Bowers Solutions

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Chapter

4.73

Erercise 4.6 This is the case of De Moivre's Law with

ar =

100

trx

100. In this case,

-.r - t

100-x

'

-&'L tp,

P,(t) =

100-x-t'

frT) = ,p,.lt*(t) =tob= Frrthermore, when De Moiwe's Law holds, we have

.l-4Ja-x =*( 0

"-5t

J-4t

da-xl

o)-x'

n.-1

At,A

d-nl @-x

= l"-"' ,i;dt 0

In particular,

(a)

O)

Z)o,za

=

W = !ffi

= {tt

-

e-t,s)

x

0.237832.

This is a more special case, because the amount of benefit is not Hence, the actuarial present value of benefit is:

1 but

rather bt =

eo'05t

.

25

l"o.os,

e-o.os,

0

id, = #

=

*

= 0.416667.

Exercise 4.7 De Moivre's Law is again assumed but this time with interest rate (not force of interest) of 10%, so that the force of interest is lnl.l, again with ar = 100. We have: (a)

4o,r,n=

ffh

=o'oe2oee'

(b) The second moment

is. ,.

271

- -

,bo,iol

d*lo='^

ldr-_#

=

0.063803.

Therefore,thevarianceis:27j6.,nr-(40,r)t = 0.063803 -0.0g20gg2

Copyright

@

ACTEX Publications 2007

s

0.055321.

Acluarial Mathematics: Solutions Manual for Exercises

o) (z*) =

dt INTEGRATION BY PARTS

, (_Ltfi+o.osr)-3)lt=r00-'r, , toor* 4a, = roo-i(-T'\rrv'wJ'i, +loo; )1,=,

=-f,oo-r), roo#(1

+ 0.0s(100

.

-'))-" .

=

-tG -0.05x)-3 .

=

-+(#)' . 46(ro -

dET

J

1 ( zP

rd;l-Ai'tt

11t=roo-x

*

o.or')-',,J1,=,

-(r * o.os(roo - 4)')

mfu(r

rolo - o s')-' )

=ffi['-(*)')-+(#=)' Exercise 4.9

(a) We have:

Mr(i

= E(e'r)

=

. [ "" ,p,. p*(t)dt.

0

Therefore,fors = -d'

Mr(s)|"=-u = I

n* ' ,P"'Pr(t)dt

= A\.

0

(b) For the gamma distribution,

.frQ) =

fr;wo"-r

"-ft

,t >0.

Therefore,

MrG)

=(*)"

Using part (a) of this problem, we obtain

A* = Copyright @ ACTEX publications 2007

Mr('=(#)"

=

[,

.fr)"

Actuarial Mathematics: Solutions Manual for Exercises

Chapter 4

|

79

Thus we have:

E(z\

=

r.1*'(r -,i)* el ; z=l

:.('-r-[+,u' )" '-*['-(:)r) *. =

e

5 -t'2 o 0.41766. -e G

Exercise 4.14 This is De Moiwe's Law with o = 100. Note that we have the following discrete versions of De Moiwe's identities proven in Exercise 4.6

A'

al-x-l

=Z ,t=0

at .rt*l ua4= a-x' ?o ,-,

a-x-l

1

rlq,''o*t = t

n-l

Ata = 2rlq,'nr*t

n-l

1.yr+l7=o@- x i

=l

&=0

a;l

(t)- x

Therefore,

(a) A+oid

=

ALxt* A*,)4

=

%

60

-,,25.35 '' 60

x

0.407159.

(b) This is a bit more worlg as it requires referring to the definition again, but we can again note the general formula fust:

rul-x-l k=0

" . rlq* =

,f'{o*r;ut+r a'-x-t ,t=0

.-l-- = o-x

o)-x "i'=

In this case,

(M)qo =

Copyright @ ACTEX Publications 2007

(Ia)an 60

t ii=^- - 60v60 -^- . out = 60 = 5.554541. 0.05

Actuarial Mathematics: Solutions Manual for Exercises

80

r

Chapter 4

Exercise 4.15 We have

.

