Solutions Manual

for

Bowers' et al.

ACTUARIAL MATHEMATICS:

LIFE CONTINGENCIES AND RUIN THEORY FOR THE ACTUARIAL STUDENT

Michael A. Gauger, Ph.D.

ACTEX Publications Winsted, Connecticut

1.

CHAPTERS

i At i = .06, we find that b = 1.0297087 and 6 = .0582689

_ 1 _ .!. A {A20 = .0652848 (a) ax = / x. Substituting Aso = .2490475

Aso = .6657528

-will produce a20 = 16.00812, aso = 12.76073, aso = 5.39685

., (b) 6' = 26 = .1165378, i' = .1236, ~, - 1.0606

{

2A2o ., 1 2 . 2-6' . Ax ,producmg ~so-

2Aso -

Substituting into }2 [2Ax - Ax 2] produces

.0151702

.1004983

.5022854

x -20 50 80

x

Var SD

3.137 1.771 10.230 3.198 9.523 3.086

2. (a) Using values from Question 1, we find directly alj.L

(b) j.L = (100) (1000) (12.761)

a2 - (100) (1000? (10.230)

a = (10)( 1000)(3 .198)

20 .111 50 .251 80 .572

Then alp 3.198

- 127.610 = .0251.

}2 [1 - 26 2a x - (1 - 6a x)2 ] }2 [1 - 26 2a x-I + 26 a x 62 a;] ~ [a x - 2a x] - a;

Chapter 5

4. COv(8a TI' v T) =COV(l- v T

, v T)

Now var{ (l - v T) + v T

] = var( 1 - v T) + var(v T

) + 2 Cov( 1 - v T, VT

)

But Var[ (l - v T) + VT

] = Var(l) = 0 Var(l) + var(v T) + var(v T )

+ 2 COY ( 1 - v T, V T) 2Var(V T

) + 2COV( 1 - v T, VT

)

5. (a) Multiply by the integrating factor and rearrange to obtain

[. ~ a, - [.0, (,,(Y) + 0) = -[. where [ = exp [- l' (,,(z) + 0) dz]

d -or dy (I· ay) = - vY yPo. Integrating over [x, w], we have

I. a Y[ = _[U: vy yPO dy or _vx xPo (ix = _vx xPo [U: vy- x y-xPx dy

which gives ax = 1U:-x v t tPx dt

= 1U:-x v t JU:-x J-LAs) sPx ds dt

= 1U:-x 1U:-x v t J-Lx(s) sPx dt ds

= 1U:-x (iST sPxJ-Lx(s) ds

(b) This time multiplication by e-oy , and rearrangement, yields

29

-oy d - -oy 1: - -oy (Y)- -oy d ( -oy -) -oy (Y)- _ e-oy. e . -a - e . u a = e . J-L a - e or - e . a = e . J-L ay dyy y Y' dy y

Integrating over (x, w) yields e-oy . (iy [ = [U: e-oy . J-L(y)(iy dy - ["" e-oy dy,

[ "" and (i = (i- - e-o(y-x) (i H(y) dy x u:-xl yr

x

ChapterS 31

- - - - - - - - -7. ax:n+!1 = ax:Tf + vpxax+!:;i'f = ax:;i'f + vn nPxax+l:TT '* ax:;i'f = vpxax+!:;i'f + ax:Tf - vn nPxax+!:TT

-Starting Value: a""-n:;i'f'

r+1 r+! 8. nlax = in VtPxdt+n+llax = in VtPxdt+vPx(nlax+l)

Starting Value n I aw- n = O.

9.

- - -= vpxCanr+n I ax+!) + anr(1-vPx)

- -vpx(ax+1:iil ) + anr(1-vPx)

- -Starting Value: a

w_ n:iil ) = anr

10. The density of TI T<l is (tPxl.l,At»lqx and the density of TI1?::! is (tPx/-Lx(t»lpx'

E[a11 10 ~ T < I]Pr(O ~ T < 1)

- -= -PxaTf + ax:TT after integration by parts

- -= px(aTf + vE[a y]) (Y = T - 1 = T(x+l»

- -= PxCaTf + vax+1)

- - -Combining these results gives ax = ax:TT + Vpxax+l

11. Var (aK1 ) = Var(iiK+11 - 1) = Var( iiK+I1)

30 Chapter 5

6. (a) From constant force Fy{t) = 1 - e-P,t, hence

1 Fy{y) = Fy{ -In(l-8y)/8) = 1 - (1-8y)pJ6 for 0 ~ y ~ "5.

(b) If Y corresponds to the n year, continuous temporary annuity then

{ (l-VT) -

Pr -- < Y 0 < Y < a-Fy{y) = Pr(Y ~ y) = 8 - _ - nl

1 a~ ~y

= {Fy( -In(1-8y)8) ~ ~ y < a~ 1 a~ ~y

= { 1 - (l-8y)pJ6 ~ ~ Y < a~ 1 a~ ~y

(c) Assuming a constant force of mortality, the graph of Y versus T for the deferred life annuity looks like

Y (0/8) - - - - - - - - - - - -

T n t

Fy(O) = Pr(Y= 0) = Pr(O ~ T~ n) = nqx = 1- e-np . If 0 <y < 0/8, then t = -In(0 - oy)/o and F(y) = Pr(Y ~ y) = Pr(T ~ t) = tqx = 1 - e-p,t = 1 - (0-8y)J.L/6.

(d) For the n year certain and life annuity, the graph of Y versus T looks like

Y 1/0 Y a~

-- -------~-

n t = -In(1-oy)/8

Fy{y) = 0 ify < a~

Fy{y) = nqx = 1 - e-p,n if y = a~

Fy(y) = Pr(Y ~ y) = Pr(T ~ t) = tqx

- 1 = 1 - e-pt = 1 - (l-8y)pJb if a~ < y < "5

T

32 ChapterS

n-I n-I n 12. (a) RHS ypx L yk kPX+1 '" k+l ~ V k+IPx L v

k kPx

t=1

LHS

(b)

k=O k=O

nlax = ax ( .. E) I-Ax I-A 1 ax:itT + n x -d~ - d

x:n

- nE:c RHS

A -7 is the value of a perpetuity of 1 annually, starting at the end of n years, or year of

death if earlier. 1- cancels the payments at the end of the year of death, and on. This com­

bination provides 1 at end of n years, if alive, which is cancelled by -nEx, leaving n I ax.

13. From 1 d . axonr + A"nr' we obtain

Ax.nr 1 - (I - v) a"nr 1 - a"nr + V· ax.nr v·a"nr - (ax.nr -1)

v . iix:1IT - ax:n=iT

14. We simply indicate the calculation of E[y2] and leave the rest to the reader;

y2 = ~ i

(

1- 0+1

i 1= v" i

(

I-VK

(i+2) P +2i

I I-,?n

i2 +2i

+ {

D 2 2 i+2 a- - ~a- + ~ea-) nl I nl I nl

2 (2) (i+2) 2 2 2 i+2 2 ) E[Y ] = j a"nr - i axnr + (anr-janr+i anr nPx

15. Under the UDD assumption each jlmPx+k . Ilmqx+k+jlm = jim Illmqx+k = ~qX+k. So the term in

brackets in the problem statement is

[I.. I·· 1..]

q~+k mS l-~j + mSI_~1 + ... + mSI_;1

_ [;[(I+OI-Ilm - I] + ; [(1+0 1-2/m

- I] + ... ] - qx+k tim)

sl[ - = (qx+k)(3(m) [

(m) 1] = qx+k tim)

Thus the sum in the problem statement reduces to

tkPxVk+I(qHk· (3(m)) = (3(m)A,. bO

,... -

Chapter 5

16. (a) a(m~ = a("'!.... + vp aIm) __ _ x:y-x! x.lj x x+ l.y-(x+ I) I

(b) a~"D = ax.l[a(m) - (3(m)(I-vpx)

a(m) - (3(m)(I-vpx) (ax.1[ = I)

a(m) - (3(m) + (3(m)vpx

17. ..(m) •. m-I ax ~a;c- 2m

( m-I)

II I ti~m) = V' nP.:Ji~;'n ~ vii nPx iix+n - 2m

- I··(m) - v"nP --(m-I)

- n ax x 2m

Subtracting these two relations gives

a~~~ "" (ax- n lax) - (~-:nl) (I-v" nPx), and the term in parentheses is ax.nr.

·(m)

18. Analogous to Ii· ax + (I +0 Ax> we have 1 jim) . a~n) + (I + 1m) A~m) ..

Equating the two RHS's and solving for a;m), we have

(m) i I [. (im») (m)] ax = l,m)aX + I,m) (I+/)A, - I+ m Ax .

. UDD (m) i A (m) Assummg ,Ax = I,m) x = S I[ Ax· Then

aIm) x

i I,m) ax

I I [i ( i(m)] + l,m)(I+I)Ax I,m) (1+1) v I,m) 1+ m) Ax

'---' s~)

II (m) 1- a~)

S I[ ax + (I + i) . -=--------::... A lm) x

'------v---' •. (m) al[

33

34 ChapterS

(

;(m)

19 As in Question 18 I i(m). a(m) + I + ~ )A(m) so • • x m.x '

a;m) I _ (I + ~)A~m)

= a<::'!.- a<::'!.(I+I'm»)A(m) a<::'!. - Ii<::'!.A(m) i(m) 001 001 m x 001 001 x

Under UDD, a~m) = I I i

I,m) - tim) . I,m) Ax

20. (a) lim ii~m) m-co

(b) a,

from Ax - d·lix - d(ax + I)

I I i I,m) - tim) . I,m) (v - d . ax) = I - d - d· ax

v - d·ax

di I - iv ltim)

,fm)im)a, + I,m) a(m) ax + -y(m)

lim f: ~ . v h/m h/mPx

l/m..() h~

1'''' v I IPX dt (by definition of the integral) = ax

/. ..(m) /. (.. m - I) . I I I lin ax "" 1m ax -"2iil = ax - 2: = ax + - 2:

m-oo

I =ax + 2:

(c) ax 1"'" v' IPX dt "" I 2: (vo oPx) + vPx + V

2 2Px + ... 1 2: +ax

21. (a) Using ii~m) "" iix - m ~ I, we have

ii(m) = ii(m) _ 1. "" ii _ m - I _ 1. x xmx2m m

m-I _ ax + 1 - "2iil - m

m-I ax + "2iil

(c) .Ia~m) (m) ( m - I) flEx ax+n ::::::: nEx ax+n + "2iil

m-I .Iax + "2iil.Ex

b (m) _ (m) I (m)_ m-I I m-I E () aX

:nJ - ax - n ax ....... ax + "2iil - n ax - "2iil n x m-I

axon[ + "2iil (I - nEx)

22. (a) •. (m) 8 25040[ ii;;)4O[' 4O~25 [a(m) 0.25040[ - (3(m) { 1- 4OE25 } 1 4O~25

.. m{-I -I} a(m) 8 250 40[ - (3() 4OE25

ChapterS 35

.(12) (3 (b) a250

4O[ = a(l2) 1i25 .4O[ - (12)(1 - 40E25)' Now

1i2504O[ 15.46630875; 4OE25 .0765778156

id a(l2) 1,12)ti I2)

(3(12) = i _ 1,12)

1,12)til2)

(.06)(.0566037736) (.0584106067)(.0581276674)

.06 - .0584106067 (.0584106067)(.0581276674)

I. 000281 006

.4681195348

Substituting in these values, we obtain

23. (a)

(b)

24. (a)

(b)

1i(12~

1i(l2~ = 15.0383835 and s(l2~ = ~ = 196.3804112 250401 250401 4OE25

y= K+(l+I)/m - -{ (li1)~1 O<K<n-I,O<J<m

(/ii)*) K 2: n

.-1

E[Y] = (lii)(mJ... = I: k I o.(~~ can be seen since the payment pattern implied by the sum is x.nl k=O x.n~kl

the same as that implied by (Iii)(mJ.... For example if m = 4 the payment pattern would be x:n1

1/4, 114, 114, 114,2/4,2/4 if x died between ages x+5/4 and x+3/2. The term olii~~~ would supply a pattern 114, 114, ... , 114 at the same ages as the above pattern, and the term II ii(m~ would supply 0, 0, 0, 0, 114, 114 at these ages. The remaining terms in the

x'n-I1

sum would supply nothing.

y= {(Dii)K+(J+I)/m l O:S K < n-I, O:S J < m o K2:n

(Da,(mJ... + (Ili)("!2.. = (n+ l)o.(mJ... x:nl .t:n] x:nl

=> (Dii)(m!... x:nl

(n+l)ii(~ - (Iii)(m!... x:nl x:nl

(

n-I ) (n+l)ii(~ - '" klii(m~

x:n1 ~ x:n-kl k~O

(problem 23(b))

n-I

o.~~ + I:( ii~~ - klii~7Lkl) k~

n-I ii(m) + '" ii(m)

x:nj" ~ x'n-k] (k = 0 term is zero)

I ~I

~~ _____ L... _, __ --t.~~_~,

36

25. (a) {

(l"")(m) a K+(l+I)/ml

y-(Iii)~) + n(ii(_m_) __ - ii~»

nl K+(J+l)/ml nl

o :5 K < n, 0 :5 J < m

n:5 K, 0 :5J<m

(b) By comparison with problem 23

E[Y] = (Iii):7~ + Il(n I ii~m»

n-l

= I> I ii:mLkl + n(n I ii;m» (See 23(b)) k~O

n-I n-l _ " ( I .. (m) + I .. (m») -" I-(m) - ~ k a,t:n_kl n ax - L-t n ax

k~O k~

Chapter 5

a--T. v T

26. b(Ia)T[+T. v T = b· TI 6 +T· yT aT[' The required expression follows directly

27.

from equating E[LHS] = E[RHS] . ,J.'-

dm:"" = a~) + nla(m)

x:nl nl x

where

_ :i.. _ + v" .. (m) - ,1m) a"l nPxax+n

i [ . (1- .. (m» ] = "m) an[ + v" nPx ,(~)aHn + (I+l)~ A ,,' ,-{m) x+n

probl~m 18 '

i I,m) [an[ + v" nPxaHn + v" nPxAHn . c]

c = (I+i) (1 - ii~») = (!) (1 - ~) = (! -~) i II d elm) delm '

I

1.

