lE(N(s)N(t)) -JE(N(s))JE(N(t)) = (A.s) 2 + A.s + A.s{A.(t- s)}- (A.s)(A.t) = A.s. In general, cov(N(s), N(t)) =A. min{s, t}. (b) N(t +h)- N(t) has the same distribution as N(h), if h > 0. Hence
which tends to 0 as h --+ 0. (c) By Markov's inequality,
lP'(IN(t+h)-N(t)l
>E).:::: E12lE({N(t+h)-N(t)} 2),
which tends to 0 as h --+ 0, if E > 0. (d) Let E > 0. For 0 < h < E- 1, lP'
(I
N(t +
I
h~- N(t) >E)
= lP'(N(t +h)- N(t):::
337
1)
= A.h + o(h),
[7.11.6]-[7.11.10]
Solutions
Convergence of random variables
which tends to 0 as h --+ 0. On the other hand,
which tends to oo as h -J, 0.
6.
By Markov's inequality, Sn
= 2:7= 1 X;
satisfies
forE > 0. Using the properties of the X's,
since E(X;) = 0 for all i. Therefore there exists a constant C such that
implying (via the first Borel--Cantelli lemma) that n- 1 Sn ~ 0.
7.
We have by Markov's inequality that <
00
forE > 0, so that Xn ~ X (via the first Borel-Cantelli lemma).
8. Either use the Skorokhod representation or characteristic functions. Following the latter route, the characteristic function of aXn + b is
where ¢>n is the characteristic function of Xn. The result follows by the continuity theorem.
9.
For any positive reals c, t, li"(X > t) -
= li"(X + c >- t + c)
< -
E{(X+c) 2 } . (t +c) 2
Set c = a- 2 jt to obtain the required inequality.
10. Note that g(u) = uj(l + u) is an increasing function on [0, oo). Therefore, forE > 0, lP' X > (I nl
E
)
=11" (
IXnl l+E > -E- ) < - .JE: ( IXnl ) l+IXnl l+E E l+IXnl
338
Solutions [7.11.11]-[7.11.14]
Problems
by Markov's inequality. If this expectation tends to 0 then Xn
~ 0.
p
Suppose conversely that Xn -+ 0. Then E(
IXnl ) E E :':: - - ·li"(IXnl :'::E)+ 1 ·lP'(IXnl >E)--+-1+1Xnl 1+E 1+E
as n --+ oo, for E > 0. However E is arbitrary, and hence the expectation has limit 0.
11. (i) The argument of the solution to Exercise (7.9.6a) shows that {Xn} converges in mean square if it is mean-square Cauchy convergent. Conversely, suppose that Xn ~ X. By Minkowski's inequality,
as m, n--+ oo, so that {Xn} is mean-square Cauchy convergent. (ii) The corresponding result is valid for convergence almost surely, in rth mean, and in probability. For a.s. convergence, it is self-evident by the properties of Cauchy-convergent sequences of real numbers. For convergence in probability, see Exercise (7.3.1). For convergence in rth mean (r ~ 1), just adapt the argument of (i) above.
12. Ifvar(Xi)::: M for all i, the variance of n- 1 Lt=1
xi is
M
1 n
2 L:var(Xi) :=:---+ 0 n i=1 n
as n--+ oo.
13. (a) We have that JP>(Mn ::0 GnX) = F(anx)n --+ H(x)
If x :=: 0 then F (an X)n --+ 0, so that H (x)
= 0.
as n--+
00.
Suppose that x > 0. Then
-logH(x) =- lim {nlog[1- (1- F(anx))]} = n----+-oo
lim {n(1- F(anx))}
n---+oo
since -y- 1 log(1 - y) --+ 1 as y t 0. Setting x = 1, we obtain n(l - F(an)) --+ -log H(l), and the second limit follows. (b) This is immediate from the fact that it is valid for all sequences {an}. (c) We have that
1- F(tex+y) 1 - F(t)
1- F(tex+y) 1- F(tex)
1-F(tex)
1 - F(t)
logH(eY)
--+ --=--='---:::c=:'-:-:-c'log H(1)
logH(ex)
log H(l)
as t --+ oo. Therefore g(x + y) = g(x)g(y). Now g is non-increasing with g(O) = 1. Therefore g(x) = e-f3x for some ,B, and hence H(u) = exp( -au-f3) for u > 0, where a= -log H(l).
14. Either use the result of Problem (7.11.13) or do the calculations directly thus. We have that JP>(Mn :=oxn/rr)
= {~ +~tan- 1
c:) r=
{1-
~tan- 1 (;J
r
if x > 0, by elementary trigonometry. Now tan- 1 y = y + o(y) as y--+ 0, and therefore as n --+ oo.
339
[7.11.15]-[7.11.16]
Solutions
Convergence of random variables
15. The characteristic function of the average satisfies as n--+ oo. By the continuity theorem, the average converges in distribution to the constant JL, and hence in probability also.
16. (a) With Un = u(xn), we have that
n
n
if /lu/1 00 :::0 1. There is equality if Un equals the sign of fn - gn. The second equality holds as in Problem (2.7.13) and Exercise (4.12.3). (b) Similarly, if /lu/1 00 :::0 I, IE(u(X))- E(u(Y))I :::0
L:
lu(x)l · lf(x)- g(x)l dx :::0
L:
lf(x)- g(x)l dx
with equality if u(x) is the sign of f(x)- g(x). Secondly, we have that llP'(X
A) -lP'(Y
E
E
A) I
=
i IE(u(X))- E(u(Y))I :::0 idTV(X, Y),
where u(x)- {
1
if X E A,
-1
ifx¢.A.
Equality holds when A= {x E ffi(: f(x) :::: g(x)}. (c) Suppose dTv(Xn, X)--+ 0. Fix a E ffi(, and let u be the indicator function of the interval ( -oo, a]. Then IE(u(Xn))- E(u(X))I = llP'(Xn :::0 a) -lP'(X :::0 a)l, and the claim follows. On the other hand, if Xn = n- 1 with probability one, then Xn ..!?.. 0. However, by part (a), dTV(Xn, 0) = 2 for all n. (d) This is tricky without a knowledge of Radon-Nikodym derivatives, and we therefore restrict ourselves to the case when X and Y are discrete. (The continuous case is analogous.) As in the solution to Exercise (4.12.4), lP'(X =I Y) ::=:: idTv(X, Y). That equality is possible was proved for Exercise (4.12.5), and we rephrase that solution here. Let J.Ln = min{fn, gn} and JL = Ln J.Ln, and note that dTV(X, Y) = L lfn- gnl = LUn + gn- 2J.Ln} = 2(1- JL). n
n
It is easy to see that 1
z:dTV(X, Y) = lP'(X
=f
Y) =
{
1 if JL = 0, ifJL=l,
O
and therefore we may assume that 0 < JL < I. Let U, V, W be random variables with mass functions lP'(u _ -
Xn
) _ J.Ln JL
lP'(V _ '
-
Xn
) _ max{fn- gn, 0} lP'(W _ ' l-JL
) _ - min{fn- gn, 0} Xn
-
1-J.L
'
and let Z be a Bernoulli variable with parameter JL, independent of (U, V, W). We now choose the pair X', Y' by (X' Y') = { (U, U) if Z = I, ' (V,W) ifZ=O.
340
Solutions [7.11.17]-[7.11.19]
Problems
It may be checked that X' andY' have the same distributions as X andY, and furthermore, li"(X' Y 1) = li"(Z = 0) = 1- fi = idTV(X, Y).
i=
(e) By part (d), we may find independent pairs (X;, Y[), 1 :::0 i :::0 n, having the same marginals as (Xi, Yi), respectively, and such thatli"(X;
:::: 211" (
i=
Y[) = idTV(Xi, Yi). Now,
n
n
L x; i= L:rf i=l
)
n ::::2 Lll"(x;
i=l
i= Y[)
i=l
n
= 2
LdTV(Xj, Yj). i=l
17. If XJ, X2, ... are independent variables having the Poisson distribution with parameter A, then Sn = X1 +X2 + · · · +Xn has the Poisson distribution with parameter An. Now n- 1sn -SA, so that IE(g(n- 1Sn)) -+ g(A) for all bounded continuous g. The result follows.
18. The characteristic function 1/Jmn of (Xn - n)- (Ym - m) ~
Umn=
vm+n
satisfies log 1/lmn (t) = n (eit!.../m+n -
1)
+ m (e-it!.../m+n -
1)
+ (m - n)it -+ _lt2 2 Jm+n
as m, n-+ oo, implying by the continuity theorem that Umn -S N(O, 1). Now Xn + Ym is Poissondistributed with parameter m + n, and therefore
_ v:mn-
J+
Xn Ym P 1 -+ m+n
asm,n-+ oo D
by the law oflarge numbers and Problem (3). It follows by Slutsky's theorem (7.2.5a) that Umn IVmn -+ N(O, 1) as required.
19. (a) The characteristic function of Xn is 1/Jn (t) = exp{i f-tnt - ia;t 2 } where fin and a; are I 2
the mean and variance of Xn. Now, limn-+oo 1/Jn(l) exists. However 1/Jn(l) has modulus e-'Ian, and therefore a 2 = limn-+oo a; exists. The remaining component eitLnt of 1/Jn (t) converges as n -+ oo, say eitLnt -+ 8(t) as n -+ oo where (t) lies on the unit circle of the complex plane. Now
e
I 2 2
1/Jn(t) -+ 8(t)e-'Ia which is required to be a characteristic function; therefore e is a continuous function oft. Of the various ways of showing that 8(t) = eitLt for some f-t, here is one. The sequence 1/Jn(t) = eitLnt is a sequence of characteristic functions whose limit 8(t) is continuous at t = 0. Therefore e is a characteristic function. However 1/1n is the characteristic function of the constant fin, which must converge in distribution as n -+ oo; it follows that the real sequence {/In} converges to some limit f-t, and 8(t) = eitLt as required. This proves that 1/Jn(t)-+ exp{ifl,t- ia 2 t 2 }, and therefore the limit X is N(fl,, a 2). 1 ,
(b) Each linear combinations Xn + t Yn converges in probability, and hence in distribution, to s X +t Y. Now s Xn + t Yn has a normal distribution, implying by part (a) that s X+ t Y is normal. Therefore the joint characteristic function of X and Y satisfies
1/Jx,r(s, t) = I/JsX+tr(1) = exp{iE(sX + tY)- i var(sX + tY)}
ap)}
= exp{i(sf-tx + t{ty)- i 0. There exists I such that lp(X;,
2:::
i,J=l
nc
2--+ 0.
Xj) ::0
n
l::Oi1 1::oi,}::On
sincecov(X;, Xj)::: lp(Xi, Xj)lyivar(X;) · var(Xj)· Therefore, 2
2/c n
2
E(Tn / n ) ::0 -
+ EC --+ EC
as n--+ oo.
This is valid for all positive E, and the result follows.
21. The integral r)O_c_dx 12 x log lxl diverges, and therefore IE( X I) does not exist. The characteristic function cp of X 1 may be expressed as ,~., )
'f'(t
= 2C
100 2
whence cp(t)- cp(O) = _ [ 2c 12
cos(tx) d X x 2 Iogx
---
00 1 -
cos(tx) dx. x 2 Iogx
Now 0::: I- cose::: min{2, 8 2 }, and therefore
Icp(t)-2c cp(O) I -< Now
11/t 2
11u
-
u
and
2
!00
t2 2 --dx+ ---dx Iogx 1/t x2Iogx '
dx ----+0
Therefore
as u--+ oo,
Iogx
2 1oo 2 dx - - dx 1uoo -x2Iogx -Iogu u x 2 I < --
if t > 0.
-
Icp(t) ~ cp(O) I = o(t)
2 = --
ulogu'
as t
t
u > I.
0.
Now cp is an even function, and hence cp' (0) exists and equals 0. Use the result of Problem (7.ll.l5) to deduce that n- 1 I:'i Xi converges in distribution to 0, and therefore in probability also, since 0 is constant. The X; do not obey the strong law since they have no mean.
342
Solutions [7.11.22]-[7.11.24]
Problems
22. If the two points are U and V then
and therefore 12 1~ 2p1 -Xn = - L)Ui -Vi) -+ n n i= 1 6
as n
oo,
--+
by the independence of the components. It follows that Xn/ ,fii ~ Problem (7 .11.3) or by the fact that
1/ .J6 either by the result of
23. The characteristic function of Yj = Xj 1 is 4J(t) = 1
. + e-ztfx) . dx = lo1 cos(tjx) dx =It I100 -cosy- dy lo 1(eztfx ltl Y2
2 0
0
by the substitution x = It II y. Therefore l/J(t) = 1- It I
100 ltl
1- cosy
y
2
as t--+ 0,
dy = 1- /It I+ o(ltl)
where, integrating by parts,
-looo 1 - cos y dy--looo -sin-u du-_ -.
