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Solution Solutions s Manual Manual to
AN INTRODUCTION TO MATHEMATICAL FINANCE: OPTIONS AND OTHER TOPICS
Sheldon M. Ross
1 1.1 (a)
1 − p0 − p1 − p2 − p3 = 0.05 (b) p (b) p 0 + p + p1 + p + p2 = 0.80
1.2 P {C ∪ ∪ 8
0 .3 − 0. 0.2 = 0.5 ∩ R} = 0.4 + 0. R} = P {C } + P {R} − P {C ∩
7
=
56
14 13
1.4 (a)
27/ 27/58 (b) 27/ 27/35
182
(b)
6
1.3 (a)
5
14 13
=
30 182
(c)
6
8
14 13
+
8
6
14 13
96
=
182
1.5
1. The probability probability that their child will develop develop cystic cystic fibrosis is the probability probability that the child receives a CF gene from each of his parents, which is 1 /4. 2. Given Given that that his sibling sibling died died of the disease, disease, each each of the parents parents much much have have exactl exactly y one CF gene. Let A denote the event that he possesses one CF gene and B that he does not have the disease (since he is 30 years old). Then P ( P (A|B ) =
1.6 Let
2/4 2 P ( P (A ∩ B ) P ( P (A) = = = 3 P ( P (B ) P ( P (B ) 3/4
A be the event that they are both aces and B the event they are of different
suits. Then
P ( P (A ∩ B ) P ( P (A) = = P ( P (A|B ) = P ( P (B ) P ( P (B )
4
3
52 51 39
=
51
1 169
1.7
(a) P ( P (AB ) = = = = c
P ( P (A) − P ( P (AB) AB ) P ( P (A) − P ( P (A)P ( P (B ) P ( P (A)(1 − P ( P (B ) P ( P (A)P ( P (B ) c
Part (b) follows from part (a) since from (a) A and B are independent, implying from (a) that so are A are A and B and B . c
c
c
the gambler loses both the bets, then X = − 3. If he wins the first bet, or loses the first bet and wins the second bet, X = X = 1. Therefore, 1.8 If
20 2 100 ) = 38 361 18 20 18 261 + = P {X = 1} = 38 38 38 361
P {X = − 3} = (
1. P {X > 0} = P {X = 1} = 2. E [X ] = 1 261 = − 3 100 361 361 1.9
−39 361
261 361
2 1. E [X ] is larger since a bus with more students is more likely to be chosen than a bus with less students. 2. 1 5882 (392 + 332 + 462 + 342 ) = 152 152 1 (39 + 33 + 46 + 34) = 38 E [Y ] Y ] = 4
E [X ] =
≈ 38. 38.697
Let N denote the numbe numberr of sets played. played. Then Then it is clear clear that P N = 2 = N denote P N = 3 = 1/2.
{
1.10
{
}
}
1. E [N ] N ] = 2.5 2. Var(N ar(N ) = 21 (2 1.11 Let
2
− 2.5)
+ 21 (3
2
− 2.5)
=
1 4
= E [X ]. µ = E Var(X Var(X ) = = = =
Let F be be 1.12 Let F
[(X µ)2 ] E [(X + µ2 ] E [X 2 2µX + E [X 2 ] 2µE [X ] + µ2 E [X 2 ] µ2
− − − −
her fee if she takes the fixed amount and X when X when she takes the contin-
gency amount. 000, E [F ] F ] = 5, 000,
SD( SD (F ) F ) = 0
25, 000(. 000(.3) + 0(. 0(.7) = 7, 7, 500 E [X ] = 25, (25, 000)2 (.3) + 0(. 0(.7) = 1. 1.875 E [X 2 ] = (25,
8
× 10
Therefore, SD( SD (X ) =
Var(X Var(X ) =
1.875
× 10 − (7, (7, 500)
1.13 n ¯ ] = 1 (a) E [X E [X i ] n i=1 1 = nµ = nµ = µ µ n
8
2
=
√
1.3125
4
× 10
3
n
¯ ) = ( 1 )2 (b) Var(X Var(X i ) n i=1 1 = ( )2 nσ2 = σ 2 /n n
n
(c)
n
¯ )2 = (X i − X
i=1
¯ + X ¯ )2 ) (X i2 − 2X i X
i=1
n
=
i=1
n
¯ X − 2X 2 i
¯ 2 X i + nX
i=1
n
=
¯ X ¯ + nX ¯ 2 X i2 − 2Xn
i=1
n
=
¯ 2 X i2 − nX
i=1
n
2
(d) E [(n − 1)S ] = E [
¯ 2 ] X i2 ] − E [nX
i=1
= = = =
¯ 2 ] nE [X 12 ] − nE [X ¯ ) + E [X ¯ ]2 ) n(Var(X 1 ) + E [X 1 ]2 ) − n(Var(X nσ 2 + nµ2 − n(σ 2 /n) − nµ2 (n − 1)σ2
1.14
Cov(X, Y ) = = = =
E [(X − E [X ])(Y − E [Y ])] E [XY − XE [Y ] − E [X ]Y + E [X ]E [Y ])] E [XY ] − E [Y ]E [X ] − E [X ]E [Y ] + E [X ]E [Y ] E [XY ] − E [Y ]E [X ]
1.15
(a) Cov(X, Y ) = E [(X − E [X ])(Y − E [Y ])] = E [(Y − E [Y ])(X − E [X ])] (b) Cov(X, X ) = E [(X − E [X ])2 ] = Var(X ) (c) Cov(cX,Y ) = E [(cX − E [cX ])(Y − E [Y ])] = cE [(X − E [X ])(Y − E [Y ])] = cCov(X, Y ) (d) Cov(c, Y ) = E [(c − E [c])(Y − E [Y ])] = 0
4
1.16
Cov(aU + bV,cU + dV ) = Cov(aU, cU + dV ) + Cov(bV,cU + dV ) = Cov(aU, cU ) + Cov(aU, dV ) + Cov(bV,cU ) + Cov(bV,dV ) = ac(1) + ad(0) + bc(0) + bd(1) = ac + bd 1.17 With
c(i, j) = Cov(X i , X j ) (a) c(1, 3) + c(1, 4) + c(2, 3) + c(2, 4) = 21 (b) 2 + 3 + 4 + 4 + 6 + 8 + 6 + 9 + 12 = 54 1.17 Let X i be
the amount it goes up in period i. Then 3
Y =
X i
i=1
and Cov(X 1 , Y ) = Cov(X 1 , X 1 ) = Var(X 1 ) = 1/4 Therefore, Corr(X, Y ) = 1.18 No,
−1 and 1.