CHAPTER 6

EI[L] = (.48544)(.2) + 0(.2) + (-.45796)(.2) + (-.89000)(.2) + (-1.29758)(.2) = -.43202

EJL2] = L(Lossd = (.48544)2(.2) + 02(.2) + (-.45796)2(.2)

k

+ (-.89000)2(.2) + (-1.29758h2) = .58424

Var/(L) = .58424 - (-.43202)2 .39760

2. The new loss function is IOv K+1 - PiiK+11

so P is the solution of

4

Lexp[.I(lOyk+l - Piik+II]2, k~ Pr(K=k)

The arithmetic needed to show P = 3.45917 is left to the reader.

3. E[u(1O - L(P)] = E[(1O - L(P» - .01(IO-L(P))2]

= E[10 - L(P) - .0I(lOO-20L(P)+L(P)2)]

= E[9 - .8L(P)-.0IL(P)2]

= 9 - .8E[L(P)]- .OIE[L(P)2] ~ var+mean2

= 9 - .8[Ao - Piio] - .01 [ (I+~f eAo-Ao 2) + (Ao-Piio)2]

To find P maximizing this expected utility, take the derivative with respect to P, set it equal to zero, and then substitute Ao = .84247, 2AO = .71457, iio =2.78298, 2AO-Ao2 = .00481. As stated this problem is ridiculous - the larger the premium the higher the satisfaction of the insurer.

The problem should have asked for the insurers indifference premium:

9 = expected utility without insurance

= expected utility with insurance = E[u(10 - L)]

= 9 - .8[Ao-Piio]- .01 [( l+~reAo-Ao 2) + (Ao-Piio)2]

L' The .. quadratic is left to the reader' P "" . 3036 whereas the ievel annual benefit premium is

Ao/an = .3027.

-- -~-~ -~~~ - -- ~- .~~~- ._---~ ---~------

38 Chapter 6

__ I 4. (a) PtA,) = '" = .02 since 50 = E[1l = Ii

5.

(b) L = e-;T - Pan is a decreasing function of T. So the 50th percentile of L, 0, is L

evaluated at the 50th percentile of T, 1~2 = 34.66 years:

0= e- J16(3466) - Pa-- '* P = ~ = .0086 34.661 S34.661

o 0- 1 . Of - 0 (c) 0 = 1- Pex '* P = ex = '" smce e- = 1 and E[an] = E[1l = ex·

P(Ax) fo'" v' ,Px"'x(t) dt This clearly shows that P(Ax) is a weighted average of /L,(t) for

6. If ",(x+t) is constant, then Ax 100

e-(~+;)'· '" dt '" t 6' and 2Ax ~ /L+26'

7.

8.

9.

- 2 2A_ Ax Then ~6ax)2

~ ~' p:+To (p+iji

;' (It+6)2

2A _ A 2

(I _ Ax;2 is given by

(",3 + 2""6 + '" 62) - 1/ - 2",2 6 62 ('" + 26)

If 6 = 0, then v' = 1, so P(Ax) foOO

,Px/Lx(t) dt

fooo ,Px dt

I o· ex

~ '" + 26

2Ax·

We wish to show that Var(vT) < Var(vT - p. an)' where P = P(Ax)' or that

~ar(v T) < (I + ~f . Var(v T). Thus we wish to show that (I + ~) > I, which it is since both

P and 6> O.

d- - d- -liX u, = (",(x)+6)ax - I and arAx = (",(x)+6) Ax - ",(x). Thus the left side is

[I + (",(x) +6) ax - I] ~ - (",(x) +6) Ax + ",(x), which is easily seen to reduce to the right ax

side.

I

I

I

Chapter 6

10. The first row of the table requires: __ A - 6A-

(i) P(A3s

,lo[) = _3S, 101 = ~,101 a 3s,1o[ I - A3s ,1o[

A - - i Al + 10 _ i [A 10 1 10 3S,101 uDO 7J 3S,1o[ v . IOP3S - 7J 3S-V . IOP3SA4S + V . IOP3S

A - '/6 [A 10 A 1 10 P(A _) = .. 3S, IO I = I 3S-v . IOP3S 4S + V • IOP3S 3S,101 a 3s,1o[ ii3S - vlO . IOP3S ii4S

(ii)

... A3s,1o[ dA3s·1O (m) P3S ,1o[ = ii _ = ~where

3S,101 I - A3s ,1o[

A3s ,1o[ = (A3S-vlO . IOP304S) + vlO . IOP3S.

11.20

pl __ pl-dOl ,,201

AI--AI-dOl ,,201

ii,,201

00

201lOAx

ii"wr 20P(,0 IIOAx) .

12. Ax f= Vk+I' k I qx L vHI (l - r) I k=O

ax

k=O

I - Ax -d-

00 I v(1 - r) "" (vr/ = v(1 - r)--

~ l-vr k=O

( I _ ~) (.!..±i) _ (_i_) (.!..±i) I+i-r i - I+i-r i

I - r

+i-r

1+ i +i-

39

Px I - r

Ax/iix T+i 2Ax will be the same as Ax, except that it is based on 6' = 26,

which implies (l + i') = (1 + i)2. Therefore, 2 Ax I .:; r . Substituting these values

• 2 Ax - Ax 2 d' I'[y' '11 d (1 - r) r mto 2 ,an sImp I mg, WI pro uce . .2 (I - Ax) 1 + 21 + 1 -

13. p<;}

Then pi;}

Aso where Aso .. (2) , aso

1210.1957 4859.30

.2490526776,

(2) 64467.45 and aso = a(2)· 4859.30 - (3(2) = 13.01224267,

where values of a(2) and (3(2) are given in Example 6.4.1.

.2490526776 .01913987. ................... "' ......

~L __ . ___ ~ ________ ~ ___ ~_ .. __ . __ ~._

Chapter 6 40

ii""-14. We really wish to show that the ratio~ can be expressed as each of the three given

a"hi

expressions.

a(m) (i) and (ii) .. "hi

a"hi

_ (3(m)(I~~x) xohl

a(m) . iidT - (3(m)(1 - hEx) _ .. - a(m) a"hi

a(m) - (3(m) [p - + d - P LI1 x:h I x.h

a(m) - (3(m) [ P!'hi + dj, which establishes (ii) . . d . ~m)

(ii) quickly becomes a(m) - (3(m)d - (3(m)· p!,hj" Since a(m) = im:Jm) and (3(m) = 'i;,im) ,

then clearly a(m) - (3(m)· d = J.,) = a(f!, which establishes (i).

To establish (iii), use a~mh = a"hi - m 2~ 1 (1- hEx). Then we have

~ _ _ m - I 1 - hEx _ _ m - 1 I • .. .. _ _ I 2 .. _ - I 2 [p .hi + dj, as m (II). _(m) ()

a"hl, m a"hl m x.

15. Obtaining the needed values from Example 6.9, we have

A50joi ~.AI In (1.06) 50,201 + 20E5O

.06 .0582689 (.13036536) + .23047353 .3647118

Then P(2) (:450,201) .3647118 11.09616

.032868.

pt(~

16. dOl pl­

dOl

~ .. (12) a,,2OT

1.032 . p.12) __ ~

,,201 - Pdol · iP!!.... dOl

(.04)(1.032) .04128

17. P < P(2) because premiums are paid later under P(2), and interest is lost. Also one-half year's premium is lost in the year of death if death occurs in the first half of the

year.

P(2) < p{4} because there is extra benefit in the year of death under p{4}, as well as further

delay in premium payment.

P(4} < p{12} Because there is greater delay in paying premiums under P{12} , hence more interest lost. The refund plus loss of unpaid premiums in year of death are

approximately equal under both.

P{12} < P because premiums are further delayed under P. The benefits are about the same.

I.

2.

3.

4.

5.

CHAPTER 7

Completely analogous to the calculation of I Yin Example 7.I.I(a).

Completely analogous to the calculation of I V in Example 7.1.I(b).

Exponential reserves have no useful properties and are never mentioned again in the text after an example in §7.l

E[u(lO-k 11)) = E[9 - .8k V + .Olk V2) = 9 - .8k V + .Olk V2

E[u(1O - kL») = E[9 - .8kL + .0IkL2)

= 9 - .8EUL) + .01ElkLf

Set these two equal and solve the quadratic in k V: example k = 3,

J = K(x + 3) = K(x) - 3IK(x)23

J 3L Probability

0 1

1.06 - .30360 = .63980 .50

1 ~ - .30360~ = .29998 1.06

.50

ElJL) = .46989, E[3L2) = .24966

Quadratic: 8.62659 = 9 - .8b 11) + .01(3 V2)

.8 ± f.8L4(.37341)(.01) = .8 ± .79061 = .46952 3 V = 2(.01) .02

In example 6.1.1 k I q, = .2 for k = 0, 1, 2, 3, 4, that is, K is the uniform discrete distribution on k = 0, I, ... ,4. Assuming the UDD, T is uniformly distributed 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 UDD :40 here is ~Ao in example 6.1.1, etc.. Exponential reserves are not

worth the effort of calculation.

42 Chapter 7

6. IL = vV - p(A"nj)aUfforO:S U < n - I,andvn-l - P'lI;Hj,U:2: n - I.

V [-]-- - - I-v P P Now IL = v v - p. aUf = v v - P ( ---r;-) = v v I + 0 - o·

[ P] 2 ( I )2 Then Var(,L) = I + 0 Var(vv) = ~ . Var(v u). 6 a"nl

Since Var(v v) = (2AxHn_11 - AXHn-11 2), 2A __ A _2

from (4.2.10), then Var(,L) = Hl'n II x~"n-II (6 . ax:nj)

Note: T - 111>1 = U, the future lifetime of (xH).

E[,L] ["-' -

io aUf o vll I - uPXH

7. UPX+I Jlx+t(u) du + an_II n-IPx+1

In-I ["-'

up,+t aUf 0 + io v" uP.t+t du + an_II n-tPx+t ax+t:n-rl

Var!.,L) I V I [2A- -A 2] 62 Var(v ) = 62 x+l:n-11 - H,.n-II

Note: T - II T>t = U, the future lifetime of (X+I).

20--8. (a) 10 V(A 3S3OI ) = A4S:WT - 20P(A3S:3OI) a4s:iOT

1 -I (b) There are no future premiums, so S V4s:iOT = Aso.5j

9, (a) "0 = -In(P(A,))1(6 + P(Ax»16 from Equation 7.2.5.

(b) 6 = In(1.06), P(A3S ) = .020266 =} Uo = T(35) - 20 = 23.25 years.

10, The minimum loss occurs when U = 100 - (35+1) = 65 - I. Setting the minimum loss to zero is

the same as setting 65 - I equal to -In(P(A3s)I(6 + P(A35»/6 = 23.25 years. Thus 1=41.75

years.

11. Analogous to the development of (7.2.9).

12. Same comment as problem 11. Nothing is done in the text with these densities except to exhibit

another formula.

Chapter 7

14. i~V(A4Q):

15. 10 V(A40:2OT):

(a) Prospective: Aso - 20P(A40) aso'\ol

(b) Retrospective: 20P(A4Q)' S40:iOT - IOk4Q

(c) From the prospective formula, we have

[I - 20P(A ) a~.iOT ] -

40 Aso Aso [ 20P(A4Q)] A-

I-~ so 10P(Aso )

(d) Alternatively, from the prospective,

[ -A~ _ 20P(A4Q)] aso:iOT [IOP(Aso) - 20P(A4Q)] asoWT aso:lol

(a) Prospective: Aso:WT P(A4Q.WT)aso:WT

(b) Retrospective: P(A4Q:WT)' s40:WT IOk4Q

[ P(A4Q:WT) ]-

(c) Analogous to 4(c): I - P(A _) Aso.iOT sO:lol

(d) Analogous to 4(d): [P(Aso:iOT) - P(A4Q.WT)]aso:WT

(e) From (a), since Aso:iOT = I - [, aso:iOT'

we have 1 - [P(A4Q.WT) + 6]aso:WT

(I) From (e), since P(A4Q:WT) + 6

we have I _ ~so: iOT a4Q:WT

(g) From (I), a40:WI. - aso:iOT a4Q:WT

a40:WT

Aso·WT -=- A4Q:WT , since a I - A40.WT

43

I - A ---r;-.

16. Retrospectively, there have been no benefits, so the reserve is just the accumulation of past . 30- - - --

premIUms: 20 Vbo I a 3S) = 30P(30 I a3S )s3S:WT'

_< L. __ ~--

44

17.

Chapter 7

Begin with the retrospective reserve formula: m V (A"m+nl)

Multiply by Px~ = .!"Ex. This produces

P(Ax:m+nl) sx:niT - mkr

. ax:mj

px.~ . m V(Axm+nl) P(Ax•m+nl ) - P(A~.mr)' since mEx . sx.mr ax:iifT - -I

and mEx . mk, = Ax.mr

This establishes (a). It is interpreted as seeing the premium P(Ax•m+nl ) in two pieces:

one which provides the coverage for those m years, P (A:.mr), and the other which provides for

Ihe 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 sx./T' and subtract ,kx

from both sides. This yields

P(Ax:m+nl) . sx:fT - ,kx = P(A:.mr) . ST./[ - ,kT + px.~· SX./[ . m V(Ax•m+nl )

,V(A,m+nl) = ,V(A:.mr) + ,Vx.~ . m V(AT.m+nl)· This establishes (b). Interpretation is totally

parallel to that for (a).

18. The given equation relates to formula (7.3.3). This equation states that the reserve at the begit\ning 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 rearrange the equation to read

20-- - -I --10 V(A 30 ) + 20P(A 30) 0.40.51 A4O.5T + 5E4O • V(A 30),

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.

19. This is totally analogous to Question 14.

20. This is totally analogous to QUestion 15.

21. This is totally analogous to Question 17.

22. Since k V _ I _ aX+k.n-kl x:nl t1x.;JT

! then aHk.n-kl ~ 6' a - 6·

.t:nl

Since" _ .. __ _ ... _ ax.nr ax+2.t.n-lkl x.nl + ax+2k.n-2kl - 2 aHk.n-kl' then a _ + .. _

.t+k:n-kl ax+k:n-kl

Thus "~+lk.n-lkl 2 _ ~ ~. Finall V _ = I _ ax+lk.n-lkl a - 5 5 y, k x+k.n-kl a-

x+k:n-kl x+k:n-kl

2.

I-~ 5

1 s·

Chapter 7

23. Fully Co.ntinuo.us:

a - _ (a) IOV(A 35 .3Of) 1- _45.