A,'4 =. frot

plq, +un . np,

,t=0

m-l

=

n-l

Ir**t

&=0

.rlq,+ Ir**t.,,\q,+rn.,p, k=m

,q!.-l+v^

.

.p*.nf'ur*t

'

rlq**^*v^ ' ^pr'v'-^ ' n-^P*+^

t=0

r^'

=,ql.A *

4-a * v^'

^P*'(lrl*^,r=a ^

P

r'

+

vn-^',-.P**-)

Ar+^,;=al.

In words: Single benefit premium for an n year endowment can be viewed as single benefit premium for z years of term life insurance protection, plus single benefit premium for an endowment for z - myears deferred by rz years' Exercise 4.16 Here we have: 19@

Ar= Ir**t . olq,* Lrr*t . olq, k=20 &=0

@

= At.,a*'20' zoPr'

I'**t'

rlq"*ro

=

/r=0

et

af

'iol'A,*ro'

and

4Zrl =

Ar,g1+ Ar,fi.

Therefore,

(a) A, = At,n* A,,)rj'Ax+2' = (A,a- A,h)*

A,,*oi.Ax+20

= = A,Id-(l- A,+2iA,,hl.

Substituting the values given we have:

0.25 so

that

=

0.55

-0.601,.h,

A,,h=i.

a a a a a a a a a

a a a a

a a a a

O) et-n = A*rol-A'ht = 0'55-0'50 = 0'05'

Copyright O ACTEX Publications 2007

a a a a a a a a a a J a a a a a a J a a a a a a

Actuarial Mathematics: Solutions Manual for Exercises

a a I

(b) Eo = 100.1000.40 = 10,248.35. Er = 1.06. to24l.3s-1s9,6ss'ljl?682 ' 9s013.79 = Ez = 1.06'

lO,7 10.36

-f

ffi

=

LI,192.34,

**{ 9s0t3.79

=

LI,694.76,

OO,OOO'

Et = 1.06'11,192.34-100,000

10,710.36,

Ec = 1.06'11,694.76 -199,969' = 12,218.04, ' 1695W 95013.79 Es = 1.06'

12,218'04-100,000';ffi = 12,762.58.

Exercise 4.19

(a) For

a death occurring

inthe mth of a year following age x + k *

the end of that m'h andits present value

(b) Consideradeathattimeswithin

ar,

*,the

benefit is payable at

is ur*# .

s.[0,]].

*'h of ayear,with

T'n"rt

I

tPl

lr,

.,0**r*1. F,*r*t(s)ds =

is the actuarial present value of benefit at age x + k +

f,*

'px+k.(u,.*n,*o**)

=

I

r'

.

*Q"***!

fi , and'

t'#

.*l*q,.r

is the actuarial present value at the beginning ofthat year. Therefore,

Ay)

=

= ir**' on,'(f,o*il*# '*l**.') t,r op,(y,r*.rl*n,*l at n ) *=o [r=o

,r=o

\.r=o

JJ

(c) Assuming Uniform Distribution of Deaths (UDD)'

,l tQ,** = mL'q,**, il^ *t

AY) = Zur kP, ,t=0

[iu.r.*

6

=

Iro "

&=0

olq*

+

[F,,'

*l

;)'I-#

Copyright @ ACTEX Publications 2007

J

*n*r)

*) =t,*.'

oln.,f) =

7 J

h}r*'*lQ,

=

j;4'

Actuarial Mathematics: Solutions Manual for Exercises

J J J J J J J

a C

Exercise 4.27 Note that for failure at time t the amount of benefit is b, = (t

-*)".

The single benefit premium

for this warranty can be expressed as:

Aa-lrn)la ="[Au

+)

(a) Under UDD

4a

-|rzr;u

=

*4'u-t;(tzri'-(+- *)4')

=

;('.i(;-+))4,

-i;

tqroa.

We know that i =109/o andthat

,

tl4o =

10.2, k =0,4, {o.L k=1,2,3.

