bowers solutions

Chapter 4.73

Erercise 4.6

This is the case of De Moivre's Law with ar = 100. In this case,

100 -.r - ttrx 100-x '

P,(t) = -&'Ltp, 100-x-t'frT) = ,p,.lt*(t) =tob=

Frrthermore, when De Moiwe's Law holds, we have

.l- =*( "-5t J-4t4Ja-x

0

n.-1At,A = l"-"' ,i;dt

0

d-a-xlo)-x'd-nl

@-x

In particular,

(a) Z)o,za = W = !ffi = {tt - e-t,s) x 0.237832.

4o,r,n= ffh =o'oe2oee'

The second moment is. 271 - - d*lo='^,. ,bo,iol - ldr-_# = 0.063803.

Therefore,thevarianceis:27j6.,nr-(40,r)t = 0.063803 -0.0g20gg2 s 0.055321.

O) This is a more special case, because the amount of benefit is not 1 but rather bt = eo'05t .Hence, the actuarial present value of benefit is:

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:

25

l"o.os, e-o.os, id, = # = * =0

(a)

(b)

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o) (z*) =

dtINTEGRATION BY PARTS

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

=-f,oo-r), . 1 ( zP 11t=roo-x

roo#(1 + 0.0s(100 -'))-" . rd;l-Ai'tt * o.or')-',,J1,=,

= -tG -0.05x)-3 . mfu(r -(r * o.os(roo - 4)')= -+(#)' . 46(ro - rolo - o s')-' )

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

(a) We have:

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

0

' ,P"'Pr(t)dt = A\.

Mr(i = E(e'r) =Therefore,fors = -d'

Mr(s)|"=-u = I n*0

(b) For the gamma distribution,

.frQ) = fr;wo"-r "-ft ,t >0.

Therefore,

MrG) =(*)"Using part (a) of this problem, we obtain

A* = Mr('=(#)" = [, .fr)"

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Chapter 4 | 79

Thus we have:

E(z\ =

z=l

e

-t'2 o 0.41766.

Exercise 4.14

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

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

:.('-r-[+,u'

'-*['-(:)r) = *.

)"

5-eG

al-x-lA' = Z rlq,''o*t

,t=0

n-lAta = 2rlq,'nr*t

&=0

a-x-l 1 a-= t t .rt*l =

ua4-?o ,-, a-x'

n-l i=l 1.yr+l- a;l

7=o@- x (t)- x

Therefore,

(a) A+oid = ALxt* A*,)4 = % -,,25.3560 '' 60

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

rul-x-l

k=0

(M)qo =

x 0.407159.

a'-x-t

,t=0 o-x o)-x" . rlq* = ,f'{o*r;ut+r .-l-- = "i'=

In this case,

(Ia)an60

t ii=^- - 60v60= -^- . out = 5.554541.60 0.05

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80 r Chapter 4

Exercise 4.15

We have

A,'4 =. frot . plq, +un . np,,t=0

m-l n-l= Ir**t .rlq,+ Ir**t.,,\q,+rn.,p,&=0 k=m

,q!.-l+v^ . .p*.nf'ur*t ' rlq**^*v^ ' ^pr'v'-^ ' n-^P*+^t=0

=,ql.A * r^' ^P*'(lrl*^,r=a + vn-^',-.P**-)

4-a * v^' ^ P r' Ar+^,;=al.

In words: Single benefit premium for an n year endowment can be viewed as single benefitpremium for z years of term life insurance protection, plus single benefit premium for anendowment for z - myears deferred by rz years'

Exercise 4.16

Here we have:

aaaaaaaaaaJaaaaaaJaaaaaaaaaaaaaaaaaaaaaaaaaI

Ar=

= At.,a*'20' zoPr' I'**t' rlq"*ro = et af/r=0

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 = 0.55 -0.601,.h,

so that A,,h=i.

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

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

@

'iol'A,*ro'

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(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 OO,OOO' ffiEt = 1.06'11,192.34-100,000 **{9s0t3.79

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

Es = 1.06' 12,218'04-100,000';ffi

Exercise 4.19(a) For a death occurring inthe mth of a year following age x + k * *,the benefit is payable at

the end of that m'h andits present value is ur*# .

(b) Consideradeathattimeswithin ar, *'h of ayear,with s.[0,]]. T'n"rtItPl I

lr, .,0**r*1. F,*r*t(s)ds = r' .

*Q"***!

is the actuarial present value of benefit at age x + k + fi , and'

f,* 'px+k.(u,.*n,*o**) = t'# .*l*q,.r

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

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

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

,l tQ,** = L'q,**,il^ m

*t kP,

" olq*

= 10,710.36,

= LI,192.34,

= LI,694.76,

= 12,218.04,

= 12,762.58.

'*l**.')