201 = .1752905 using al = a(oo)a 1- (3(00), . a

35:ilf .1.11 x n

ax.nr = ax - v" "p.,ii.,+n and the values of ax in the table

Semi-Co.ntinuo.us:

- i i a45 (b) 10 V(A35 ) = 0(10 V35 ) = 0(1 - (35) = .08566

Fully-Discrete:

(c) 10 V:5•3Of = A~5.wr - p:53Ofa45.wr = .03273 using A~5wr = A45 - ,l°20P45A65 = .08846

AI _ '0 I _ 35.301 _ A35 - V 30P35A65 = 004815

P--- - .. 30 ... 35.3°1 0.35.301 0.35 - V 30P35a65

a45.wr = a45 - ,l0 2oP45a65 = 11.575

24. (a) No.. Recall that A,.nr -I I

A,·nr + A,.nr i AI o x.nr i

I i + Axnr 'I 0 A,.nr

25.

(b) Yes. k V(A,)

(c) -I

Yes. k V(A,.nr)

At+k - peA x) . iix+k o ·Ax+k i P .. o' .t - a.t+k

i [Ax+k - px · ax+d = ~. k Vx - I -1

At+k.n-k! - P(A,.nr)· "x+h-kl

lAI_ ipl_. a _ "0 x+k:n-kl '0 x"nl x+k:n-kl

i [A I pl ... _]_ i o x+k.n-k! x.nr O,+k-n_kl - 0 kVI­x:nl

v4) 5 30wr A35j5f

p(4) _ . a(4) _ 30201 35.151

_ _ A30.wr . "l4) A35.151 "l4)_ 35.j5f

5V4)_ - 5V -30.201 30.201

A30.wr . a35·m a30.wr

[a35•m

A30.wr a30.wr

30·2°1

A3O.wr . al4) _ al40~ 35.151

.. (.~201 ] °35.j5f .. (4)

°30.wr

45

46 Chapter 7

25. (continued)

Similarly, 20 V4) £,;4) .. (4) A30 .. (4) 5 30 A35 - 20 30 . a35 ,l5j = A35 - a(4) _ a35,l5j

30.201

20 V4) 20 A30 A30 .. (4) 5 30 - 5 V30 il -' a35 l5j - -::(4)' a35 ,l5j

302[01

,,(4:30,2]01

A a35,l5j _ a35 ,l5j 30 a - .. (4)

30201 a30.2O[

Thus

v4) - v - A-

5 30,20[ 5 30,201 = 30.201 20.;4) 20V A30 5 Vio - 5 30

Note that this is true without any assumption, so it is true under UDD as well.

26 () V (m) A ..!m) .. (m) (..!m) ..!m)) .. (m) . a 15 40 55 1'40 a55 1';5 - 1'40 ass' so yes.

27.

(m) m) .. (m) m) Ass (fI-;') ) (b) IS V 40 A55 ~o ass A55 - fl-40 . [f;m) = 1 - [f;m) Ass, so yes.

55 55

() V(m) ..!m) .. (m) k b . d fi ..

C 15 4{) 1'40 S 4{),l5j - IS 4{), Y retrospectIve e ImtlOn, so yes.

(m) m) .. (m) .• m) .. (m) ii~~) (a) 15 V 4{) A55 - P4{) a55 = I - da55 - fl-4{) a55 # I - .. (m) , so no.

a4{)

Formula (7.7.2) shows that apportionable reserves are equal to fully continuous reserves, regardless of the premium-paying mode, as long as we assume immediate payment of claims.

Thus the given expression is equal to 10 V(A4{) , item (a). Furthermore, all of (c) thru (f) are

expressions for IOV(A40), so they are all correct. Checking out item (b), [p{4}<A55 ) - p{4}(A4{)]

al~} = A55 - p(4)(Al4{)' al~}, the prospective definition of 1504)(A40). Therefore all are correct.

1

I.

I I

2.

I 3.

I 4.

CHAPTERS

(a) The numerator is A, under a constant force assumption and the denominator is a., under the same assumption:

Ax = I)I)(vj+I)(rj)(l - r) = v(l - r)f,0rj

j~ j~O

oc

ax = l:Yrj

j~O

P (I ) I-px q, '* x=V -r=T+T=m

1f = .£':obI1PxJLx(x)dt

.£':0 W, v' rPx dt

(a) From the gamma density equation 1 = Jooo f3"~(a)I"-le-Pt dt it follows that

,B°r(a) = JoOOI"-le-PI dt; r(a) = (a-I)! if a is a positive integer. By constant force

lAx = E[TvTl = J;'te--"p.e-P' dt = JLJo""r-le-PI dt where,B = JL + 6. Thus - - 1 lAx = JL,B2r(2) = JL(JL + 6)2(J!). With constant force we also know ax = JL + 6' Note that

lAx JL(JL + 6)2 .." 3 both of these values are independent of age. So 1f = ~ = 1/( + 6) = JL(JL + 6)

ax JL

(b) ,V = tAxH + lAw - 1faw = t[JL ~ 6] + JL(JL+W - JL(JL+6)3/(JL+6) = t[JL ~ 6]

Notice that if j < h then 0Ch = -1fCh. Thus

Cov(0, Ch) = E[0Chl- E[01E[Chl

= -1fjE[Chl- E[01E[Chl

= -E[Chl(E[Cj + 1fjl)

- [(vbh+1 qx+h - 1fh)hP,] [(vb1+1qaj - 1f)jPx + 1fll

= [(1fh - vbh+1qx+h)hPxl [vb1+lqx+jjP, + 1fjjq,l

using 1 - jPx = jqx in the final step.

J ~

:1i i

48

5.

ChapterS

(a) Using the result in question 4 and the assumptions of the example (bh = 1 for all h,

1r0 = AI _, 1rj = PHI for i = I, 2, ... ), we obtain Cov(Co, Ch) = 0 since the first term in cli .

brackets in the solution to question 4 is (A:.lT

- vqx)oPx = (0)(1) = 0

(b) if 0 <i < h then the expression in problem 4 becomes

Cov(C" Ch) = [(PHI - vq'+hhp,] [vvl qx) + PxVqx)] (c) The second factor is always positive so their covariance is negative if

PHI - vq,+h = PHI - P:+h,lT < 0, i.e., if P,+I < P:+h•lT

6. If i < h ::; /I we see Ulat CjCh = -1rCh similar to question 4. Thus

Cov(£;, Ch) = E[£;Chl- E[qE[Chl

= E[-1rChl- E[qE[Chl

= -(E[Ch])(E[£;1 + 1r)

Now E[£;I = sJ[1rvlq,) -1rV+IP,) after a little work on the £; expression:

{o K=O, ... ,j-I

Cj = ~1rsJ+Tf)-1r K:i 1r K -}+I, ...

The covariance factor E[£;I + 1r is always positive:

E[£;I + 1r = "J[1rvlq,) + 1r(l-,+IP,) = [sJ[vlqx) + j+lqxl1r .

So the covariance expression is negative if E[Chl > O.

0< 1r[Shf(h Iq'}- h+lqxl

{o} "hf > (h+lqx)/(hlq,)

7. Replacing h by h + I, (8.3,9) becomes h V + 1rh

h+1 V· V· PHh = (h V + 1rh) - bh+l' V· qHh

h+IV ~V + 1rh)(I + i)

Px+h bh+1 . q,+h

Px+h

The interpretation is most easily seen if we write it as

bh+1 . V· q,+h + HI V· V· Px+h' Then

(h V + 1rh)(1 + i) = bh+l · qHh + h+l V· PHh

Now the old reserve plus the premium, with interest to the end of the year, is sufficient to provide bh+1 if the policy dies (with probability qHh), or to provide the new reserve if the policy lives

(with probability Px+h)'

Chapter S 49

8. (a)

k~l

k-I \ L,,' ,~,}PX I:vh+lhlq, AI-'" Px - V· qx+h _ h=O h=O _" _ P ~ - V f;o k-hEx+h - vII. kPt vk kP:c - sx.kl x - kE.t - II. t

(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 (hV, + P,)(1 + i) = qHh (1- h+1 V,) + h+1 V" then Px - V· qx+h (I - h+ 1 V,) = v . h+ I Vx - h V,. Thus the given summation becomes k-I L[V'h+lV, - hV,](1 +i)k-h. h=O This is a telescoping series which is easily seen to reduce to k V,. Interpretation: k V, 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 insurance amount

over the reserve.

9. From(8.3.14),1rh_1 = (bh-hV)vqHh-1 + V'hV - h-1V, Here7rh= 7r,andbh =,y,so we have 1r = V· hV - h-I V, or (h-I V+ 7f)(1 + I) = hV, Then with oV = 0, we find

IV = 7f(l +i)

2V [7f(1 + i) + 7f)(1 + i) 7fS2[' etc.

Thus kV 7f'" kf

t an_hi v" h-II q, = ~t (l - rh) I' h-II q, h=1 h=1

to. (a) 1r . ax.Rl = PVB

1 n I I d L (v" h-d qx - v". h-dqx) = d [AlnT - v" nql]

h=1

~ [ I - d a,nT - nEx - v" + v" ,,p,] 1 - v" " " " -d- - ax.nT = anT - ax.nT

Thus 7f = anT.-:- ax.nT aX:n!

(b) From part (a) we see that, at time k, the PVB 1r'" a--a -a

x

+

k

.

n

-

k1

' Thus k V (a- _ " ) n-kl '+k.n-kl' Clearly the PVP is n-kl ax+k.n-kl - 7f' a -x+k.n-kl·

I. i Ii

Ii

50 Chapter 8

11. (8.4.3) says k+' V = bk+1 yl-, I-,qx+k+, + k+1 V· yl-, I-,Px+k+,' Multiplying by sPx+k, we

obtain sPx+k . k+' V bk+J vl-

s s II-sqx+k + k+l V· v1- s Pr+k

bk+l yl-, (qx+k - ,qx+k) + HI V· yl-, Px+k

sPx+k . k+' V + yl-, ,qx+k bk+l yl-'(bk+1 . qx+k + k+1 V· Px+k)

yl-'(k V + 7Tk)(l + I)

(1 + i)'0. V + 7Tk)

Interpretation: The old reserve plus premium, with interest to time s, will provide the reserve at time s if (x + k) has survived to that time, or provide for the then present value of death benefit (bk+1 to be paid at year-end) if (x + k) has died.

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 terminal reserves are easily obtained.

Since r is the fraction of the year beyond the last premium payment point, then (~ - 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.

13. Note: (Ax.4O[) is omitted in the following for simplicity.

14.

(a) 201/2V ~ . 20 V + ~. 21 V + . p

(b) 201/2 V ~ . 20 V + ~. 21 V + 0 (no unearned premium when paid continuously)

(c) 201/2 V 2)

(d) 202/3 V2)

~ . 20 V2) + ~. 21 V2) + 0 (no u.p. since 20~ is premium payment point)

i . 20 V2) + ~. 21 V2) + i· pO) (u.p. for i year)

(e) 201/2 y{2} ~ . 20 y{2} + ~. 21 y{2} + 0 (same reason as (c»

(I) 202(3 y{2} i . 20 y{2) + ~. 21 y{2} + i' p{2} (same as (a»

10 1/6 y{4}(A25 ) "" ~. 10 y{4} + i'" y{4} + fi' P{4} (u.p. for month)

~. IOV(A25 ) + i' ,,11(:425 ) + il P{4}(A25 ),

since apportionable r~serves are equal to fully continuous ones.

IO V(A25 ) 1 - ~35 .0529169; "V(A25

) a25

- - / .. {4} cf4) A25 P{4}(A 25 ) = A25 a25 = --y-' ~ = .0053099. U Q2S

These values produce 10 1/6 y{4} (A25 ) "" .0544801.

a36

a25

.0596409;

Chapter 8 51

15. (a) This is a special case of (b) with k = O.

(b) Var 0.L) = f VU' hPx+dv2(bh+k+l - k+h+l \1)2. Px+k+h . qx+k+h]' h=O

For discrete whole life insurance of I, bk+h+1 = I, so we have

Var(;.L) = f: VU' hPx+k [v2(1 - k+h+1 \1)2. Px+k+h . qx+k+h]

h=O

00 ii.Hh+k+l

[ ( )

2 ] £; VU' hPx+k v2 ~ Px+hH qx+h+k

( )

2 oc iix+k+h+l (h+l) L --a-- v2 hPx+k . Px+h+k . qx+h+k'

h=O x

K

16. First let us review this matter for a whole life policy. The loss is L = bk+1 yHI - L 7Th yh, h=O

where K is the discrete time to death (i.e., death in policy year K + I). Then

Ah { ~ . bh+l - ~ V + 7Th)

v . h+l V - ~ V + 7Th)

if K :':: h if K = h ifK2:h+l.

That is, if death occurs prior to time h, there is no loss allocated to year h. If death occurs in year h, the loss allocated to that year is y . bh+1 - ~ V + 7Th), valued at the beginning of that year with benefit paid out at year-end. If death occurs after year h, the loss allocated to year h is only the reserve increase, valued at the beginning of that year, which is v . h+ I V - (h V + 'ift):

(a) For the life annuity-due, the net single premium is ax, and the loss is L = a K+1I - ax for

death in policy year K. The loss allocated to year h is

{

o Ah = -(ax+h - 1) = -ypx+h ax+h+1

vax+h+l - (ax+h - 1) = vqx+h ax+h+1

ifK:':: h -ifK = h

ifK2::h+l.

That is, if death occurs prior to year h, there is no loss allocated to year h. If death occurs in year h, the loss allocated to year h is the actuarial gain equal to the a.p.v. of the remaining payments, which will now not be paid. The loss is - VPx+h . ax+h+1 = -(ax+h - 1). Remember, if death occurs in year h, the dollar

has already been paid to the annuitant, since it is an annuity-due.

If death occurs after year h, the loss in year h is the dollar annuity payment, plus the reserve "increase," which is negative since it is really a decrease. Thus the total loss for that year is

1 + v· iix+h+l - iix+h = vqx+h iix+h+l-

52 Chapter 8

x

(b) (i) L v h A" h=O

k-l x

L vh A" + yk Ak + L yh Ah

" h=O h=k+1

k-I

= L y" [v. ax+h+1 - aHh + 1] + yk ( - ax+k+ I) {

Ah = 0 for

+ 0 h ~ k + 1, since k is h=O the year of death

k-I k-I "(h+1.. h·· )+" h k k·· ~ v a,+h+1 - y ax+h ~ y + y - y ax+k h=O h=O

(v k iiHk - yO . d,) + ak[ + vk - yk aHk ak+11 ax L.