Therefore,

4A = 0.2v+0.Iv2

+0.1v3+0.lva+ 0.2vs

o 0.53207989,

(u)toa = 0.2v+0.2v2 +0.3v3 + 0.4va +vs x 1.4666285, and

;('. +(} -*))4' -I*,,n iu 0,t9 *!( - rn1.10[ [, ' - -a]]0., lrl,r.to 5[t-* ))

3207s8s-

I

0'10 l. 4666z's = 0.3072t5. s ln1.l0

(b) If the warranted return is the reduction in the price of a new product, the answer to part (a) would not change, as the customer could take a cash refrmd and apply it toward the purchase ofa new product in any case.

Copyright @ ACTEX Publications 2007

Actuarial Mathematics: Solutions Manual for Exercises

_I

I

Erercise 4.28

Irt

O

be the cumulative distribution function of the standard normal distribution. Then:

Eek)

=l"u.onu.ft;* o,

-f* 0

=

."-+(h.+n\ o,

!s.'l2x

o

rz _1(rt+zit+.(*f))dt

= l-L'e'e' l-

i542n

t =5 "*'g!."-i?a*t)'

41

l",l2r

t=o + ,=! = d" = z"*(t 2 "-i" +"+"!h' ll\dz=dt, t+oo=) z-+6 ,

lr=L*k. l-102'

@)

-t(r)

v, = 2e1t8(t-t(r) x o.6ee2.

b) 'A- = 2e4t8[t-t(;)

= 2et/2(r-oOl) x 0.5232.

@)'A*-(4)' = 2ett2(r-o1r;)

-(r"'''[t-.(;))' '

o.oror.

(d) Let q2s b"themedian of Z.T\ensrnceZand7areinverselyrelated, it 6l't isthemedianof 2i we have: 0.5

= Pdz.€2'5) = Pr(z' €|'t)

r{ 'I

= f,@o' =

+@ .

t2

J#"-hd,

I

--_t-, t=d.t =l.riit=,,, ,-* = "=!9'tl ,-El

=

INTEGRATION BY SUBSTITUTION

=2f#,=,[,'[#)) l0

This implies that

= '-r[4]) [10J Copynght @ ACTEX Publications 2007

I

0.2s,

or .[#)

=

0 75.

Achtarial Mathenatics: Solutions Manual for Exercises

92 a C\aoter 4

The 756 percentile of the standard normal distribution is approximately 0.675. Thus 20.5

+ But

()s =

e{.os'f's

-

= 0.675,

e4'os6;s

x

or

€l5 = 6.1s.

0.713.

(e) We have

6

Z. = [r.fr?) dt

o

=*l 2t "-t dtoJzr

o,

f u=m, /=0+ U=0 zodu=ff, 't,o+ u-> INTEGRATION BY SI'BSTITUTION

=w"-udtt u2,

-

=

"--o.os.i.s7ss

(-#*)r x

0.6710

=

fto-o)

< 0.6992 =

=

#"

7-s788.

4.

Exercise 4.29

Let u(w) =

-"-6*.

By Jensen's inequality,

E(-"-6,) = -E(e-6r) = -7rn.

Actuarial Mathematics: Solutions Manual for Exercises

106 o Chapter 5

(b) We have

(Diiln+(rg(f; = @+\ffi, so ttrat, using the result from the previous problem,

(DiDn = @+L)ii(1-(r,iln (n+t)ii(1-_i__ i;(1 +

lr:*

Zt,# -

rlr$n) =an .Eo:%

n

=

Ia(4. L't

k=l

a:.nl

Exercise 5.25

(a) We have

II rn>9^,

Y=4 ':

"-;l

0 p(x). Exercise 6.6

If the force is constant, then

Z, =\n*+6)t . pdt = 0

#g,

and

')-

U .,, - p+26.

Copynght O ACTEX Publications 2007

Actuarial Mathematics: Solutions Manual for Exercises

Chapter

Tberefore

'.q,-Z:' _'Z-_ei

(6a*)'

6. lll

equals

(l-A,)'

pp2

Vop-

(p3

+2p26+p621-2p26 62 (p+26)

_p

p+2

_

'L.