I-
11:
0
y
2
0
u
2
It follows that Tn = n- 1 '£,)= 1 Xj 1 has characteristic function
as t--+ oo, whence 2Tn/7r is asymptotically Cauchy-distributed. In particular, 1P'(I2Tn/rrl >
2100
1)--+ -
11:
1
du 1+u
1 2
as t
--2 =-
--+
oo.
24. Let mn be a non-decreasing sequence of integers satisfying 1 ::: mn < n, mn
noting that Ynk takes the value ±1 each with probability '£Z= 1 Ynk· Then n
lP'(Un-:/= Zn) :"':
LlP'(IXkl ~
i whenever mn
--+
< k < n. Let Zn
n
mn) :"':
L k=mn
k=1
343
k2
--+
0
oo, and define
as n--+ oo,
[7.11.25]-[7.11.26]
Solutions
Convergence of random variables
from V.:hich it follows that Un/ Jn
-S N(O, 1) if and only if Zn/ Jn -S N(O, 1). Now mn
L
Zn =
Ynk
+ Bn-mn
k=1
where Bn-mn is the sum of n- mn independent summands each of which takes the values ±1, each possibility having probability Furthermore
1·
I
1
1 mn 2 -2:Ynk ::S mn Jn k=1 Jn
which tends to 0 if mn is chosen to be mn = n- 1 Bn-mn
Ln 1/ 5J; with this choice for mn, we have that
-S N(O, 1), and the result follows.
Finally, var(Un)
=
t (2 - ~)
k=1
k
so that 1 n 1 var(Un/ Jn) = 2- 2 --+ 2. n k=1 k
L
25. (i) Let c/Jn and cp be the characteristic functions of Xn and X. The characteristic function !frk of XNk is 00
!frk(t)
= 2::: cfJ1 (t)f'(Nk = J) }=1
whence 00
11/fk(t)-
cp(t)l:::::
L lc/Jj(t)- cp(t)llP'(Nk = j). }=1
Let E > 0. We have that ifJ} (t) --+ cp (t) as j --+ oo, and hence for any T > 0, there exists J (T) such that lifJ} (t) - cp(t)l < E if j :::: J (T) and It I ::::: T. Finally, there exists K(T) such that lP'(Nk ::::: J (T)) ::::: E if k :::: K (T). It follows that
if It I ::::: T and k :::: K(T); therefore !frk(t) --+ cp(t) ask--+ oo. (ii) Let Yn = SUPm;:::n IXm -XI. ForE > 0, n :0:: 1, lP'(IXNk- XI>
E)::::: lP'(Nk::::: n) + lP'(IXNk- XI> E, Nk ::::: lP'(Nk ::::: n)
+ lP'(Yn
> n)
> E) --+ lP'(Yn > E)
Now take the limit as n --+ oo and use the fact that Yn ~ 0.
26. (a) We have that ( a(n
k
)
an- -,n - =
+ 1, n)
ITk ( 1 i=O
"'l
k .) -1. ) < exp ( - L..J - . n i=O n
344
ask --+
00.
Solutions
Problems
e
(b) The expectation is
L_g
En=
Jnn)
J
where the sum is over all j satisfying n - M Jn
(j-
g
:::
(nj+l _
Jn
j!
nl;~n
j ::: n. For such a value of j,
n) nle-n =e-n
Jn
[7.11.27]-[7.11.28]
nl ) (j- l)! ,
j!
whence En has the form given. (c) Now g is continuous on the interval [ -M, 0], and it follows by the central limit theorem that
1
I I 2 g(x)--e-'1.x dx
0
En--+
.J2ii
- M
I
x I 2 --e-'1.x dx
!oM
=
=
.J2ii
0
1-e-'J.M
2
------.,=-
.J2ii
Also,
where k
= LM JnJ.
Take the limits as n --+ oo and M --+ oo in that order to obtain 1
.
-- I-E fort~ T. We may approximate SM(t) by the random variable sk(t) as follows. With Aj = {ISj - sk(t) I >
E.Jk(t)}, IP'(AM(t)) ::0 IP'(AM(t)• lmn(t) k(t)-1
= 1) + IP'(AM(t)• lmn(t) = 0) n(t)-1
:::lP'( U Aj)+lP'( U j=m(t) < -
Aj)+IP'(lmn(t)=O)
j=k(t)
{k(t) - m(t) }a 2 E2k(t)
+ if t
345
{ n(t)- k(t) }a 2
E2k(t) ~
T,
+E
[7.11.29]-[7.11.32]
Solutions
Convergence of random variables
by Kolmogorov's inequality (Exercise (7.8.1) and Problem (7.11.29)). Send t --+ oo to find that as t --+ oo.
Now Sk(t)/ .Jk(i)
~
N(O, a 2 ) as t --+ oo, by the usual central limit theorem. Therefore
which implies the first claim, since k(t)j(et) --+ 1 (see Exercise (7.2.7)). The second part follows by Slutsky's theorem (7.2.5a).
29. We have that Sn = Sk
+ (Sn -
Sk), and so, for n :::: k,
Now S'fJAk :::: c 2 IAk; the second term on the right side is 0, by the independence of the X's, and the third term is non-negative. The first inequality of the question follows. Summing over k, we obtain E(S;) :::: c 2 1P'(Mn > c) as required.
30. (i) With Sn = Lt=1 xj, we have by Kolmogorov's inequality that 1 m+n lP' ( max ISm+k- Sml > E) ::0 2 E(X~) 1~k~n E k=m
L
forE > 0. Take the limit as m, n --+ oo to obtain in the usual way that {Sr : r :::: 0} is a.s. Cauchy convergent, and therefore a.s. convergent, if 2::\"' E(X~) < oo. It is shorter to use the martingale convergence theorem, noting that Sn is a martingale with uniformly bounded second moments. (ii) Apply part (i) to the sequence Yk = Xkfbk to deduce that 2::~ 1 Xkfbk converges a.s. The claim now follows by Kronecker's lemma (see Exercise (7.8.2)).
31. (a) This is immediate by the observation that ei.(P) =
f Xo IJ PN_ij lJ · i,j
I
(b) Clearly Lj Pij = 1 for each i, and we introduce Lagrange multipliers {JL; : i E S} and write V = A.(P) + Li JLi Lj Pij. Differentiating V with respect to each Pij yields a stationary (maximum) value when (N;j / Pij) + JLi = 0. Hence Lk N;k = - JLb and
z=E\
(c) We have that Nij = Nik Ir where Iris the indicator function oftheeventthatthe rth transition out ofi is to j. By the Markov property, the Ir are independent with constant mean Pij. Using the strong law oflarge numbers and the fact that Lk N;k ~ oo as n --+ oo, Pij ~ E(/t) = Pij. 32. (a) If X is transient then V;(n) < oo a.s., and JLi = oo, whence V;(n)/n ~ 0 = ttj 1. If X is persistent, then without loss of generality we may assume Xo = i. LetT (r) be the duration of the rth
346
Solutions [7.11.33]-[7.11.34]
Problems
excursion from i. By the strong Markov property, the T (r) are independent and identically distributed with mean J.Li . Furthermore, V;(n)-J
::r
V;(n)
-
n
1
V;(n)
::r ·
T(r) < < " V;(n) - V;(n)
"
T(r)
By the strong law of large numbers and the fact that V; (n) ~ oo as n --+ oo, the two outer terms sandwich the central term, and the result follows. (b) Note that ~~,;;;J f(Xr) = ~iES f(i)V;(n). With Q a finite subset of S, and n; = J.Lj 1, the unique stationary distribution,
::":
{"I
V;(n) 1 L.t ------:-
iEQ
n
J.Lz
I
1 )} 11/lloo, + "(V;(n) L.t - - +---:i(X < Y) = 'A/('A + J-t). (a) In this case, 1{
p
=2
'A 1-t 1-t } 1 { 1-t 'A 'A } 'A + J-t . 'A + J-t + 'A + J-t + 2 'A + J-t . 'A + J-t + 'A + J-t
(b) If 'A< f.-t, and you pick the quicker server, p
= 1-
(-1-t-)
1
= 2+
2'A~-t ('A +
~-t)2.
2
"A+~-t
2'A~-t
(c) And finally, p = ('A+ J-t) 2 .
2. The given event occurs if the time X to the next arrival is less than t, and also less than the time Y of service of the customer present. Now,
3.
By conditioning on the time of passage of the first vehicle, IE(T) =loa (x
+ IE(T) )'Ae-1.x dx + ae-1.a,
and the result follows. If it takes a time b to cross the other lane, and so a + b to cross both, then, with an obvious notation, (a) (b)
eal.- 1 ebJ.L- 1 IE(Ta) +IE(n) = - - + - - , 'A 1-t ek{M(-v)k _ M(-v)k+I} = 1- M(-v). k=O 1- sM(-v) (b) In this case, lE(sN(T)) = lE(eAT(s-l)) = MT(A.(s -1)). When T has the given gamma distribution, MT(e) = {vj(v- e)}b, and
lE(sN(T)) =
(-v-) (1 - ~) b
v+A.
v+A.
The coefficient of sk may be found by use of the binomial theorem.
372
b
Excess life
Solutions
[10.3.1]-[10.3.3]
10.3 Solutions. Excess life 1. Let g (y) = lP'( E (t) > y), assumed notto depend on t. By the integral equation for the distribution of E(t),
l
+ y) + g(y) dF(x). Write h(x) = 1- F(x) to obtain g(y)h(t) = h(t + y), for y, t :=:: 0. With t = 0, we have that g(y)h(O) = h(y), whence g(y) = h(y)/ h(O) satisfies g(t + y) = g(t)g(y), for y, t :=:: 0. Now g is g(y)
= 1-
F(t
left-continuous, and we deduce as usual that g(t) the renewal process is a Poisson process.
= e-At for some A..
Hence F(t)
=
1 -e-At, and
2.
(a) Examine a sample path of E. If E(t) = x, then the sample path decreases (with slope -1) until it reaches the value 0, at which point it jumps to a height X, where X is the next interarrival time. Since X is independent of all previous interarrival times, the process is Markovian. (b) In contrast, C has sample paths which increase (with slope 1) until a renewal occurs, at which they drop to 0. If C (s) = x and, in addition, we know the entire history of the process up to time s, the time of the next renewal depends only on the length ofthe spent period (i.e., x) of the interarrival time in process. Hence C is Markovian.
3.
(a) We have that lP'(E(t) _:::: y) = F(t
+ y)
-l
G(t
+ y- x)dm(x)
where G(u) = 1- F(u). Check the conditions of the key renewal theorem (10.2.7): g(t) = G(t + y) satisfies: (i) g(t) :::: 0, (ii) f0 g(t) dt .:::= fcF[1 - F(u)] du = E(X 1) < oo, (iii) g is non-increasing. We conclude, by that theorem, that
00
1 lim lP'(E(t) _:::: y) = 1 - -
t-+00
f1
100 g(x)dx = 0
loy -[11 F(x)]dx. 0 f1
(b) Integrating by parts,
i oo o
xr 1 -[1-F(x)]dx=11
11
laoo
xr+1 E(xr+1) --dF(x)= 1 . M(r + 1)
o r+1
See Exercise (4.3.3). (c) As in Exercise (4.3.3), we have that E(E(tY) = J0 ryr- 11P'(E(t) > y) dy, implying by(*) that
00
E(E(tn = E({CX1- t)+n
+ i:ol~ 0 ryr- 1 1P'(X 1
> t
+ y -x)dm(x)dy,
whence the given integral equation is valid with
Now h satisfies the conditions of the key renewal theorem, whence lim E(E(tY)
t-+oo
= .!._ { 00 h(u) du = .!._ { { yr dF(u + y) du 11 Jo 11 JJoyjC(t)=x)=lP'(Xt>y+xiXt>x)=
1- F(y +x)
1- F(x)
,
whence E(E(t)jC(t)=x)= fool-F(y+x)dy=E{(Xt-x)+}_ Jo 1- F(x) 1- F(x)
5. (a) Apply Exercise (10.2.2) to the sequence X;- f-L, 1 :::= i < oo, to obtain var(TM(t)- t-LM(t)) = a 2 E(M(t)). (b) Clearly TM(t) = t + E(t), where E is excess lifetime, and hence t-LM(t) = (t + E(t))- (TM(t)t-LM(t)), implying in tum that f-L 2 var(M(t)) =
where
SM(t)
var(E(t)) + var(SM(t))- 2cov(E(t),
SM(t)),
= TM(t)- {LM(t). Now
as t --+ oo
ifE(XI) < oo (see Exercise (10.3.3c)), implying that 1 - var(E(t)) --+ 0 t
as t --+ oo.