1/4
(1/4)(3/4)
√
= 1/ 3
since for such a pair Corr(X, Y ) = 2, and correlations are always between
1.18
Let X i be the amount it goes up in period i. Then 3
Y =
X i
i=1
and Cov(X 1 , Y ) = Cov(X 1 , X 1 ) = Var(X 1 ) = 1/4 Therefore, Corr(X 1 , Y ) =
1.19 No,
√
1/4 = 1/ 3 (1/4)(3/4)
since for such a pair Corr(X, Y ) = 2 and correlations are always between
−1 and 1.
1.20
h(y)P (Y = y) =
y
h(y)P (Y = y)
i y:h(y)=hi
=
hi P (Y = y)
i y:h(y)=hi
=
hi
i
=
P (X = i) = F (i)
y:h(y)=hi
hi P (h(Y ) = h i )
i
1.21
P (Y = y)
− F (i − 1)
2
5
2.1
1. P (Z <
−.66) = P (Z > .66) = 1 − P (Z < .66) = 1 − Φ(.66) = 1 − .7454 = .2546
2. P ( Z x) + P (Z < −x) = 2P (Z > x)
where the last equality comes from the fact that Z is symmetric. 2.4
a = 2µ,
b =
−1. Cov(X, Y ) =
−Var(X ) = −σ
2
2.5
(a) 127.7 (b) 127.7 (c) 127.7
± 19.2 ± (1.96)(19.2) ± 57.6
Let X 1 and X 2 denote the life of the first and the second battery respectively. It is given that X 1 and X 2 are both normal random variables with mean 400 and standard deviation 50. Let Z denote a standard normal random variable. 2.6
√
1. X 1 + X 2 is a normal random variable with mean 800 and standard deviation 50 2. P (X 1 + X 2 > 760) =
X + X − 800 760 − 800 √ > √ P 1
2
50 2 50 2 P (Z > .5657) = P (Z < .5657) = Φ(.5657) 0.7142
≈ 2. X 2
− ≈
− X is a normal random variable with mean 0 and standard deviation 50√ 2. 25 X − X √ > 50√ 2 P (X − X > 25) = P 50 2 ≈ P (Z > .3536) = 1 − Φ(.3536) ≈ 1 − .6382 = .3618 1
2
2
1
1
6 3. P ( X 1
| − X | > 25) = 2P (X − X > 25) ≈ .7236 2
2
1
the time the that it takes to develop the i th print. Then, the time that it takes to develop 100 prints, call it X , can be expressed as 2.7 Let X i be
100
X X =
i
i=1
It follows from the central limit theorem that X approximately has a normal distribution with mean 1800 and standard deviation 10. Therefore,
{
− 1800 } { X −101800 > 1710 10 1 − Φ(−9) Φ(9) ≈ 1
= P
P X > 1710
}
= = The probability for part (b) is 0
Let X i be the mileage for person i, i = 1, . . . , 30. From central limit theorem, X i is approximately a normal random variable with mean 25000 30 and standard deviation 12000 30. 2.8
30 i=1
×
× √
30
1. P
i=1
> 25000 = P
X i
30
30
X i
−25000 √ ×30 > 0 = Φ(0) = 0.5 12000× 30
i=1
2. 30 i=1
X P 23000 < log 1.3 > 1.3 = P X > 1.3 = P i
i=0
i
i=0
7 We can use the central limit theorem to approximate variable Y with the same mean and variance.
999 i=0
log X i with a normal random
E [Y ] = 1000 E [log X i ] = 1000 ( p log u + (1 p)log d) Var(Y ) = 1000 Var(log X i ) = 1000 ( p(log u)2 + (1 p)(log d)2 .00137872 )
−
−
−
≈ 1.3787 ≈ 0.1206
Therefore 999
log X > log 1.3 P i
i=0
≈ =
≈
P Y > log 1.3 Y − 1.3787 log1.3 − 1.3787 √ P √ > .1206 .1206 P (Z > 3.2146) = Φ(3.2146)
−
≈ .9993
where Z stands for a standard normal random variable. 2.10 Let X i be
the movement in period i. Then we can approximate normal random variable Y with the same mean and variance
700 i=1
X i with a
700
E [Y ] =
E X = 700(−.39 + .41) = 14 Var X = 700(.39 × 1.02 + .20 × .02 + .41 × .98 ) = 559.72 i
i=1
700
Var(Y ) =
2
2
2
i
i=1
Therefore, 700
P X > 10 i
i=1
≈
10.5 − 14 Y − 14 ≈ Φ(.1479) ≈ .5588 P (Y > 10.5) = P √ > √ 559.72
559.72
3.1 This
follows since X (t + y) ( X (y)) = (X (t + y) X (y)) is the negative of a normal with mean µt and variance σ 2 t, and so has mean µt and variance σ2 t.
3.2 (a)
−
−−
− −
−
16; (b) 18
√
X (2) 16 > 20 16 18) 18 = P (Z > .9428) .173
(c) P (X (2) > 20) = P (
√ −
−
≈
X (.5) 11.5 10 11.5 > ) 14.5 14.5 = P (Z > .3939)
(d) P (X (.5) > 10) = P (
√ − −
= P (Z < .3939)
√ −
≈ .655
where Z is a standard normal.