(ii) E[Ahl [Ah I K = h] . Pr (K = h) + [Ah I K> h] . Pr (K > h)

(a,+h - 1)· IIPx qHh + (v· aHh+1 - d,+h + l)hPx . Px+h

(aHh - 1) IIPx + (ax+h - 1) IIPx . PHh

+ y aX+HI . IIPx . Px+h (iix+h - I)IIPx . Px+h (d,+h - I) hPx + (iiHh - 1) hPx = O.

(iii) Rewrite the expressions for Ah . We see that Ah - aHh for K = h, and, for K > h, . .. h A aHh aHh (1 - PHh) qHh SInce ax+h = vpx+ha,+h+l> we aye h = -- - ax+h = = a,+h . --.

PHh Px+h Px+h

Thus Ah 0 for K < h; - aHh for K = h; aHh . qHh for K > h. PHn

ThenE[A~l o . Pr (K < h) + (a'+h)2. Pr (K = h) + (aH h)2 (qHh) 2 • Pr (K > h) Px+h

2 2 qHh ( )

2

aHh . hPxqHh + ax+h Px+h . IIPx . Px+h·

Using a,+h VPHh iiHh+l , we have

Var(Ah) E[A~l yZ2 .. 2 yZ .. 2 2 PHh aHh+ I . IIPx . qHh + . a,+h+ I . qHh . IIPx . Px+h

yZ .. 2 [ 2 2] aHh+1 . IIPx PHh qHh + qHh Px+h

yZ ii;+h+1 . IIPx . PHh . qx+h [Px+h I qx+h].

I

17. From Ex. 8.3.2, bh = I + h V. Thus ,-1

(a) Var(L) = L ;h IIPx [; (bh+1 - h+1 V)2 PHh qx+h] h=O n-I

L ;h l,px[ ; Px+h qHh] h=O

n-I

L ;(h+l) IIPx PHh qx+h. h=O

Chapter 8

(b) If iLx(t) = .01, then ,p, = e- Ol', q,+, = 1 - e-.Dl for all t. 19

18. (a)

(b)

(c)

(d)

(e)

Var(L) = L ;(h+l) e-OI(h+l) (1 - e- 01)

h=O

; = e=26 (I - e-·ol )(; e- Ol + y4 e- 02 + ... + y40 e-·20 )

[

Note: 1 - e .1

(

- 11 (1- e- ol ) _e_· ___ )( 1 - e-22 )

1 - e .11 .076090

A25 .0067994 20P25 a25201

~~V25 = A44 - 20P25 = .1926071 - .0067994 = .1858077

;~V25 = A45 = .2012024

Var(20L ) = 2A45 - A45 2 = .0680193 - (.2012024)2 = .0275369

Var(lsL) is the sum of the terms in Column I:

Column 1

yZ(l - 19 V)2 P43 q43

+ 0(1 - 20 V)2 2P43q44

+ v6(1 - 21 V)2 3P43q45

+ v8(1 - 22 ~2 4P43q46

Column 2

y4 2P43 . yZ(1 - 21 V)2 P45q45

v4 2P43 . y4(1 - 22 V)2 2P4Sq46

The sum of the terms in Column 2 is v42P43 . Var(20L). Thus

Var(18L ) yZ(1 - 19 V)2 P43q43 + y4(1 - 20 V)2 2P43q44 + 0 2P43· Var(zoL)

(1.0W2 (1 - .1858077h9965573)(.0034427)

+ (1.06)-4 (1 - .2012024)2(.9965573)(.9962930)(.0037070)

+ (1.06)-4 (.992863)(.0275369) = .0255406

19. (a) The rate of change in the reserve is made up of three components: 7f, the instantaneous rate of premium income

, V . (.5 + iLx(t»: the instantaneous rate of reserve increase, due to interest and

survivorship (i.e., reserve forfeiture by those who die) - b, iLx(t): the instantaneous rate of benefit outgo

(b) Again there are three components to the rate of change of the reserve: 7f, the instantaneous rate of premium income

.5 ·,V

-(b, - ,V)iL,(t)

the instantaneous rate of reserve increase due only to interest

(i.e., no growth in reserve due to forfeiture of those who die)

the instantaneous rate of reserve drain to pay cost of insurance

based on net amount at risk

54

d - --20. We know that Iii ,V'" + 0', V - (b, - ,v) J.l.(t). If b, ,v, and

d -'" = 7r, then Iii ,V ,,+ o·,V.

d Recall that ~v = - 0 v. Thus Iii (v· ,v)

d - -v'Iii'V+,V, -0,1.

So if we multiply our equation ~ ,V " + 0', V by V, we obtain

d -v'Iii'V = ,,·v + o·v·,V

d - - d or v· Iii ,V - o· V ,V = 7r V = Iii(V . ,V).

Integrating we have v· ,V = J" v' = -~ v + c

Att=O,wehave 0 =-~ + c,soc=~.

Then v,V = ~ (I - v) "ai[' so,V = 7r,si['

21. Since, V(A)

d Now Iii tPx

Thus !i tPx aXH dt ---

ax

I _ a.::+" then we seek !i tPx_aH , ax dt ax

d- -- tPx J.lx(t) and Iii ax+, = (J.lx(t) + 0) aXH - I.

1- [ ,Px { (J.l.(t)+ 0) £1.+, - I} - aw ' tPxJ.lx(t) j ax I [ - j [I -0 aH ,] =- {;. IPX . aXH - ,Px = - IPX ---- .

~ ~

Since 1- = Ji(:4x) + 0, and a.::+t = I - ,V(Ax )' ax ax

we have -,Px [li(A) + 0 - 6 { 1-,V(Ax)} j = -tPx [P(A.) + 6 . ,V(Axlj.

d 22. (a) Iii ,Px ·,V

d (b) Iii v ,V

d -(c) liivtPx"V

- d -- ,Pxll.(t) ·,V + tPx' Iii ,V

- tPxJ.lx(t)· ,V + tPx [", + (0 + J.lx(t)) .,V - b,J.lx(t)j

tPx [", + 6· ,V - b,J.lx(t)j - d -

- ov·,V+ v'Iii'V

- 6 v "V + V [", + (6 + J.l.(t)) ,V - b,J.l,(t)]

v [", + J.l..(t) ·,V - b,J.lx(t)j d - - d -Iii ,E, ·,V = -,Ex(J.lx(t) + 0) ',V + ,Ex' Iii ,V

- ,Ex (J.lx(t) + 6) ,V + ,Ex ['" + (0 + J.lx(t)) ,V - b,J.lx(t)]

,Ex ('" - b,J.l,(t)) = v ,Px ('" - b,J.lx(t))

Chapter 8 Chapter 8

23. (8.4.3) is k+' V bk+ 1 v1- s l-sQx+k+s + k+l V· \11-.1' l-sPx+k+s

bk+1 yl-' I-,qx+k+, + HI V· yl-' (l - I-,qx+k+')

bk+1 yl-, (I - s) qx+k + k+1 V· yl-'(l - (l - s) qt+k), under Balducci

yl-' [bk+1 (I - s) qx+k + HI V (1 - (I - s)(l - PHk)) j yl-' [bk+1 (l - s) qx+k + HI V (I - s) Px+k + s· k+l V]

yl-' [(I - S)("k + k V)(I + i) + S· HI V]

I roo - 2 1 24. The RHS is =2 Jo (V· ax+,) . tPxJ.lx(t) dt =2 . I.

~ 0 ~

Now 1

so 1

rOO (v ax+i d( - ,Px) dt

Jo 2(v aw ) . d(lax+,) - ,Px

_ tPx (v ax+i 1

00

+ roo tP.· 2(1 ax+,) d(1 aw ) dt o Jo '----v---"

1 aXH J.lx(t) - 1

a~ + 2 [L'" (I ax+/ tPxllx(t) dt - [' v, ,Px aH , dt]"

a~ + 2 [I - f' v, tPx ax+, dr].

2 f' v' tPx ax+, dt - a~ 2 fl. tI ax dt - a~

rOO 1"" 2 2 Jo 1 , y'sPx ds dt - ax

rOO y' _ V' -2 = 2 Jo ~. sPx ds - ax

210'" y'sPx l' 1 dtds

"-.:,..-' as(

2 (- 2- ) li ax - ax -2

- ax'

. .2 aT-aX (- 2-) Then the enllre R.H.S. IS ~ -' -6- - I.

-2 - a x

Substituting ax I - Ax 2- I - 2Ax -2 ---,,-, ax -2-6-' ax

1 - lA., + A,2

62

this last expression for R.H.S. becomes

021£12 ( 2Ax -Ax 2), which is Var [L(Ax)j, which is also given by the L.H.S. Q.E.D.

x

55

56

25.

Chapter 8

Note: J = K - klK2k is the curtate future lifetime of (x+k).

This problem is a generalization of the allocation of loss situation described in the chapter, where an insured is alive k years after issue and has a reserve (savings account) of k V, which can be viewed as a lump sum premium at the time. The insured can subject the fund to a loss if death occurs in the next m years. If the insured survives the m years, it can be settled for the then reserve (savings account) of k+m V and no further loss will occur. The proof proceeds in the same

manner as the proof in the text.

(a) There are two cases: (i) O:s J :S m - 1 and (ii) J 2: m.

m-l ~l m-I

(i) L 0Ak+h L 0 Ak+h + VAk+1 + L 0 Ak+h h~O ,,= 0 1+l

In the first sum, J 2: ,,+ 1 and in the second sum, J :S h - I, so substituting from the definition of Ak+h, we have

In-I

L o Ak+h h~O

1-1

L (v'HI k+lo+1 V - 0 k+h V - 01fk+h) h~O

+ v'+1 bk+1+l - v' kHV - v'1fk+1 + o.

Upon summing and cancelling terms, this last right hand side becomes

1

bk+l+1 - k V - L 0 "k+h, as required. h~O

(ii) In this case, J 2: h + I for all h 0,1 .. ··, m - I. Therefore,

m-I

L 0 Ak+h h~O

m-I

L (0+1 k+h+1 V - 0 k+h V - 0 "k+h)'

h~O

m-I

which easily becomes yn k+m V - k V - L 0 "k+h, as required. h~O

(b) This follows exactly as in the standard situation, from variance of a sum, where the covariance of any pair is zero.

Chapter 8

26. (a) In Example 7.4.4, we have 5-year endowments issued at age 50, for $1000.

1000 P,05[ 170.803

{ 1000 ,1+1 - 170.083 iij+II' j = 0,1,2, Outcomej 2L = , .. = 3,4 .. ·· 1000 v - 170.083 aT[,

0 773.31

I 559.46

2 357.71

2:3 357.71

EIzL] (773.31)(.0069724) + etc. 362.12543

Var(2L) = EIzL2] - (EIzL]f

(773.31)2(.0069724) + etc. - (362.12543j2

132,625.83 - 131,134.83 = 1491.

(b) One-year term variances

h v2(1ooo - 1000· Hh+1 V5o,5[ . P5Hh . q52+h)

o 1187.1419 I 343.8443 2 0.0

Conditional Probability

.0069724

.0075227

.0081170

.9773879

Var(zL) 1187.1419 + (106)-2 (343.8443)P52 + 0 1491

(c) In like manner, Var(,L) 343.84 is found.

(d) Var (4L) = 0, since the same result occurs whether the policy lives or dies. There is no

uncertainty regarding the loss.

27. (a) E[Z] = [375(1) + 375(3)] (362.12) + [250(1) + 250(3)] (561.08)

57

+ [125(1) + (125)(3)] (773.31) = 1,490,915.

(b) Var(Z) = [375(1) + 375(9)] (1491.03) + [250(1) + 250(9)] (343.84) + 0 6,450,962.

Thus SD = 2539.8745

Now if C ~5319~704~15 = 1.645, then c = 1,495,093 which is 1.0028 times the reserve.

(c) Var(ZI) = [375(1) + 375(9)] (1187.14) + [250(1) + 250(9)] (343.84) + 0 = 5,311,375

CI = 1.645 )5,311 ,375 = 3791.14, which is .00254 times the reserve. I .~

I I

58 Chapter 8

(d) E[Z] = 149,091,500

Var(Z) = 645,096,250

c = 14~,091,500 + 1.645)645,096,250 = 149,133,280, which is .00025 times the reserve

Var(Zl) = 531,137,500

C1 = 1.645 )531,137,500 = 37,911.36, which is .00025 times the reserve.

28. (a) E[Z] = [375(1) + 375(3)] (362.12) + [250(1) + 250(3)] (561.08) = 1,104,260

30.

(b) Var(Z) = 6,450,962 as before C = 1.645)6,450,962 + 1,104,260 = 1,108,483, which is 1.00378 times the reserve.

(c) Var(Zl) = [375(1) + 375(9)](1187.14) + [250(1) + 250(9)](343.84) = 5,311,375 as before

C1 = 1.645) 5,311,375 = 3791.14, which is .00343 times the reverse.

(d) E[Z] = 110,426,000

Var(Z) = 645,096,250

C = 110,426,000 + 1.645)645,096,250 = 110,467,780, which is 1.000378 times

the aggregate reserve

Var(Zl)

Since Px:3T

Since 1 Vx:3T

531,137,500

1.645 )531,137,500 = 37,911, which is .00034 times the reverse.

~ 10,000 [i . 10 v{1}(A30 ) + i· 11 vf1}(A30 ) + i' p{l} (A30 ) ]

= 5,000 [1OV(A30) + 11V(.A30) + P{1}(A30 )].

1 d, then iix:3T

1 1 2.083. = - =

Px:3T d .94 .2 -iix:3T + ""3 + T.2

1 iiX+1:2j .66

then iix+l:2j .78 iix:3T 1.625 = - = 3 = = iix:3T

(a) iix:3T = 1 + vPx iix+1:2j' so qx 1 - (1.2i.~2~83) = .2

(b) iiX+l:2j = 1 + VPx+1, so qx+l = 1 - (1.2)(.625) = .25

..