Erercise 6.7

Ifd=0,thenyt=land @

I t p,p,(t) dt

Ff+l=s-

I tP, dt

=*. €a

0

Erercise 6.8 We would like to show that

var(vu) < Var(vr -F.dT), where F

=FU),

or, equivalently, that

var(v?) But this follows directly from the fact that

. (t.5)'

.var(vr).

(t. #) > 1, as both F and d are posifive.

Exercise 6.9 We have d-a* dn

"= (u@)+6)a,-t

and

dZ, dx

-,(p(*)*6\4-pG).

(r +

(p(x)

Based on this

(,. *)uG,)

-

*

=

+a)a - r)F1 a) - ((pr.1 + d) a - p@)

= (t + (p(x) + a)a, -r)-f; -@@) + d)7" + p(x) = (p(x)+ a)a -(p(xy+ a)A + pG) = P(x).

Copyright @ ACTEX Publications 2007

Actuarial Mathematics: Solutions Manual for Exercises

Exercise 6.10

calculations in the first row are based on the following identities: z:s"iol

= ,,io *n:r^*toErs l('hs-rc'Ers)+

rcEts'

i(,4ts- rc Ets' /as) + 5'

Asid = (4s-rcEx'&s)+

__ 5s:iol : D

rcEzs,

#(4s-rcE$'&s)+ d' =mt;, =ffi l-(4s-rcEts'4s

P(zrr,rr)

rcEE

rcEgs

+ roErs) '

: ilr*roi E*n - T:4m Ats.n-=i

ns'&s) + vro' to pts) . 4Ar_vto.topts.,4cr)+rto.rofrs)' d'((,hs-v\o'

ro

Calculations in the other rows follows analogous formulas.

Exercise 6.11 We have

,o4u-P.1a

=HH=W=#=

zoP(zolrc4).

Exercise 6.12 We calculate

4

@@6

= iv**t . *lq, = iro*t1t- r)rk = v(l-r)l(vr)& = v(l-r) k=0

t=0

&=0

#

:- l-r l+i-r'

'd.=+=('-ff|) +=#+=#, #=

P* = ax Furthermore

,'A"

\-.r,. l+,

is calculated the same way as A, ,butthat it is based on doubled force

interest, resulting in

Based on these values

Copyright @ ACTEX Publications 2007

of

24=& 24-L2 _ (l-r)r (1-A,)2 l+2i+i2 -r Actuarial Mathematics: Solutions Manual for Exercises

l16 o CtrapterT

Exercise 7.5

In example 6.I.1 elq, : .2 for k : 0, 1,2,3,4, futrs, f is fte uniform discrete distribution on 0,1, ..., 4. Assuming the UDD, Iis rmiformly dishibuted on [0, 5]. Thus this problem is a repeat of example 6.1.1 to the situation where the variables are now continuous instead of discrete. By the IDD a here is 6.1.1, etc.. Exponential reseryes are not worth ino ^exaryle the effort of calculation.

k:

Exercise 7.6

& : vu - PZ,,;)a6ifor0 ( U < n - t,andvn-t - F.a;4,U ) n Nowrz : vu -P.uq: vu-P(#) : ,u[r + #] - f

:

rhen

var(,e

since

var(vu)

from (4.2.10),

Note:

. f]'uuruur: (4)'.

(r7,*,,;4

-

var(vu).

7,*,nn1r),

2Z "+t"n-d then Var(rz) - "x+tin-i --/ ' ar,;)z -2 16

Z- tlr>t: U, the future lifetime of (x*l).

Exercise 7.7

f'-t _aA uPx+t ltx+t(u) du * d;4 n-tp*+t J, vu - uPxtt

El'Ll :

Var(1L) Note:

:

[t

upx*rurrl,'o-'

* Io' ' vu upx+tdu *

u) : $ v"'t, .fi|r7,*,,v

:

d;4n-tpx*t

:

dx+t:Frr

- V,*,,42f

?- tlr>,: U, the future lifetime of (x*/).

