Chapter
4.73
Erercise 4.6 This is the case of De Moivre's Law with
ar =
100
trx
100. In this case,
-.r - t
100-x
'
-&'L tp,
P,(t) =
100-x-t'
frT) = ,p,.lt*(t) =tob= Frrthermore, when De Moiwe's Law holds, we have
.l-4Ja-x =*( 0
"-5t
J-4t
da-xl
o)-x'
n.-1
At,A
d-nl @-x
= l"-"' ,i;dt 0
In particular,
(a)
O)
Z)o,za
=
W = !ffi
= {tt
-
e-t,s)
x
0.237832.
This is a more special case, because the amount of benefit is not Hence, the actuarial present value of benefit is:
1 but
rather bt =
eo'05t
.
25
l"o.os,
e-o.os,
0
id, = #
=
*
= 0.416667.
Exercise 4.7 De Moivre's Law is again assumed but this time with interest rate (not force of interest) of 10%, so that the force of interest is lnl.l, again with ar = 100. We have: (a)
4o,r,n=
ffh
=o'oe2oee'
(b) The second moment
is. ,.
271
- -
,bo,iol
d*lo='^
ldr-_#
=
0.063803.
Therefore,thevarianceis:27j6.,nr-(40,r)t = 0.063803 -0.0g20gg2
Copyright
@
ACTEX Publications 2007
s
0.055321.
Acluarial Mathematics: Solutions Manual for Exercises
o) (z*) =
dt INTEGRATION BY PARTS
, (_Ltfi+o.osr)-3)lt=r00-'r, , toor* 4a, = roo-i(-T'\rrv'wJ'i, +loo; )1,=,
=-f,oo-r), roo#(1
+ 0.0s(100
.
-'))-" .
=
-tG -0.05x)-3 .
=
-+(#)' . 46(ro -
dET
J
1 ( zP
rd;l-Ai'tt
11t=roo-x
*
o.or')-',,J1,=,
-(r * o.os(roo - 4)')
mfu(r
rolo - o s')-' )
=ffi['-(*)')-+(#=)' Exercise 4.9
(a) We have:
Mr(i
= E(e'r)
=
. [ "" ,p,. p*(t)dt.
0
Therefore,fors = -d'
Mr(s)|"=-u = I
n* ' ,P"'Pr(t)dt
= A\.
0
(b) For the gamma distribution,
.frQ) =
fr;wo"-r
"-ft
,t >0.
Therefore,
MrG)
=(*)"
Using part (a) of this problem, we obtain
A* = Copyright @ ACTEX publications 2007
Mr('=(#)"
=
[,
.fr)"
Actuarial Mathematics: Solutions Manual for Exercises
Chapter 4
|
79
Thus we have:
E(z\
=
r.1*'(r -,i)* el ; z=l
:.('-r-[+,u' )" '-*['-(:)r) *. =
e
5 -t'2 o 0.41766. -e G
Exercise 4.14 This is De Moiwe's Law with o = 100. Note that we have the following discrete versions of De Moiwe's identities proven in Exercise 4.6
A'
al-x-l
=Z ,t=0
at .rt*l ua4= a-x' ?o ,-,
a-x-l
1
rlq,''o*t = t
n-l
Ata = 2rlq,'nr*t
n-l
1.yr+l7=o@- x i
=l
&=0
a;l
(t)- x
Therefore,
(a) A+oid
=
ALxt* A*,)4
=
%
60
-,,25.35 '' 60
x
0.407159.
(b) This is a bit more worlg as it requires referring to the definition again, but we can again note the general formula fust:
rul-x-l k=0
" . rlq* =
,f'{o*r;ut+r a'-x-t ,t=0
.-l-- = o-x
o)-x "i'=
In this case,
(M)qo =
Copyright @ ACTEX Publications 2007
(Ia)an 60
t ii=^- - 60v60 -^- . out = 60 = 5.554541. 0.05
Actuarial Mathematics: Solutions Manual for Exercises
80
r
Chapter 4
Exercise 4.15 We have
.
A,'4 =. frot
plq, +un . np,
,t=0
m-l
=
n-l
Ir**t
&=0
.rlq,+ Ir**t.,,\q,+rn.,p, k=m
,q!.-l+v^
.
