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Actuarial Mathematics :It is the solution manual
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 . 26' . 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 xI + 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 yxPx dy
which gives ax = 1U:x v t tPx dt
= 1U:x v t JU:x JLAs) sPx ds dt
= 1U:x 1U:x v t JLx(s) sPx dt ds
= 1U:x (iST sPxJLx(s) ds
(b) This time multiplication by eoy , and rearrangement, yields
29
oy d  oy 1:  oy (Y) oy d ( oy ) oy (Y) _ eoy. e . a  e . u a = e . JL a  e or  e . a = e . JL ay dyy y Y' dy y
Integrating over (x, w) yields eoy . (iy [ = [U: eoy . JL(y)(iy dy  ["" eoy dy,
[ "" and (i = (i  eo(yx) (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(1vPx)
 vpx(ax+1:iil ) + anr(1vPx)
 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  eP,t, hence
1 Fy{y) = Fy{ In(l8y)/8) = 1  (18y)pJ6 for 0 ~ y ~ "5.
(b) If Y corresponds to the n year, continuous temporary annuity then
{ (lVT) 
Pr  < Y 0 < Y < aFy{y) = Pr(Y ~ y) = 8  _  nl
1 a~ ~y
= {Fy( In(18y)8) ~ ~ y < a~ 1 a~ ~y
= { 1  (l8y)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 enp . If 0 <y < 0/8, then t = In(0  oy)/o and F(y) = Pr(Y ~ y) = Pr(T ~ t) = tqx = 1  ep,t = 1  (08y)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(1oy)/8
Fy{y) = 0 ify < a~
Fy{y) = nqx = 1  ep,n if y = a~
Fy(y) = Pr(Y ~ y) = Pr(T ~ t) = tqx
 1 = 1  ept = 1  (l8y)pJb if a~ < y < "5
T
32 ChapterS
nI nI 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) IAx IA 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
(
IVK
(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 + (anrjanr+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+OIIlm  I] + ; [(1+0 12/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:yx! x.lj x x+ l.y(x+ I) I
(b) a~"D = ax.l[a(m)  (3(m)(Ivpx)
a(m)  (3(m)(Ivpx) (ax.1[ = I)
a(m)  (3(m) + (3(m)vpx
17. ..(m) •. mI ax ~a;c 2m
( mI)
II I ti~m) = V' nP.:Ji~;'n ~ vii nPx iix+n  2m
 I··(m)  v"nP (mI)
 n ax x 2m
Subtracting these two relations gives
a~~~ "" (ax n lax)  (~:nl) (Iv" 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) mco
(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:
moo
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
mI _ ax + 1  "2iil  m
mI ax + "2iil
(c) .Ia~m) (m) ( m  I) flEx ax+n ::::::: nEx ax+n + "2iil
mI .Iax + "2iil.Ex
b (m) _ (m) I (m)_ mI I mI E () aX
:nJ  ax  n ax ....... ax + "2iil  n ax  "2iil n x mI
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<nI,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'nI1
sum would supply nothing.
y= {(Dii)K+(J+I)/m l O:S K < nI, 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
(
nI ) (n+l)ii(~  '" klii(m~
x:n1 ~ x:nkl k~O
(problem 23(b))
nI
o.~~ + I:( ii~~  klii~7Lkl) k~
nI ii(m) + '" ii(m)
x:nj" ~ x'nk] (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»
nl
= I> I ii:mLkl + n(n I ii;m» (See 23(b)) k~O
nI nl _ " ( I .. (m) + I .. (m») " I(m)  ~ k a,t:n_kl n ax  Lt n ax
k~O k~
Chapter 5
aT. 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(IOL(P))2]
= E[10  L(P)  .0I(lOO20L(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 eAoAo 2) + (AoPiio)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, 2AOAo2 = .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[AoPiio] .01 [( l+~reAoAo 2) + (AoPiio)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 3SV . IOP3SA4S + V . IOP3S
A  '/6 [A 10 A 1 10 P(A _) = .. 3S, IO I = I 3Sv . 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[ = (A3SvlO . IOP304S) + vlO . IOP3S.