This is valid under the weaker assumption that E(XI) < oo, as the following argument shows. By Exercise (10.3.3c), E(E(t) 2 ) = a(t) +lot a(t- u) dm(u), where a(u) = JE:({(Xt- u)+} 2 ). Now use the key renewal theorem together with the fact that a(t):::: E(XIl(x 1 >tJ)--+ Oast--+ oo. Using the Cauchy-Schwarz inequality,
as t --+ oo, by part (a) and (** ). Returning to (*), we have that f-L2 -var(M(t))--+ lim t t--'>00
{a2
-(m(t)+l) } =a2 -. t {L
374
Solutions [10.4.1]-[10.5.3]
Renewal-reward processes
10.4 Solution. Applications 1. Visualize a renewal as arriving after two stages, type 1 stages being exponential parameter A and type 2 stages being exponential parameter J-t. The 'stage' process is the flip-flop two-state Markov process of Exercise (6.9.1). With an obvious notation, Pn(t)
= _A_e-(A+JL)t + _1-t_. A+J-t
A+J-t
Hence the excess lifetime distribution is a mixture of the exponential distribution with parameter J-t, and the distribution of the sum of two exponential random variables, thus,
where g(x) is the density function of a typical interarrival time. By Wald's equation, E(t
+ E(t)) = E(SN(t)+1) = E(X1)E(N(t) + 1) =
We substitute
(± + ~)
(m(t)
+ 1).
p 11 (t) (-A1+-1-t1) + (1- Pn(t))-1-t1=1-t-1 + A
E(E(t)) = Pn(t) to obtain the required expression.
10.5 Solutions. Renewal-reward processes 1. Suppose, at time s, you are paid a reward at rate u(X(s)). By Theorem (10.5.10), equation (10.5.7), and Exercise (6.9.11b),
lot
1 t 0
-
Suppose lu (i) I .::; K < oo for all i
E
a.s.
1 J-tjgj
l{X(s)=j)dS ~ - - = lfj.
S, and let F be a finite subset of the state space. Then
where T1 (F) is the total time spent in F up to time t. Take the limit as t F t S, to obtain the required result.
~
oo using (*), and then as
2. Suppose you are paid a reward at unit rate during every interarrival time of type X, i.e., at all times tat which M(t) is even. By the renewal-reward theorem (10.5.1), 1
t
lot I
a.s. E(reward during interarrival time) EX 1 = . E(length of interarrival time) EX 1 + EY1
M s is even ds ~
o { ()
}
3.
Suppose, at timet, you are paid a reward at rate C(t). The expected reward during an interval (cycle) of length X is fox s ds = ~ X 2 , since the age C is the same at the time s into the interval. The
375
[10.5.4]-[10.6.3]
Solutions
Renewals
result follows by the renewal-reward theorem (10.5.1) and equation (10.5.7). The same conclusion is valid for the excess lifetime E(s), the integral in this case being fox (X- s) ds = ~ X 2 . 4. Suppose Xo = j. Let V1 = min{n ::: 1 : Xn = j, Xm = k for some 1 :S m < n}, the first visit to j subsequent to a visit to k, and let Vr+1 = min{n ::: Vr : Xn = j, Xm = k for some Vr + 1 :S m < n}. The Vr are the times of a renewal process. Suppose a reward of one ecu is paid at every visit to k. By the renewal-reward theorem and equation (10.5.7),
By considering the time of the first visit to k, E(V1
I Xo = j) = E(Tk I Xo = j) +E(1) I Xo = k).
The latter expectation in (*) is the mean of a random variable N having the geometric distribution f'(N = n) = p(l - p)n- 1 for n ::: 1, where p = f'(1) < Tk I Xo = k). Since E(N) = p- 1, we deduce as required that 1/f'(1) < Tk I Xo = k) lfk = - - - - " - - - - - - - - -
E(n
I Xo
= j) + E(1j
I Xo
= k)
10.6 Solutions to problems 1. (a) For any n, JFl'(N(t) < n) :S JFl'(Tn > t)--+ 0 as t--+ oo. (b) Either use Exercise (10.1.1), or argue as follows. Since JL > 0, there exists E (> 0) such that f'(X1 >E) > 0. For all n, f'(Tn :S nE) = 1- f'(Tn > nE) :S 1- f'(X1 > E)n < 1,
so that, if t > 0, there exists n = n(t) such that f'(Tn :S t) < 1. Fix t and let n be chosen accordingly. Any positive integer k may be expressed in the form k =an+ fj where 0 :S fj < n. Now JFl'(Tk :S t) :S f'(Tn :S t)a for an :S k < (a+ 1)n, and hence 00
00
k=1
a=O
L JFl'(Tk :S t) :S L nf'(Tn :S t)a <
m(t) =
00.
(c) It is easiest to use Exercise (10.1.1), which implies the stronger conclusion that the moment generating function of N (t) is finite in a neighbourhood of the origin. 2.
(i) Condition on X 1 to obtain
r
,, 2 v(t)= Jo E{(N(t-u)+1) }dF(u)= Jo {v(t-u)+2m(t-u)+1}dF(u).
Take Laplace-Stieltjes transforms to find that v* = (v* +2m*+ 1)F*, where m* = F* + m* F* as usual. Therefore v* = m*(l +2m*), which may be inverted to obtain the required integral equation. (ii) If N is a Poisson process with intensity A., then m(t) =At, and therefore v(t) 3.
Fix x
E
lit Then
376
= (A.t) 2 + A.t.
Solutions [10.6.4]-[10.6. 7]
Problems wherea(t) = l y)--+ lim lP'(E(u) > y) U---+00
as t--+ oo
= {oo !._[1- F(x)] dx
}y
1-i
by Exercise (10.3.3a). The current and excess lifetimes have the same asymptotic distributions.
5. Using the lack-of-memory property of the Poisson process, the current lifetime C(t) is independent of the excess lifetime E(t), the latter being exponentially distributed with parameter J..... To derive the density function of C(t) either solve (without difficulty in this case) the relevant integral equation, or argue as follows. Looking backwards in time from t, the arrival process looks like a Poisson process up to distance t (at the origin) where it stops. Therefore C(t) may be expressed as min{Z, t} where Z is exponential with parameter J....; hence J....e-J..s
!qt)(s) = { 0
and lP'(C(t) = t) = e-At. Now D(t) convolution formula) to be as given.
=
C(t)
if s < t ·f - , s > t,
1
+ E(t),
whose distribution is easily found (by the
6. The ith interarrival time may be expressed in the form T + Zi where Zi is exponential with parameter J..... In addition, Z1, Zz, ... are independent, by the lack-of-memory property. Now
1- F(x)
= lP'(T + zl
> x)
= lP'(Zl
>X- T)
= e-A(x-T)'
X~
T.
Taking into account the (conventional) dead period beginning at time 0, we have that
lP'(N(t)
~ k)
k
= lP'(kT
+ ~ Zi
.::0 t) = lP'(N(t- kT)
~ k),
t
~
kT,
l=l
where N is a Poisson process. 7. We have that X1 = L + E(L) where Lis the length of the dead period beginning at 0, and E(L) is the excess lifetime at L. Therefore, conditioning on L,
377
[10.6.8]-[10.6.10]
Solutions
Renewals
We have that lP'(E(t) s y) = F(t
+ y)
-lot {1-
F(t
+ y -x)}dm(x).
By the renewal equation,
m(t
+ y) =
F(t
rr+y F(t + y- x)dm(x),
+ y) + Jo
whence, by subtraction,
i
t+y
{1-F(t+y-x)}dm(x).
lP'(E(t)sy)= t It follows that
1P'(X 1 sx)=
{x
{x {1-F(x-y)}dm(y)dFL(l)
}z=O }y=l
= [FL(l) {x{1-F(x-y)}dm(y)]x
~0
h
+
r FL(l){1-F(x-l)}dm(l)
k
using integration by parts. The term in square brackets equals 0.
8. (a) Each interarrival time has the same distribution as the sum of two independent random variables with the exponential distribution. Therefore N(t) has the same distribution as L~ M(t)J where M is a Poisson process with intensity A. Therefore m(t) = ~JE(M(t))- ilP'(M(t) is odd). Now JE(M(t)) = At, and oo (At)Zn+le-At lP'(M(t) is odd)="'"' = ~e-A.t(eA.t- e-At). ~ (2n + 1)!
n=O
With more work, one may establish the probability generating function of N(t). (b) Doing part (a) as above, one may see that iii(t) = m(t).
9.
Clearly C(t) and E(t) are independent if the process N is a Poisson process. Conversely, suppose that C(t) and E(t) are independent, for each fixed choice oft. The event {C(t) 0:: y} n {E(t) 0:: x} occurs if and only if E (t - y) ::=: x + y. Therefore lP'(C(t) ::=: y)lP'(E(t) ::=: x) = lP'(E(t- y) ::=: x
+ y).
Take the limit as t ~ oo, remembering Exercise (10.3.3) and Problem (10.6.4), to obtain that G(y)G(x) = G(x + y) if x, y ::=: 0, where
1
00
G(u) =
u
1
-[1- F(v)] dv. f.-i
Now 1- G is a distribution function, and hence has the lack-of-memory property (Problem (4.14.5)), implying that G(u) = e-Jcu for some A. This implies in tum that [1- F(u)]/ f.-i = -G'(u) = Ae-Jcu, whence f.-i = 1/A and F(u) = 1- e-Jcu.
10. Clearly N is a renewal process if Nz is Poisson. Suppose that N is a renewal process, and write A for the intensity of N 1 , and Fz for the interarrival time distribution of Nz. By considering the time X 1 to the first arrival of N,
378
Solutions [10.6.11]-(10.6.12]
Problems
Writing E, Ei for the excess lifetimes of N, Ni, we have that
Take the limit as t
~
oo, using Exercise (10.3.3), to find that
1 100 -[1-
F(u)] du = e-Ax
xf.L
100 -[11 F2(u)]
du,
xf.L2
where f.L2 is the mean of F2. Differentiate, and use(*), to obtain
which simplifies to give 1- F2(x) = c equation has solution F2 (x) = 1 - e -ex.
fx
00
[1- F2(u)] du where c = AJL/(JL2- J.L); this integral
11. (i) Taking transforms of the renewal equation in the usual way, we find that m* e -
F*(e)
< ) - 1- F*(e)
where as
-----1 1- F*(e)
F*(e) = E(e-ex,) = 1- elL+ ~e 2 (JL 2
+a 2 ) +o(e 2 )
e ~ 0. Substitute this into the above expression to obtain
and expand to obtain the given expression. A formal inversion yields the expression form. (ii) The transform of the right-hand side of the integral equation is
By Exercise (10.3.3), Ff:(e) = [1 - F*(e)]/(J.Le), and(*) simplifies to m*(e) - (m* - m* F* F*)/(J.Le), which equals m*(e) since the quotient is 0 (by the renewal equation). Using the key renewal theorem, as t ~ oo,
lo [1t
0
1
looo [1- FE(u)]du = --JE(X2) a2 + JL2 = 2
FE(t -x)]dm(x) ~JL 0
2JL
by Exercise (10.3.3b). Therefore,
12. (i) Conditioning on X 1, we obtain
379
2JL
[10.6.13]-[10.6.16]
Solutions
Renewals
Thereforemd* = Fd* +m*Fd*. Alsom* = F* +m*F*,sothat
whence md* = Fd* + md* F*, the transform of the given integral equation. (ii)ArguingasinProblem(10.6.2), vd* = Fd*+2m*Fd*+v*Fd*wherev* = F*(l+2m*)/(l-F*) is the corresponding object in the ordinary renewal process. We eliminate v* to find that
by(*). Now invert.
13. Taking into account the structure of the process, it suffices to deal with the case I = 1. Refer to Example (10.4.22) for the basic notation and analysis. It is easily seen that fJ = (v - 1)A.. Now F(t) = 1- e-vA.t. Solve the renewal equation (10.4.24) to obtain
+lot
g(t) = h(t)
h(t- x) diii(x)
where iii(x) = vA.x is the renewal function associated with the interarrival time distribution Therefore g(t) = 1, andm(t) = ef3t.
F.
14. We have from Lemma (10.4.5) that p* = 1 - Fz + p* F*, where F* = Fy Fz. Solve to obtain
* p
=
1- Fz 1- F*F* y z
15. The first locked period begins at the time of arrival of the first particle. Since all future events may be timed relative to this arrival time, we may take this time to be 0. We shall therefore assume that a particle arrives at 0; call this the Oth particle, with locking time Yo. We shall condition on the time X 1 of the arrival of the next particle. Now lP'(L > t
I X1
= u) = {
lP'(Yo>t)
ifu>t,
lP'(Yo > u)lP'(L' > t - u)
ifu ~ t,
where L' has the same distribution as L; the second part is a consequence of the fact that the process 'restarts' at each arrival. Therefore lP'(L > t) = (1- G(t))lP'(X1 > t)
+lot
lP'(L > t - u)(1- G(u))fx 1 (u)du,
the required integral equation. If G(x) = 1 - e-11-x, the solution is lP'(L > t) = e-11-t, so that L has the same distribution as the locking times of individual particles. This striking fact may be attributed to the lack-of-memory property of the exponential distribution.