AY) = Zur,t=0

6

= Iro&=0

7JJJJJJJJJJJaC

[iu.r.* *l *n*r)

[F,,' + ;)'I-# *) =t,*.' oln.,f) = h}r*'*lQ, = j;4'


Aa-lrn)la ="[Au +)4a -|rzr;u

Exercise 4.27

Note that for failure at time t the amount of benefit is b, = (t -*)". The single benefit premiumfor this warranty can be expressed as:

(a) Under UDD

= *4'u-t;(tzri'-(+-= ;('.i(;-+))4, -i;

*)4')tqroa.

We know that i =109/o andthat

Therefore,

and

, 10.2, k =0,4,tl4o = {o.L k=1,2,3.

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,

;('. +(} -*))4' -I*,,n iu

- 0,t9 [, *!( ' - -a]]0., 3207s8s- I 0'10 l. 4666z's = 0.3072t5.rn1.10[ 5[t-* lrl,r.to )) 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 purchaseofa new product in any case.


_II

Erercise 4.28

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

Eek) =l"u.onu.ft;* o,0

= -f*

."-+(h.+n\ o,!s.'l2xo rz _1(rt+zit+.(*f)

= l-L'e'e' )dtl-

i542n

= t "*'g!."-i?a*t)' 415 l",l2r

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

= +"+"!h' "-i" d" = z"*(t -t(r),

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 isthemedianof2i we have:

0.5 = Pdz.€2'5)

This implies that

= Pr(z' €|'t)r{ +@ . t2 I --_t-, t=d.t = "=!9'tl= 'I

f,@o' = J#"-hd, =l.riit=,,, ,-* = ,-ElINTEGRATION BY SUBSTITUTION

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

= 0.2s, or .[#) = 0 75.'-r[4])[10J

lr=L*k. t=o + ,=!l-102' 2ll\dz=dt, t+oo=) z-+6

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I

92 a C\aoter 4

The 756 percentile of the standard normal distribution is approximately 0.675. Thus20.5

+ = 0.675, or €l5 = 6.1s.

But ()s = e{.os'f's - e4'os6;s x 0.713.

(e) We have

6 dt =*l 2t

"-t o,Z. = [r.fr?)o dtoJzrfu=m, /=0+ U=0

zodu=ff, 't,o+ u->

E(-"-6,) = -E(e-6r) = -7r<-e-6EQ) = -e-6E(r) =-e-52'

-o o -e-o"'=v"t 3Ar.Hence,

Exercise 4.30

(a) v, = *r[-lO*)

b) z=brvr=D,oo[-h-)

(c) E(zi) = ,lu,["-[-i--)J'] = "["*,[-ir--)] = E(z)@ro,cei6,

INTEGRATION BY SI'BSTITUTION

=w"-udtt = (-#*)r = fto-o) = # " 7-s788.

u2, - "--o.os.i.s7ss

x 0.6710 < 0.6992 = 4.

Exercise 4.29

Let u(w) = -"-6*. By Jensen's inequality,

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94 o Chapter 5

(b) We have

The coefficient of variation is

Exercise 5.3

We have

lr = 100.1000'12.76I,o2 = !00.10002.10.230,

o = ,[7 rv 10.1000.3.198.

,JPu.* x o.o25t.g*p

Exercise 5.4

We have

Note that

But

Therefore

var(a71) = #('r, -'2,)= #(r-262a,-g-da,)')= #(t -262a" -t + 26a, - d'al)

= ?(u. -' a- -)-a:.

cov(Aa71,vt) = Con(t -u',r').

var (6-vz ) + vr ) = Var(l-vr ) + Var(vu ) + 2Cov(l-v',r' ).

var({t-v?)+vr) = Var(l) = 6

= var(l) + var(vr) + Var(v?) + 2Cov(I-vr,vr )= zYar(vr)+ 2Cov(l-vr ,vr).

Cov(l-vr,vt) = -V-(rt).

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Cbapter 5 I 99

Exercise 5.11

We have

var(cq) = var(h-l) = Var(4;r) = u.[L#) = #trrt ".'1.

Exercise 5.12

(a) We haven-l n-l

= Iru*t k+lPx = lPr'lvk kPx+t = rEr'dr*r,il'&=0 fr=O

(b) hthis case

,la, = ii,-(d,,a+,8,) =+ +- nE, = #- nE*.

Recall that], isthepresentvalueofaunitlevelperpetuitydue.Basedonthat f it,tr"value of a perpetuity of I annually, starting at the end of n years, or year of death if earlier.

$ cao"etsthe payments at the end of the year of death, and on. This combination provides

1 at end of n years, if alive, which is cancelled by - nE,, leaing ,la*.

Exercise 5.13

From| = dii -*A -.

recalling tl:mit d = 1-v, we obtain

A*A = | - (L-v)ii *s = | - d*A * r' r.,^ = v' d,,,-,- (or,r-r) = v' d ril- ti,;11.