Exercise 7.8 1a) [email protected])

:

V+s,x:1

-

2sP(A35,,s1) z+s,i6i

(b) There are no future premiums, so 5 Z+1,O

:

Vlost

Exercise 7.9

: -lnCP(A,Xd + P(,aS1ti from Equati on 7.2.5. o) 6 : ln(1.06),P(Vd : .020266 =+ uo : T\35) - 20 :23.25 years. (a)

uo

t.

Chapter 7

. ll7

Exercise 7.10

(35J0 :65 t.-Setting the minimum loss to 100 The minimum loss occurs when zero is the same as setting 65 t equal to -ln(P(As)/(6 + P(135))/6 23.25 years. Thus

U:

t:

47.75 years.

-

-

-

:

Erercise 7.11 Analogous to the development of (7.2.9).

Exercise 7.12 Same comment as problem 11. Nothing is done in the text vtdth these densities except to €xhibit another formula.

Erercise 7.14 fl-v(.1'aq1:

(a) Prospective: Zso - 2sF(Vai att,Tdi (b) Retrospective: zoP@+i.s+o,i6i - ,oEoo (c) From the prospective formula, we have

It_^rce,,r$]z*:lr-fu*urn^ln* (d) Alternatively, from the prospective,

l*

- roP(tdfdso,6i :

I roF(Zso)

-

roF('n^1] aro3q

Exercise 7.15

,o-tr(A*.6):

-

(a)

Prospective: Vso,Tot

O)

Rehospective: [email protected]+0,I0

(c)

Anarogous to

(d)

Analogous to

4(c): I

t

J

-

P(A+o.ro)]aso,O

: | - 6 Z56.iq,

- p@ao,ro1) + a]aro.t

From (e), since F(7*.61)

,\ (g)

.From(rr,

air,t-z1y-

*6

dn;tg

---v4ofrl

sinceZ

roEoo

'

4(d): F(7ro,ioj)

(0

,^

-

t -'+#4lr,,t t'"so:lol/

L

(e) From (a), since Vro5ol we have

P1Ao*5paro,6i

| -V : ---T-'

:

wehave

#^,

Ato,Tol

Aoo,^l

- - l=z;{,

-

t - 3t'g u4otzol

Exercise 7.16 Retrospectively, there have been no benefits, so the rreserve is just the accumulation of past soF($l zrsFssfrl.

premiums: tz|V(tola34

:

Exercise 7.17 Begrn with the retrospective resewe formula:

Multiply AyP,,*t : ! P,,;t' and

-/-

\

^r(,,^*;1) : Vlr,A

ffi.

:

^T @,.-*,t)

: FF,*-)

s,.6

-

^8,.

n"rproduces \

-/P\A,,^*,1)

-

--r P(ai,a),since,E,'s".4

:

d,d

^E*, ^8,

This establishes (a). It is interpreted as seeing the premium

PF,,^ ) in two pieces: one

which provides the coverage for those m years,n@ia),and the other which provides for the reserve after m years, if alive. Thus the reserve is in the nature of a pure endowment benefit, so the premium for it is a P.E. premium. Now multiply equation (a) by Fr"1, and subtract ,E* fromboth sides. This yields

totally parallel to that for (a).

Exercise 7.18 The given equation relates to formula (7.3.3). This equation states that the reserve at the beginning of the interval (at time 10, interval length 5) is the a.p.v. of benefits payable during the interval plus the a.p.v. of a P.E. for the amount of the reserve at the end of the interval, less the a.p.v. of net premiums to be received during the interval. If we rea:range the equation to read

--jt

6r-i t= ilV@ro) +' z"nP(7n)d1,6.r: ZloA + 2OTtr-7 t

sE+o.V(An),

we show that the a.p.v. of all resources available to the insurer at the beginning of the interval is equal to a.p.v. of the uses of those resources.

Exercise 7.19 This is totally analogous to Question 14. Exercise 7.20 This is totally analogous to Question 15. Exercise 7.21 This is totally analogous to Question 17.

Irrlrei,*7.22

.