11.20
pl __ pldOl ,,201
AIAIdOl ,,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)
~ lvr k=O
( I _ ~) (.!..±i) _ (_i_) (.!..±i) I+ir i  I+ir i
I  r
+ir
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 onehalf 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(lOk 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,andvnl  P'lI;Hj,U:2: n  I.
V []   Iv 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  AXHn11 2), 2A __ A _2
from (4.2.10), then Var(,L) = Hl'n II x~"nII (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 nIPx+1
InI ["'
up,+t aUf 0 + io v" uP.t+t du + an_II ntPx+t ax+t:nrl
Var!.,L) I V I [2A A 2] 62 Var(v ) = 62 x+l:n11  H,.nII
Note: T  II T>t = U, the future lifetime of (X+I).
208. (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.nkl x:nl t1x.;JT
! then aHk.nkl ~ 6' a  6·
.t:nl
Since" _ .. __ _ ... _ ax.nr ax+2.t.nlkl x.nl + ax+2k.n2kl  2 aHk.nkl' then a _ + .. _
.t+k:nkl ax+k:nkl
Thus "~+lk.nlkl 2 _ ~ ~. Finall V _ = I _ ax+lk.nlkl a  5 5 y, k x+k.nkl a
x+k:nkl x+k:nkl
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
SemiCo.ntinuo.us:
 i i a45 (b) 10 V(A35 ) = 0(10 V35 ) = 0(1  (35) = .08566
FullyDiscrete:
(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.nk!  P(A,.nr)· "x+hkl
lAI_ ipl_. a _ "0 x+k:nkl '0 x"nl x+k:nkl
i [A I pl ... _]_ i o x+k.nk! x.nr O,+kn_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  fl40 . [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  fl4{) a55 # I  .. (m) , so no.
a4{)
Formula (7.7.2) shows that apportionable reserves are equal to fully continuous reserves, regardless of the premiumpaying 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 ) Ipx 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"lePt dt it follows that
,B°r(a) = JoOOI"lePI dt; r(a) = (aI)! if a is a positive integer. By constant force
lAx = E[TvTl = J;'te"p.eP' dt = JLJo""rlePI 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, ... ,jI
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
kI \ L,,' ,~,}PX I:vh+lhlq, AI'" Px  V· qx+h _ h=O h=O _" _ P ~  V f;o khEx+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 kI L[V'h+lV,  hV,](1 +i)kh. 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 = (bhhV)vqHh1 + V'hV  h1V, Here7rh= 7r,andbh =,y,so we have 1r = V· hV  hI V, or (hI 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" hII q, = ~t (l  rh) I' hII q, h=1 h=1
to. (a) 1r . ax.Rl = PVB
1 n I I d L (v" hd qx  v". hdqx) = 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'" aa a
x
+
k
.
n

k1
' Thus k V (a _ " ) nkl '+k.nkl' Clearly the PVP is nkl ax+k.nkl  7f' a x+k.nkl·
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 IIsqx+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 yearend) if (x + k) has died.
12. Interpretation for both (a) and (b): The reserve is sought at a duration between two consecutive premiumpayment 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 yearend. 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 annuitydue, 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 annuitydue.
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
kl x
L vh A" + yk Ak + L yh Ah
" h=O h=k+1
kI
= 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
kI kI "(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 nI
L ;h l,px[ ; Px+h qHh] h=O
nI
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) eOI(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  e22 )
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 lsQx+k+s + k+l V· \11.1' lsPx+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 aTaX ( 2) Then the enllre R.H.S. IS ~ ' 6  I.
2  a x
Substituting ax I  Ax 2 I  2Ax 2 ,,, ax 26' 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.