Exercise 5.14

We begin with the calculation of E (f') :

lz r-r**' i+2 r-vzKl;.: . ----.:r---_, K=0,I,2,...,r-I,y2 =)t

, t ,-+ztl z r-r' i+2 r-v2' , 2 i+2 .t

-' "-+ a1-=a- +-''a71, K = n,n*l,....Lt i i i'+2i nt i nt i '

dr.il = Zur rP"k=l

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102 r g63p1"r 5

Exercise 5.18

Recall the formula

and its analogue

By equating the two we obtain

i'a" * (1+i),4,

and from this

d...............:n)

AssumingUDD,

,= go,*

1 = f .ar+(l+i)A",

t = i@).oY) *(r.#)^f,

= i@).*,.(r.#)nr,,

#(n.',n"-(,.*)^r,)

AY ,;,g+ =,{()'q*.

Therefore

*t =# r.

"ff'

= tf) 'o,

= tf' 'o"

= ,f, .o,

= tf)o**

.h e+i) A.-#[' .#)nr,r,,

.# e+i)..4, #['.#) ,f,.,e.

.n(#-# [' .+)"r-,)

.^"#(-[,.#) ,f,)t,fi (t-"i-,)

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Chapter 5 r 105

p(tz)=ffi_ t-tz.(e+i)t/t2 _t))

tz. (1t+i1t t t2 -t). tz. (r - 1r*4-t,t, ;

- 0'06 - 12'(t.Oetrtz - ')= l: 0.4681195348.

By substituting these values, we obtain A[?.h"15.0383835 and

3111-+= HxD6.3to4n2.

Exercise 5.23

(a) We have

(b) Based on part (a)

E(D = @)ff) =f,Oagil,k=0

and this can be seen since the payment pattern implied by the sum is the same as that impliedAV (Q!]. Forexampleif m= 5 thepaymentpatternwouldbe

+.+,*, 1].,?,2,if x dies between ages at age x +1.2I.

Exercise 5.24

(a) We have0<K< n-1,03J <m,

K>n.

l<ra>9-n, o<K< n-l,o3J <m,Y=l ^-;l

ftrlf, , K 2n.

, = {Y'F*EI,

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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__ lr:*i;(1 + Zt,# - rlr$n) =an .Eo:% n

= Ia(4.L't a:.nlk=l

Exercise 5.25

(a) We have

(b) Again using results from problem 23

I rn>9^,I "-;lY=4 /' : lt'lf '."(a3*-rff'),

0<K< n,O<J <m,

n<K,0<J <m.

E(Y) = (Iiln*"Qaff )n-l

= l*l a%+"(Srff')&=0

' = ;(-' ii(\+,iaff')n-l

= >ovf).,t=0

Exercise 5.26

We have

d(h)n+T-vr = u.u^-{'' +T.vr = af,.

We take the expected value of both sides and obtain

6(Id).+(IA)* = q*.

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I l0 r Chapter 6

Exercise 6.4

(a) We have F(.a,) = p=0.02, as E(I) =i=SO.

(b) We have L = e6r -Pail and this is a decreasing function of 2' The 50ft percentile of I isfound at the value corresponding to the 50-th percentile of Z, which is ln2 times the mean, or

ry=ffix34.66 years. Based on this

0 = ,-o.o6"ue -p f -r=t=oj#0.02

P = =L az 0.0086..rb'?

o.s2 I

so that

(c) We have

0=l-PZ* = P-

. since with zero force of intere st e8t =t and f (aV)

Exercise 6.5

We have

I v' . 1p,p"(t) dtFru = g _ .

Ivt ' ,p" dt0

This shows that FQe) is weighted averige of p,(t) for r > 0. As all values of p,(t) exceedp(x), we conclude that F(A,)> p(x).

Exercise 6.6

If the force is constant, then

| =uo,€x

= E(T\ = Z*.

and

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

')- U.,, - p+26.

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Chapter 6. lll

Tberefore '.q,-Z:' _'Z-_ei equals(6a*)' (l-A,)'pp2

Vop- (p3 +2p26+p621-2p2662 (p+26)

_p _p+2 'L.

Erercise 6.7

Ifd=0,thenyt=land @

I t p,p,(t) dtFf+l=s- =*.I tP, dt €a0

Erercise 6.8

We would like to show that

var(vu) < Var(vr -F.dT),where F =FU), or, equivalently, that

var(v?) . (t.5)' .var(vr).

But this follows directly from the fact that (t. #) > 1, as both F and d are posifive.

Exercise 6.9

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

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

d-a*dn

dZ, -dx

Based on this

(,. *)uG,) - * = (r + (p(x) +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).


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'

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

P(zrr,rr) =ffi =mt;, #(4s-rcE$'&s)+ d' rcEgsl-(4s-rcEts'4s + roErs) '

Calculations in the other rows follows analogous formulas.

Exercise 6.11

We have

,o4u-P.1a =HH=W=#=

l-r:- l+i-r'

Based on these values

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

D _ _ Ats.n-=i : ilr*roi - d'((,hs-v\o' ro ns'&s) + vro' to pts) .5s:iol : E*n - T:4m - 4Ar_vto.topts.,4cr)+rto.rofrs)'

zoP(zolrc4).