Chapter 8

(c) Ah v2(bh+1 - h+l V) Px+h . qx+h . hPx

Ao = (/2r (3 - .66)2 (.2)(.8) = .6084

(1~2r (3 -1.56)2 (.8)(.75)(.25) = .216

Var (oL) = Ao + v2 Al = .6084 + .69444 (.216)

(d) Var (IL) Al = .216 = .27 Px .8

.7584

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

31. (a) From h V + 7rh = bh+1 vqx+h + h+l Vv(l-qx+h) it follows that:

Ak(K = h) = vbh+I-7rh-h V = vbh+1 - [bh+1 vqx+h + h+l Vv(l-qx+h)]

= (bh+1- h V)v - (bh+l-h V)vqx+h; and

Ak(K = k > h) = -7r + Vh+l V - h V

= vh+l V - [bh+1 vqx+h-h+l Vv(l-qx+h)]

= (bh+1- h+1 V)v[ (1 - qx+h )hPxqx+h - qx+h . h+lPx] '-v---'"

Px+h

So the term in brackets is h+lPx . qx+h - qx+h . h+lPx = 0

(c) Since E[Ah] = 0, Var(Ah) = E[A~]. Now

(*) E[A~] = E[A~ IK < h]Pr(K < h] + E[A~ IK 2 h]Pr(K 2 h)

-------------zero

59

and Ah = bI + c where I is a Bernoulli variable «Pr(I = 1) = Pr(K = hi K 2 h) = qx+h, b = (bh+1 - h+l V)v). Also E[Ah I K 2 h] = 0 so

E[A~ I K 2 h] = Var(Ah I K 2 h) = b2 Var(l) (Bernoulli above)

= «bh+1 - h+l V)v)2(qx+h)(1 - qx+h)

Substituting into (*) above

CHAPTER 9

1. Totally analogous to Example 9.2.1. A series of tedious integration tricks:

(i) (Xi 1 ds = _1_ if n > 1 =? Is(s) = n - 1 is a density on [0,00) Jo (1 + s)n n - 1 (1 + s)n

(ii) r c (JO 1 d d 1 (n - l)(n - 2)

Jo Jo (1 + s + tt t s = (n - l)(n - 2) if n > 2 =? Is. rts,t) = (1 + s + tt is a

joint density for 0 ::; s, t < 00

(iii) From (i), E[(l + S)m] = (n - 1) (:xJ ds n-m = n - 1 1 if m < n + 1. Jo (1 + s) n - m - Thus

E[l + S] = n - 21 = 1 + E[S] =? E[S] = -..L2

. Similarly n- n-

~ =: i = E[(l + S)2] = 1 + 2E[S] + E[S2] can be used with E[S] to find E[S2]

(.) S' '1 1 f ( .. ) E[(l S T)m] (n - l)(n - 2) W' h 2 d E[S] IV lIlli ar y, rom 11, + + = (n _ m _ l)(n _ m _ 2)' It m = an ,

E[S2] from (iii) you can calculate E[S11 for the covariance.

2. 100100 (n - l)(n - 2) dx = 1

s t (l+x+y)n dy (1+s+t)n-2

3. Analogous to Example 9.2.3.

4. (a) Pr(T> n) = nPxy = nPx' nPy, by independence.

(b) Pr [T(x) > nand T(y) ::; n, or T(y) > nand T(x) ::; n] = nPx (1 - nPy) + nPy (1 - nPx) = nPx + nPy - 2· nPx . nPy-

(c) Pr [at least one survives] = 1 - Pr[neither survives]

= 1 - Pr{ max [T(x) , T(y)] ::; n}

1 - nq;:y = nP;:y = nPx + nPy - nPx.,fy

(d) Pr [T ::; n] = nqxy 1 - nPxy = 1 - nPx . nPy

(e) Pr [at least one fails] 1 - Pr[both survive] = 1 - nPx' nPy.

(f) Pr [T(x) ::; nand T(y) ::; n] = nqx' nqy = (1 - nPx) (1 - nPy)

= 1 - nPx - nPy + nPx· nPy·

5. We seek nPx . n-IPy, which is Px . n-IPx+ 1 . n-IPy, or Px' n-IPx+ 1:y

Al . 1 - h - nPy-1 d' / ternatIve y, nPy-I - Py-l' n-IPy, so t at n-IPy - --, pro ucmg nPx:y-l Py-l . Py-l

Chapter 9 61

6. Intuitively, tPxx J.i.xx(t) is the p.d.f. of the R.V. T = T(x,x).

7.

Thus the integral is Pr(T S n) = nqxx. An interesting algebraic approach is to note that J.i.xx(t) = 2IlAt), and tPxx = tPx . tPx· Then the integral becomes

r 1 21n

. d 2 io tPx (tPxJ.i.x(t» dt 2( - 2 tPx ), SInce tPxJ.i.At) = - dt tPx·

Then we have 1 - nP; = 1 - nPxx = nqxx.

1 If T = T(xy), Fr(t) = 1 - ST{x)TIy)(t, t) = 1 - n 2 from problem 2 above

(l+2t) -

(a) F-'t) - F'(t) _ 2(n - 2) )1\ - T - (1 + 2tt-1

(b) 1 Sr(t) = 1 - F(t) = n 2

(1 + 2t) -

{'XJ (CO dt (c) E[T(xy)] = io tpxydt = io (1 + 2t)n-2 (s = 2t, ds = 2dt)

lCO 1 ds 1 1 . = 2 n-2 = -2 --3 (see solutIOn to problem 1)

o (1 + s) n- ..

8. Analogus to the given example and equation (9.3.8).

9. tP;'j 1 - tq~ = 1 - tqx . tqy = 1 - (1 - tPx)(1 - tPy) = 1 - (1 - tPx - tPy + tPx . tPy)

tPx + tPy - tPx· tPy = tPx + tPy - 2 tPx . tPy + tPxy = tPx (1 - tPy) + tPy (1 - tPx) + tPxy·

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

10. Pr [at least one dies in (n + 1)] = 1 - Pr[neither dies in (n + 1)]

1 {I - Pr[x dies in (n + I)]} {1 - Pr[y dies in (n + I)]}

1 - (1 - nPx + n+lPx)(1 - nPy + n+lPy) - 1 - (1 - n I qx - n I qy + n I qx . n I qy)

n I qx + n I qy - n I qx . n I qy n I qxy is the probability that the second death out of x and y occurs in (n + 1), which is not the

same event as above. Algebraically, n I qxy = n I qx + n I qy - n I qxy. Clearly n I qxy i= n I qx . n I qy.

62

11. (a)

(b)

(c)

FT(xy)(t) = rq-xy = rqx rqy = FT(x)(t)FT(y)(t).

problem 1. Now fT(-xy/t) = F~-xy)(t)

Now plug

Chapter 9

in F (t) - 1 1 from T(x) - - (1 + t)n-2

Calculate E[T(xy)] as E[T(x)] + E[T(y)] - E[T(xy)]. Use problems #7 and #1.

density J.t-xy(t) = survival: use results from (a)

12. We seek 3SP40. We note that 2SP2S:S0 = 2SP2S· 2SPSO = SOP2S. Furthermore,

SOP2S lSP2S ·3SP40 = SOP2S, so that 3SP40 = -­

lSP2S

.2 2 = = ~ g.

13. We will need rPx = exp [ - l (1000 - x - S)-1 ds]

t exp [In(IOO-X-S) 1:] = 1 100 -x·

We will also make use of rPxJ.tAt)

1

1 t = 100 -x· Then rP40 = 1- 60 and rPSO -

and rP40J.t40(t) - 60 and rPSOJ.tSO(t) 1

(a)

(b)

lOP40:S0

= 50·

= lOP40· lOPSO = (1 - !~) (1 - ;~) = ~~. ~~ = ~ 50 40 2 29

= lOP40 + lOPSO - lOP40:S0 = 60 + 50 - 3 = 30

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

rP40:S0 J.t40:S0(t) = rP40· rPso(J.t40(t) + J.tso(t»

t 1- 50'

60 - t 50 - t (1 1) 55 - t ="60 ""50 60 - t + 50 _ t = 1500' 0 S t S 50

(Note it is still t S 50, not t S 55). Then

o e40:S0 = E[1]

o (d) e40:S0

000 = e40 + eSO - e40:S0 160

(1 - ;0) dt + lS0

(1 - ;0) dt - 18.06

(60 - (60)2) + (50 _ (50)2) - 18 06 = 36.94 = 120 100'

1 (50 (e) E[T2] = 1500 Jo r(55 - t) dt = 1 [55 1 150 ] . 1500 "3 r3 - 4 t 0 = 486.11111

Then Var(1) = E[T2] = { E[1] } 2

= (486.11) - (18.06)2 = 160.11.

jiiP

Chapter 9

(t) E[T2] = 160 r-160 dt + 150

r-150 dt - 486.1111 i

(60)3 + (50)3 _ 486 11111 = 1547.22222. = 180 150 .

Then Var(]) = 1547.2222

(g) Cov [T(40:50), T(40:50)] =

(36.94)2 = 182.66.

o 0 e40:50 e 40:50

160 t 150 t = (1 - -) dt . (1- -) dt - (18.06)(36.94)

o 60 0 50

(30)(25) - (18.06)(36.94) 82.86.

63

(h) r T(40:50),T(40:50) Cov[T(40:50), T(40:50)]

= = VVar[T(40:50)]' Var[T(40:50)]

82.86 =

y/(160.11)(182.66) .4845.

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

d 0 14. dx exx

d ('Xl = dxJo tPxx dt

= foX! 2tpx [tPx(J.L(x) - J.Lx(t))] dt, from Ex. 20(a), Chapter 3

= 2J.L(x) foco tPxx dt - 2 foco tPx tPxJ.Lx(t) dt

(c.f. Exercise 3)

= J.L(xx) ~xx - 1

15. The probability that both die in year t is t-I!Q30' r-I!Q40 = Cr-IP30 - rP30)(r-IP40 - rP40), ,:xl

and the overall probability is 2:: (r-IP30:40 - tP30 . r-IP40 - tP40' r-IP30 + tP30:40) t=1

00

= L (t-IP30:40 - P30 . r-IP31:40 - P40 . t-IP30:41 + P30:40 . r-IP31:41) t= 1

00 ac

Now L t-IPxy = 1 + L rPxy = 1 + exy. Thus we have r= 1 t= 1

1 + e30:40 - P30 (1 + e31:40) - P40 (1 + e30:41) + P30:40 (1 + e31:41)

64

16. The probability that both will die at age 40 + t last birthday is (1O+tP30 - 1l+tP30)(tP40 - t+lP40),

and the overall probability is

:xJ

L (1O+tP30 -11+tP30)(tP40 - t+lP40) t=O

oc

L (lOP30 . tP40:40 - llP30' tP41:40 - lOP30' P40 . tP40:41 + llP30' P40 . tP41:41) t=O ""----v-'

llP30

2 . UP30 (1 + e40:41) + UP30' P40 (1 + e41:41)

17. For T = T(l,l), we seek Pr(2 < T < 4). Now

and tPxII = exp [- fot (10 - x - S)-l dS] = 1 _ _ t_ lo 10 -x'

63 3645 = = 100 - 10,000

18. (a)

(b) Differentiate the answer above.

1 [ (ea(.OS) _1)(ea(.03) -1)] 19. FT(x)T(y)(5, 5) = CiJn 1 + ea- 1

20. (a) As a --t 0 T(x) and Try) are independent so

(b) If a = 3, from problem 19 sq-xy = .000266

(c) If a = -3, from problem 19 sq-xy = .004232

2655 10,000 - 53112000

Chapter 9

21. In general, am, = au + ay - aUY , so a(xy): liT = axy + a liT - axy:1iT = aliT + n \ axy.

This annuity will pay until the first failure out of x and y, or until time n, whichever Thus, it pays for n years for certain, and beyond that as long as the joint status (xy) survivi

22. This insurance will pay at the death of x, or at time n, whichever is later.

Ax:1iT = Ax + AliT - Ax:IiT' where AliT = v n.

{

V n T < n

Alternatively, let Z = v T T> n

23. COy [vT(,Xy), vT(XY)] _ E[vT(,Xy). vT(XY)] - E[vT(,Xy)]E[vT(XY)]

_ E[v T(x) . v T(y)] - E[V T(,Xy)] E[ v T(xy)]

- E [ v T(x)] E [v T(Y)] - E [v T(,Xy) ] E [v T(XY)] due t; independer

= Ax ·Ay - (Ax +Ay -Axy)

= (Ax - Axy)( Ay - Axy) ;JI''' '""

24. For 0 < t < 20, the annuity will pay if either is alive, since both are under age 20 < t < 25, the annuity will pay only if (25) is alive. By the current payment tee) Chapter 5,

apv = 120

vt

tP25:30 dt + 1:5

vt

tP25 dt

= fo25vttP25 dt + fo20 vt tP30 dt - fo20vttP25:30 dt

25. In this case, the annuity will pay for k = 21,.··, 25 only if (30) is alive, and for k = 2 either is alive.

25 00

Thus apv = L v k kP30 + L v k kP25:30 k=21 k=26

00 00

= L v k kP30 + L v k kP25 k=21 k=26

= 20 \ a30 + 25\ a25 - 25\ a25:30

00

L vk

kP25:30 k=26

66 Chapter 9

26. By current payment techniques,

apv ~ ~ v' [kP" + i kPy (1- kP,) + ~ kPx (I - ,py)]

= n-I k 1 1 1 LV (2 kPy + 3 kPx + 6 kPXY) k=O

27. A payment will be made at time k if (1) x is then alive, (2) y was alive at time k - n, so that time k is during y's lifetime, or within n years after the

death of y, and (3) k:S; m.

m

Then apv = L v k kPx k-nPy

k=1

n m

= ""' v k ,_nx + ""' v k Ln nP sl'nce P ~ K.t' ~ K.t'X k- y, k-n y k=1 k=n+1

m-n

+ ""' j+n h' ax:iiT ~ v j+nPx . jpy, were J j=1

m-n

ax:iiT + nEx L vj

jPx . jpy j=1

a + E·a -x:iiT n x x+n:y:m-nl

k-n

1 for k :s; n

28. The annuity is payable for the following ranges of t, under the stated conditions: t < 5: not at all, since neither is over age 60; 5 < t < 15: only if (55) is alive and (40) is dead; 15 < t < 20: if (55) is alive; t > 20: if either is alive

Then apv = {IS Vl

lP55 (1 - IP40) dt + {20 Vi rP55 dt + (Xl v r rP55:40 dt i5 il5 i20

1°C V r rP55 dt + 1~ v t rP40 dt - 115

V r rP40:55 dt - 1~ v t rP40:55 dt

= 5 I a5S + 20 I a40 20 I a40:55

Chapter 9

29. (a) Payment is at 1 per year while x lives, and P per year if y is alive with x dead. This i~ same as 1 for lifetime of x, plus p for lifetime of y, minus p for the joint lifetime.