.p*.nf'ur*t
'
rlq**^*v^ ' ^pr'v'-^ ' n-^P*+^
t=0
r^'
=,ql.A *
4-a * v^'
^P*'(lrl*^,r=a ^
P
r'
+
vn-^',-.P**-)
Ar+^,;=al.
In words: Single benefit premium for an n year endowment can be viewed as single benefit premium for z years of term life insurance protection, plus single benefit premium for an endowment for z - myears deferred by rz years' Exercise 4.16 Here we have: 19@
Ar= Ir**t . olq,* Lrr*t . olq, k=20 &=0
@
= At.,a*'20' zoPr'
I'**t'
rlq"*ro
=
/r=0
et
af
'iol'A,*ro'
and
4Zrl =
Ar,g1+ Ar,fi.
Therefore,
(a) A, = At,n* A,,)rj'Ax+2' = (A,a- A,h)*
A,,*oi.Ax+20
= = A,Id-(l- A,+2iA,,hl.
Substituting the values given we have:
0.25 so
that
=
0.55
-0.601,.h,
A,,h=i.
a a a a a a a a a
a a a a
a a a a
O) et-n = A*rol-A'ht = 0'55-0'50 = 0'05'
Copyright O ACTEX Publications 2007
a a a a a a a a a a J a a a a a a J a a a a a a
Actuarial Mathematics: Solutions Manual for Exercises
a a I
(b) Eo = 100.1000.40 = 10,248.35. Er = 1.06. to24l.3s-1s9,6ss'ljl?682 ' 9s013.79 = Ez = 1.06'
lO,7 10.36
-f
ffi
=
LI,192.34,
**{ 9s0t3.79
=
LI,694.76,
OO,OOO'
Et = 1.06'11,192.34-100,000
10,710.36,
Ec = 1.06'11,694.76 -199,969' = 12,218.04, ' 1695W 95013.79 Es = 1.06'
12,218'04-100,000';ffi = 12,762.58.
Exercise 4.19
(a) For
a death occurring
inthe mth of a year following age x + k *
the end of that m'h andits present value
(b) Consideradeathattimeswithin
ar,
*,the
benefit is payable at
is ur*# .
s.[0,]].
*'h of ayear,with
T'n"rt
I
tPl
lr,
.,0**r*1. F,*r*t(s)ds =
is the actuarial present value of benefit at age x + k +
f,*
'px+k.(u,.*n,*o**)
=
I
r'
.
*Q"***!
fi , and'
t'#
.*l*q,.r
is the actuarial present value at the beginning ofthat year. Therefore,
Ay)
=
= ir**' on,'(f,o*il*# '*l**.') t,r op,(y,r*.rl*n,*l at n ) *=o [r=o
,r=o
\.r=o
JJ
(c) Assuming Uniform Distribution of Deaths (UDD)'
,l tQ,** = mL'q,**, il^ *t
AY) = Zur kP, ,t=0
[iu.r.*
6
=
Iro "
&=0
olq*
+
[F,,'
*l
;)'I-#
Copyright @ ACTEX Publications 2007
J
*n*r)
*) =t,*.'
oln.,f) =
7 J
h}r*'*lQ,
=
j;4'
Actuarial Mathematics: Solutions Manual for Exercises
J J J J J J J
a C
Exercise 4.27 Note that for failure at time t the amount of benefit is b, = (t
-*)".
The single benefit premium
for this warranty can be expressed as:
Aa-lrn)la ="[Au
+)
(a) Under UDD
4a
-|rzr;u
=
*4'u-t;(tzri'-(+- *)4')
=
;('.i(;-+))4,
-i;
tqroa.
We know that i =109/o andthat
,
tl4o =
10.2, k =0,4, {o.L k=1,2,3.
Therefore,
4A = 0.2v+0.Iv2
+0.1v3+0.lva+ 0.2vs
o 0.53207989,
(u)toa = 0.2v+0.2v2 +0.3v3 + 0.4va +vs x 1.4666285, and
;('. +(} -*))4' -I*,,n iu 0,t9 *!( - rn1.10[ [, ' - -a]]0., lrl,r.to 5[t-* ))
3207s8s-
I
0'10 l. 4666z's = 0.3072t5. s ln1.l0
(b) If the warranted return is the reduction in the price of a new product, the answer to part (a) would not change, as the customer could take a cash refrmd and apply it toward the purchase ofa new product in any case.