16. (a) It is clear that M(tp) is a renewal process whose interarrival times are distributed as X1 + Xz + · · · + XR where lP'(R = r) = pqr- 1 for r:::: 1. It follows that M(t) is a renewal process whose first interarrival time X(p) = inf{t : M(t) = 1} = p inf{t : M(tp) = 1}
380
Solutions [10.6.17]-[10.6.19]
Problems
has distribution function 00
lP'(X(p)::::: x) = LlP'(R = r)Fr(xjp). r=1
(b) The characteristic function QJp of Fp is given by QJp(t)
=
f
pqr-1100 eixt dFr(tfp)
r=1
=
-oo
f
pqr-14J(pt{
=
pqy(pt)
1 - qqy(pt)
r=1
where qy is the characteristic function of F. Now qy (pt) = 1 + i Jlpt + o(p) as p ,), 0, so that QJ (t)
=
P
1 + ijlpt + o(p) 1-itLt+o(1)
=
1 + o(l) 1-itLt
as p ,), 0. The limit is the characteristic function of the exponential distribution with mean fL, and the continuity theorem tells us that the process M converges in distribution as p ,), 0 to a Poisson process with intensity 1I fL (in the sense that the interarrival time distribution converges to the appropriate limit). (c) If M and N have the same fdds, then c/Jp(t) = qy(t), which implies that qy(pt) = qy(t)f(p + qqy(t)). Hence 1/f(t) = c/J(t)- 1 satisfies 1/f(pt) = q + pl/f(t) fort E JR. Now 1/f is continuous, and it follows as in the solution toProblem(5.12.15) that 1/f has the form 1/f(t) = 1+,Bt, implyingthatqy(t) = (1 +.Bt)- 1 for some .B E «:::. The only characteristic function of this form is that of an exponential distribution, and the claim follows.
17. (a) Let N(t) be the number of times the sequence has been typed up to the tth keystroke. Then N is a renewal process whose interarrival times have the required mean fL; we have that JE(N(t))/t --+ fL- 1 as t --+ oo. Now each epoch of time marks the completion of such a sequence with probability ( 1 0 ) 14 , so that
6
~lE(N(t)) = ~ t
t
t (-1-) (-1-) 14--+
n= 14
100
100
14
as t--+ oo,
implying that fL = 1028 . The problem with 'omo' is 'omomo' (i.e., appearances may overlap). Let us call an epoch of time a 'renewal point' if it marks the completion of the word 'omo', disjoint from the words completed at previous renewal points. In each appearance of 'omo', either the first 'o' or the second 'o' (but not both) is a renewal point. Therefore the probability un, that n is a renewal point, satisfies ( 1 0 ) 3 = un + un-2( 1 0 ) 2 . Average this over n to obtain
6
6
C~or = n~moo ~ j; {Un +un-2 C~or} = ~ + ~ C~or, and therefore fL = 106 + 102 . (b) (i) Arguing as for 'omo', we obtain p 3 = un + pun-1 + p 2un-2, whence p 3 = (1 + p + p 2 )/ fL. (if) Similarly, p 2q = Un + pqun-2, so that fL = (1 + pq)f(p 2q). 18. The fdds of {N(u) - N(t) : u 2:: t} depend on the distributions of E(t) and of the interarrival times. In a stationary renewal process, the distribution of E(t) does not depend on the value oft, whence {N(u) - N(t) : u 2:: t} has the same fdds as {N(u) : u 2:: 0}, implying that X is strongly stationary.
19. We use the renewal-reward theorem. The mean time between expeditions is B fL, and this is the mean length of a cycle of the process. The mean cost of keeping the bears during a cycle is B(B - 1)cfL, whence the long-run average cost is {d + B(B - 1)ctL/2}/(B fL).
i
381
11
Queues
11.2 Solutions. M/M/1 1. The stationary distribution satisfies 1f = equations
1f P
when it exists, where P is the transition matrix. The Plfn-1 lfn = - 1+p
lfn+1 +1+p
for n > 2, -
with l::~o :rri = 1, have the given solution. If p ;::: 1, no such solution exists. It is slightly shorter to use the fact that such a walk is reversible in equilibrium, from which it follows that 1f satisfies for n ;::: 1. 2. (i) This continuous-time walk is a Markov chain with generator given by g01 = Bo, gn,n+ 1 = Bnp/(1 + p) and gn,n-1 = Bn/(1 + p) for n ;::: 1, other off-diagonal terms being 0. Such a process is reversible in equilibrium (see Problem (6.15.16)), and its stationary distribution v must satisfy Vngn,n+1 = Vn+1gn+1,n· These equations may be written as for n ;::: 1. These are identical to the equations labelled(*) in the previous solution, with :rrn replaced by vnBn. It follows that Vn = C:rrn/Bn for some positive constant C. (ii) If Bo = A., Bn =A.+ f.-i for n ;::: 1, we have that
whence C = 2A. and the result follows. 3. Let Q be the number of people ahead of the arriving customer at the time of his arrival. Using the lack -of-memory property of the exponential distribution, the customer in service has residual servicetime with the exponential distribution, parameter f.-t, whence W may be expressed as S 1 + S2 + · · · + S Q, the sum of independent exponential variables, parameter f.-t. The characteristic function of W is w(t) =E{lE(eitW
I Q)}
=E{ (f.-t~it)Q}
-c--_1_-,...,---P--,- = (1 _ p) 1-pf.-t/({t-it)
382
+p
(
1-i - A. ) . {t-A.-it
Solutions
M/M/1
[11.2.4]-[11.2.7]
This is the characteristic function of the given distribution. The atom at 0 corresponds to the possibility that Q = 0. 4. We prove this by induction on the value of i + j. If i + j = 0 then i = j = 0, and it is easy to check that rr(O; 0, 0) = 1 and A(O; 0, 0) = 1, A(n; 0, 0) = 0 for n ::: 1. Suppose then that K ::: 1, and that the claim is valid for all pairs (i, j) satisfying i + j = K. Let i and j satisfy i + j = K + 1. The last ball picked has probability i /(i + j) of being red; conditioning on the colour of the last ball, we have that rr(n;i,j)= "+i .rr(n-1;i-1,j)+ "+j .rr(n+1;i,j-1). l 1 l 1 Now (i - 1) + j = K = i + ( j - 1). Applying the induction hypothesis, we find that rr(n; i, j) = i
~j
{ A(n- 1; i - 1, j)- A(n; i - 1, j)}
+ i
~j
{ A(n + 1; i, j - 1)- A(n + 2; i, j - 1) }·
Substitute to obtain the required answer, after a little cancellation and collection of terms. Can you see a more natural way?
5.
Let A and B be independent Poisson process with intensities J.... and p, respectively. These processes generate a queue-process as follows. At each arrival time of A, a customer arrives in the shop. At each arrival-time of B, the customer being served completes his service and leaves; if the queue is empty at this moment, then nothing happens. It is not difficult to see that this queue-process is M(J....)/M(p,)/1. Suppose that A(t) = i and B(t) = j. During the time-interval [0, t], the order of arrivals and departures follows the schedule of Exercise (11.2.4), arrivals being marked as red balls and departures as lemon balls. The imbedded chain has the same distributions as the random walk of that exercise, and it follows that JP>(Q(t) = n I A(t) = i, B(t) = j) = rr(n; i, j). Therefore Pn (t) = 'E.i,j rr(n; i, j)JP>(A(t) = i)JP>(B(t) = j). 6.
With p
= A/ p,, the stationary distribution of the imbedded chain is, as in Exercise (11.2.1 ),
{ io - p) Jrn = iO - p2)pn-1 ~
if n = 0, if n::: 1.
In the usual notation of continuous-time Markov chains, go= by Exercise (6.10.11), there exists a constant c such that rro
Now'£.; rr;
=
;J.... (1- p),
Jrn
=
J....
and gn =
J....
+ p, for n :::
1, whence,
2 1 for n::: 1. 2 (/...: p,) (1- p )pn-
= 1, and therefore c = 2/... and Jrn = (1 -
p)pn as required. The working is reversible.
7. (a) Let Q; (t) be the number of people in the ith queue at timet, including any currently in service. The process Q1 is reversible in equilibrium, and departures in the original process correspond to arrivals in the reversed process. It follows that the departure process of the first queue is a Poisson process with intensity J...., and that the departure process of Q 1 is independent of the current value of Q1· (b) We have from part (a) that, for any given t, the random variables Q1 (t), Q2(t) are independent. Consider an arriving customer when the queues are in equilibrium, and let W; be his waiting time (before service) in the ith queue. With T the time of arrival, and recalling Exercise (11.2.3), JP>(W1
= 0,
W2
= 0)
> JP>(Q;(T)
= (1 -
= 0 fori= 1, 2) = lP'(Q1 (T) = O)JP>(Q2(T) = 0) = lP'(W1 = O)JP>(W2 = 0).
P1)(1 - P2)
Therefore W1 and W2 are not independent. There is a slight complication arising from the fact that T is a random variable. However, T is independent of everybody who has gone before, and in particular of the earlier values of the queue processes Q;.
383
[11.3.1]-[11.4.1]
Solutions
Queues
11.3 Solutions. M/G/1 1.
In equilibrium, the queue-length Qn just after the nth departure satisfies
where An is the number of arrivals during the (n + l)th service period, and h(m) = 1 - 8mo· ·Now Qn and Qn+l have the same distribution. Take expectations to obtain
0 = lE(An) - IP'(Qn > 0), where lE (An) = A.d, the mean number of arrivals in an interval of length d. Next, square (*) and take expectations:
Use the facts that An is independent of Qn, and that Qnh(Qn) 0
= {(A.d) 2 + A.d} + IP'(Qn
> 0)
=
Qn, to find that
+ 2{ (A.d- l)lE(Qn)- A.dlP'(Qn >
0)}
and therefore, by (** ), 2p- p2 lE(Qn) = 2 (1 _ p)
From the standard theory, Ms satisfies Ms(s) = Ms(s - A. + A.Ms(s)), where Ms(e) ~-t/(J-t- e). Substitute to find that x = Ms (s) is a root of the quadratic A.x 2 - x(A. + 1-t- s) + 1-t = 0. For some small positive s, M B (s) is smooth and non-decreasing. Therefore M B (s) is the root given.
2.
3. Let Tn be the instant of time at which the server is freed for the nth time. By the lack-of-memory property of the exponential distribution, the time of the first arrival after Tn is independent of all arrivals prior to Tn, whence Tn is a 'regeneration point' of the queue (so to say). It follows that the times which elapse between such regeneration points are independent, and it is easily seen that they have the same distribution.
11.4 Solutions. G/M/1 1. The transition matrix of the imbedded chain obtained by observing queue-lengths just before arrivals is 0 ... 0 ...
ao
)
The equation :rr = :rr P A may be written as 00
Trn
= L ai:rrn+i-1
for n 2:: 1.
i=O
It is easily seen, by adding, that the first equation is a consequence of the remaining equations, taken in conjunction with L:Q" :rri = 1. Therefore :rr is specified by the equation for :rrn, n 2:: 1.
384
Solutions [11.4.2]-[11.5.2]
G/G/1
The indicated substitution gives 00
en= en-1 l.:a;ei i=O
which is satisfied whenever e satisfies
It is easily seen that A(8) = Mx(~-t(8 - I)) is convex and non-decreasing on [0, 1], and satisfies A(O) > 0, A(l) = 1. Now A'(l) = ~-tlE(X) = p- 1 > 1, implying that there is a unique 'f/ E (0, 1) such that A('f/) = 'f'J. With this value of 'f'J, the vector 1r given by Trj = (1 - TJ)'f/1, j ;:: 0, is a stationary distribution of the imbedded chain. This 1r is the unique such distribution because the chain is irreducible.
2. (i) The equilibrium distribution is Trn = (1- TJ)'f'Jn for n 2: 0, with mean l:~o nnn = TJ/0- TJ). (ii) Using the lack-of-memory property of the service time in progress at the time of the arrival, we see that the waiting time may be expressed as W = S1 + S2 + · · · + S Q where Q has distribution 1r, given above, and the Sn are service times independent of Q. Therefore JE(W)
= lE(S1)lE(Q) = ___!}_j£__. (1- 'f/)
We have that Q(n+) = 1 + Q(n-) a.s. foreachintegern, whencelim 1 _, 00 lP'(Q(t) = m) cannot exist. Since the traffic intensity is less than 1, the imbedded chain is ergodic with stationary distribution as in Exercise (11. 4.1).
3.