Since

pZ,;

Since

drA

Ttu"

m

*

: , - d.ffi- : *,** W

5

6'

tt_

, vzl?-kn-2H : 2.ii**p,4,A* +ffir_O nxlkn-lcl ti: )'-g : t' FinallY' kY*asT: . wr*?-kn-2A qx*kn-kl

iira2la@

l--4

4 -r - r-5

1

5'

kdn723 FtI$r Continuous:

: |#^:

(a) ro-K4ss-i)

dA:

dt

-f

'l752905usingz'';

nPxdx+n and the values

fud4,ontinuous: (b) rcY6s): i(rovzs;

: a(x)ii,q-

of d" in the table

: itr - ffi): .08566

FuIIy-Disoete:

(c)

rc%lssT

pt

:

Arts.^1l-

_-;3s'301 - 3s301 atssTi

Eisf4:

dos

-

0(m),

Pt

r.o.1aor.m1:

Ats ii3s

-

-

.03273

usingl|r.,

:

A4s

-

f0

zopesAos

flgon$Aes: .oo4gl5

v"u soPssdes

- fo zoPtsiies: ll'575

hdrc724

:

: i n'.a + A,,k # *n.a : i.e*r - i ' Px' iix+k {tCL) : : I |,l.*r - P,. d*+kl : t. rr, (c) Yes. {(1,;1) - V.i*a=A ' e(/.;1).dnp.4 : t' A|**a - t P'",a' d,+*i4 (a) (b)

7',a * A,,h 7,+t - P(A,).a**e

No. Recall thatV*6 Yes.

: tlni.-^-

pta.ii,*r.frf

:

t.ov);t

:

.08846

bdc&7 Xcptacinghbyh+1,(8.3.9)becomes nV * nn Then6'1Y'v'Pr+n : (nV * ri - bn+r 'v'1x+n

*+tv:

@#P

-bn+t

Thc interpretation is most easily seen GV

+

nn\Q

i) :

+

:

bn+r'v'Qx+h

*

n+rV'v'Px+h.

ffi

if we write it

bn+r ' Qx+t

*

as

r+t Y'P*+t

Now the old reserve plus the premium, with interest to the end of the year, is sufficient to povide b6.1 if the policy dies (with probability Qx+n), or to provide the new reserve if the policy lives (with probability P,+n).

kercise

8.8

(a)

k-l

Dr'*' ,lq*

fi'w:

At :i,'APr-#:kv*

-+re,

(retrospective form). The reserve is just the accumulated value of all of the premium income, less the accumulated value of all death benefits paid out, taking account of the benefit of survivorship in the accumulations.

(b)

Since (r,Z' +

t :

PrXl +

P, - r.e,+h(I-n+rVr) k-l

Dt"' h4

n+rv,

(l - n+tV') * l+r l/*,then v.n+rV, - tVr. Thusthegivensummationbecomes

e*+r,

:

- nv,l(o*l*-n.

a telescoping series which is easily seen to reduce to kl/x. Interpretationz rY, is the accumulated value of past premiums without benefit of survivorship, less the accumulated value of past benefits without benefit of survivorship, such benefits being only the excess of the

This is

insurance amount over the reserve.

Exercise 8.9 From (8.3.14),

rn-r :

IJercrl: fr,andfuwehave

Tr

:

@n

1rV,sO

v. pV - 1,-1V,or 1'-rY *zr)(1

tV : zr(l+t) 2V : [zr(1 +i) + Tlrl:.s

pV

-

- nV) v Qx+h-r *

n.

6

n

n](1

+;) :

v ' 1V

* i) :

r'i1,etc.

-

n-tY.

1V. Then with

sV:0,

we find

Exercise 8.10

: Du^i h=t

(a) v'ii*,4: PYB

n-rlq,

: hDO-r-\l h=l

It(,/ ,-,1a"- r',-te,) if, - dii*d #-ii"fr: Thus

(b)

uo

n:

;:'a

ux:nl

-f

nE*

: hViu - ,"

rqrf

+ tr ,Pr)

dit-d*;t

.