Exercise 6.12

We calculate@@64 = iv**t . *lq, = iro*t1t- r)rk = v(l-r)l(vr)& = v(l-r) #t=0 k=0 &=0

'd.=+=('-ff|) +=#+=#,P* = #= \-.r,.ax l+,

Furthermore ,'A" is calculated the same way as A, ,butthat it is based on doubled force ofinterest, resulting in 24=&

24-L2 _ (l-r)r(1-A,)2 l+2i+i2 -r


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 onk: 0,1, ..., 4. Assuming the UDD, Iis rmiformly dishibuted on [0, 5]. Thus this problem is arepeat of example 6.1.1 to the situation where the variables are now continuous instead ofdiscrete. By the IDD a here is ino ^exaryle

6.1.1, etc.. Exponential reseryes are not worththe effort of calculation.

Exercise 7.6

& : vu - PZ,,;)a6ifor0 ( U < n - t,andvn-t - F.a;4,U ) n - t.

Nowrz : vu -P.uq: vu-P(#) : ,u[r + #] - frhen var(,e : [t . f]'uuruur: (4)'. var(vu).

since var(vu) : (r7,*,,;4 - 7,*,nn1r),2Z

--/ -2from (4.2.10), then Var(rz) - "x+tin-i "+t"n-d16 ' ar,;)z

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

f'-t _J, aA uPx+t ltx+t(u) du * d;4 n-tp*+t

vu - uPxtt

- upx*rurrl,'o-' * Io' ' vu upx+tdu * d;4n-tpx*t : dx+t:Frr

$ v"'t, u) : .fi|r7,*,,v - V,*,,42f

Note: ?- 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

(a) uo : -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.

Exercise 7.7

El'Ll :

Var(1L) :

Chapter 7 . ll7

Exercise 7.10

The minimum loss occurs when U: 100 - (35J0 :65 - t.-Setting the minimum loss tozero is the same as setting 65 - t equal to -ln(P(As)/(6 + P(135))/6 : 23.25 years. Thust : 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 - P1Ao*5paro,6i

O) Rehospective: [email protected]+0,I0 - roEoo

(c) Anarogous to 4(c): I t -'+#4lr,,tL ' t'"so:lol/ J(d) Analogous to 4(d): F(7ro,ioj) - P(A+o.ro)]aso,O

(e) From (a), since Vro5ol : | - 6 Z56.iq,

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

(0 From (e), since F(7*.61) * 6 : #^, wehave t - 3t'gu4otzol

, \ ,^ air,t-z1y- dn;tg - Ato,Tol - Aoo,^l -(g) .From(rr, ---v4ofrl - - l=z;{,

sinceZ : | -V---T-'

Exercise 7.16

Retrospectively, there have been no benefits, so the rreserve is just the accumulation of pastpremiums: tz|V(tola34 : soF($l zrsFssfrl.

Exercise 7.17Begrn with the retrospective resewe formula: ^T @,.-*,t) : FF,*-) s,.6 - ^8,.Multiply AyP,,*t : ffi. n"rproduces- !

-/- \ -/- \ --rP,,;t' ^r(,,^*;1) : P\A,,^*,1) - P(ai,a),since,E,'s".4 : d,d

and ^E*, ^8, : Vlr,A

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 forthe reserve after m years, if alive. Thus the reserve is in the nature of a pure endowmentbenefit, so the premium for it is a P.E. premium. Now multiply equation (a) by Fr"1, andsubtract ,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 thebeginning of the interval (at time 10, interval length 5) is the a.p.v. of benefits payableduring the interval plus the a.p.v. of a P.E. for the amount of the reserve at the end of theinterval, less the a.p.v. of net premiums to be received during the interval. If we rea:rangethe equation to read

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

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

Exercise 7.19This is totally analogous to Question 14.



Irrlrei,*7.22. Since pZ,; : , - d.ffi- : *,** WSince drA * iira2la@ : 2.ii**p,4,A* ffir_OTtu" m : )'-g : t' FinallY' kY*asT:

kdn723FtI$r Continuous:

Ats - flgon$Aes: .oo4gl5

56'tt_, vzl?-kn-2H+- nxlkn-lcl

ti-. wr*?-kn-2Al--4 qx*kn-kl -r 4 1- r-5 5'

(a) ro-K4ss-i) : | - #^: 'l752905usingz''; : a(x)ii,q- 0(m),dA: dt - f nPxdx+n and the values of d" in the table

fud4,ontinuous:(b) rcY6s): i(rovzs; : itr - ffi): .08566

FuIIy-Disoete:(c) rc%lssT : Arts.^- Pt r.o.1aor.m1: .03273 usingl|r., : A4s - f0 zopesAos : .08846

1l-pt _-;3s'301 -- 3s301 atssTi ii3s - v"u soPssdes

Eisf4: dos - fo zoPtsiies: ll'575

hdrc724(a) No.