Copyright @ ACTEX Publications 2007
Actuarial Mathematics: Solutions Manual for Exercises
_I
I
Erercise 4.28
Irt
O
be the cumulative distribution function of the standard normal distribution. Then:
Eek)
=l"u.onu.ft;* o,
-f* 0
=
."-+(h.+n\ o,
!s.'l2x
o
rz _1(rt+zit+.(*f))dt
= l-L'e'e' l-
i542n
t =5 "*'g!."-i?a*t)'
41
l",l2r
t=o + ,=! = d" = z"*(t 2 "-i" +"+"!h' ll\dz=dt, t+oo=) z-+6 ,
lr=L*k. l-102'
@)
-t(r)
v, = 2e1t8(t-t(r) x o.6ee2.
b) 'A- = 2e4t8[t-t(;)
= 2et/2(r-oOl) x 0.5232.
@)'A*-(4)' = 2ett2(r-o1r;)
-(r"'''[t-.(;))' '
o.oror.
(d) Let q2s b"themedian of Z.T\ensrnceZand7areinverselyrelated, it 6l't isthemedianof 2i we have: 0.5
= Pdz.€2'5) = Pr(z' €|'t)
r{ 'I
= f,@o' =
+@ .
t2
J#"-hd,
I
--_t-, t=d.t =l.riit=,,, ,-* = "=!9'tl ,-El
=
INTEGRATION BY SUBSTITUTION
=2f#,=,[,'[#)) l0
This implies that
= '-r[4]) [10J Copynght @ ACTEX Publications 2007
I
0.2s,
or .[#)
=
0 75.
Achtarial Mathenatics: Solutions Manual for Exercises
92 a C\aoter 4
The 756 percentile of the standard normal distribution is approximately 0.675. Thus 20.5
+ But
()s =
e{.os'f's
-
= 0.675,
e4'os6;s
x
or
€l5 = 6.1s.
0.713.
(e) We have
6
Z. = [r.fr?) dt
o
=*l 2t "-t dtoJzr
o,
f u=m, /=0+ U=0 zodu=ff, 't,o+ u-> INTEGRATION BY SI'BSTITUTION
=w"-udtt u2,
-
=
"--o.os.i.s7ss
(-#*)r x
0.6710
=
fto-o)
< 0.6992 =
=
#"
7-s788.
4.
Exercise 4.29
Let u(w) =
-"-6*.
By Jensen's inequality,
E(-"-6,) = -E(e-6r) = -7r<-e-6EQ) = -e-6E(r) =-e-52' Hence,
-o
e-o"'=v"t
o
3Ar.
Exercise 4.30
(a) v, =
*r[-lO*)
b) z=brvr=D,oo[-h-) (c) E(zi) =
,lu,["-[-i--)J']
Copyneht O ACTEX Publications 2007
=
"["*,[-ir--)]
= E(z)@ro,cei6,
Actuarial Mathematics: Solutions Manual for Exercises
94 o Chapter 5
(b) We have
lr = 100.1000'12.76I, o2 =
!00.10002.10.230,
o = ,[7 The coefficient of variation is
g* p
rv 10.1000.3.198.
,JPu.*
x o.o25t.
Exercise 5.3 We have
var(a71) =
#('r, -'2,)
=
#(r-262a,-g-da,)') = #(t -262a" -t + 26a, - d'al) = ?(u.
-'
a-
-)-a:.
Exercise 5.4 We have
cov(Aa71,vt) = Con(t
-u',r').
Note that
var (6-vz ) + vr
) = Var(l-vr ) + Var(vu ) + 2Cov(l-v',r'
).
But
var({t-v?)+vr) = Var(l) =
6
= var(l) + var(vr)
=
+ Var(v?) +
2Cov(I-vr,vr )
zYar(vr)+ 2Cov(l-vr ,vr).
Therefore
Cov(l-vr,vt) = -V-(rt).
Copyright O ACTEX Publications 2007
Actuarial Mathematics: Solutions Manual for Exercises
Cbapter 5
I
99
Exercise 5.11 We have
var(cq) =
var(h-l)
= Var(4;r) =
u.[L#)
=
".'1.
#trrt
Exercise 5.12
(a) We have n-l
n-l
dr.il = Zur rP" = k=l
(b) hthis
Iru*t &=0
k+lPx
= lPr'lvk
kPx+t
= rEr'dr*r,il'
fr=O
case
,la,
= ii,-(d,,a+,8,) =+
+-
nE, =
#-
nE*.