3.3 (a)
With 2 p 1 = .3162, E [X (1)] = 10 + 10(.9487)(.3168) = 13.005 (b) Var(X (1)) = 9[1 .09998] 8.10018 (c) P (X (.5) > 10) = (.6581)5 + 5(.6581)4 (.3419) + 10(.6581)3 (.3419)2 = .7773
3.4 (a)
−
−
≈
With W being normal with mean .1 and variance .04,
P (S (1) > S (0)) = P (eW > 1) = P (W > 0) = P (Z >
−.1/.2) = P (Z < .5) = .6915
(b) (.6915)2 (c) P (S (3) < S (1) > S (0)) = P (S (1) > S (0))P (S (3) < S (1) S (1) > S (0))
|
= P (S (1) > S (0))P (S (3) < S (1)) = .6915P (Z <
3.6 S (t)/s is
√ −.2/ .08)
distributed as eW , where W is normal with mean µt and variance tσ2 . Hence, E [S (t)] = sE [eW ] = seµt+tσ
2
/2
and E [S 2 (t)] = s 2 E [e2W ] = s 2 e2µt+2tσ 3
2
Hence, Var(S (t)) = s 2 e2µt+2tσ
2
− s2e2µt+tσ
2
3.7 This
2
= s 2 e2µt+tσ (etσ
2
− 1)
follows directly from the formula for P (T y t) given in the text, upon using that limx→∞ ¯ Φ(x) = 0 and limx→∞ ¯ Φ( x) = 1. Hence, when µ y ) = P (T y <
2
∞) = e2yµ/σ ,
y > 0
the representation that S (t) = seX (t) , where X (t), t X (0) = 0, gives 3.8 Using
P ( max S (v) 0 v t
≤≤
≥ y) = P
max X (v)
0 v t
≤≤
≥ 0 is Brownian motion with
≥ log(y/s)
Now use Corollary 1, setting t = log(y/s).
3.9 The
desired probability is P (M (t)
34 1+r .
6 5 + = 6.7444 − (1.05) 3 (1.05)4
That is, when r > .2.
4.20
104e−r = 110e−2r giving that er = 110/104 or r = log(110/104) = .0561
4.21
Ae−rs +
4.22 (a)
∞
−rs
−r(s+nt) = Ae−rs ∞ −rt )n = Ae n=0 (e 1−e
n=1 Ae
−rt
The interest earned by time t + h on the interest earned in (t, t + h) is of smaller order
than h. (b) From (a), we have D(t + h) D(t) rD(t) h 0, the approximation becomes exact and we obtain that
−
Letting h
→
≈
D (t) = rD(t) (c) Integrating both sides of
D (t) D(t)
= r, yields that log(D(t)) = rt + C 5
or D(t) = K ert for some constant K . Evaluating at t = 0 gives D = D(0) = K .
4.23 By
Proposition 4.2.1, the cash flow 100, 140, 131 is preferable for any positive interest rate.
110/(1 + r)2 = 100 or r = .0488 (b) If R is the rate of return, then R is equally likely to be 4.24 (a)
4.25 1000e
4.26 100
√ 1.2 − 1 or 0. Hence, E [R] = .0477.
−.8 = 449.33
= 40(1 + r)−1 + 70(1 + r)−2 yielding that r = .0498
4.27 (a)
No, it is greater than 10 percent if and only if 8 16 110 (b) yes because 1.11 + (1.11) + (1.11) = 100.624 > 100. 2
3
i
xi /(1.1)i is greater than 1.
4.28
P (
X 1 X 2 + > 100) = P (1.1X 1 + X 2 > 121) 1.1 (1.1)2 121 126 = P (Z > ) 55.25 = P (Z < .6727) = .780
√ −
where the preceding used that 1.1X 1 + X 2 is normal with mean 126 and variance 55.25
6
The present value of your net return is 10e−.12 10
5.1 (a)
(b)
−
5.2 (a)
− 10 = −1.1308
-5; (b)2e .03
− − 5 = −3.059
5.3 Because
the call option will be exercised, purchasing it costs C at time 0 and then costs K at time 1 with the result being owning the security at time 1. Another investment that yields the security at time 1 is to purchase it at time 0 for its initial price s. By the law of one price s = C + Ke −r giving that C = s
5.4 If C
− Ke −r .
> S an arbitrage is effected by simultaneously selling the call and buying the security.
≥ 0, the put call option parity implies that Ke −rt ≥ S − C
5.5 Because P
5.6 Use
≥ S − Ke −rt = 30 − 28e−.05/3
Exercise 5.5 to obtain C
5.7 (a)
is not necessarily true (to see this, let K be exceedingly large); (b) is true for if P > K an arbitrage can be effected by selling the put.
5.8 This
follows from the put call option parity formula.
5.10 Buying
the security, buying the put, and selling the call has an initial cost of S + P C and no matter what the price at time t yields K at that time. (If S (t) K then the sold call is worthless and we exercise the put to sell the security for K . If S (t) > K then the bought put is worthless but the purchaser of the call will exercise and we will receive K for the security.) A second investment that yields K at time t is to ;put Ke −rt in the bank at time 0. The parity formula now follows from the law of one price.
≤
7
−
(a) K ; (b) If P is cost of put then law of one price yields s + P = Ke −rt , giving that P = Ke−rt s. (Note that if K e−rt < s then se rt > K > S (t), showing that selling the security with the intention to purchase it at time t yields an arbitrage.) 5.11
−
both yields 1 at time t, as would putting e−rt in the bank at time 0. Hence, the law of one price gives C 1 + C 2 = e −rt 5.12 Buying
25 = S +P C > Ke−rt = 20e−.1/4 , an arbitrage is effected by selling the security, selling the put, and buying the call. This yields 25 and will cost you 20 at time t. 5.13 Because
−
If C and P are the costs for the European versions, then C a = C and P a call parity formula yields S + P a C a Ke −rt 5.14
≥ P .