(b) Yes.

Recall thatV*6 : 7',a * A,,h : i n'.a + A,,k

{tCL) : 7,+t - P(A,).a**e : i.e*r - i# *n.a' Px' iix+k

: 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: tlni.-^- pta.ii,*r.frf : t.ov);t

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

*+tv: @#P -bn+t ffiThc interpretation is most easily seen if we write it as

GV + nn\Q + i) : bn+r ' Qx+t * r+t Y'P*+t

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

kercise 8.8

(a) fi'w:k-lDr'*' ,lq* At --+re, :i,'APr-#:kv*

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

(b) Since (r,Z' + PrXl + t : e*+r, (l - n+tV') * l+r l/*,thenP, - r.e,+h(I-n+rVr) : v.n+rV, - tVr. Thusthegivensummationbecomesk-lDt"' n+rv, - nv,l(o*l*-n.h4This is 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 withoutbenefit of survivorship, such benefits being only the excess of theinsurance amount over the reserve.

Exercise 8.9

From (8.3.14), rn-r : @n - nV) v Qx+h-r * v ' 1V -IJercrl: fr,andfu- 1rV,sOwehave Tr : v. pV - 1,-1V,or 1'-rY *zr)(1 * i) : 1V.

tV : zr(l+t)2V : [zr(1 +i) + n](1 +;) : r'i1,etc.

Tlrl:.s pV - n. 6 n

n-tY.

Then with sV:0, we find

Exercise 8.10

(a) v'ii*,4: PYB : Du^i n-rlq, : hDO-r-\l n-ilq*h=t h=l

It(,/ ,-,1a"- r',-te,)if, - dii*d - nE* - f +

#-ii"fr: dit-d*;t

- ," rqrf

Thus n : uo ;:'a .ux:nl

(b) From part (a) we see that, attime ft, the PW - d;:A - d"apffi. Clearly the PYP is

n .d,*1,,;4.Thus 1z : (o^ - a-**.4) - n'ii"*p.4.

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 : b*+t vr-t r[-"qr+,t * t+r V'vl-t pr+t

: bk+t vr-t (qr+* - sQx+k) * t+r V'r'-' p**r

sPx+k'*+sV * vr-t ,qt+kbk+t : 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 thereserve at time s if (x * &) has suruived to that time, or provide for the then present value ofdeath 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 twoconsecutive premium-payment points. This reserve is approximated by interpolatinglinearly between the two adjacent policy year terminal reserves, and adding the unearnedpremium for the current premium period. The interpolation coefficients on the twoierminal reserves are easily obtained. Since r is the fraction of the year beyond the lastpremium payment point, then (j - r) is the fraction of the year remaining to the nextpremium payment point, so that is the appropriate fraction of annual premium unearned.Note that this fraction multiplies the annual premium, not the fractional premium actuallypaid at each premium payment point.

: hViutr ,Pr)

ffit2t(e) ElT -t375(l) + 375(3)l (362.12) + t2s0(1) + 2s0(3)l(561.08):t,104,260

(b) Yt(4 : 6,450,962 as before

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

vu(z) = [37s(l) + 37s(9)X1187.14) + [250(l) + 2s0(9)](343.84) : s,3rt,37s as before

ct : 1.645 \ffi : 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) : 531,137,500c1 : 1.645 \@ : 37,gll,whichis.o0o34timesthereverse.

Erercise 8.29

We seek 10,000 rcrpY{tl(4')

10,000 ll.rcv{t}(Vto) + }.rrv{r}(Ato) + t.p{t}(7ro)]

s,000 [ toV (Ail * n-vGzi + P{t}(7ro)].

(c)

(d)

Exercise 8.30

SinceP,a : *, - 4 then d",Tt" : ffi : T-iESince12".1 : t-W: f,**d,+r't: .78ii*.4:(a) d*,Tt : l*uprd,*1a,soQx:1-(1'?'*q83) :

: 2.083.

t.625

.2

(b) dx+ril1 : L * vPx+r, so4r+t : 1 - (1'2X'625) :'25

(c) It'r, : f (b^*t - r,+rY) p"+h' Qx+h' hPx

As : (#)' G-.66)2(.2X.8) : .6084

A1 : (i=)' (3 - l.s6)2 (.8x.7s)(.2s) : .216Var (62) : Ao * I Ar : .6084 + .69M4 (.216) : .7584

(d) Var(12) : X : # : -27

(Note that there is no risk in the final year of an annual premium endowment.)

186 . Chapter 15

Exercise 15.5

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

Therefore,

Commissions of 0.40G + 0.0sG\",l,sl = 0.35G + 0.O5Gajao1iol.

Premium tax of 0.02G\+o1lil.

Maintenance expenses of 12.50+ 4o4oj = 8.50 + 4\+01,A.