(b) Let the initial payment for the joint and survivor be 1, as above. Let the payment rate the life annuity be r Then r. (j(m) = (j(m) + p . (j(m) _ p. (j(m)

. x x Y xy' .. (m)

d th d . d .. 1 ax an e eSlre ratIo IS r = .. (m) + .. (m) .. (m)

ax p . ay - p . axy

30. (a)

(b) USing! tPx = tPx(J.L(x) - J.LAt» and differentiating under the integral sign with re

x gives

31. From (9.8.4) and (9.8.5), we have

J.Lx(S) + j.Ly(s) = 2A + BCS(c' + eY)

and 2J.Lw(s) = 2A + BCS(2eW).

Then c' + eY 2ew, or eX- Y + 1 = 2ew- y, so log (eLl + 1) log 2 + (w-

w-y log (eLl + 1) log 2

log e

68

32. In either case, iiSO:60:lOI = iiSO:6O - IOESO:6O ii60:70'

10 £70 IOE 50:60 = V 10 lOP SO . IOP60 = V 20PSO' (1.06)10 £SO = .4127435

(a) In Example 9.8.1 we found that W = 66.11276 replaces (60:70) Then by uniform seniority, W = 56.11276 replaces (50:60) By interpolation, iiSO:6O = .88724 iiS6 :S6 + .11276 iiS7 :S7 = 10.19419 From Example 9.8.1 we have ii60:70 = 7.55637. Then iisO:6O:

lOI = 10.19419 - (.41274351)(7.55637) = 7.075349

(b) More directly, iisO:60:lOI = 10.19438 - (.41274351)(7.55633) = 7.075554

33. (a) Since (ww) == (xy), then tPww = (tPwf = tPxy = tPx' tPy tPw = (tPx' tpy)1!2, the geometric mean.

(b) Consider the quantity [tP~!2 - tP~!2J 2 > 0, due to squaring.

Then [tPx - 2(tPx' tpy)1!2 + tPy] > 0, so tPx + tPy > 2(tpxy)I!2 = 2(tPww)1!2 = 2· tPw

(c) Since tPx + tPy > 2· tPw, and tPxy = tPww,

then tPx + tPy - tPxy > tPw + tPw - tPww. so tPxy > t~, and aX;; > aww

34. axy = 100

v t tPxy dt = 100

e-t6 . e - J~ /l·x(S) + J1y(s) ds dt

= 100

e-t6 e - J~ 2A + Bc'(c'+cY) ds dt.

Now we let CW = c< + cY , and obtain

axy = 100

e-t6 e-tA e - J~ A+BcW+Sds dt

= 100

e-(HA)t tPw dt = 100

V It tPw dt = a~ ,where 8' = 8 + A

Chapter 9

Further, Axy = 1-8a = xy -I -I -/-/

1 - 8aw = 1 - (8' - A) aw = 1 - 8/ aw + A· aw -I -I

= Aw + A -aw

35. We wish to have axy = aww , which will be true if /l(XY) = /l(ww), or if /lM(x) + /IF(y) = /lM(W) + /IF(w).

Th 3 3bx b -_ 3a + 32bw b en a + 2 + a + y + a + W

4a + b(~ x + Y) 4a + b (~ w) 5 3 2 w = 2 x + y

3 2 W = SX+SY

l

Chapter 9

37. Since qx = qy = 1, then ~Xy = 101

tPx· tPy dt = 101

(1 - t)(l - t) dt

38. If T(xy) were uniform, given T(x) < 1 and T(y) < 1, then this conditional F[T(xy) I T(x) < 1 n T(y) < 1] would be t.

Prob. of the condition

69

-----Now Pr[T(xy) ~ t I T(x) < 1 n T(y) < 1] qx . qy = unconditional probability. We note

39.

that the unconditional probability we desire is the probability that the first death is before t, with both deaths before time 1. This can be written as

t . q . q + t· qy . qx - t· qx . t . qy __ 2t _.:1 ..J. t Then our conditional c.d.f. = ---'=x----"'y'--_~'---=-__ =-_.:...:.. £- "/ qx· qy

1 [i - P .3 -4 ] -1

b = [In(l + i)r1 = "2 +J-~+ ...

{i [1 -. ·2 ·3 -4

H)lf' = (~-J+~-3 + '"

v

Call this part 1.

70 Chapter 9

40. j-I/mPxy - j/mPxy = j-I/mPx' j-I/mPy - j/mPx . j/mPy

( 1 - j ~ 1 . qx) ( 1 - j ~ 1 . qy) - (1 - ~ . qx) ( 1 - ~ . qy )

j - 1 j - 1 (j - 1)2 j j l = 1 - 1i1 . qx - 1i1 . qy + m2 . qx . qy - 1 + m . qx + m . qy - m2 . qx . qy

1 1 l-2j + 1 -l 1 1 m . qx + m' qy + m2 qx . qy + m' qx . qy - m' qx . qy

1( ) m+1-2j = m qx + qy - qx . qy + m2 qx . qy

1 + m+ 1- 2j = m' qxy m2 qx . qy

Now replace x, y with x + k, y + k, and substitute into (9.8.13), obtaining

~ k ~ jim ( 1 m + 1 - 2j ) = ~ v k.Pxy ~ v m . qx+k:y+k + 2 qx+k . qy+k k=O j=I m

00 00 m

= " k a(m) +" vkkPrv qx+k . qy+k " vj/m. m + :r -2j ~ v kPxy' qx+k:y+k IT ~ -J ~ k=O k=O j=I

i 00 [ m . (1 1 2')] 00 . = i(m) L vk+

I kl qxy + L vV/m)-I m + d - J L v k

+I etc.

k=O J=I k=O

i A + [(1 + i) a'!!:.) + i(m) xy 11

i(m)

i(~)Axy + i(~) [1 + ~ - tf:) + ~]L etc. = (9.8.14)

CHAPTER 10

m

J-L(r) (x) = L J-LV)(x) is constant. ThenfT.J(t,)) = Ilv)(x) e-t.~(T) (x)

j=l

(b) h(j) = 100

fS,J(s,J) ds = J-LV'ex) 100

e-s.p.(T)(x) ds

= _ J-LV'ex) e-s.~(T)(x) 100

J-L(r\x) 0

m m (c) fr<t) = L fT.J(t,)) = e-t. p.(x) (T) L J-LV\X) = J-L(r\x) e-t· p'<T)(X)

j=l j=l T and J are independent, since the joint distribution is the product of the marginais:

hAt, J) = fr<t)· fTV)

2. First we obtain ,p~~ ~ exp [ -I,' 503_ s d!-] ~ e~~ t) l

(r) v) (50 - t)3 ( j) j(50 - t)2 (a) fT.J(tJ) = tP50 J-L50(t) = 50 50 - t = (50)3

2 3(50 - t)2

(b) fr<t) = LfT.J(t,)) = (50)3 j=l

150 . 150

..L [ _ ! (50 _ S)315~] (c) h(j) = o fT.J(s, J) ds = ~ (50 - s)2 ds = 50 0 503 3 0

(d) fT.J(t, )) = h.r<.i I t) . fr<t) , so h.r<.i I t) fT.J(tJ) 1 . = fr<t) = 3J

3. (a) fr<t) = fT,At, 1) + fT.J(t,2)

= P(ul+vl)e-(u\+v\)t + (1-p)(u2+v2)e-(u2+v2)t

h(l) = roo fT.J(t, l)dt = ~ + (l-P)U2 Jt=o ul +Vl U2 + v2

h(2) = J!!J..- + (1-p)v2 Ul+Vl U2+ v2

"

=

72

4.

I (1) 3 q65

(r) (r) (r) P65 . P66 . P67

= [1 - qrs>][ 1 - q~][ 1 - q~i] = [1 - (.02 + .05)][1 - (.03 + .06)][1 - (.04 + .07)]

(.93)(.91)(.89) = .753207

(r) (1) 3P65 . q68 = (.753207)(.05) .03766035

J:2) + J:2) + J:2)

Chapter 10

65 66 67 £(r)

165.04 1000 .16504, using the table of Example 10.3.1

65

5. (a) Probability of graduation is 4Pbr) = .3024. Then the number of graduates, G, is a binomial

6.

R. V., with n = 1000, P = .3024.

E[G] = np = 302.4

V(G) = np(1 - p) = 210.95424

(b) Similarly, number of failures, F, is binomial with n = 1000, and

P = 4qg) = .15 + (.60)(.10) + (.60)(.70)(.05) = .231

Then E[F] = np = 231, V(E) = np(1 - p) = 177.639

k

0 1 2 3

(a) fAl)

fA3)

iT) k

1.1) k = ir) (1)

k qk

1000 150 600 60 420 21 336 0

231 total failures

.231 1000 = total others withdrawals = .4666

1000

1.2) k

graduates _ 1000 - 231 - 466.6 1000 1000

(b) fA} I k = 2) . Pr(k = 2) = Pr(termination at k =

ThenfAI I k = 2) = :~~ = .25 and.fJ(2 I k = 2)

Of course .fJ(3 I k = 2) == o.

= ir) (2) k . qk

250 120

63 33.6

466.6

= .3024

2 via mode))

~ = 75 .20 .

Chapter 10

7.

8.

9.

(a) From (10.4.1), e;"' ~ a· exp [-1' J.!{l}(y) + J.!(2)(y) dy] ~ a . exp [ -1' ( 1

= a. exp [ - x + In a ~ x] = (a - x)e-X

(b) ~1) = 101

R.;1tJ.L;1)(t) dt = 101

e-x- t dt = e-x _ e-x- 1

(c) Jx2) = tiT) (2)(t) dt = t (a - x - t) e-

x-

t dt

Jo xH J.Lx Jo - dt - e-x- t

Again iT) , x

(a)

(c)

= - (a - x - t) e-x- t I: - 101

e-x-

t dt

_ (a - x) e-x - (a - x - 1) e-x- 1 - e-x + e-x- 1

= (a - x - 1) e-x - (a - x - 2) e-x- 1

= 1000 exp [- r e+ ~ dY] Jo a-y

= 1000 exp [-ex + In (a - x2) -In aJ

1000 e-cx . a - x"-

=

iT) d x+t dx iT)

X

a

=

R.;1t J.L;T)(t) - J.L(T)(X)

iT) x

d R.x(f) - R.(f)

x+t = dx iT) X

=

[R.;T)f

= _1 [_ iT) I/.(f)(x) iT) x r- + R.;1tJ.L~) (t) + (R.~) - R.~~t) J.L(T) (x) J x

= tp;T) J.L~\t) + tq~) J.L(T) (x) - J.L(f)(x)

d R.(f) - R.(f) x x+t

dt iT) x

1 [(T) (f) ] = iT) R.x+tJ.Lx (t)

x

+ 1 a-y

74

10.

11.

12.

Chapter 10

k = O:qoV) = qo (T) [ 0) 1 1 - exp q~T) log Po

tic (I) -o - 1 exp [ :!~ IOg.60] = .17433

tic (2) -o - 1 exp [ :~~ IOg.60] = .27332

Similarly,

q, (I) _ 1 - 1 exp [ :j~ log. 70] = .11210

q, (2) _ 1 - 1 exp [ :;~ log. 70] = .21163

cI (1) -2 - 1 exp [ :~~ IOg.80] = .05426

cI(2) -2 - 1 exp [ :~~ IOg.80] = .15410

cf (1) - (1) 0 3 - q3

cf (2) - (2) 3 - % = .10

(a) rIx (1) = 1 - p~(1) = 1 - J/J1.(I\t) dt - e 0 x = 1 e-c

II (T) (1\t) dt II (T) d (b) m;l) o tPx J..Lx c 0 tPx t

= II (T) dt = II (T) dt = c o tPx o tPx

(c) q;l) = 101

tp;T) J..L;l)(t) dt = c 101

tp;T) dt

Let each decrement be uniformly distributed. tp;T) J..L;T)(t) = q;T), and

Then the total is so distributed, so

(T) _ q;T) W m = x - r11-t.q(T)dt

Jo x

(b) Likewise, tp;T) J..L~\t) = q~), leading to m~) = 1

~) (e) qxV) = x =

.e(T)

~) m~ x if all decrements are uniform, so q~) = _-:-x_

x L(T) + lefT) , 1 + lm(T) x 2 x 2 x

Assuming a uniform distribution for decrement (;) only, (i. e., in the single decrement model of Chapter 3),

Chapter 10

13.

cf') (c) mlV) = ~

x LV) x R(}) - l cf')

x 2 x

(t) Then m~ (}) [1 - i ifx V)] = ifx (}), so

1 l n1 (}) 2 '1x

ml V) = cI, V) 1 + l ml V) and cI, V) = _----,-x __ [ ]

ml V)

x x 2 x' x 1 + l ml V) 2 x

As shown by (l0.5.3), ifx(}) 2:: q~). N IV) ow mx fx in the single decrement model for =

x

decrement (j), whereas ifx V) dx Since J.L~) (t) 2:: 0, Lx ~ Rx, so m~V) 2:: ifx V) . Rx·

Therefore ml V) , x > ifx(}) 2:: q~)

14. P~o(l) = 1 - ~o(l) = .98, andp~(2) = 1 - ~0(2) = .96.

Then q<;d = 1 - prd = 1 - p~o (1) . p~o (2) = 1 - (.98)(.96) = .0592

(r) (r) m40 .20 2 d (r) 9 I (1) ; (2) -.2.-. pl- (2)

15. (a) q40 = 1 + l (r) = TI = II' (un er UDD), so P40 = II = P40 . P40 - 10 40

2 m40

I (2) 10 1(2) ThenP40 = II' so q40 =

1 II = .09091 1(1)

(b) By assuming UDD in the single decrement tables, J.L~(t) = 1 _q:~ ql(I) - 1 ~1.1t' and 40

1(2)