ml ~l mI
(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
InI
L o Ak+h h~O
11
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,
mI
L 0 Ak+h h~O
mI
L (0+1 k+h+1 V  0 k+h V  0 "k+h)'
h~O
mI
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 5year 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) Oneyear 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(lqx+h) it follows that:
Ak(K = h) = vbh+I7rhh V = vbh+1  [bh+1 vqx+h + h+l Vv(lqx+h)]
= (bh+1 h V)v  (bh+lh V)vqx+h; and
Ak(K = k > h) = 7r + Vh+l V  h V
= vh+l V  [bh+1 vqx+hh+l Vv(lqx+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 nm = 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)n2
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 . nIPy, which is Px . nIPx+ 1 . nIPy, or Px' nIPx+ 1:y
Al . 1  h  nPy1 d' / ternatIve y, nPyI  Pyl' nIPy, so t at nIPy  , pro ucmg nPx:yl Pyl . Pyl
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 + 2tt1
(b) 1 Sr(t) = 1  F(t) = n 2
(1 + 2t) 
{'XJ (CO dt (c) E[T(xy)] = io tpxydt = io (1 + 2t)n2 (s = 2t, ds = 2dt)
lCO 1 ds 1 1 . = 2 n2 = 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) = rqxy = 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)n2
Calculate E[T(xy)] as E[T(x)] + E[T(y)]  E[T(xy)]. Use problems #7 and #1.
density J.txy(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(IOOXS) 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 r160 dt + 150
r150 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 tI!Q30' rI!Q40 = CrIP30  rP30)(rIP40  rP40), ,:xl
and the overall probability is 2:: (rIP30:40  tP30 . rIP40  tP40' rIP30 + tP30:40) t=1
00
= L (tIP30:40  P30 . rIP31:40  P40 . tIP30:41 + P30:40 . rIP31:41) t= 1
00 ac
Now L tIPxy = 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 sqxy = .000266
(c) If a = 3, from problem 19 sqxy = .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)]
= nI 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 knPy
k=1
n m
= ""' v k ,_nx + ""' v k Ln nP sl'nce P ~ K.t' ~ K.t'X k y, kn y k=1 k=n+1
mn
+ ""' j+n h' ax:iiT ~ v j+nPx . jpy, were J j=1
mn
ax:iiT + nEx L vj
jPx . jpy j=1
a + E·a x:iiT n x x+n:y:mnl
kn
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
wy 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
et6 . e  J~ /l·x(S) + J1y(s) ds dt
= 100
et6 e  J~ 2A + Bc'(c'+cY) ds dt.
Now we let CW = c< + cY , and obtain
axy = 100
et6 etA 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 = 18a = 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. jI/mPxy  j/mPxy = jI/mPx' jI/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 l2j + 1 l 1 1 m . qx + m' qy + m2 qx . qy + m' qx . qy  m' qx . qy
1( ) m+12j = 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
JL(r) (x) = L JLV)(x) is constant. ThenfT.J(t,)) = Ilv)(x) et.~(T) (x)
j=l
(b) h(j) = 100
fS,J(s,J) ds = JLV'ex) 100
es.p.(T)(x) ds
= _ JLV'ex) es.~(T)(x) 100
JL(r\x) 0
m m (c) fr<t) = L fT.J(t,)) = et. p.(x) (T) L JLV\X) = JL(r\x) et· 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 JL50(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 + (1p)(u2+v2)e(u2+v2)t
h(l) = roo fT.J(t, l)dt = ~ + (lP)U2 Jt=o ul +Vl U2 + v2
h(2) = J!!J.. + (1p)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)eX
(b) ~1) = 101
R.;1tJ.L;1)(t) dt = 101
ex t dt = ex _ ex 1
(c) Jx2) = tiT) (2)(t) dt = t (a  x  t) e
x
t dt
Jo xH J.Lx Jo  dt  ex t
Again iT) , x
(a)
(c)
=  (a  x  t) ex t I:  101
ex
t dt
_ (a  x) ex  (a  x  1) ex 1  ex + ex 1
= (a  x  1) ex  (a  x  2) ex 1
= 1000 exp [ r e+ ~ dY] Jo ay
= 1000 exp [ex + In (a  x2) In aJ
1000 ecx . 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 ay
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 ec
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  r11t.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 It. 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 KI' 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) tsqx+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(jl) 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 a1 (3s ds
(T) Jt s e tPx = roo a1 (3s ds
Jo s e
(a) Now ir.J(t, j) tP~T) /L~\t)
= L 100
sa1 e(3s ds. r(a) t
= e j3a ~1 (3t fOf]' = 1 r(a) e,
= (1  e)j3a ra1 e(3t for)' = 2. Next, r(a) ,
Then w~;l ?,••
fA)) = f(sJ) ds =  sa1 e(3s ds = ·130. 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(~) = Lj3o.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 (38 r(8) = L (3a2 r(a + 2) r(a) r(a)
= r\:)(3a2 (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 Unl must have been 0 so that Un = 1.