Death benefit of 1000{+01F.

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

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

8.50+ a\*l4+1000{+o[H 1 ooo{+o1B + 4d1+o16i+ 8'50

or

and hence

G-

Exercise 15.6

This policy has a single premium fI that pays for all benefits and expanses, so that premium iscalculated as the actuarial present value at issue of all benefits and expenses. The premium paysfor 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,

0.98i1+o16 -0.05&1+olrol-0.35 0'93d1+ol,E +0.05'roE1+01'A[+o]+ro:i3 -0.35

0.935n = 2.50 + 2.50ii,s+ 10001- .r,

2.50+2.50A 1+10007 -xint x.:nt

0.93s

so that

and


Chapter 15 o 187

Exercise 15.7

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

- An initial expense of eo.

- Annual expenses of (e, + ezPr)iir.- The cost of claim settlement of q.A,.- Benefit of A*

Therefore,(aPr+c)ii" = €0 * (q+qPr)ii, + er. A* + Ar.

Recallthat ^(, =#,h"n""(recalltlrat 1 = .4r+diir)' 4x'

("+ * ")r, = "o *(",. rt)r" + q. A, + A,,

aA* + cii* = €o * 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) " = Bvr + a B a7 + 0 aa + p (B n + f)a7 - (B n + f)a7.

We assume that T(x) andB are independent.

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


il

Exercise 9.3Analogous to Example 9.2.3.

Exercise 9.4

(a) Pr(T > n) = npry : ,px . npy,by independence.(b) Pr[T(x) ) n and W) S ,,.?, \y) ) n and Kx) < n]: np*(l-np) * ,py(l'-,pr)' : npx * npy _ 2.np".npy.(c) Pr [at least one survives] : I - prfneither survivesl: I - pr{max[t\r),W)]<n]

: l-ngl,_:nPfi: ,p, 1rp, npx.npy(d) Pr[T<n] : ,ery : 1 - npxt: r-npr.npy(e) Prfatleastone failsJ : 1 _ pr[bothsurvive] : I _,p*. npy.(0 Pr[I(x) 4 nand W)3n] : nQx.nQy: .(l_,p,)(t_,py)

/,* /,*'a', l,*r# dvdx -n * F

Chapter 9

Exercise 9.1

(iv)

Exercise 9.2

Totaty anarogous to Exampre g.2.r. Aseries of tedious integration tricks:(i) /-# *= *irnlr =+ fi(s): #isadensiryon[0,oo)(ii) I- [- qirrdtds : r';ri6,=T irn > 2 + fs,r6r) = @, ,t\(n,-=?)isajointdensityfor0(s,t(oo -' ' (l+s+Dr(iii) From(i),Et(r+s)'l= ,r-tll c# = vffiirm<.n *l.rhus

EU + sl: FL: I +EF] =+ E[^SJ : T5 simitarlyH : EKt +,S)2J : | + 2Etsl+ E[^r2J can be used with E[^iJ to find E[^s2]

: I - ,P" - ,py'* npx.npy.

I 16 o Chanter 9

Exercise 9.5

We seek nPx' n-tPy, which is Px' n-tPx+t ' n-rPy, 91 P*_n-lPxltyAlternatively, npy-r : Py-r' n-rPytso that n-rPy : #,producing nPx:y-r/py-t'

Exercise 9.6

Intuitively, tpo PoQ) is the p.d.f' of the R.V. T : T\xl).Thus the integral is PdT I n) - ,qo. Aninteresting algebraic approach is to note thatp*(t) :2pr(t),and po: tPx.lpr. Then the integral becomes

P / 1 ^l'r,lo" *-(,p'p,'(t)) dt : ,(-rdl|r), since P,F,Q) : -ft,P,'Thenwehave 1 - nP? : | - rPu : nQn.

Exercise 9.7

tf T: IW),FrU): 1- S4,yr6,y(t, t): L - # fromproblem2 above

(a) fr4): F[(t): ffiO) S1(t):l-F(r):dF(c) Elr@y)l:

Io* tpxydt : lr* & (s :2t, ds : 2dt)

: Ir* ,61s" : th (see solution to problem l)

Exercise 9.8

Analogus to the given example and equation (9.3.8).

Exercise 9.9.

1-tQ7 - 1- tQx.tQy - 1 - (1 -rpJ(t -,P) : I - (1 -tP,-tPy*tP"'tPy)tP* * p, - P*'Py : tP, * tp, - ZtP''tPv I tP'vtp, (l - ,py) * ,P, (l - tP) * tP"y.

Reasoningly, the event of at least one out of x and y surviving t years is obtained if xsurvives andy does not, or ify survives and x does not, or ifboth survive.

tP-ry =::

Erercise 9.10

Pr [at least onedies in (n + l) : I - pr[neither dies in (n + 1): I - J t - pr[xdiesintr+ r1]'{t _ rrgai"rirl"i rX}t: I - (1 -npx*,+e)(l-,py*,*ipr): I - (l _n.lq,_,1q, I ,lq,.,lqr): nlq, + nlq, - nlq,.,lq,y l ea is the probability that the second death out of x and y occurs in (z * l), which is notthe same event as above.