The put
− ≥
5.15 First
note that P 1 P 2 for if P 1 < P 2 an arbitrage is effected by buying the P 1 put and selling the P 2 put. So assume that P 1 P 2 . If K 1 K 2 < P 1 P 2 then an arbitrage is effected by selling the K 1 , P 1 put and buying the K 2 , P 2 put. This has an immediate return of P 1 P 2 . If the sold put is exercised at time t then if you also exercise the bought put you will have to pay K 1 K 2 , which is less than your immediate return.
≥
≥
−
−
−
−
5.16 If
it were less than could buy the exercise time t put and sell the exercise time s < t put. If the sold put is ever exercised then you should exercise the bought put at that time. These latter transactions cancel each other and you have the initial difference in prices as your arbitrage.
5.17 (a)
True because it is clearly true for an American call option and the European is worth the same amount. (b) This is only true if the domestic interest rate is at least as large as the foreign rate. (c) This need not be true since it is sometimes optimal to exercise early and so being forced to continue can be detrimential.
8
5.18 (a)
5.19
s
If the security has a lot of volatility.
−d
(S (t) K )+ + (K S (t))+ = S (t) (b) (S (t) K 1 )+ (K 2 S (t))+ (c) 2(S (t) K )+ S (t) (d) S (t) (S (t) K )+ 5.20 (a)
− − −
−
− − −
−
−
|
− K |.
5.21 If
not then an arbitrage would result from buying the one with lower strike and selling the one with higher strike.
5.22 (a)
negative
5.23 Suppose
you buy the K = 110 call and sell the K = 100 call. This would give you 20 C at time 0 and would cost at most (if the sold option is exercised then you should exercise the other option) 10 at exercise time t. So there would be an arbitrage if 20 C 10e−rt . Hence, C 20 10e−rt .
−
− ≥
≥ −
5.24 Convexity
follows from the analogous result for call options upon using the put call option
parity formula.
5.25 yes,
to show that having λ put options with strike K 1 and 1 λ put options with strike K 2 is better than having one put option with strike K = λK 1 + (1 λ)K 2 exercise the put pair at the same time that the single put is exercised, taking (K ∗ s)+ as the return from exercising a K ∗ strike put when the security price is s.
−
− −
5.26 If s is
the price at time t 1 than better than exercising is to sell the security at that time and then exercise and give back the security at time t2 . This follows because exercising at time t1 gives a time t1 return of s K 1 , whereas the latter policy gives a time t1 return of s K 2 e−r(t −t ) .
−
−
9
2
1
5.27 This
follows because
S (t)
− max(K, S (t) − A) = S (t) + min(−K, −S (t) + A) = min(S (t) − K, A) where we used that − max(a, b) = min(−a, −b). Hence, (S (t) − max(K, S (t) − A))+ = max{0, min(S (t) − K, A)} = min{(S (t) − K )+ , A)} 5.28 A
function is concave if the curve obtained when it is plotted is such that the straight line connecting any two of its points lies below or on the curve.
5.29 An
arbitrage is a sure win, so neither is necessarily true.
10
17
6.1 We
need to see whether we can find a probability vector ( p1 , p2 , p3 ) for which all bets are fair. In order to have all bets fair, pi = 1/(1 + oi ). Therefore, p1 = 1/2
p2 = 1/3
p3 = 1/6
Since the p i ’s sum up to 1, ( p1 , p2 , p3 ) is indeed a probability vector which makes all bets fair. Therefore, no arbitrage is present. 6.2 To
rule out the arbitrage opportunity, o 4 must satisfy the equation, 1 1 1 1 + + + =1 1 + 2 1 + 3 1 + 4 1 + o4
Therefore, o 4 = 47/13. 6.3 No
arbitrage is present since 1 1 1 1 1 1 + + = + + = 1 1 + o1 1 + o2 1 + o3 3 3 3
6.4 If
no arbitrage is present, then ( p1 , p2 , p3 ) = (1/2, 1/3, 1/6) has to be the probability vector which makes all bets fair. Therefore o12 ( p1 + p2 ) − p3 = 0 o23 ( p2 + p3 ) − p1 = 0 o13 ( p1 + p3 ) − p2 = 0 6.5 If
⇒ ⇒ ⇒
o12 = 1/5 o23 = 1 o13 = 1/2
the outcome is j , then the betting scheme x i , i = 1, . . . , m, gives me m
oj xj
−
x
i
=
m
x + x = (1 + o )x x ox (1 + o ) (1 + o )(1 + o ) =1 (1 + o ) 1 j
j −
j
j
j
j −
i=1
i =j
=
j
i=1
j
−
6.6 From
−1 − j m i=1
m i=1 −1 i
i
−1
Example 6.1b, p = (1 + 2r)/3. The payoff of the put option is 0 if the stock price goes up, and 100 if the stock price goes down. To rule out arbitrage, the expected return of buying one put option under the probability distribution has to be zero. That is, p0 + (1 − p)100 − P = 0 1 + r Therefore 1 + 2r 100 200(1 − r) = P = 1 − 3 1 + r 3(1 + r)
18 The put-call parity says
K 1 + r In this example, one can check the following indeed holds. S + P − C =
6.7
Since the call option expires in period 2 and the strike price K = 150, it is clear that C uu = 250
C ud = 0
C dd = 0
Let p denote the risk neutral probability that the price of the security goes up, then p =
1 + r − d 1 − 0.5 1 = = 2 − 0.5 3 u−d
where we assume r = 0. Then we can find C by computing the expected return of the call option in the risk neutral world. 1 3 = pC ud + (1 − p)C dd = 0
C u = pC uu + (1 − p)C ud = C d 1 250 C = p 2 C uu + 2 p(1 − p)C ud + (1 − p)2 C dd = 2 × 250 = 3 9 6.8 See
Example 8.1a for details.