11.5 Solutions. G/G/1 1. Let Tn be the starting time of the nth busy period. Then Tn is an arrival time, and also the beginning of a service period. Conditional on the value of Tn, the future evolution of the queue is independent of the past, whence the random variables {Tn+l - Tn : n :::: 1} are independent. It is easily seen that they are identically distributed. 2. If the server is freed at time T, the time I until the next arrival has the exponential distribution with parameter J-t (since arrivals form a Poisson process). By the duality theory of queues, the waiting time in question has moment generating function Mw(s) = (1- 0/(1- M1(s)) where M1(s) = ~-t/(J-t- s) and = lP'(W > 0). Therefore,
s
s
Mw(s)=
s~-t( 1 -0 +(1-S),
~-tO-n-s
the moment generating function of a mixture of an atom at 0 and an exponential distribution with parameter ~-t(1 - S). If G is the probability generating function of the equilibrium queue-length, then, using the lackof-memory property of the exponential distribution, we have that Mw(s) = G(~-t/(J-t- s)), since W is the sum of the (residual) service times of the customers already present. Set u = ~-t/(J-t- s) to find that G(u) = (1- 0/(1- su), the generating function of the mass function f(k) = (1- Osk fork::: 0.
385
[11.5.3]-(11.7.2]
Solutions
Queues
It may of course be shown that ~ is the smallest positive root of the equation x = M x (~-t(x where X is a typical interarrival time. 3.
1)),
We have that 1- G(y)
= lP'(S- X>
y)
=
lx:;
lP'(S > u
+ y)dFx(u),
y
E
JR,
where Sand X are typical (independent) service and interarrival times. Hence, formally, dG(y) = - {
lo
00
dlP'(S > u
+ y) dFx(u)
= dy
1-y ~-te-JJ,(u+y) 00
dFx(u),
since fs(u + y) = e-JJ,(u+y) if u > -y, and is 0 otherwise. With F as given,
1
x F(x- y)dG(y) = {{ {I -oo }}-oo- 1 as x -->- oo, since the queue is stable, and therefore A= 1. Using the equation for F(O), we find that B = ~(1-
JS).
9. Q is a M('A)/M(JL)/oo queue, otherwise known as an immigration-death process (see Exercise (6.11.3) and Problem (6.15.18)). As found in (6.15.18), Q(t) has probability generating function G(s, t) = { 1 + (s- l)e-JLt} I exp{p(s- 1)(1 - e-JLt)}
where p ='A/ IL· Hence JE.(Q(t)) = Ie-JLt
+ p(l- e-JLt),
lP'(Q(t) =0) = (1-e-JLti exp{-p(1-e-ILt)}, as t -->- oo. If JE.(J) and JE.(B) denote the mean lengths of an idle period and a busy period in equilibrium, we have that the proportion of time spent idle is JE.(/)/{lE.(J) + JE.(B)}. This equals limt-H>o lP'(Q(t) = 0) = e-P. Now JE.(J) = A- 1, by the lack-of-memory property of the arrival process, so that JE.(B) = (eP- 1)/'A.
10. We have in the usual way that Q(t
+ 1) =At+ Q(t)- min{l, Q(t)}
where At has the Poisson distribution with parameter A. When the queue is in equilibrium, JE.(Q(t)) = + 1)), and hence
JE.(Q(t
lP'(Q(t) > 0) = JE.(min{l, Q(t)}) = JE.(At) ='A. We have from (*)that the probability generating function G(s) of the equilibrium distribution of Q(t) (= Q) is
G(s) = JE.(sAt)JE.(sQ-min{1,Q}) = e).(s-1){lE.(sQ-1 l(Q:::l))
391
+ lP'(Q =
0) }.
[11.8.11]-[11.8.13]
Solutions
Queues
Also, G(s) = JE.(sQ I{Q:::1J) + JP>(Q = 0), and hence G(s) = whence
e).(s- 1) {
~G(s) +
( 1-
~) (1- A)}
(1 - s)(l -A) G(s) = 1- se -).(s -1).
The mean queue length is G' (1) = ~A(2 - A)/(1 -A). Since service times are of unit length, and arrivals form a Poisson process, the mean residual service time of the customer in service at an arrival time is ~, so long as the queue is non-empty. Hence A
JE.(W) = JE.(Q)- ~JP>(Q > 0) = - - 2(1- A)
11. The length B of a typical busy period has moment generating function satisfying M B (s) = exp{s- A+ AMs(s)}; this fact may be deduced from the standard theory ofM/G/1, or alternatively by a random-walk approach. Now T may be expressed as T = I + B where I is the length of the first idle period, a random variable with the exponential distribution, parameter A. It follows that MT(s) = AMs(s)/(A- s). Therefore, as required, (A- s)MT(s) = Aexp{s- A+ (A- s)MT(s) }. If A :::: 1, the queue~length at moments of departure is either null persistent or transient, and it follows that JE.(T) = oo. If A < 1, we differentiate(*) and sets = 0 to obtain AlE.(T)- 1 =A 2 JE.(T), whence JE.(T) = {A(l - A)} -1. 12. (a) Q is a birth-death process with parameters Ai = A, /-[i = {[, and is therefore reversible in equilibrium; see Problems (6.15.16) and (11.8.3). (b) The equilibrium distribution satisfies krci = {[7ri+1 fori :::: 0, whence Jri = (1- p)pi where p = A/ f.-[. A typical waiting time W is the sum of Q independent service times, so that
Mw(s)=GQ(Ms(s))=
1-p
1- P{[/({[- s)
=
(1-p)(f.-[-S) . f.-[(1- p)- s
(c) See the solution to Problem (11.8.3). (d) Follow the solution to Problem (11.8.3) (either method) to find that, at any timet in equilibrium, the queue lengths are independent, the jth having the equilibrium distribution ofM(A)IM({[j )/1. The joint mass function is therefore
where
Pj
=A/{[J.
13. The size of the queue is a birth-death process with rates Ai =A, /-[i = {[ min{i, k}. Either solve the equilibrium equations in order to find a stationary distribution 7r, or argue as follows. The process is reversible in equilibrium (see Problem (6.15.16)), and therefore Ailri = /-[i+17ri+1 for all i. These 'balance equations' become ifO~i n) ----+ 0 as n ----+ oo, so that the final two terms tend to 0 (by the uniform integrability of the Yi and the finiteness of lEI Yr I respectively). Therefore Yr An standard theorem (7.10.3).
~
Yr, and the claim follows by the
3. By uniform integrability, Y00 = limn-+oo Yn exists a.s. and in mean, and Yn = lE(Y00 I :Fn). (a) On the event {T = n} it is the case that Yr = Yn and JE(Y00 I :Fr) = lE(Y00 I :Fn); for the latter statement, use the definition of conditional expectation. It follows that Yr = JE(Y00 I :Fr ), irrespective of the value of T. (b) We have from Exercise (12.4.7) that :Fs ~ :Fr. Now Ys = JE(Yoo I :Fs) = lE{lE(Y00 I :Fr) I :Fs)
= JE(Yr I :Fs).
4.
Let T be the time until absorption, and note that {Sn} is a bounded, and therefore uniformly integrable, martingale. Also IP'(T < oo) = 1 since T is no larger than the waiting time for N consecutive steps in the same direction. It follows that JE(So) = lE(Sr) = NIP'(Sr = N), so that IP'(Sr = N) = JE(So)/ N. Secondly, {S~- n : n 2':: 0} is a martingale (see Exercise (12.1.4)), and the optional stopping theorem (if it may be applied) gives that lE(S6)
= JE(Sf -
T)
= N 21P'(Sr = N) -lE(T),
and hence JE(T) = NlE(So) - lECS6) as required. It remains to check the conditions of the optional stopping theorem. Certainly IP'(T < oo) = 1, and in addition lE(T 2) < oo by the argument above. We have that lEI Sf- Tl ::0 N 2 + JE(T) < oo. Finally, lE{ (S~ - n)l(T>n}} ::0 (N 2 + n)IP'(T > n) ----+ 0 as n ----+ oo, since JE(T 2) < oo.
5. Let :Fn = cr(SJ, S2, ... , Sn). It is immediate from the identity cos(A + ).) + cos(A- ).) = 2 cos A cos). that lE(Yn+l I :Fn)
=
cos[,J,.(Sn + 1- ~(b- a))]+ cos[,J,.(Sn- 1- ~(b- a))]
2(cos ,J,.)n+l
= Yn,
and therefore Y is a martingale (it is easy to see that lEI Yn I < oo for all n ). SupposethatO 0, then lP' (max Yk > 1~k~n
x) ::=: lP' (max (Yk + c) > (x + c) 2
1~k~n
2 ).
Now (Yk + c) 2 is a convex function of Yk, and therefore defines a submartingale (Exercise (12.1.7)). Applying the maximalinequality to this submartingale, we obtain an upper bound ofE{ (Yn +c) 2 }/ (x + c) 2 for the right-hand side of(*). We set c = E(Y1)/x to obtain the result.
4. (a) Note that Zn = Zn-1 + Cn{Xn- E(Xn I :Fn_I)}, so that (Z, !F) is a martingale. LetT be the stopping timeT= min{k: qYk 2:: x}. ThenE(ZTAn) = E(Zo) = 0, so that
since the final term in the definition of Zn is non-negative. Therefore n
xlP'(T ::S n) ::S E{cTAnyT/\n} ::S LckE{E(Xk I :Fk-d},
k=1 where we have used the facts that Yn 2:: 0 and E(Xk I :Fk-d 2:: 0. The claim follows. (b) Let X 1, X 2, ... be independent random variables, with zero means and finite variances, and let Yj = 1 X;. Then Y/ defines a non-negative submartingale, whence
L;{=
5. The function h(u) = lulr is convex, and therefore Y;(m) = IS;- Smlr, i 2:: m, defines a submartingale with respect to the filtration :F; = a ({x1 : 1 :::: j :::: i}). Apply the HRC inequality of Problem (12.9.4), with q = 1, to obtain the required inequality. If r = 1, we have that
E(ISm+n- Sml) ::S
m+n L EIZkl k=m+1
by the triangle inequality. Let m, n --+ oo to find, in the usual way, that the sequence {Sn} converges a.s.; Kronecker's lemma (see Exercise (7.8.2)) then yields the final claim. Suppose 1 < r ::=: 2, in which case a little more work is required. The function h is differentiable, and therefore
h(v)- h(u) = (v- u)h 1 (u)
+ fov-u {h'(u +x)- h 1(u)} dx.
Now h'(y) = rlylr- 1sign(y) has a derivative decreasing in IYI· It follows (draw a picture) that h 1 (u + x)- h 1(u) ::=: 2h 1 ( ix) if x 2:: 0, and therefore the above integral is no larger than 2h(i(v- u)). Apply this with v = Sm+k+1 - Sm and u = Sm+k- Sm, to obtain
404
Solutions [12.9.6]-[12.9.10]
Problems
Sum over k and use the fact that
to deduce that E(ISm+n- smn
~ 2 2 -r
m+n
L
E(lzkn·
k=m+l
The argument is completed as after (*). 6.
With
h
= l{Yk=O)• we have that
+ nYn-tiXn 1(1 - ln-t) I.Tn-1) = ln-tE(Xn) + nYn-1 (1- ln-t)EIXnl = Yn-t
E(Yn I .Tn-t) = E( Xnln-1
since E(Xn) = 0, EIXnl = n- 1. Also EIYnl ~ E{IXniO El Yn I < oo. Therefore (Y, !F) is a martingale. Now Yn
=
0 if and only if Xn
=
0. Therefore lP'(Yn
+ niYn-tl)} =
0)
=
and EIYtl < oo, whence
lP'(Xn
=
0)
=
1 - n- 1 ---+ 1
as n ---+ oo, implying that Yn ~ 0. On the other hand, Ln lP'(Xn =1- 0) = oo, and therefore lP'(Yn =1- 0 i.o.) = 1 by the second Borel-Cantelli lemma. However, Yn takes only integer values, and therefore Yn does not converge to 0 a.s. The martingale convergence theorem is inapplicable since supn EIYnl = 00. 7. Assume that t > 0 and M(t) = 1. Then Yn = e 1Sn defines a positive martingale (with mean 1) with respect to .Tn = a(Xt, X2, ... , Xn). By the maximal inequality,
and the result follows by taking the limit as n ---+ oo.
8. The sequence Yn = ~Zn defines a martingale; this may be seen easily, as in the solution to Exercise (12.1.15). Now {Yn} is uniformly bounded, and therefore Y 00 = limn---+oo Yn exists a.s. and satisfies E(Y00 ) = E(Yo) = ~. Suppose 0 < ~ < 1. In this case Zt is not a.s. zero, so that Zn cannot converge a.s. to a constant c unless c E {0, oo}. Therefore the a.s. convergence of Yn entails the a.s. convergence of Zn to a limit random variable taking values 0 and oo. In this case, E(Y00 ) = 1 · lP'(Zn ---+ 0) + 0 · lP'(Zn ---+ oo ), implying that lP'(Zn ---+ 0) = ~, and therefore lP'(Zn ---+ oo) = 1 - ~.