Algebraically, nlqy : nlq, + ,lqy - ,lq,y. Clearly ,lq,y * ,lq,.nlqy.

Exercise 9.11

(a) FrwG): tQo: tQxtQy: F4aQ)F49Q).fromproblem 1. NowfslG): F[oot!)

o) calculate Etw)l as E[(-r)] + EVU)I - nv@v)1. use problem s #7 and,#t.

(c) u4Q) : ##' use results from (a)

Exercise 9.12

We seek tsp+0. We note that zspzs:so : zspzs . zspso : 5ep25. Furthennore,

tsPzs' 35p40 : soP2s,so that 35pas : #, : 3 : ?.

Exercise 9.13

Now plug in FaaQ) - 1-

t- T0o=.

tPso : 1- #,

23

We will need p, : e-[ip'6)as - eXp [- Ir'(1000 -, - "l-' a"l

: ex' lr,t*-'-", lr] : ;

We will also make use of 1p*1t,(r) : TO#-. Then rpqo : f _ f6 andand,pastta(t) : # and, 75str5s!) : ,,10.'

(a) rcpqo:so : rcpq.rcpso : (t - *B)(t _ *B) : eg # :(b) roPa*o: toPco* ropso- ropto:so:;8+f8-?:H

I

(c) The p.d.f. of T : Z(40:50) is

tp+o:so ttq,so(t) : tpqo . tpso?-tq(t) + pn?))

: k k(#=7*#) :T5#,0<'<so(Note it is still r ( 50, not t < 55).

B+o,so : EtTf : r#',/"nrs-t)dt: r+00 lf r-+rlr'] : ,r.ou

(d) 8ao;m : 9qo + 8ro - 8ro,ro

: fo* (t - .6) o, *

fo'o (t - #) dt - t8.06

: (uo - fS}3) + (so - qf}l) - 18.06 : 36.e4

(e) Err2t: tbo/" rlss-t1 at - "hf fft'-inl,'] : +se.rrrri

Then Var(Z) : E[72] : {ttn}' : (486.1r; - 1ra.Oo;2 : 160.11.

(D Elr2f : lo* ,rtaoo, * loto

f no at - +se.rrrri: tf$ * Gr+# - 486.1r1n : ts47.22222.

Then Var(Z) : 1547.2222 - pe.S+72 : n2.66.

(g) cov lr1+o:so), (40s0)] : 8* 8ro - goo,ro gae,

: louo

{, - dnlo, . fo'o {r- s6l dt - (t8.06)(36.s4)

'" : (30x2s)-(18.06x36.94) : 82.86.

Cov[(40:50), I(40:50)] : 82.86 _ jcj<var[(40:50)J. va444651] y'(t6o.llxl82.66)

(h) r4+oso;,44613n-q :

Note: These answers differ slightly from the text answers since we have roundedprior answers for use in later calculations.

140 o Chanter9

Exercise 9.17

ForZ:

and ,p{

(1,1), we seekPr(2 < T < 4). Now

- expl- I,' r(tf) *] : ($)-' : (s),,

- eXpl- Ir' (10-x-"1-'*] : t - #.we seek 2pr,r - 4pt:t: ef (8) - (.e)4(;) : ,%t - frtrE : ,?$fo :531/2000

Exercise 9.18

(a) FrwQ) : Fq,rft) + FrelQ) - Fao{t)

:t*t- (2,-[n[t.ss]):[rnlt.$$]

O) Differentiate the answer above.

Exercise 9.19

F4*1ror(5,s1:[hlt.W]

Exercise 9.20

(a) As o + 0 (x) and7|O)) are independent so

sQ4 : sqx. sQy: (.05X.03) : .0015

If a: 3, fromproblem 19 sQd: .000266

If a : -3, fromproblem 19 srv: .004232

Exercise 9.21

In general, dfr : au * du - auv,so o@A : ctry * ai - a"y.nl : a4 * rlar.This aruruity will pay until the first failure out of x andy, or until time n, whichever is later.Thus, it pays for nyearc for certain, and beyond that as long as the joint status (ry) survives.

,pl

o)(c)

Exercise 9.22

This insurance will pay atthe death ofx, or at time z, whichever is later.V*r=, : 7* +V;t -Vr,;1, where ZA : u n.