6.9 We
need to find a betting strategy which gives a weak arbitrage if (a) C = 0 and (b) C = 50/3. (a) C = 0. In this case it is clear that buying one share of stock is a weak arbitrage. At time 0, one does not have to pay out anything. At time 1, the profit is 50 if the stock goes up to 200, and 0 if the stock price is either 100 or 50. (b) C = 50/3. In this case it is not that clear how a weak arbitrage can be established. But since the price of the call option is high, it’s intuitive that we want to sell it. So, let’s consider a portfolio consisting of selling one share of the call option and buying x share(s) of the stock. Our return depends on the price of stock at time 1, which is tabulated as follows. stock price at time 1 200 100 50
balance value of the portfolio at time 0 at time 1 50/3 − 100x −50 + 200x 50/3 − 100x 0 + 100x 50/3 − 100x 0 + 50x
profit (r = 0) 100x − 100/3 50/3 −50x + 50/3
From the above table, if we choose x = 1/3, then the profit is 50/3 if the stock price is 100 at time 1, and 0 otherwise, which is a weak arbitrage.
×
250 =
250 3
6.3 (a)
There is no arbitrage if there are probabilities p1 , p2 , p3 such that 4 p1 + 8 p2
− 10 p3 = 0
i pi =
1 and
and 6 p1 + 12 p2
− 16 p3 = 0 Because the first equation implies that 6 p1 + 12 p2 − 15 p3 = 0, any solution must have p3 = 0. Consequently, p 1 , p2 would need to satisfy p 1 + 2 p2 = 0 which is impossible because the p i must be nonnegative. Hence, an arbitrage is possible. One such is to let x1 = 1.5, x2 = −1. (b) No arbitrage if there are probabilities p 1 , p2 , p3 such that 6 p1
− 3 p2
−2 p1 + 6 p3
i pi =
1 and
= 0 = 0
10 p1 + 10 p2 + xp3 = 0 Hence, p2 = 2 p1 , and p1 = 3 p3 Using that i pi = 1, this yields that p1 = .3, p2 = .6, p3 = .1. Therefore, no arbitrage is possible if x = 90.
−
Let S (0) = s and suppose that us > K > ds. If you purchase y shares of the security by borrowing x and investing the remaining ys x then your payoff at time t is 6.10
−
payoff =
yus yds
− (1 + r)x, − (1 + r)x,
if S (1) = us if S (1) = ds
Hence, the payoff from the option is replicated if we choose x, y so that yus
− (1 + r)x = us − K
and yds Setting y =
us K (u d)s ,
− −
x =
ds(us K ) (1+r)(u d)s does
− −
the no arbitrage cost of the option is ys
− (1 + r)x = 0
the trick. It now follows, by the law of one price, that ds(us−K ) − x = usu−−dK − (1+r)(u −d)s .
−d .3 2 3 With p = 1+r u−d = .425 = 12/17, the expected payoff is 25(1 p) p = .7606 , so the no arbitrage cost is .7606(1.1)−5 = .4723 (b) yes (c) 25(1/2)5 = 25/32 = .78125 6.11 (a)
−
11
−d arbitrage is possible if new bet is fair when the up probability is p = 1+r u−d = .7380. The bet will pay off if at least 2 of the first 3 moves are up moves. Hence, no arbitrage if 6.12 No
C = e−.15 100 (.7380)3 + 3(.7380)2 (.2620)
12
19
7.1 (a) .33/
√
√
225,
(b).33/ 12
7.2 Since
the unit of time is one year, t = 4/12 = 1/3. The probability that the call option will be exercised at t = 1/3 is the probability that the stock price at t = 1/3 is greater than the strike price K = 42, which is P (S (1/3) > 42) =
S (1/3) 42 P > S (0) S (1/3) 40
42 = P log > log 40 S (0) = P (X > log 1.05)
where X is a√ normal random variable with mean µt = .12/3 = .04 and standard deviation √ σ t = .24/ 3. Therefore, the above probability is equal to
− .04 = 1 − Φ log1.05√ − .04 √ > log1.05√ P .24/ 3 .24/ 3 .24/ 3 ≈ 1 − Φ(.0634) ≈ 1 − .5253 = .4747
X − .04
7.3 The
parameters are t = 1/3
so we have that
r = .08
K = 42
σ = .24
.08/3 + .242 /6 − log(42/40) √ ω = .24/ 3
S = 40
≈ −.0904
Therefore, C
≈ 40Φ(−.0904) − 42e−. / Φ(−.2289) ≈ 40 × .4639 − 42e−. / × .4094 ≈ 1.8137 08 3
08 3
7.4 From
the put-call parity, we can derive the no-arbitrage cost to a put option P = C − S (0) + Ke−rt √ = S (0)Φ(ω) − Ke−rt Φ(ω − σ t) − S (0) + K −rt √ = S ( 0) (Φ(ω) − 1) + Ke−rt (1 − Φ(ω − σ t))
where ω is defined in equation (7.7) in text (page 87). The parameters are K = 100
S (0) = 105
ω ≈ .571767 √ ω − σ t ≈ .359635
r = .1
σ = .30
Φ(.571767) ≈ .7163 Φ(.359635) ≈ .6404
t = 1/2
7.5 (a)
With W being a normal random variable with mean .03 and variance .045 P (S (.5)/S (0) < .9) = P (W log(40/38)) = e −.03 100(1
7.8 e
− (.32)2/2)/2 = .0044 and variance
− Φ(.207)) ≈ 40.55
−.03 100(1 − Φ( log(40/38) √ ))
7.9 No,
.0512
you also need to know the drift of the geometric Brownian motion.