9. It is a consequence of the maximal inequality that lP'(Y; =:: x) ~ x- 1E(Ynl{Y,i::::x)) for x > 0. Therefore E(Y;) = fooo lP'(Y; =:: x) dx
~ 1 + E { Yn
1
00
~ 1+
1
00
lP'(Y;
=:: x) dx
x-l I(l,Y,t](x) dx}
= 1 + E(Yn log+ Y;) ~ 1 + E(Yn log+ Yn)
+ E(Y;)fe.
10. (a) We have, as in Exercise (12.7.1), that E(h(X(t)) I B,X(s)
=i) =
LPij(t)h(j) j
405
fors < t,
[12.9.11]-[12.9.13]
Solutions
Martingales
for any event B defined in terms of {X (u) : u ::=: s}. The derivative of this expression, with respect to t, is (P1 Gh')i, where P1 is the transition semigroup, G is the generator, and h = (h(j) : j :::: 0). In this case,
(Gh1)j
=L
gjkh(k)
= Aj { h(j + 1)- h(j)}- IJ-j { h(j)- h(j- 1)} = 0
k
for all j. Therefore the left side of (*) is constant for t ::=: s, and is equal to its value at time s, i.e. X (s). Hence h(X(t)) defines a martingale. (b) We apply the optional stopping theorem with T = min{t: X(t) E {0, n}} toobtainE(h(X(T))) = E(h(X(O))), and therefore (1 - n(m))h(n) = h(m) as required. It is necessary but not difficult to check the conditions of the optional stopping theorem.
11. (a) Since Y is a submartingale, so is y+ (see Exercise (12.1.6)). Now
Therefore {E(Y:+m I :Fn) : m :::: 0} is (a.s.) non-decreasing, and therefore converges (a.s.) to a limit Mn. Also, by monotone convergence of conditional expectation,
E(Mn+l
I :Fn) = m--+oo lim E{E(Y:+m+l I :Fn+I) I:Fn} = m--+00 lim E(Y:+m+l I :Fn) = Mn,
and furthermore E(Mn) = limm---+oo E(Y:+n) :::0 M. It is the case that Mn is :Fn-measurable, and therefore it is a martingale. (b) We have that Zn = Mn - Yn is the difference of a martingale and a submartingale, and is therefore a supermauingale. Also Mn ::=: Y: ::=: 0, and the decomposition for Yn follows. (c) In thi~ case Zn is a martingale, being the difference of two martingales. Also Mn :::: E(Y: I :Fn) = Y: ::=: Yn a.s., and the claim follows.
12. We may as well assume that JJ- < P since the inequality is trivial otherwise. The moment generating function of P- Cr is M(t) = iCP-JLH5;a 212 , and we choose t such that M(t) = l, i.e., t = -2(P- J.J-)/a 2. Now define Zn = min{etYn, 1} and :Fn = a(Cr, C2, ... , Cn). Certainly EIZnl < oo; also E(Zn+l I :Fn) :::0 E(etYn+l I :Fn) = etYn M(t) = etYn and E(Zn+l I :Fn) :S 1, implying that E(Zn+l I :Fn) :S Zn. Therefore (Zn, :Fn) is a positive supermartingale. LetT = inf{n : Yn ::=: 0} = inf{n : Zn = 1}. Then T 1\ m is a bounded stopping time, whence E(Zo) ::=: E(ZT Am) ::=: JP>(T ::=: m). Let m ---+ oo to obtain the result.
13. Let :Fn = a(Rr, R2, ... , Rn). (a) 0 :S Yn :S 1, and Yn is :Fn-measurable. Also
E(Rn+l
I Rn)
= Rn
Rn
+ n+r+ b'
whence Yn satisfies E(Yn+I I :Fn) = Yn. Therefore {Yn : n ::=: 0} is a uniformly integrable martingale, and therefore converges a.s. and in mean. (b) In order to apply the optional stopping theorem, it suffices that JP>(T < oo) = l (since Y is uniformly integrable). However JP>(T > n) = ~ n~I = (n + l)- 1 ---+ 0. Using that theorem,
1· ··· E(YT) = E(Yo), which is to say that E{T I (T + 2)} = 1, and the result follows. (c) Apply the maximal inequality.
406
Solutions [12.9.14]-[12.9.17]
Problems
14. As in the previous solution, with fi,n the a-field generated by A1, A2, ... and :Fn, E(Yn+1
I 13 (J>n) =
Rn + An ) ( Rn ) Rn + Bn +An Rn + Bn Rn ---=Yn, Rn +Bn (
+(
Rn
Rn
+ Bn + An
) (
Bn Rn
+ Bn
)
sothatE(Yn+1 I :Fn) = E{E(Yn+1 I fi,n) I :Fn} = Yn. Also IYn I ::=: 1, and therefore Yn is a martingale. We need to show that lP'(T < oo) = 1. Letln be the indicator function of the event {T > n}. We have by conditioning on the An that
E(In
I A)
IT
= n-1 ( 1 - -1- ) --+
J=O
2 + SJ
IT 00
J=O
(
1 - -1- ) 2 + S1
L-{=
as n --+ oo, where s1 = 1 Ai. The infinite product equals 0 a.s. if and only if"L,1 (2+ SJ )- 1 = oo a.s. By monotone convergence, lP'(T < oo) = 1 under this condition. If this holds, we may apply the optional stopping theorem to obtain that E(YT) = E(Yo), which is to say that
15. At each stage k, let Lk be the length of the sequence 'in play', and let Yk be the sum of its entries, so that Lo = n, Yo = L,j= 1 Xi. If you lose the (k + 1)th gamble, then Lk+ 1 = Lk + 1 and Yk+1 = Yk + Zk where Zk is the stake on that play, whereas if you win, then Lk+1 = Lk - 2 and Yk+1 = Yk- Zk; we have assumed that Lk :::: 2, similar relations being valid if Lk = 1. Note that Lk is a random walk with mean step size -1, implying that the first-passage time T to 0 is a.s. finite, and has all moments finite. Your profits at time k amount to Yo - Yb whence your profit at time T is Yo, since YT = 0. Since the games are fair, Yk constitutes a martingale. Therefore E(YT 1\m) = E(Yo) =j:. 0 for all m. However T 1\ m --+ T a.s. as m --+ oo, so that YT 1\m --+ YT a.s. Now E(YT) = 0 =j:. Iimm---+ooE(YT/\m), and it follows that {YT/\m : m :0:: 1} is not uniformly integrable. Therefore E(supm YT 1\m) = oo; see Exercise (7.10.6). 16. Since the game is fair, E(Sn+1 I Sn) = Sn. Also ISnl ::=: 1 + 2 + · · · + n < oo. Therefore Sn is a martingale. The occurrence of the event {N = n} depends only on the outcomes of the coin-tosses up to and including the nth; therefore N is a stopping time. A tail appeared at time N- 3, followed by three heads. Therefore the gamblers G1, G2, ... , G N -3 have forfeited their initial capital by time N, while G N -i has had i + 1 successful rounds for 0 ::=: i ::=: 2. Therefore SN = N- (p- 1 + p- 2 + p- 3 ), after a little calculation. It is easy to check that N satisfies the conditions of the optional stopping theorem, and it follows that E(SN) = E(So) = 0, which is to saythatE(N) = p-1 + p-2 + p-3.
In order to deal with HTH, the gamblers are re-programmed to act as follows. If they win on their first bet, they bet their current fortune on tails, returning to heads thereafter. In this case, SN = N- (p- 1 + p- 2q- 1) where q = 1- p (remember that the game is fair), and therefore E(N) = p-1 + p-2q-1.
17. Let :Fn = a ({Xi , Yi : 1 ::=: i ::=: n}), and note that T is a stopping time with respect to this filtration. Furthermore lP'(T < oo) = 1 since Tis no larger than the first-passage time to 0 of either of the two single-coordinate random walks, each of which has mean 0 and is therefore persistent. Let a'f = var(X1) and ai = var(YI). We have that Un - Uo and Vn - Vo are sums of independent summands with means 0 and variances a'f and ai respectively. It follows by considering
407
[12.9.18]-[12.9.19]
Solutions
Martingales
the martingales (Un- Uo) 2 -no} and (Vn- Vo) 2 -no} (see equation (12.5.14) and Exercise (10.2.2)) that E{(UT - Uo) 2} = o}E(T), E{(VT - Vo) 2} = o}E(T). Applying the same argument to (Un E { (UT
+ VT- Uo-
+ Vn)- (Uo + Vo), we obtain
Vo) 2} = E(T)E{(Xr
+ Yr) 2} =
E(T)(a12 + 2c + a 22).
Subtract the two earlier equations to obtain E{ (UT - Uo)(VT - Vo)} = cE(T) if E(T) < oo. Now UT VT = 0, and in addition E(UT) = Uo, E(VT) = Vo, by Wald's equation and the fact that E(X r) = E(Yr) = 0. It follows that -E(Uo Vo) = cE(T) if E(T) < oo, in which case c < 0. Suppose conversely that c < 0. Then(*) is valid with T replaced throughout by the bounded stopping timeT 1\ m, and hence
0 :S E(UT Am VTAm) = E(Uo Vo)
+ cE(T 1\ m).
Therefore E(T 1\ m) :S E(Uo Vo)/(21cl) for all m, implying that E(T) = limm-+oo E(T and so E(T) = -E(Uo Vo)/c as before.
1\
m) < oo,
18. Certainly 0 :S Xn :S 1, and in addition Xn is measurable with respect to the a-field :Fn = a(Rr, R2, ... , Rn). Also E(Rn+l I Rn) = Rn- Rn/(52- n), whence E(Xn+l I :Fn) = Xn. Therefore Xn is a martingale. A strategy corresponds to a stopping time. If the player decides to call at the stopping time T, he wins with (conditional) probability X T· and therefore IP'(wins) = E(XT ), which equals E(Xo) (= by the optional stopping theorem. Here is a trivial solution to the problem. It may be seen that the chance of winning is the same for a player who, after calling "Red Now", picks the card placed at the bottom of the pack rather than that at the top. The bottom card is red with probability irrespective of the strategy of the player.
i)
i,
19. (a) A sums of money in weekt is equivalentto a sums /(1 +a)t in weekO, since the latter sum may be invested now to yield s in week t. If he sells in week t, his discounted costs are :L~= 1 cI (1 + a )n and his discounted profit is Xt/(1 + a)t. He wishes to find a stopping time for which his mean discounted gain is a maximum. Now
so that tt(T) = E{ (1
+ a)-T ZT} -
(c/a).
(b) The function h(y) = ay - fy IP'(Zn > y) dy is continuous and strictly increasing on [0, 100 ), with h(O) = -E(Zn) < 0 and h(y) --+ oo as y --+ oo. Therefore there exists a unique y (> 0) such that h (y) = 0, and we choose y accordingly. (c) Let :Fn = a(Zr, Z2, ... , Zn). We have that 00
E(max{Zn, y}) = y
+
l
00
[1- G(y)]dy = (1 +a)y
where G(y) = IP'(Zn :S y). Therefore E(Vn+l I :Fn) = (1 non-negative supermartingale.
408
+ a)-ny
:S Vn, so that CVn, :Fn) is a
Solutions [12.9.20]-[12.9.21]
Problems
Let ~-t(r) be the mean gain of following the strategy 'accept the first offer exceeding r - (cja)'. The corresponding stopping timeT satisfies JP>(T = n) = G(r)n(l- G(r)), and therefore 00
~-t(r)
+ (cja)
=
L E{ (1 + a)-T Zr l{T=n)} n=O 00
= L(l
+ a)-nG(r)n(l- G(r))E(Z, I z,
n=O
=
1 +a { r(l - G(r)) 1 +a- G(r)
+
1
00
,
> r)
(1 - G(y)) dy } .
Differentiate with care to find that the only value of r lying in the support of Z 1 such that ~-t' (r) = 0 is the value r = y. Furthermore this value gives a maximum for ~-t( r ). Therefore, amongst strategies of the above sort, the best is that with r = y. Note that ~-t(Y) = y(l +a) - (cja). Consider now a general strategy with corresponding stopping time T, where JP>(T < oo) = 1. For anypositiveintegerm, T /\misaboundedstoppingtime, whenceE(VTAm):::: E(Vo) = y(l+a). Now IVT Am I :::: l::~o IVil, and l::~o EIVi I < oo. Therefore {Vr Am : m :=::: 0} is uniformly integrable. Also Vr Am --+ Vr a.s. as m --+ oo, and it follows that E(Vr Am) --+ E(Vr ). We conclude that J-t(T) = E(Vr)- (cja) :::: y(l +a)- (cja) = J-t(y). Therefore the strategy given above is optimal. (d) In the special case, JP>(ZI > y) = (y - 1)- 2 for y ::::, 2, whence y = 10. The target price is therefore 9, and the mean number of weeks before selling is G(y)/(1 - G(y)) = 80.