Then J.L40(r) (t) = J.L4(l0) (t) + J.L40(2) (t) = q40 +.1 I-t. q/(2) 1-.1t

1(2) (2) q40

J.L40 (t) = 1(2) . 1 - t· Q40

rPx(r) = exp - fot q40 +

[

1(2)

io 1 - r· ql;;/

m(r) = 20 x . fol tP~~ J.L~~(t) dt

rl (r) d Jo tP40 t

= fol [.1 + q~~) - .2 q~~) t] dt

fol [1- (.1 + q~~») t + .lq~~) p] dt

1(2) .1 + .9 Q40

7 1(2) .95 - 15 Q40

1(2) 27 .20, so Q40 = 298

40

.0906

75

76

16. Under the assumption, m~)

(I) (I) qo'

m - = .15

1 _ ! q(r) 2 k

.15 -1 _ ! q(r) I .80 o -

1 - Z (.40) 2 0

(1) .10 .11765;

(2) .20 m -

.85 m l = .85 = I -

Chapter J(J

(2)

.1875 (2) % .25

.3125 = mo = = .80 =

1 I (r)

- Z qo

.23529

(I) .05 .05556;

(2) .15 .16667 m(l) = o· (2) - J.Q - 10526 m 2 .90

m 2 .90 = m3 - .95 - . 3 '

17. First we find p;r) = [1 - ct (1)] [1 - ct (2)] [1 - ct (3)]

Th (r) us, P62 = .76048, (r)

q62 = .23952

(r) P63 = .85027,

(r) q63 = .14973

(r) P64 .82115,

(r) q64 .17885

(r)

Then qV) q62 I 'V) 'V) = - .87478166 InP62 62 [Tri' nP62 nP62

(1) q62 = .01767; (2)

q62 .02665; (3)

q62 = .19520

Similarly, (1)

q63 = .02054; (2) q63 = .03193;

(3) q63 = .09726

(1) q64 = .02578;

(2) q64 = .03705; (3)

q64 .11603

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

19. (a) m~ V) ;::::: m~) is justified by the constant force assumption. ,/ (j) V)

A . h I V) V) h '1x qx (b) cceptmg t at mx ;::::: mx ' t en I v);::::: I (r)' if decrements are uniformly 1 - Z . ct 1 - z . qx

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

q' V) [ 1 _ ! . q(r)]

(c) Clearly this leads to q~);::::: x I 2 V; ,or to 1- z 'ct

(d) ct V) [1 - ~ . q;r)] = q.~) [1 - ~ . ct (j) ] , which in turn implies

V) d (j) [ 1 (r) 1 V)] _ V) d (j) _ qx '1x 1 - 2" . qx + 2"' qx - qx ' or '1x - I [ (r) (j)]

1 - z qx - qx

Chapter 10

20. mV) q~)

(1) .02 = 1 1 (r) . E.g., m65 = 1 x

-:2. qx 1 - :2 (.07) .02073, etc.

m'v) q~ v)

'(1) .02052 = 1 - ! . q', v) . E.g., m65

1 - ~ (.02052) x

2 x .02073, etc.

mV) 21. (a) There is no justification for this relationship. qV) ~ ; (r) , assuming UDD ir

x 1+-.m 2 x

mUltiple decrement model.

(b) t i r) dt = L(r) ~ i r) - -21 . ~r) (if (T) is UDD) Jo x+t x x

so L (r) [1 + 1.. m(r)] ~ C(r) and L (r) ~ x 2 x x' x

1 + 1. .m(r) 2 x

c(r) _ 1.. L (r) . m(r) x 2 x x'

(ACCEPTABLE)

(c) We know q;1) = 11 tP;r) J.£;l)(t) dt = 11 tP~ (1) . tP~ (2) J.£;1\t) dt

= 11 [1 - t . ciY)] cfx (1) dt , under the stated assumptions

= q;l) 11 [1 - t . cfx (2)] dt

= q~(l) [1 - ~ . i (2)] (ACCEPTABLE)

22. Ii::::} iii: If tq~) = Kj. tq;r), then 1t sp;r) J.£~\s) ds = Kj ·1t sp~r) J.£;r)(s) ds

Differentiating both sides with respect to t, we find tP~r) J.£~\t) = Kj. tP;r) J.£;r) (t), or J.£~\t) = Kj. J.£;r) (t), as required

Iii ::::} iiil: L.H.S. of (iii) 1 - q', v) - It J1('\s)ds -K-I' J1(T)(s)ds (f .")) = = e 0 x - e 1 0 'x rom (u t x

{ (r)} Kj

tPx = { 1 _ tq;r) } AJ = R.H.S. of (iii)

P/V)

t X = {tP~r)} AJ from (iii);

d ,v) d { (r)}Kj

- dt tPx - dt tPx

P' v)

t x { (r)}Kj

tPx K·-1

1 • (r). (r)(t) tPx J.£x K{ (r)

j tPx

{ (r)}AJ

tPx

10J.£;r)(t), as required

78

Iii => il: If f.L~\t) = ~f.L~r)(t), then tP~r) . f.L~\t) = ~ tP~r) f.L~r)(t) for all t

Thus it sp~r) f.L~\s) ds = ~ it sp~r) f.L~r)(s) ds,

or 'q' (j) = K-. q(r) as required t x :J t X ' •

23. (a) T and J are independent if and only iffr,J(tJ) = h{t)!J(;) , that is, if and only if

Cancel tP~r) from both sides above to obtain the desired result.

-I.' 11(1)(s) ds = e 0 x •

Now replace f.L~l)(s) by ~f.L~r)(s) from part (a) and the result is finished.

25. All results are direct, so no "solution" is illustrated.

26. If each decrement is UD D, then so is the total (1").

and f.L~\1/2)

!{ t· q(j) dt x

1 t (r) - . qx

= (r) , 1 - t· qx

= m(j) as established in Exercise 12. x '

Chapter 10

27. (a) Equation (10.6.3) allows us to express q~3) in terms of the three ct(j), from which ct(3) can

be obtained. Construction of the table from a set of ct (J) has already been explored.

q~l) q~l) (b) Here we would solve for q(l), using c/,x(l) = -----,.,.....:;....:-----,

X 1 - Hq~r) - q~l)] = 1 - Hq~2) + q~3)r [This relationship was developed in Exercise 19.]

Having all q~) establishes the multiple decrement table.

-Chapter 10

28. Only decrement 3 has been changed from Example 10.6.2, so we still have q~~ = .037:

q~d = .01843 from Example 10.6.2. In this problem, however, q~~ = 1.

29.

(3) (1) (2) Thus q69 = 1 - q69 - q69 = .94434.

1000 -+ 800

I 50 ~OO 51

Since ~~ = q;~ . £;~ = 200 withdraw immediately following age 50, only 800 persons bl

interval (50, 51]. Since ~~ = .06 ~~ = 12 die, then elso (1) = 8~0 = .015. Note that, 1

one can withdraw during the year of age, we really have a single decrement (death only) II

30. We seek the probability of termination for cause 1 or 3, or no termination at all (Le., survi age 65).

31. (a)

(b)

cix (f) . One way is to use the relations m~) ~ m~ (j) ~ 1 _ 1.. if, (j) , J 1,2.

2 x 4 m(j)

Then m (T) = '"' m(j) and q(j) ~ x J' 1 2 3 Co er ly s x L...J x' x 1 (T)' = , ,. nv se ,u e j=1 1 + 2:' mx

m,(j)

~ , (j) , j = 3, 4. 1 + 2:' mx

4

i (j) is to produce p;T) from p;T) = IT (1 - i (j)) .

)=1

i (j) by assuming UDD for each single decrement.

= q(T) co y

The advantage of getting all fou

Then the q~) are obtained from

80 Chapter 10

32. Reasoningly, q~) is smaller that the "real rate" (or net probability) because of the operation of the

other decrements. i Ij) is the probability of falling to cause (j), if only it could occur, up to age

x + 1. It is reduced by the probability of falling to (j) before (x + 1) after leaving for another cause. This nets to the probability of falling to (j) while in the group, which is the probability

represented by q~). More formally, the given identity is the case t = 1 of the more general

relation

qlj) _ rI Ij) '" lot (r) (k) 11j) ds t x t'1x + ~ sPx J..Lx (s) t-sqx+s

k#j 0 0.

Let F(t) denote the LHS of this equation. The student can verify that

which can be seen to reduce to J..L~) (t) . - F(t), so that J..L~) (t) F(t) + F' (t) = 0, for t ~ 0. It is also

clear that F(O) = 0. Thus we have an initial value problem in differential equations, whose unique solution is F(t) = 0, for t ~ 0, which establishes the desired relationship.

33. (a) (i) tP~) ::; tp~Ij), for 0< t < 1, and equality cannot hold throughout (0,1)

due to the positive forces of decrement. Thus we fmd

w(r)(o) = 1 > 1 = wlj) (0) r1 (r)dt r1 IIj)dt

Jo tPx Jo tPx

(1·1·) 1(1) l(j-l) 101

11j) dt l(j+l) I(m) Px ..... Px . tPx . Px ..... Px o

(iii)

< 101 tp~Ij)· .... tp~lj) ..... tP~(m) dt,

m since tP~(k) decreases on (0, 1). Recalling that tp;T) = II tp~(I),

i=l (r) 11j)

we have w(r)(1) = Px <];1 Px,lj) = wlj\l). r 1 (r) dt d

JO ~x 0 ~x t

W(T)(t) Let F(t) = ~. Then F(O) > 1 and F(1) < 1 by parts (i) and (ii). Also F is

w (t)

continuous and decreasing (since F(t) = k· II tp~(I)). i#j

Thus, by the Intermediate Value Theorem, there is a unique r E (0,1) with F(r) = 1, i.e., w(T)(r) = wlj\r).

Chapter 10

(b) From part (a), the graphs of W(T) and wV) are shown below, and we want to show th W(T)(t)

1

o r 1

shaded areas are the same. But this is equiva

fo1 [W(T)(t) - wV\t)] dt = 0, which is ob

t true.

(c) This is an immediate application of the M.V.T. for integrals.

(Editor'S Note: These solutions to Exercises 32 and 33 were contributed by Davi Mathematics Department, University of Bridgeport.)

34. First we will need some pieces:

Thus roo a-1 -(3s ds

(T) Jt s e tPx = roo a-1 -(3s ds

Jo s e

(a) Now ir.J(t, j) tP~T) /L~\t)

= L 100

sa-1 e-(3s ds. r(a) t

= e j3a ~-1 -(3t fOf]' = 1 r(a) e,

= (1 - e)j3a ra-1 e-(3t for)' = 2. Next, r(a) ,

Then w~;l ?,-••

fA)) = f(sJ) ds = -- sa-1 e-(3s ds = --·13-0. r(a) = e, 100 e j3a 100 e j3a o r(a) 0 (a)

2

Likewise, h(J) = 1 - e, fori = 2. Finally,f:r(t) = L fT.J(tJ) j=l

(b) E[1] = fooo t· h{t) dt = rq:) fooo ro e-(3t dt

= L roo ~-1 e-(3t dt where ~ = a + 1 r(a) Jo '

= Lj3-~ f(~) = Lj3-o.-1 r(a + 1) r(a) r(a)

£ = r(a) a r(a) =

82

Similarly, E[T2] = 1"00 ? fr(t)dt = r\:) 100

fk+l e-(3t dt

Then Var(1) =

L roo f+l e-(3t dt where 8 = a + 2 r(a) Jo '

= L (3-8 r(8) = L (3-a-2 r(a + 2) r(a) r(a)

= r\:)(3-a-2 (a + 1) r (a + 1)

E[T2] - {E[1] r a(a,6~ 1)

= p(:) (a + 1)(a) r(a)

(~r = %2

Chapter 10

a(a + 1) = 2 ,6

(c) In general jj(j! T = t) = f.£~ I f.£<;) (t). From the given formulas for the forces we see that

jj(l!T= t) = e, jj(2!T= t) = 1- e. In (a) we saw jj(1) = e, fA2) = 1- e. Since the

marginal and conditional distributions are identical T and J are independent.

CHAPTER 13

1. (a) c - W = ± 1 so Un = Un- 1 ± 1. Since u is an integer we see that if ruin occurs at· then Un-l must have been 0 so that Un = -1.

(b) ;j;(u) = expS -!-u) = exp(-~u) = e-R(u+l) E[exp(R)IT < 00] exp(R)

(c) w-c

(d)

0-1 P 2-1 q

~ qe2r - er + p = 0, a quadratic in er •

Hence e _ 1 ± !1-4pq

- 2q , which is < 1 when you select the - option. Now r > 0 so er > 1 means

r _ 1 + ! 1-4pq h "'R _ I (1 + ! 1-4Pq ) e - 2q ,ence - n 2q

'IjJ(u) = e-R(u+l) = q _ _ ( 2 ) u+l

1 + !1-4pq

2. (a) Sn,m = W n+1 + Wn+2 + ... + Wn+m f

= (Yn+1+aWn) + (Yn+2+aYn+l+a2Wn) + ... + (Yn+m+aYn+m-l+ .. + am-;-)¥~~

m m-i m

= LYn+iLaj + WnLa

j,

i=l j=O j=l

or t ( 1 -1 a::i+l) Yn+i + ( a "1 ~:+l) Wn by evaluating the sums of powers of a.

(b) use lim d' = 0 since -1 < a < 1

(c) Take the expected value of both sides in (a). Minor algebra results in the desired relati<

84 Chapter 13

3. Pr(claim in [t,t + dt] I N(t) = i, Ti = S, S < t)

time

= Pr(V;+1 E [t-s,t-s+dt] I V;+I > t - s)

= Pr(Vi+1 E [t-s,t-s+dtD/Pr(V;+1 > t - s)

= f(t-s)dt 1 - F(t-s)

• o

no claim ,.-----''''----.,

• s = Ti

(ithclaim)

• • t t+dt

The infonnation says Ti+1 E [t,Hdt] given Ti+1 > t. Then Vi = Ti+1 - T; = Ti+1 - s since it is given Ti = s. Hence

{T;+I E[t,Hdt]ITi+1 >t} = {Vi+1 E[t-s,t-s+dt]IVi+1 >t-s}.

4. Since N(t) is Poisson with parameter At

Pn(t) = Pr(N(t) = n) = e-At(At)n In!

(a) ]Jo(t) = e-At is the probability of no claims in [0, t]

p~(t) = _Ae-At (i.e., decreases proportional to Po(t»

(b) p~(t) = [-Ae-At(At)n + e-Atn(At)n-1 . A]/n!