13
The risk neutral geometric Brownian motion has drift r σ 2 /2 = .06 .08 = .02. No arbitrage if under the risk neutral geometric Brownian motion. Under this process W log(S (1)/S (0)) is normal with mean .02 and variance .16. Hence, no arbitrage if 7.10 (a)
−
−
−
−
≡
10 = e−.06 (5P (W log(1.1))) Now, P (W log(1.1)) = P (W > .0953) = 1 Hence, x
) = 1 − Φ(.28825) ≈ .386 − Φ( .1153 .4
.06 − 5(.469) ≈ 10e .386 ≈ 21.433
(b) Using that the actual mean of W is .05 yields that P (S (1) 760) =
X + X − 800 760 − 800 √ > √ P 1
2
50 2 50 2 P (Z > .5657) = P (Z < .5657) = Φ(.5657) 0.7142
≈ 2. X 2
− ≈
− X is a normal random variable with mean 0 and standard deviation 50√ 2. 25 X − X √ > 50√ 2 P (X − X > 25) = P 50 2 ≈ P (Z > .3536) = 1 − Φ(.3536) ≈ 1 − .6382 = .3618 1
2
1
2
1
6 3. P ( X 1
| − X | > 25) = 2P (X − X > 25) ≈ .7236 2
2
1
2.7 Let X i be the time the that it takes to develop the i th print. Then, the time that
it takes to develop 100 prints, call it X , can be expressed as 100
X X =
i
i=1
It follows from the central limit theorem that X approximately has a normal distribution with mean 1800 and standard deviation 10. Therefore,
{
− 1800 } { X −101800 > 1710 10 1 − Φ(−9) Φ(9) ≈ 1
= P
P X > 1710
}
= = The probability for part (b) is 0
2.8 Let X i be the mileage for person i, i = 1, . . . , 30. From central limit theorem,
30 i=1 X i is
approximately a normal random variable with mean 25000 deviation 12000 30.
× √
30 X i i=1
1. P
> 25000 = P
30
30 Xi i=1
× 30 and standard
−25000 ×30 √ > 0 = Φ(0) = 0.5 12000× 30
2. 30 i=1
X P 23000 < log 1.3 > 1.3 = P X > 1.3 = P i
i=0
i
i=0
7 We can use the central limit theorem to approximate variable Y variable Y with with the same mean and variance.
999 i=0 log X i with
[log X i ] = 1000 ( p ( p log u + (1 p)log E [Y ] Y ] = 1000 E [log p)log d) Var(Y Var(Y )) = 1000 1000 Var(l ar(log og X i ) = 1000 ( p(log p(log u)2 + (1 p)(log p)(log d)2 .00137872 )
−
−
−
a normal random
≈ 1.3787 ≈ 0.1206
Therefore 999
log X > l og 1.3 P > log i
i=0
≈ =
≈
log 1.3 P Y > log 1.3787 log1. log1.3 − 1.3787 − Y − √ P √ > .1206 .1206 Φ(3.2146) P ( P (Z > 3.2146) = Φ(3.
−
≈ .9993
where Z where Z stands stands for a standard normal random variable. 2.10 Let X i be the movement in period i. i . Then we can approximate
normal random variable Y variable Y with with the same mean and variance
700 i=1 X i with
700
E [Y ] Y ] =
E X = 700(−.39 + .41) = 14 Var 700(.39 × 1.02 + .20 × .02 + .41 × .98 ) = 559. 559.72 X = 700(. i
i=1 700
Var(Y Var(Y )) =
i
2
2
2
i=1
Therefore, 700
P X > 10 > 10 i
i=1
≈
− 14 10. 10.5 − 14 Y − 10.5) = P √ Φ(.1479) ≈ .5588 ≈ Φ(. P ( P (Y > 10. > √ 559. 559.72
559. 559.72
a
8
3.1
(a) (a) 100 100e.1+.2 = 100e 100e.3 For parts (b) and (c), use that Y .4. This gives
log(S (10)/ (10)/100) ≡ log(S
is normal normal with with mean mean . .11 and variance
(b) P S (10) > (10) > 100 100
}
= P Y > 0 Y .1 .1 = P > .4 .4 .1 = Φ( ) = Φ(. Φ(.158) = . = .5628 5628 .4
(b) P S (10) < (10) S . In this case, one can buy one share of stock and sell a call option today, which gives C S dollars. When the call option expires, the value of the portfolio is
≥
≥
≤
−
vs
− v = min{S (t), K } > 0 c
Therefore it is an arbitrage. 5.4 (a) is not true. (b) is true. The easiest way to see this is to use the put call parity and the result from Exercise 5.3. rt
− Ke−
P
= C
− S ≤ 0
Therefore rt
≤ Ke − ≤ K
P
Also, one can argue that an arbitrage opportunity exists if P > K . 5.5 From the put call parity,
− Ke −
P
rt
+ S = C
≥ 0
Therefore rt
≥ Ke− − S
P
15 Also, one can argue that an arbitrage opportunity exists if the inequality doesn’t hold. 5.7 Let P (t) denote the price of an American put option having exercise time t. Given
s < t, we want to show that P (s) P (t). We prove this by showing that an arbitrage is present if the statement is not true. Suppose P (s) > P (t), then we buy one share of the put option having exercise time t and sell one share of the put option having exercise time s, which gives us P (s) P (t) dollars today. Consider the strategy that whenever the put option we sold is exercised by the buyer, we exercise our put option at the same time. This strategy guarantees us no cash flow in the time period (0, s]. On the other hand, if the put option we sold is not exercised by the buyer, we then still have our put option at time s, whose value is always nonnegative. Either way, it is a sure win situation (remember we receive P (s) P (t) today), therefore an arbitrage.