20. Since G is convex on [0, oo) wherever it is finite, and since G(l) =land G'(l) < 1, there exists a unique value of IJ (> 1) such that G(l]) = IJ. Furthermore, Yn = IJZn defines a martingale with mean E(Yo) = IJ. Using the maximal inequality (12.6.6),
n
1
k
IJ) ::S
JP>(supZn::::, k) = lP' ( supYn :=::: n
k-l I]
for positive integers k. Therefore 00
E(supZn) ::S n
1
L - .1
k=I I] -
21. LetMn bethenumberpresentafterthenthround, soMo = K, andMn+I = Mn -Xn+I· n::::, 1, where Xn is the number of matches in the nth round. By the result of Problem (3.11.17), EXn = 1 for all n, whence E(Mn+I + n + 1 I :Fn) = Mn + n, where :Fn is the a-field generated by Mo, M1, ... , Mn. Thus the sequence {Mn + n} is a martingale. Now, N is clearly a stopping time, and therefore K = Mo + 0 = E(MN + N) =EN. We have that
+ n + 1) 2 + Mn+I I Fn} = (Mn + n) 2 - 2(Mn + n)E(Xn+I ::S (Mn + n) 2 + Mn,
E{ (Mn+I
- 1) + Mn
+ E{ (Xn+I
where we have used the fact that
1 if Mn > 1, if Mn = 1.
var(Xn+l I :Fn) = { 0
409
I
- 1) 2 - Xn+I :Fn}
[12.9.22]-[12.9.24]
Solutions
Martingales
+ n) 2 + Mn}
Hence the sequence {(Mn supermartingales,
is a supermartingale. By an optional stopping theorem for
and therefore var(N) _:::: K.
22. In the usual notation, E(M(s
+ t) I.'Fs)
= E(fos W(u)du
+ 1s+t W(u)du-
HW(s
= M(s) + tW(s)- W(s)E([W(s + t)-
+ t)- W(s) + W(s)} 3 1.'Fs)
W(s)l 2
I.'Fs) = M(s)
as required. We apply the optional stopping theorem (12.7.12) with the stopping timeT= inf{ u : W ( u) E {a, b}}. The hypotheses of the theorem follow easily from the boundedness of the process fort E [0, T], and it follows that
· Hence the required area A has mean E(A) = E( (T W(u)du) =
h
3 ) = ~a 3 (~) + ~b 3 (-a-). ~E(W(T) 3 3 a-b 3 a-b
[We have used the optional stopping theorem twice actually, in that E(W(T)) = 0 and therefore JP>(W(T) =a)= -bj(a- b).]
23. With .'Fs = a(W(u) : 0 _:::: u _:::: s), we have for s < t that E(R(t) 2 I .'Fs) = E(IW(s)l 2 + IW(t)- W(s)l 2
+ 2W(s) · (W(t)-
W(s)) I:Fs) = R(s) 2
+ (t- s),
and the first claim follows. We apply the optional stopping theorem (12.7.12) with T = inf{u : IW(u)l =a}, as in Problem (12.9.22), to find that 0 = E(R(T) 2 - T) = a 2 - E(T).
24. We apply the optional stopping theorem to the martingale W(t) with the stopping timeT to find that E(W(T)) = -a(1 - Pb) + bpb = 0, where Pb = JP>(W(T) = b). By Example (12.7.10), W (t) 2 - t is a martingale, and therefore, by the optional stopping theorem again,
whence E(T) = ab. For the final part, we take a = band apply the optional stopping theorem to the 2 t] to obtain martingale exp[eW(t)-
ie
on noting that the conditional distribution ofT given W (T) = b is the same as that given W (T) = -b. Therefore, E(e-i 02 T)
=
1/ cosh(be), and the answer follows by substituting s
410
=
ie 2.
13 Diffusion processes
13.3 Solutions. Diffusion processes 1.
It is easily seen that
I
E{ X(t +h)- X(t) X(t)} =(A.- p,)X(t)h + o(h), E( {X (t +h) - X (t)} 2 1 X (t)) = (A.+ p,)X (t)h + o(h),
which suggest a diffusion approximation with instantaneous mean a(t, x) = (A.- p,)x and instantaneous variance b(t, x) =(A.+ p,)x. 2. The following method is not entirely rigorous (it is an argument of the following well-known type: it is valid when it works, and not otherwise). We have that
-aM = at
1oo e -oo
0Yaf -
at
dy
=
1oo {ea(t, y) + ~e b(t, y) }e -oo 2
0Y f
dy,
by using the forward equation and integrating by parts. Assume that a(t, y) = I:n an (t)yn, b(t, y) = I:n f3n (t)yn. The required expression follows from the 'fact' that
3.
Using Exercise (13.3.2) or otherwise, we obtain the equation aM= emM + le 2M 2 at
with boundary condition M(O, e) = l. The solution is M(t) = exp{ ~e (2m + e)t}.
4.
Using Exercise (13.3.2) or otherwise, we obtain the equation aM aM 1 2 -=-e-+-e M 2 ae at
with boundary condition M(O, e) = l. The characteristics of the equation are given by dt 1
de e
2dM e2M'
with solution M(t, e) = e! 02 g(ee- 1) where g is a function satisfying 1 = e! 02 g(e). Therefore M = exp{~e 2 (1- e- 21 )}.
411
[13.3.5]-[13.3.10]
Solutions
Diffusion processes
5. Fix t > 0. Suppose we are given W1 (s), W2(s), W3(s), for 0:::;: s :::;: t. By Pythagoras's theorem, R(t + u) 2 =Xi+ X~+ X~ where the Xi are independent N(Wi (t), u) variables. Using the result of Exercise (5.7.7), the conditional distribution of R(t + u) 2 (and hence of R(t + u) also) depends only on the value of the non-centrality parameter e = R(t) 2 of the relevant non-central x2 distribution. It follows that R satisfies the Markov property. This argument is valid for the n-dimensional Bessel process. 6. By the spherical symmetry of the process, the conditional distribution of R(s +a) given R(s) = x is the same as that given W(s) = (x, 0, 0). Therefore, recalling the solution to Exercise (13.3.5),
lP'(R(s +a):::;: y I R(s) =x) =
=
I
i { (u-x) 2 +v 2 +w 2 } (u,v,w): dudvdw 2a 2:rra 312 exp u2+v2+w2:sy2 ( )
y 1 2:n: l:n: 1 { 1p=O 312 exp ¢=0 0=0 (2:rra)
=loy~ { exp (
(p
~ax)2) -
p 2 - 2px cos e + x 2 } p 2 sm8d8d¢dp . 2a exp (
(p ;ax)2)} dp,
and the result follows by differentiating with respect to y. 7. Continuous functions of continuous functions are continuous. The Markov property is preserved because gO is single-valued with a unique inverse. I 2
8. (a) Since E(eaW(t)) = e2a t, this is not a martingale. (b) This is a Wiener process (see Problem (13.12.1)), and is certainly a martingale. (c) With :Ft = a(W(s) : 0:::;: s :::;: t) and t, u > 0, E{ (t + u)W(t + u)- lot+u W(s)ds
I:Ft} =
(t + u)W(t)- lot W(s)ds -lt+u W(t)ds
= tW(t)- lot W(s)ds,
whence this is a martingale. [The integrability condition is easily verified.]
9. (a) With s < t, S(t) = S(s) exp{a(t- s) + b(W(t)- W(s)) }. Now W(t)- W(s) is independent of {W(u) : 0:::;: u :::;: s}, and the claim follows. (b) S (t) is clearly integrable and adapted to the filtration !F = (:Ft) so that, for s < t, E(S(t) I Fs) = S(s)E(exp{a(t- s) + b(W(t)- W(s))} I Fs) = S(s)exp{a(t- s) + ~b 2 (t- s)}, which equals S(s) if and only if a+ ~b 2 = 0. In this case, E(S(t)) = E(S(O)) = 1.
10. Either find the instantaneous mean and variance, and solve the forward equation, or argue directly as follows. With s < t, lP'(S(t):::;: y I S(s) = x) = lP'(bW(t):::;: -at+ logy I bW(s) =-as+ logx). Now b(W(t)- W(s)) is independent of W(s) and is distributed as N(O, b2(t- s)), and we obtain on differentiating with respect to y that
f (t, y 1 s, x) -_
1 ( (log(y/x)-a(t-s)) 2 ) exp , 2 yy2:rrb2(t- s) 2b (t- s)
412
x,y > 0.
Solutions
Excursions and the Brownian bridge
[13.4.1]-[13.6.1]
13.4 Solutions. First passage times 1.
Certainly X has continuous sample paths, and in addition EIX (t) I < oo. Also, if s < t,
as required, where we have used the fact that W(t)- W(s) is N(O, t - s) and is independent of :Fs.
2.
Apply the optional stopping theorem to the martingale X of Exercise (13.4.1), with the stopping time T, to obtain E(X(T)) = 1. Now W(T) = aT + b, and therefore E(elfrT+ilib) = l where 1/J = i ae + 2 . Solve to find that
1e
E(elfrT)
= e-ilib = exp { -b( Jaz- 21/1 +a)}
is the solution which gives a moment generating function. We have that T :::: u if and only if there is no zero in (u, t], an event with probability l (2/n) cos- 1 { .JU?t}, and the claim follows on drawing a triangle.
3.
13.5 Solution. Barriers 1. Solving the forward equation subject to the appropriate boundary conditions, we obtain as usual that F(t, y) = g(t, y I d)+ e-Zmd g(t, y I -d)-
j_:
2me 2 mx g(t, y I x) dx
I
where g(t, y I x) = (2nt) -2 exp{ -(y - x - mt) 2 j(2t)}. The first two terms tend to 0 as t -+ oo, regardless of the sign of m. As for the integral, make the substitution u = (x - y- mt) I ..fi to obtain, as t-+ oo, -
1
-(d+y+mt)/../i
-00
I 2
e-zu { 21mle-21mly 2me 2my - - du -+ ..,fiir 0
ifm < 0, ifm:::: 0.
13.6 Solutions. Excursions and the Brownian bridge 1.
I
2
Let f(t, x) = (2nt)-ze-x /(Zt). It may be seen that
IP'(W(t)
>xI Z, W(O) =
0)
= lim IP'(W(t) w-!-0
>xI Z, W(O) =
w)
where Z = {no zeros in (0, t]}; the small missing step here may be filled by conditioning instead on the event {W(E) = w, no zeros in (E, t]}, and taking the limit as E .j, 0. Now, if w > 0,
IP'(W(t) > x, Z I W(O)
= w) =
1
00
{J(t, y- w)- f(t, y
by the reflection principle, and
IP'(Z I W(O) =
w)
= l- 2
Loo f(t, y)dy 413
=
i:
+ w)} dy
f(t, y)dy
[13.6.2]-[13.6.5]
Solutions
Diffusion processes
by a consideration of the minimum value of Won (0, t]. It follows that the density function of W(t), conditional on Z n {W(O) = w}, where w > 0, is
h ( ) _ f(t, x- w)- f(t, x X -w f(t, y) dy
w
Jw
+ w)
'
X> 0.
Divide top and bottom by 2w, and take the limit as w {. 0: lim hw(x) = _ _1_ 8f = ::_e-x2/(2t) f(t, 0) ax t ,
X> 0.
w-),0
2.
It is a standard exercise that, for a Wiener process W,
C=;),
E{W(t) I W(s) =a, W(1) =
0}
=a
E{W(s) 2 W(O) = W(l) =
0}
= s(l- s),
1
if 0 ::::: s ::::: t ::::: 1. Therefore the Brownian bridge B satisfies, for 0 ::::: s ::::: t ::::: 1, E(B(s)B(t)) = E{ B(s)E(B(t) I B(s))} = 1 - t E(B(s) 2 ) = s(l- t) 1-s as required. Certainly E(B(s)) = 0 for all s, by symmetry. 3. W is a zero-mean Gaussian process on [0, 1] with continuous sample paths, and also W(O) = W(l) = 0. Therefore W is a Brownian bridge if it has the same autocovariance function as the Brownian bridge, that is, c(s, t) = min{s, t}- st. For s < t, cov(W(s), W(t)) = cov(W(s)- sW(1), W(t)- tW(l)) = s- ts- st
+ st =s-st
since cov(W(u), W(v)) = min{u, v}. The claim follows. 4. Either calculate the instantaneous mean and variance of W, or repeat the argument in the solution to Exercise (13.6.3). The only complication in this case is the necessity to show that W(t) is a.s. continuous at t = l, i.e., that u- 1 W(u- l) ---+ 0 a.s. as u---+ oo. There are various ways to show this. Certainly it is true in the limit as u---+ oo through the integers, since, for integral u, W(u- l) may be expressed as the sumofu -1 independent N(O, 1) variables (use the strong law). It remains to fill in the gaps. Let n be a positive integer, let x > 0, and write Mn = max{IW(u)- W(n)l : n::::: u ::::: n + 1}. We have by the stationarity of the increments that oo
oo
E~)
n=O
n=O
X
L)"CMn ::=: nx) = ~)P'CM1 ::=: nx)::::: 1 + - - < oo, implying by the Borel-Cantelli lemma that n- 1 Mn ::::: x for all but finitely many values of n, a.s. Therefore n - 1 Mn ---+ 0 a.s. as n ---+ oo, implying that
. 1 . 1 hm --IW(u)l::::: hm -{IW(n)l +1 n-+oo n
u-+oo u
+ Mn}---+ 0
5.
a.s.