Note:

= -APn(t) + APn-l(t) = -A(Pn(t) - Pn-I(t»

(i.e., decreases proportional to the difference)

1 +0 = ,c means that as 0 - 0 we must have c - API and as 0 - 00 we must have "PI

c- 00.

5. These limits can be reasoned from the graph below

1 + (l+O)p,r

R 'Y

As c - API one has 1 +0 = ,c - 1, that is, the slope of the line approaches PI, which is also "PI

Mx(O). The point of intersection slides toward the left to a limiting position of (0,1). That means R - O. On the other hand c - 00 means e - 00. The slope of the line increases and the intersection point moves upward and to the right. That means R - 'Y.

Chapter 13

6. If R is the adjustment coefficient then

1 + (l+8)PIR = Mx(R) = 1 + PIR + pzR2 /2! + .... The last part follows from Taylor's theorem and the fact that Mf)(O) = E[Xk] = Pk. S X ~ 0 each Pk ~ 0, and we know R > 0 so

1 + (1 +8)pI R 1 + PI R + pzR2 /2 + (pos. terms)

=* 1 + (l+8)PIR > 1 + PIR + pzR2 /2

7. The given density

p(x) = !(3e-3X ) + !(7e-7X ) x> 0

is a weighted average of exponentials with parameters 3 and 7. Thus

Mx(t) = ! (3~t) +! (6) ~ '-v-'

and'Y = min{3,7} = 3.

The equation for R,

t<3 t<7

1+"(l+,,2/5),,,(!·1+!·+),R = J~+!~, ' 1+8 Pi M;(R)

is equivalent (common denominators, then cross multiply) to

126 - 30R = 126 - lSR - 14R2 + 2R3 or

o = R· 2· (R2 -7R + 6) = 2R(R - l)(R - 6). ,~-1P-

But 0 < R < 'Y so obviously R = 1 is the choice of the 3 roots of the above equation.

S. P(1) = .25, p(2) = .75 means PI = E[X] = l.75 and

Mx(t) = E[etx] = Letxp(x) = .25et + .75e2t . x

R = log(2) satisfies

1 + (l+8)(l.75)R = .25eR + .75e2R = (.25)(2) + (.75)(4)

so

86 Chapter 13

9. R is the solution for 0 < r < "f of

I + Xr = 100

eTX

. p(x)dx

since the integraris E[eTX ] = Mx(r). Use integration by parts with u = eTX , dv = p(x)dx and

du = reTxdx, v = P(x) - 1 = -(1 - P(x»

to rewrite the integral as

The first term above is

-limeT(x)(1-P(x» + eO. (1-0) = -0 + 1 x-oo

Substituting into the first equation gives

1 + Xr = 1 + r 100

eTX(l-P(x»dx,

or, after rearrangement,

10. Ifu = 0 then U(T)IT<oo is -L1• Thus -RO

1jJ(0) = E[e~R(-LI)] =

Now (Mx(t) - l)/Plt ML1(t) and

1 + (l+O)PIR = Mx(R)

is equivalent to

1+0

Substituting into the above gives

1jJ(0) = ML~ (R)

as desired.

1 - 1+0

11. (a) If ruin occurs surplus must drop below the initial level and (perhaps at the first time this happens) eventually also drops below zero. If LI = Y < u then u - Y > 0 and the subse­quent probability of ruin is 1jJ(u - y). On the other hand, if LI = y > u then ruin has already occurred. Thus 1jJ( u) is given by

(1!0) i:o,-#i(1-P(Y»dY," ~ + P-----b d P (L E v[ +d]) ruin occurs ro . rops r 1 Y,Y Y at later time

, below u #

ruin occu~s after the

\, (l~O\Jy=~ "k(l-P(Y»dY#

Prob. drops Pr(Lv> u) \, below u .--_1 ___ -'

ruin occ~s at first first record low record low

where 1 - .d. (1 + 8)pl - C"

Chapter 13

11. (continued)

12.

(b) Using the standard calculus result

d [l b(U) 1 l b

(U) du f(u,y)dy = (Juf(u,y»)dy + f(u,b(u»· b'(u) - f(u,a(u»· a'( a(u) a(u)

we have

'lj;'(u) = ~ [lu

[l - P(y)lzP'(u - y)dy + [l-P(U)lzP(O)] - ~[l-P(u)]

= ~[-[l-P(Y)lzP(U-Y)I: -lu

p(y)'lj;(U- y )dY +[l-P(U)]'lj;(O)] - ~[1

= ~ [-[1 -P(u)]'lj;(O) + ~~'lj;(U) -lu 'lj;(u - y)P(y)dy + [l-P(u)]

AI

= ~['lj;(U)-lu'lj;(U-Y)P(Y)dY-[l-P(U)]]. - c!

(Note eror in textbook; the inside ~ should be deleted.)

ML1(t) = Mx(t)-l (1+Ptt+P2t2/2!+P3t3/3!+··)-1 ~t - ~t

= 1+(~)t+(~)t2/2!+ .. .

( = I) 1 + E[Ldt + E[LDt2/2! + .. . genera

=? E[LtJ = ~,E[LD = ~

Var(L t ) = ..a. _ (A)2 3pt 2pt

13. (a) For a negative binomial with parameters r,p

E[N] = '0lP

' Var(N) = '0l p2 .

The case r = 1 is the geometric. The number of record lows, N, is geom

q = 'lj;(O) = I! 8 (i.e., "failure" means setting another record low which is the sa

with 0 initial capital):

E[N] = ~ = (I! 8) / (I! e) = b Var(N) = ~ = 17;8

p 8

(b) L = L t + ... + LN where each Li is distributed as L t and the Li are independeJ from basic facts about random sums

E[L] = E[LdE[N] = (~) (b) Var(L) = E[N]Var(L t ) + (E[Ld)2Var(N) = (b) (3~t -(~ r) + (~)

88

14. In general

If X = 2 then PI = 2, Mx(t) = e2t . Substitute into the above.

15. (a) ¢(O) = .3e-o + .2e-o + .le-o = .6 and in general ¢(O) = I! o. Thus 0

(b) When X is distributed like a mixture of exponentials and n

¢(u) = 2:= Cie-riu where rl < r2 < ... ;=1

the text shows R = rl. Here it means R = 2 since rl = 2, r2 = 4, r3 = 7

n

Chapter 13

16. If p(x) = 2:=Ai/J;e-,8iX, a weighted average of exponentials, then moments about the origin of X i=1

are similar weighted averages: n

E[X] = 2:=Ai . J. i=1 Z

Note: Ji = E[X;J ifpi(x) = (Jie-,8iX.

17. (a) p(x) = te- 3X + ~6 e-6x = ~(3e-3X) + ~~ (6e-6x)

~ E[X] = ~(t) + ~(i) = ~~ = -f, = PI

(b) 1 + 0 - c - 1 _ 27 - ~ - 3(5/27) - 15

(c) Mx(r) = ~ (3~r) + ~ (6~r) d 1 O[Mx(r) - 1]

( ) 1 +0 . 1 +(1+0)Plr-Mx (r)

0 -.u-_1 - 15 5

54 - 17r 3(3 - r)(6 - r)

= 15 . 4/5(lOr - 3r2 )/3(3 - r)(6 - r) 27 (r3 - 6r2 + 8r)/3(3 - r)(6 - r)

= 4 r(10-3r) 4 [-1 -2] 9 r(r - 2)(r - 4) = 9· r-4 + r-2

= 1._4_+1._2_ 9 4-r 9 2-r

(e) above and (13.6.12) and (13.6.13) imply

¢(u) = ~e-4u + ~e-2U

18. (a) The given density is a gamma with a = 2 and {J = 3/5 so PI = E[X] = a/ (J = 10/3

(b) 1+0 = A~l = 1. (\~/3) = 3 ~ 0 = 2

(c) Mx(t) = (1 - t/ {J)-a = (1 - 5t/3)-2 = (3! 5ti

""'" 1

I !

\

I

Chapter 13

18. (continued)

(d) 1 e[Mx(r) - 1] _ 2 (30r - 25r2)

1 + e 1 + (1+e)Plr - Mx(r) - 3" . 5r(5r - 4)(10r - 3)

2 (6 - 5r) = 3"' (5r-4)(10r-3)

_ 2 [ 2/5 9/5] _ 1 ( .8) 2 ( . - 3" 5r-4 - 10r-3 - -15 .8-r + 3' 3' (e) 'Ij;(u) = - 115 e-·8u + ~e-.3U

19. (a) L = 0 means surplus never drops below its initial level. Thus

Pr(L=O) = 1 - (l!e) = (l!e) and ~ = Pr(L=O)

The given distribution function relation means similar linear relations for moments origin:

E[L] = l!e' 0 + l!e . ~ = (1:e),8

E[L2] = l!e .02 + l!e [~ + (~r] =

(b) 'Ij;(u) = Pr(L > u) = 1 - Pr(L ~ u)

~ 1- (I(u)l!e +G(u:a,,8)I!e)

1- I!o- G(u: a,/l)lio (f(U) = { ~ = l!e P - G(u: a,,8)]

20. (a) If p(x) = I:A;,8;e-/3;X then P(x) = I:A;(1 - e-/3;X). Hence

1 - P(y) = 1 - I:A;(1 - e-/3;X) = 1 - (I:A;- I:A;e-/3;X)

a(a+l) (I +e),82

u ~ 0) u<O

= 1 - 1 + D;e-,B,x = D;e-,B,x. Finally,

!L1(Y) = J...[1 - P(y)] = 1 . I:A;e-,B;X = t ( :;1,8; ) ,8;e-f3/x , PI 2:,A;I ,8;_ "'A ./,8.

I-I L.. J '] j=1

a weighted average of the same exponentials with new weights.

(b) New weights: AI/(b A2/(h I:Aj/,8j' I:A),8j' ...

(c) E[Ld = t ( A;I,8; ) l . ;=1 I:A),8j ,81

=

n

I:A;I(3; ;=1

n

I:A),8j j=1

90 Chapter 13

21. (a) p(x) = ~e-3X + ie-7X = ~(3e-3X) + ~(7e-7X), a weighted average of two exponentials with

A -A - 1 . bl 20· 4L - 3 A., - 7 ",di - 5 ] - 2 - 2 as In pro em . f3] - 2' f32 - 2' L.....t f3i -

iL,(Y) = ?0(3e-3X ) + ?O(7e-7x)

(b) E[L]] = ?O . t + ?O . ~ = ?O (c) 2 _ 3 2 7 2

E[Ld - TO . f3? + 10 . f3i _ 2 2 _ 200 - 30 + 70 - 2100

_ 2 1 _ 29 Var(L]) - 21 - 25 - (21)(25)

22. (Un+I1Un=x) = x + Gn+] so ifu(w) = _e-nw

E[u(x+Gn+])] = E[_e-n(x+Gn+,)] = -e-nxMG(-a.).

Thus E[u(Un+])IUn=x] 2: u(x) is equivalent to

which in turn is equivalent to M G ( -a.) :::; 1

The graph below and MaC - R) = 1 finish the proof

MG(-r)

r

23. (a) If A claims are expected in 1 unit of time then X = Af claims are expected in f units of time. Similarlyc = cf and :;j;(u,t) = 'lj;(u,ft)

(b) Af = 1 ::::} f = 1/ A

24. Since u = 0, U(T) = -L I • Since claims are uniform on (0,10) we have

p(x) = 110 , P(x) = to for 0 :::; x :::; 10,

hence

iL,(Y) =! [l-W] forO:::;y:::; 10.

rio (400/12) E[Ld can be computed as Jo ~(1 - fo)dy = 130 from the L]-density or as P2 /2pI = 2.5

from the X-density. Finally

E[U(T - )IT < 00] = E[X] - E[Ld = 5 - 130 = j.

CHAPTER 14

r-.J _ O-§a r-.J ~ '0 ~ 16. Let 0 (3 = 1 so that R = -rv-. The max occurs at a = ° if -I-a' -a 1+0

~ > '073 _ ~ . O-§a . 1 - a (all a) 1 +0 - 1 +'0 - I-a I-a (l-a)+O-~a

..<-->.. _0 _ > 0 - § a (II ) r7 1 + 0 - (1-a)[(1-a) + (O-~a)] a a

Multiply out and arrange terms to arrive at the inequality

a 2:: 20 is (all a in [0, 1]).

Thus ~ 2:: 20 will insure that R is maximal when a = °

17. With no reinsurance (i.e. a = 0) E[L] = O~' With proportional reinsurance '0

r-.J(3 - (3 d E[L] - 1 - (1 - ai Th .. - ° 'f - 1 _ a an - '0 ~ - (0 _ ~a)j3' e mmlmum occurs at a-I

_1 _ (1- ai 0(3 - (0 - ~a)(3

The inequality is equivalent to

(all a in [0, 1]).

20i S :S a (all a in [0, 1].

Thus ~ ~ 20 is the desired relation.

18. Since X is exponential with parameter 1

=> c' = retained premium = (1 +O)E[.X] - (1 +OE[I,6]

= (l +0) - (1 +Oe-,6

lJ' ( lJ ( ,6/ ,6 ret. premo => 1 + u = 1 + u - 1 +Oe- (l-e-) = E[ret. claim]

The final ingredient needed for the equation determining the new adjustment coeffici moment generating function of X' = X - I,6(X), the retained claim amount:

X' = {x x < (3 (3 x 2:: (3

92 Chapter 14

The equation for.the new adjustment coefficient is

or l_e-(I-r)fJ -(I-r)fJ

1 + e . -r

()' (l+()-(1+0e- fJ _ 1 = ()-Se- fJ

1 - e- fJ 1 - e-fJ

19. (a) With a normal W we saw in Chapter 13 that R = 2(c -; fl). We need only adapt this to . (J

(b)

retained claims since (1 - a)W is normal with mean (1 - a)E[W] = (1 - a)10 and variance

(l - aiVar(W) = (l - ai4. The retained premium is

1.25E[W] - 1.40E[aW] = 12.50 - 14a.

Thus, after reinsurance, the new adjustment coeffiecient is

rv ,

dR da

rv ,

R 2((l2.50-14a) - (l-a)10) = 4(1 - ai

o ::::} 0 = numerator

= (1-ai(-2)+(1.25-2a)2(1-a)

= 2(1 - a)[-1 + a + 1.25 - 2aJ

= 2(1 - a)(.25 - a) ::::} a = .25

1.25 - 2a (l-ai