≤
−
−
5.8 No, it is not valid for European puts. Suppose we buy the put option having
exercise time t and sell the put option having exercise time s, where s < t. If the sold put option is exercised at time s, we are forced to pay K for one share of stock. At time t, we have a debt of Ker(t−s) and our put option guarantees us to sell that stock for at least K , which is not enough to pay off the debt. Therefore, we can not guarantee a sure win. 5.9 s
−d
5.10
(a) If S (t) > K , the call option is worth S (t) K and the put option is worthless. On the other hand if S (t) < K , the put option is worth K S (t) and the call option is worthless. Therefore, the payoff is S (t) K .
− − |
|
−
(b) Consider the following two cases. K 1 K 2 , then
≥
S (t) − K payoff = 0 S (t) − K
, S (t) > K 1 , K 2 S (t) , S (t) < K 2
1
≤
2
≤ K
1
K 1 < K 2 , then
S (t) − K payoff = 2S (t) − K − K S (t) − K 1
1
2
2
, S (t) > K 2 , K 1 S (t) , S (t) < K 1
≤
≤ K
2
5.7 Let P (t) denote the price of an American put option having exercise time t. Given
s < t, we want to show that P (s) P (t). We prove this by showing that an arbitrage is present if the statement is not true. Suppose P (s) > P (t), then we buy one share of the put option having exercise time t and sell one share of the put option having exercise time s, which gives us P (s) P (t)
≤
−
16 dollars today. Consider the strategy that whenever the put option we sold is exercised by the buyer, we exercise our put option at the same time. This strategy guarantees us no cash flow in the time period (0, s]. On the other hand, if the put option we sold is not exercised by the buyer, we then still have our put option at time s, whose value is always nonnegative. Either way, it is a sure win situation (remember we receive P (s) P (t) today), therefore an arbitrage.
−
5.8 No, it is not valid for European puts. Suppose we buy the put option having
exercise time t and sell the put option having exercise time s, where s < t. If the sold put option is exercised at time s, we are forced to pay K for one share of stock. At time t, we have a debt of Ker(t−s) and our put option guarantees us to sell that stock for at least K , which is not enough to pay off the debt. Therefore, we can not guarantee a sure win. 5.10
(a) If S (t) > K , the call option is worth S (t) K and the put option is worthless. On the other hand if S (t) < K , the put option is worth K S (t) and the call option is worthless. Therefore, the payoff is S (t) K .
− − |
|
−
(b) Consider the following two cases. K 1 K 2 , then
≥
S (t) − K payoff = 0 S (t) − K
, S (t) > K 1 , K 2 S (t) , S (t) < K 2
1
≤
2
≤ K
1
K 1 < K 2 , then
S (t) − K payoff = 2S (t) − K − K S (t) − K 1
1
2
2
, S (t) > K 2 , K 1 S (t) , S (t) < K 1
≤
≤ K
2
17
6.1 We need to see whether we can find a probability vector ( p1 , p2 , p3 ) for which all
bets are fair. In order to have all bets fair, pi = 1/(1 + oi ). Therefore, p1 = 1/2
p2 = 1/3
p3 = 1/6
Since the p i ’s sum up to 1, ( p1 , p2 , p3 ) is indeed a probability vector which makes all bets fair. Therefore, no arbitrage is present. 6.2 To rule out the arbitrage opportunity, o 4 must satisfy the equation,
1 1 1 1 + + + =1 1 + 2 1 + 3 1 + 4 1 + o4 Therefore, o 4 = 47/13. 6.3 No arbitrage is present since
1 1 1 1 1 1 + + = + + = 1 1 + o1 1 + o2 1 + o3 3 3 3 6.4 If no arbitrage is present, then ( p1 , p2 , p3 ) = (1/2, 1/3, 1/6) has to be the proba-
bility vector which makes all bets fair. Therefore o12 ( p1 + p2 ) o23 ( p2 + p3 ) o13 ( p1 + p3 )
− p = 0 ⇒ − p = 0 ⇒ − p = 0 ⇒ 3 1 2
o12 = 1/5 o23 = 1 o13 = 1/2
6.5 If the outcome is j , then the betting scheme x i , i = 1, . . . , m, gives me m
x ox − j j
i
=
m
x + x = (1 + o )x − x ox − (1 + o )− (1 + o )(1 + o )− − =1 (1 + o )− 1− j j
j
j
j
j
j
i=1
i=j
=
1 j m i=1
j
m i=1 i
i
i=1 1
1
6.6 From Example 6.1b, p = (1 + 2r)/3. The payoff of the put option is 0 if the stock
price goes up, and 100 if the stock price goes down. To rule out arbitrage, the expected return of buying one put option under the probability distribution has to be zero. That is, p0 + (1 p)100 P = 0 1+r Therefore 1 + 2r 100 200(1 r) = P = 1 3 1+r 3(1 + r)
−
−
−
−
18 The put-call parity says
− C = 1 K +r
S + P
In this example, one can check the following indeed holds. 6.7
Since the call option expires in period 2 and the strike price K = 150, it is clear that C uu = 250
C ud = 0
C dd = 0
Let p denote the risk neutral probability that the price of the security goes up, then p =
1+r d 1 = 2 u d
−
−
− 0.5 = 1 − 0.5 3
where we assume r = 0. Then we can find C by computing the expected return of the call option in the risk neutral world. C u = pC uu + (1 C = p 2 C uu + 2 p(1
ud + (1
− p)C
2
dd =
− p) C
1 32
×
C d = pC ud 250 250 = 9
1 3 dd = 0 ud =
− p)C + (1 − p)C
× 250 = 250 3
6.8 See Example 8.1a for details. 6.9 We need to find a betting strategy which gives a weak arbitrage if (a) C = 0 and
(b) C = 50/3. (a) C = 0. In this case it is clear that buying one share of stock is a weak arbitrage. At time 0, one does not have to pay out anything. At time 1, the profit is 50 if the stock goes up to 200, and 0 if the stock price is either 100 or 50. (b) C = 50/3. In this case it is not that clear how a weak arbitrage can be established. But since the price of the call option is high, it’s intuitive that we want to sell it. So, let’s consider a portfolio consisting of selling one share of the call option and buying x share(s) of the stock. Our return depends on the price of stock at time 1, which is tabulated as follows. stock price at time 1 200 100 50
balance value of the portfolio at time 0 at time 1 50/3 100x 50 + 200x 50/3 100x 0 + 100x 50/3 100x 0 + 50x
− − −
−
profit (r = 0) 100x 100/3 50/3 50x + 50/3
−
−
From the above table, if we choose x = 1/3, then the profit is 50/3 if the stock price is 100 at time 1, and 0 otherwise, which is a weak arbitrage.