In the notation of Exercise (13.6.4), we are asked to calculate the probability that W has no zeros in the time interval between s /(1 - s) and t /(1 - t). By Theorem (13.4.8), this equals 1 2 1 - - cos-
:rr
s(1-y) 2 -1 t(1-s) =;cos
414
~
V~·
Solutions
Stochastic calculus
[13.7.1]-[13.7.5]
13.7 Solutions. Stochastic calculus 1. Let :Fs = u(Wu : 0:::;: u :::;: s). Fix n :::0: 1 and define Xn(k) = IWktjznl for 0:::;: k:::;: 2n. By Jensen's inequality, the sequence {Xn (k) : 0 :::;: k :::;: 2n} is a non-negative submartingale with respect to the filtration :Fktf2n, with finite variance. Hence, by Exercise (4.3.3) and equation (12.6.2), X~= max{Xn(k): 0 ::=:: k ::=:: 2n} satisfies
E(X~2 ) = 2lo xlP'(X~ > x) dx :::;: 2lo 00
00
E(Wt
= 2E(Wt X~) ::=:: 2VECWhE(Xj?)
/{X~:o:x}) dx =
2E{ wt lox:; dx}
by the Cauchy-Schwarz inequality.
Hence E(X~ 2 ) :::;: 4E(Wh. Now X~ 2 is monotone increasing inn, and W has continuous sample paths. By monotone convergence,
2.
See the solution to Exercise (8.5.4).
3.
(a) We have that
w?,
The first summation equals by successive concellation, and the mean-square limit of the second it in mean square. summation is t, by Exercise (8.5.4). Hence limn--+oo Ir (n) = i Likewise, we obtain the mean-square limits: lim /z(n) = i
n--+oo
4.
w?-
w? +it,
lim /3(n) = lim h(n) =
n--+oo
n--+oo
iw?.
Clearly E(U (t)) = 0. The process U is Gaussian with autocovariance function
Thus U is a stationary Gaussian Markov process, namely the Omstein-Uhlenbeck process. [See Example (9.6.10).] 5.
Clearly E(Ut) = 0. For s < t, E(UsUs+t) = E(Ws Wt)
+ ,B 2 E (1~ 0 1~ 0 e-.B(s-u)Wue-.B(t-v)Wv du dv)
- E ( Wt.B los e-.B(s-u)Wu du) - E ( Ws lot e-.B(t-v)Wv dv) =s+,B 2e-.B(s+t)1s
1 1
e.B(u+v)min{u,v}dudv
u=O v=O
- ,B los e-.B(s-u) min{u, t} du- lot e-,B(t-v) min{s, v} dv e2.Bs - 1 -,B(s+t) 2,8 e
415
[13.8.1]-[13.8.1]
Solutions
Diffusion processes
after prolonged integration. By the linearity of the definition of U, it is a Gaussian process. From the calculation above, it has autocovariance function c(s, s +t) = (e-f3(t-s) -e-f3(t+sl)j(2(3). From this we may calculate the instantaneous mean and variance, and thus we recognize an Omstein-Uhlenbeck process. See also Exercise (13.3.4) and Problem (13.12.4).
13.8 Solutions. The Ito integral 1. (a) Fix t > 0 and let n :=:: I and 8 = tjn. We write tj = jtjn and Vj = Wtr By the absence of correlation of Wiener increments, and the Cauchy-Schwarz inequality,
~ ~{(tj+1- tj) ltj+l E(IVH1- Wsl 2)ds} j=O
=
tl
n-1 1 -(tj+1 -tj) 3 . 0 2
L
=
n-1 1 (t)3
L. 2 n
--+ 0
asn--+ oo .
]= 0
]=
Therefore,
=
lim ( tWtn--'>00
n-1
L vj+1 (tj+1- tj) )
= tWt-
j=O
lor Ws ds. t
(b) As n --+ oo, n-1 n-1 2:: V/(Vi+1- Vj) = 2:: {v/+ 1 - V/- 3Vj(Vj+1 - Vj) 2 - --oo
M(t, B)= exp {
2adeat
t
}
{3(1 - ea )
which converges to e- 2adffJ as t--+ oo.
6.
The equilibrium density function g(y) satisfies the (reduced) forward equation d --(ag) dy
where a(y) are
=
-fJy and b(y)
2
1 d + --(bg) = 2 dy2
0
= a 2 are the instantaneous mean and variance.
The boundary conditions
y = -c,d. Integrate (*) from -c to y, using the boundary conditions, to obtain
-c
~
y
~d.
Integrate again to obtain g(y) = Ae-f3Y 2fa 2 . The constant A is given by the fact that f~c g(y) dy = 1.
7.
First we show that the series converges uniformly (along a subsequence), implying that the limit exists and is a continuous function of t. Set n-1 . (k ) ~ sm t Zmn(t) = ~ -k-Xk>
Mmn = sup{IZmn(t)l: 0 ~ t ~
7r }.
k=m
We have that M2 < mn -
su
n-1 eikt
~ -X P l~ k k
O::or:::,:rr k=m
12
n-1 x2 < ~ ___k_ ~ k2
-
k=m
422
+2
n-m-11n-1-1 X·X· I 1 J+l ~ ~ ~ ~ · ( · + l) · 1=1
J=m 1 1
Solutions
Problems
[13.12.8]-[13.12.8]
The mean value of the final term is, by the Cauchy-Schwarz inequality, no larger than n-m-1
2
1
n-l-1
L:
"
+ [)2
L.J j2(j }=m
1=1
< 2(n - m) -
F.t--
m m4 ·
--
Combine this with (*) to obtain
It follows that
implying that .2:::~ 1 M2n-I 2n < oo a.s. Therefore the series which defines W converges uniformly with probability 1 (along a ;ubsequence), and hence W has (a.s.) continuous sample paths. Certainly W is a Gaussian process since W(t) is the sum of normal variables (see Problem (7.11.19)). Furthermore E(W(t)) = 0, and st
cov ( W(s), W(t) ) = -
2 ~ sin(ks) sin(kt)
+- L.J
T(
T(
2
k
k=1
since the Xi are independent with zero means and unit variances. It is an exercise in Fourier analysis to deduce that cov(W(s), W(t)) = min{s, t}.
8.
We wish to find a solution g(t, y) to the equation
IYI <
b,
satisfying the boundary conditions g(O, y) = 8yo
if IYI :S b,
g(t, y) =
0 if IYI =b.
Let g(t, y I d) be the N(d, t) density function, and note that g(-, · I d) satisfies (*) for any 'source' d. Let 00
L:
g(t, y) =
c-1)kg(t, y 12kb),
k=-00
a series which converges absolutely and is differentiable term by term. Since each summand satisfies (*),so does the sum. Now g(O, y) is a combination of Dirac delta functions, one at each multiple of2b. Only one such multiple lies in [-b, b], and hence g(y, 0) = 8ao· Also, setting y = b, the contributions from the sources at -2(k- 1)b and 2kb cancel, so that g(t, b) = 0. Similarly g(t, -b) = 0, and therefore g is the required solution. Here is an alternative method. Look for the solution to(*) of the form e-l.nt sin{inn(y + b)jb}; such a sine function vanishes when IYI =b. Substitute into(*) to obtain An = n 2 n 2 /(8b 2 ). A linear combination of such functions has the form 00 g(t,y) =Lane
-Ant .
sm
n=1
423
(
nn(y +b) ) . 2b
[13.12.9]-[13.12.11]
Solutions
Diffusion processes
We choose the constants an such that g(O, y) = 8yo for IYI < b. With the aid of a little Fourier analysis, one finds that an = b- 1 sin(~mr). Finally, the required probability equals the probability that wa has been absorbed by time t, a probability expressible as 1- f~b fa(t, y) dy. Using the second expression for this yields
r,
9. Recall that U(t) = e- 2mD(t) is a martingale. LetT be the time of absorption, and assume that the conditions of the optional stopping theorem are satisfied. Then E(U(O)) = E(U(T)), which is to say that 1 = e2ma Pa + e-2mb(l - Pa).
10. (a) We may assume that a, b > 0. With Pt(b)
= lP'(W(t)
> b, F(O, t) I W(O) =a),
we have by the reflection principle that Pt(b) = lP'(W(t) > b I W(O) =a) -lP'(W(t) < -b I W(O) =a) = lP'(b- a< W(t) < b +a I W(O) =
o),
giving that apr (b) ---az;=
f(t, b +a)- f(t, b- a)
where f(t, x) is the N(O, t) density function. Now, using conditional probabilities, lP'(F(O, t) I W(O) =a, W(t) =b)=-
1 apr(b) = 1 _ e-2abft. f(t, b- a) ab
(b) We know that
2
lP'(F(s, t)) = 1 - - cos- 1 { T(
VS/t} = -2 sin- 1 { VS/t} T(
if 0 < s < t. The claim follows since F (to, t2) s; F (t1, t2). (c) Remember that sinx = x + o(x) as x 0. Take the limit in part (b) as to
+
+0 to obtain ,fiJ!i2.
11. Let M(t) = sup{W(s): 0:::; s:::; t} and recall that M(t) has the same distribution as IW(t)l. By symmetry, lP'( sup IW(s)l::::
w) :::; 2lP'(M(t):::: w)
= 2lP'(IW(t)l::::
w).
O~s~t
By Chebyshov's inequality,
Fix E > 0, and let An(E) = {I W(s)l/s > E for somes satisfying 2n- 1 < s :::; 2n }. Note that
424
Solutions [13.12.12]-[13.12.13]
Problems
for all large n, and also
Therefore L:n lP'(An(E)) < oo, implying by the Borel-Cantelli lemma that (a.s.) only finitely many of the An(E) occur. Therefore t- 1W(t)--+ 0 a.s. as t--+ oo. Compare with the solution to the relevant part of Exercise (13.6.4).
12. We require the solution to Laplace's equation p(w)
=
v2 p
= 0, subject to the boundary condition
0 ifwEH, { 1 'f G. 1 W E
Look for a solution in polar coordinates of the form 00
p(r, e) =
L rn {an sin(ne) + bn cos(ne)}. n=O
Certainly combinations having this form satisfy Laplace's equation, and the boundary condition gives that 00
H(e) = bo
+L
{an sin(ne)
+ bn cos(ne) },
1e1
< n,
n=1 where H(e) = {
~
if - n <
e < o,
ifO = PoR satisfies e = tan- 1(D/d), whence lP'(E>.::; e)= lP'(D.::; dtane) =
425
1
e
2 + -;·
1e1
< ~n.
+ d2)
,
u E ffi:.,
[13.12.14]-[13.12.18]
Solutions
Diffusion processes
14. By an extension of Ito's formula to functions of two Wiener processes, U = u(Wr, W2 ) and V = v(Wr, W2) satisfy dU = Ux dWr dV = Vx dWr
+ uy dW2 + J-Cuxx + Uyy) dt, + vy dW2 + J-Cvxx + Vyy) dt,
where Ux, Vyy, etc, denote partial derivatives of u and v. Since ¢ is analytic, u and v satisfy the Cauchy-Riemann equations Ux = vy, uy = -vx, whence u and v are harmonic in that Uxx + uyy = Vxx + Vyy = 0. Therefore,
The matrix (
~xUy
uy) is an orthogonal rotation oflH:2 when u~ +u~ = 1. Since the joint distribution Ux
of the pair (Wr, W2) is invariant under such rotations, the claim follows. 15. One method of solution uses the fact that the reversed Wiener process {W (t- s)- W (t) : 0 ::: s ::: t} has the same distribution as {W(s) : 0 ::: s ::: t}. Thus M(t)- W (t) = maxo~s~dW(s)- W(t)} has the same distribution as maxo~u~dW(u)- W(O)} = M(t). Alternatively, by the reflection principle, lP'(M(t) 2':: x, W(t)::: y) = lP'(W(t) 2':: 2x- y)
for x 2':: max{O, y}.
By differentiation, the pair M(t), W(t) has joint density function -2¢' (2x - y) for y ::: x, x 2':: 0, where¢ is the density function of the N(O, t) distribution. Hence M(t) and M(t)- W(t) have the joint density function -2¢'(x + y). Since this function is symmetric in its arguments, M(t) and M(t)- W(t) have the same marginal distribution. 16. The Lebesgue measure A(Z) is given by
A(Z)
= fooo l{W(t)=u) du,
whence by Fubini's theorem (cf. equation (5.6.13)), JE:(A(Z))
= fooo lP'(W(t) = u) dt = 0.
17. Let 0
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