Altematively,

Then 7.:x:nl

(vn TlntetZ- J -trt T) n

EI4 : vn nQ, * nl7,

* - 9"'a-v'nP') *Ar-A*4*vn

: ," - Zt,A * v, (l - np")yn - Vn npx

Exercise 9.23Cov [v?@), ,r@Df : [email protected])] - n[vw]n[vw1

: ElvT\n . urrDl - z[vzwtlnlvxvtl: nfvx,tlafvnt)] _ ZfvrootlnlrW)l due to independence: 7: 7r-- Z7'VZ ,: 7*'7, - (7, *V, -7r)7,: 17, -71117, -Vr) r

Exercise 9.24

For 0 < t < 20, the annuity will p.ay.if either is alive, since both are under age 50. For':n;r'";:' the annuitv wili pav onrv ir (25) i; .ii*. iy tr,. .uo.nt p"y-r* technique of

f2o r25aPv : J, vt ,pE,Ta O, * Jro vt p25 dr

: l" r, pzs dt * I'o r: pso dt _ I'o

r, *2sso dt: Azs,4 * dzo'fr1 - dzs,to,q

Exercise 9.25

kr this case, the annyltf will pay for /r : 21,...,25 onlyif (30) is alive,and for k : 26,... if either is aiive.

25Thus apv : tru kho + iro *pxsob2l tol;

oo oo: Dur kPn * f ,o kpzs -F2t n:ie: zolazo * zslazs - zslazs,to

t'r oorr,rolF26

148 o CamF l0

p{l' p9' p[)

f'-nsl l-*)ll,-oslll - (.02 +.os)l[1 - (.03 +.06)][1 - (.04 +.07)]

: (.93X.91X.89) : .7s3207

,k['] : ,p{l .qtl : (.7s3207)(.0s) : .0376603s

,qtl : et#--e : if# : Jrso4,usingthetableof Example 10.3.1

Exercise 10.5

(a) Probability of graduation ir op[") : .3024.

Then the number of graduates, G, is a binomial

R.IZ., with n : 1000, p : .3024.

ElGl:nP:302.4V(G) : np(I - p) : 210.95424

(b) Similarly, number of failures,4 is binomial with n : 1000, andp: 4qt): .15 * (.60X.10) + (.60X.70)(.05) : .231ThenE[F] : np:231, V(F): np(l-p):177.639

Exercise 10.4

,pg :

Exercise 10.6

k0I23

al' : 4)ql])150

"602l0

BT

d? : $ .q?)

2501206333.6res

_a_1000600420336

(a) fie) : aE#rl* : .231

fi(z) : total others withdrawalsTfi(3): q#: W: so24

(b) fi(i lk - 2) . Pr(k : 2) : Pr(termination at k : 2 iamodeT)Thenf(l lk : 2) : # : .25 andfl(Zlk : 2) : # : .75

Of coursef (3lk:2) : 0.

: .4666

Excrcin f0.?

(a) From (10.4.1),

*) :' * -l- [ p'U) * rrot ar]

-''*[-1,t, * &)dvf :;.*[-'+ k+]: (a-x1s-,

(b) 4D : fo' tJ,r:r<,1o, : lo' "*-, at : "-x - "-x-l

@) 8) : I'c{J,p!r)@at: fo'%lL:ry: - (a - x - t) "--,l,to - Irt e-*-t dt

:'-' -? :"; .:-; -:':l-"':'"-x I "-x-'

Erercise 10.8

Aga^, *)

Erercise 10.9

(a) *,n9'

roooexp[-1,'"* &*]1000exp [-* * n (' - *) - rn | : 1000 e-d -t

: -4&dx F):-

: *l,lry,ri', ,o,r.ff') : og,[ug,a> - r<,t1gf

(b) t,u9 :

(c) #,nP : #W: fi[4],rgatl: ,pg)pga>

Lry - 0.,clx ltr)

bt- *)pa(,) * *,ryff'l* (c-\*,)l,tatl,pP p?al * ,oyt ro@) - pa(r)

I {? r Chanter l0

Exercise 10.17

Firstwe findpf) : [r - n't"1 [t - o'o'] [t - f o']

rhus,pf]pgp{}

rlter_q]|/2

qt)

Similarly, qt]

: - .87478166 tnp'S

: .76048,

: .85027,

: .821,1,5,

qr) :h\qi; :

q{} :

.23952

.t4973

.17885

: ffi'mP',: .01767; qg)

qgqzl :q?:

.02054;

.02578;

: .02665 qg) : .Igszo

.031e3; q?l : .0s726

.03705; q?]:.11603

Exercise 10.18

The result is direct, so no "solution" need be illustrated. The purpose of the exercise is toshow the closeness of results to those of Exercise 16.

Exercise 10.19

(a) *',a on9) is justified by the constant force assumption.

(b) Accepting that d.a x m9, *"n i d"r?:i6 = ; f :O. if decrements arer-l'lx r-2'Yxuniformly

distributed in both the single decrement tables and the multiple decrement model.

(c) Clearly this leads to 4f) = ,orto

(d) l,'1, - L nl'l: oglt - L n:,'1, which in turn implies

t - i'qf)

d,alr - * n9' + t' og) : q9, or d,a : # -qExercise 10.20

*9 : 4* v.e.,-tl:

m'a:4cfu : '02073'etc'

bowers solutions

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