19
√
√
7.1 (a) .33/ 225,
(b).33/ 12
7.2 Since the unit of time is one year, t = 4/12 = 1/3. The probability that the call
option will be exercised at t = 1/3 is the probability that the stock price at t = 1/3 is greater than the strike price K = 42, which is
S (1/3) 42 P > 40 S (0) 42 S (1/3) P log > log
P (S (1/3) > 42) = =
S (0) = P (X > log 1.05)
40
where X is a normal random variable with mean µt = .12/3 = .04 and standard deviation σ t = .24/ 3. Therefore, the above probability is equal to
√
√
log1.05 − .04 X − .04 log1.05 − .04 √ > √ = 1 − Φ √ P .24/ 3 1 Φ(.0634)
≈ −
.24/ 3 1 .5253 = .4747
.24/ 3
≈ −
7.3 The parameters are
t = 1/3 so we have that
r = .08
K = 42
σ = .24
.08/3 + .242 /6 log(42/40) ω = .24/ 3
− √
S = 40
≈ −.0904
Therefore, C
.08/3
≈ 40Φ(−.0904) − 42e− Φ(−.2289) ≈ 40 × .4639 − 42e− × .4094 ≈ 1.8137 .08/3
7.4 From the put-call parity, we can derive the no-arbitrage cost to a put option
P = C S (0) + Ke−rt = S (0)Φ(ω) Ke−rt Φ(ω σ t) S (0) + K −rt = S ( 0) (Φ(ω) 1) + Ke−rt (1 Φ(ω σ t))
−
− −
−
√ − √ − −
where ω is defined in equation (7.7) in text (page 87). The parameters are K = 100
S (0) = 105 ω
ω
≈√ .571767 − ≈ σ t
.359635
r = .1
σ = .30
Φ(.571767) Φ(.359635)
≈ .7163 ≈ .6404
t = 1/2
20 Therefore, P = 105(.7163
.05
− 1) + 100e−
(1
− .6404) ≈ 4.418
7.5 A call option with a strike price equal to zero is equivalent to a stock, since the
payoff is max 0, S (t) K = max 0, S (t) 0 = S (t). Therefore, the price of such a call option is equal to the price of the stock S (0). The same conclusion can be easily verified by the B-S formula by plugging in K = 0.
{
− }
{
−}
7.6 From the Black-Scholes formula, ω
C = S (0)Φ(ω)
− Ke −
rt
→ ∞ as t → ∞. Therefore, as t → ∞, √ Φ(ω − σ t) → S (0) × 1 − K × 0 = S (0)
7.7 The payoff is F if S (t) > K and 0 otherwise. So, the risk-neutral valuation (or
the unique no-arbitrage cost) of such a call is equal to C a = e −rt F
W
× P {S (0)e
> K
}
where W is normal with mean (r σ2 /2)t and variance σ 2 t (see page 88 in text). Therefore
−
C a = e−rt F = e−rt = e−rt
× P {W > log(K/S (0))} W − (r − σ /2)t log(K/S (0)) − (r − σ /2)t √ √ F × P > σ t σ t (r − σ /2)t − log(K/S (0)) √ F × P Z < σ t √ Φ(ω − 2
2
2
= e−rt F
σ t)
where Z stands for a standard normal random variable and (r + σ2 /2)t log(K/S (0)) ω = σ t
−√
as defined in equation (7.7) in text (page 87). The parameters F = 100
K = 40
So
√ − σ t ≈ −.207242 × 100 × .4179 = 40.55.
ω
≈ e−
Therefore, C a
.03
S (0) = 38
σ = .32 Φ( .207242)
−
r = .06
≈ .4179
t = 1/2
21
8.1 If there is no arbitrage, then there exists p = ( p50 , p175, p200) such that both buying
stocks and buying call options are fair. This means we are able to solve ( p50, p175, p200) from the following linear equations.
−C + 25 p + 50 p = 0 −50 p + 75 p + 100 p = 0 175
50
(0.1) (0.2) (0.3) (0.4)
200
175
200
p50 + p175 + p200 = 1 0 p50, p175, p200 1
≤
≤
By letting p 50 = x, we can solve p 175 and p 200 in terms of x from (2) and (3), which gives p175 = 4
− 6x
p200 = 5x
−3
Together with the constraint in (4), the solution from (2)–(4) can be written as ( p50, p175, p200) = (x, 4
− 6x, 5x − 3) 3/5 ≤ x ≤ 2/3 Therefore, if C = 25 p + 50 p = 100x − 50 where 3/5 ≤ x ≤ 2/3, then we are able to solve the linear equations (1)–(4). In other words, if 10 ≤ C ≤ 50/3, there is no arbitrage 175
200
opportunity. 8.2
(a) u (x) = 1/x
u(x) = log x
u (x) =
2
−1/x
Therefore a(x) = 1/x. (b) u(x) = 1
x
− e−
u (x) = e −x
u (x) =
−e−
x
Therefore a(x) = 1. 8.3 Using the notation defined in Example 8.2a, let f (α) denote the expected utility
of the final fortune, then f (α) = log x + p log(1 + α) + (1 1 p p f (α) = 1+α 1 α
− p)log(1 − α)
− −− Since p < 1/2, f (α)
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