PW - d;:A - d"apffi. a-**.4) - n'ii"*p.4.

From part (a) we see that, attime ft, the

n .d,*1,,;4.Thus

n-ilq*

1z : (o^ -

Clearly the PYP is

Exercise 8.11 (8.4.3) sals pasZ

:

bk+t

rr-s

t-sQx+k+"

*

r+t

V'vr-t

l-,rPr+/c+s'

Multiplying bY ,P*+k,we obtain spx+k.

k+sv :

:

sPx+k'*+sV

*

vr-t

b*+t vr-t r[-"qr+,t bk+t vr-t (qr+*

,qt+kbk+t :

*

t+r

V'vl-t pr+t

- sQx+k) *

t+r

V'r'-'

p**r

vr-t(b*t' Qx+* * *tY'P*+t) ,t-sQ,V + zrrXl + t)

: : (t+i)GVqri

Interpretation: The old reserve plus premium, \nith interest to time s, will provide the reserve at time s if (x * &) has suruived to that time, or provide for the then present value of death benefrt(bwr to be paid a?year-end) if (x * &) has died.

Exercise 8.12

Interpretation for both (a) and (b): The reserve is sought at a duration between two consecutive premium-payment points. This reserve is approximated by interpolating linearly between the two adjacent policy year terminal reserves, and adding the unearned premium for the current premium period. The interpolation coefficients on the two ierminal reserves are easily obtained. Since r is the fraction of the year beyond the last premium payment point, then (j - r) is the fraction of the year remaining to the next premium payment point, so that is the appropriate fraction of annual premium unearned. Note that this fraction multiplies the annual premium, not the fractional premium actually paid at each premium payment point.

ffit2t (e) ElT

-t375(l) +

(b) Yt(4 :

375(3)l (362.12)

6,450,962

as

+

t2s0(1)

+

2s0(3)l(561.08):t,104,260

before

c:l.&5l@+l,lo4,260:l,108,483,whichis1.00378timesthereserve. (c)

vu(z) = [37s(l) + 37s(9)X1187.14) + [250(l) + 2s0(9)](343.84) : ct :

(d)

:

\ffi

1.645

s,3rt,37s

as

before

37gl.l4,which is .00343 times the reverse'

44 : 110,426,000 Ya{Q : 645,096'250 c : tto,426,ooo+ 1.645\@250 : 110,467,780, which is 1.000378 times the aggregate reserve

Yar(Z) : c1 :

531,137,500 1.645

:

\@

37,gll,whichis.o0o34timesthereverse.

Erercise 8.29 We seek 10,000

rcrpY{tl(4') 10,000

ll.rcv{t}(Vto)

s,000 [ toV (Ail

*

+ }.rrv{r}(Ato) +

t.p{t}(7ro)]

n-vGzi + P{t}(7ro)].

Exercise 8.30

SinceP,a

: *, - 4

Since12".1

then d",Tt"

: t-W:

:

: ffi

f,**d,+r't:

T-iE

.78ii*.4:

(a) d*,Tt : l*uprd,*1a,soQx:1-(1'?'*q83) : (b)

dx+ril1

:

L

(c) It'r, : f

*

(b^*t

vPx+r,

-

so4r+t

r,+rY)

p"+h'

:

1

-

(1'2X'625)

(d)

2.083.

t.625 .2

:'25

Qx+h' hPx

As : (#)' G-.66)2(.2X.8) : .6084 A1 : (i=)' (3 - l.s6)2 (.8x.7s)(.2s) : * I Ar :

:

.216

+ .69M4 (.216) :

Var (62)

:

Var(12)

: X : # : is no risk in the final year of an annual premium endowment.)

Ao

(Note that there

.6084 -27

.7584

186

.

Chapter 15

Exercise 15.5 Since a select-and-ultimate table is used the life insured is [a0]. Let G be the expense-loaded premium. Then this premium pays for the following items Qisted in terms of their actuarial present value at issue):

Commissions Premium tax

of

0.40G + 0.0sG\",l,sl

0.35G + 0.O5Gajao1iol.

of 0.02G\+o1lil.