(b) ;j;(u) = expS !u) = exp(~u) = eR(u+l) E[exp(R)IT < 00] exp(R)
(c) wc
(d)
01 P 21 q
~ qe2r  er + p = 0, a quadratic in er •
Hence e _ 1 ± !14pq
 2q , which is < 1 when you select the  option. Now r > 0 so er > 1 means
r _ 1 + ! 14pq h "'R _ I (1 + ! 14Pq ) e  2q ,ence  n 2q
'IjJ(u) = eR(u+l) = q _ _ ( 2 ) u+l
1 + !14pq
2. (a) Sn,m = W n+1 + Wn+2 + ... + Wn+m f
= (Yn+1+aWn) + (Yn+2+aYn+l+a2Wn) + ... + (Yn+m+aYn+ml+ .. + am;)¥~~
m mi 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 [ts,ts+dt] I V;+I > t  s)
= Pr(Vi+1 E [ts,ts+dtD/Pr(V;+1 > t  s)
= f(ts)dt 1  F(ts)
• 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[ts,ts+dt]IVi+1 >ts}.
4. Since N(t) is Poisson with parameter At
Pn(t) = Pr(N(t) = n) = eAt(At)n In!
(a) ]Jo(t) = eAt is the probability of no claims in [0, t]
p~(t) = _AeAt (i.e., decreases proportional to Po(t»
(b) p~(t) = [AeAt(At)n + eAtn(At)n1 . A]/n!
Note:
= APn(t) + APnl(t) = A(Pn(t)  PnI(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) = !(3e3X ) + !(7e7X ) 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)(1P(x» + eO. (10) = 0 + 1 xoo
Substituting into the first equation gives
1 + Xr = 1 + r 100
eTX(lP(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(1P(Y»dY," ~ + Pb 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(lP(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 + [lP(U)lzP(O)]  ~[lP(u)]
= ~[[lP(Y)lzP(UY)I: lu
p(y)'lj;(U y )dY +[lP(U)]'lj;(O)]  ~[1
= ~ [[1 P(u)]'lj;(O) + ~~'lj;(U) lu 'lj;(u  y)P(y)dy + [lP(u)]
AI
= ~['lj;(U)lu'lj;(UY)P(Y)dY[lP(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) = .3eo + .2eo + .leo = .6 and in general ¢(O) = I! o. Thus 0
(b) When X is distributed like a mixture of exponentials and n
¢(u) = 2:= Cieriu 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 e6x = ~(3e3X) + ~~ (6e6x)
~ 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)PlrMx (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(103r) 4 [1 2] 9 r(r  2)(r  4) = 9· r4 + r2
= 1._4_+1._2_ 9 4r 9 2r
(e) above and (13.6.12) and (13.6.13) imply
¢(u) = ~e4u + ~e2U
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"' (5r4)(10r3)
_ 2 [ 2/5 9/5] _ 1 ( .8) 2 ( .  3" 5r4  10r3  15 .8r + 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;ef3/x , PI 2:,A;I ,8;_ "'A ./,8.
II 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) = ~e3X + ie7X = ~(3e3X) + ~(7e7X), 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(3e3X ) + ?O(7e7x)
(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) = _enw
E[u(x+Gn+])] = E[_en(x+Gn+,)] = enxMG(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) =! [lW] 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 Xdensity. 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 Ia' a 1+0
~ > '073 _ ~ . O§a . 1  a (all a) 1 +0  1 +'0  Ia Ia (la)+O~a
..<>.. _0 _ > 0  § a (II ) r7 1 + 0  (1a)[(1a) + (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 aI
_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 (le) = 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(Ir)fJ (Ir)fJ
1 + e . r
()' (l+()(1+0e fJ _ 1 = ()Se fJ
1  e fJ 1  efJ
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((l2.5014a)  (la)10) = 4(1  ai
o ::::} 0 = numerator
= (1ai(2)+(1.252a)2(1a)
= 2(1  a)[1 + a + 1.25  2aJ
= 2(1  a)(.25  a) ::::} a = .25
1.25  2a (lai