Maintenance expenses Death benefit

=

of 12.50+

4o4oj = 8.50 + 4\+01,A.

of 1000{+01F.

Therefore,

GAt*lul=0.35G+0.05G41+ol,iol +0.02Giip1.;.1+8.50+4\ao1a+1000{+ol,zs,t,

or c(o'laa1+01.- - 0.05d1+01,a - 0.35) = 8.50 + 4a1*1,Tit+ 1000{+o;,rsl, and hence

G-

8.50+

a\*l4+1000{+o[H

0.98i1+o16

1

-0.05&1+olrol-0.35

ooo{+o1B + 4d1+o16i+ 8'50

0'93d1+ol,E +0.05'roE1+01'A[+o]+ro:i3

-0.35

Exercise 15.6 This policy has a single premium

fI

that pays for all benefits and expanses, so that premium is calculated as the actuarial present value at issue of all benefits and expenses. The premium pays for the following items (listed in terms of their actuarial present value at issue):

-

Taxes

of 0.025II.

Commissions

of 0.04n"

Other expenses

of

5

+ 2.50a*--t

=

2.50 +2.50ii".;1.

Benefits of 10001,;1.

Therefore,

fI

=

0.025fI + 0.04fI + 2.50 + 2.5Oii,.A+ I 0001- .r,

so that

0.935n =

2.50 +

2.50ii,s+

10001-

.r,

and

2.50+2.50A xint 1+10007x.:nt0.93s

Copyright @ ACTEX Publications 2007

Actuarial Mathematics: Solutions Manual for Exercises

Chapter 15 o 187

Exercise 15.7

ThelevelannualconfractpremiumG=aPr+cpaysforthefollowing(listedintermsoftheir actuarial present value at issue):

-

An initial expense of eo. Annual expenses of (e, + ezPr)iir.

of q.A,.

The cost of claim settlement

Benefit of

A*

Therefore,

(aPr+c)ii" = Recallthat ^(, '

=#,h"n""(recalltlrat 4x'

("+ * ")r, aA* +

cii* =

€o

€0

1

=

* (q+qPr)ii,

+ er. A* + Ar.

= .4r+diir)

"o

*(",. rt)r"

+

q. A, + A,,

* erii, + erA, + qA, + A, = Ar(l+eo+er+er)

+ iir(er+deo),

and we conclude that

a = l+eo+e2+%, c = e1+dq. Exercise 15.8 We have, for

Z>

0,

t(r 61, a) We assume that

(a)

"

T(x) andB

= Bvr

+ a B a7 +

0

aa + p (B n + f)a7 - (B n + f)a7.

are independent.

under the conditional equivalence principle ,

n(t(r{u),8)"lB - b) 0. Therefore,

o = n(r(r6y,a)"ln=t)

t(atf + aBaa + laa+ p(ar + f)an -(ao + y)aola = t) = s(t ' + abaa+ ilaa+ p(tn + f)afl-(tr + f)dn) =

=

b-4 + abd* + 0d* + p(bn+

Copyright @ ACTEX Publications 2007

f)a. - (br+ f)a,.

Actuarial Mathematics: Solutions Manual for Exercises

Chapter 9 Exercise 9.1

Totaty anarogous to Exampre g.2.r. Aseries of tedious integration tricks:

(i)

(ii)

/-#

*= *irnlr

fi(s):

#isadensiryon[0,oo)

qirrdtds : r';ri6,=T-' irn > 2 I[isajointdensityfor0(s,t(oo

(iii) From(i),Et(r+s)'l=

,r-tll c#

+ fs,r6r)= @, ,t\(n,-=?) ' (l+s+Dr

vffiirm n) = npry : ,px . npy,by independence. (b) Pr[T(x) ) n and W) S ,,.?, \y) ) n and Kx) < n] : i l

(c)

np*(l-np) * ,py(l'-,pr)' :

Pr [at least one survives]

npx

: I - prfneither survivesl : I - pr{max[t\r),W)]
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