Recall that], isthepresentvalueofaunitlevelperpetuitydue.Basedonthat value of a perpetuity of
$
I annually, starting
f
at the end of n years, or year of death
it,tr" if earlier.
cao"etsthe payments at the end of the year of death, and on. This combination provides
1 at end of n years,
if alive, which is cancelled by - nE,, leaing ,la*.
Exercise 5.13 From
| = dii -*A -.
recalling tl:mit d
= 1-v,
we obtain
A*A = | - (L-v)ii *s = | - d*A * r' r.,^ = v' d,,,-,- (or,r-r) = v' d
ril- ti,;11.
Exercise 5.14 We begin with the calculation
of E (f')
:
lz r-r**' i+2 r-vzK l;.: ,. ----.:r---_, t ,-+zt
y2 =)t lz
K=0,I,2,...,r-I,
i+2 r-r' i+2 r-v2' a1-=a, 2 +-''a71, nt nt i i ' Lt i -'i i'+2i "-+
Copynght @ ACTEX Publications 2007
.t
K = n,n*l,....
Actuarial Mathematics: Solutions Manual for Exercises
102
r
g63p1"r 5
Exercise 5.18 Recall the formula 1
= f .ar+(l+i)A",
and its analogue
t = i@).oY) *(r.#)^f, By equating the two we obtain
i'a" * (1+i),4, =
i@).*,.(r.#)nr,,
and from this
d...............:n)
=
,
go,* #(n.',n"-(,.*)^r,)
AssumingUDD,
AY
,;,g+
=,{()'q*.
Therefore
*t =# r. .h
e+i)
A.-#[' .#)nr,r,,
"ff'
.# e+i)..4, #['.#) ,f,.,e. = tf' 'o" .n(#-# .+)"r-,) = tf) 'o,
['
=
,f,
.o,
= tf)o**
Copyright @ ACTEX Publicarions 2007
.^"#(-[,.#)
t,fi
,f,)
(t-"i-,)
Actuarial Mathematics: Solutions Manual for Exercises
Chapter 5
r
105
p(tz)=ffi _
t-tz.(e+i)t/t2 _t)) tz. (1t+i1t t t2 -t). tz. (r - 1r*4-t,t, ; 0'06
-=
- 12'(t.Oetrtz - ')
l:
0.4681195348.
By substituting these values, we obtain A[?.h"15.0383835 and
3111-+=
HxD6.3to4n2.
Exercise 5.23
(a) We have
Y=l
l
9-n, o
^-;l
ftrlf,
,
K 2n.
(b) Based on part (a)
E(D = @)ff)
=f,Oagil, k=0
and this can be seen since the payment pattern implied by the sum is the same as that implied
AV
(Q!].
Forexampleif m=
5
thepaymentpatternwouldbe
+.+,*, 1].,?,2, if x dies between ages at age x +1.2I.
Exercise 5.24
(a) We have
,=
Copyright @ ACTEX Publicarions 2007
0
{Y'F*EI, K>n.
Actuarial Mathematics: Solutions Manual for Exercises
106 o Chapter 5
(b) We have
(Diiln+(rg(f; = @+\ffi, so ttrat, using the result from the previous problem,
(DiDn = @+L)ii(1-(r,iln (n+t)ii(1-_i__ i;(1 +
lr:*
Zt,# -
rlr$n) =an .Eo:%
n
=
Ia(4. L't
k=l
a:.nl
Exercise 5.25
(a) We have
II rn>9^,
Y=4 ':
"-;l
0
lt'lf '."(a3*-rff'),
(b) Again using results from problem
n
23
E(Y) =
(Iiln*"Qaff
)
n-l
=
l*l a%+"(Srff')
&=0
' = ;(-' ii(\+,iaff') n-l
=
>ovf).
,t=0
Exercise 5.26 We have
d(h)n+T-vr = u.u^-{''
+T.vr = af,.
We take the expected value of both sides and obtain
6(Id).+(IA)* =
Copyrieht @ ACTEX Publications 2007
q*.
Actuarial Mathematics: Solufions Manual for Exercises
I
l0 r
Chapter 6
Exercise 6.4
(a) We have F(.a,) = p=0.02, as E(I) =i=SO. of 2' The 50ft percentile of I is found at the value corresponding to the 50-th percentile of Z, which is ln2 times the mean, or
(b) We have L = e6r -Pail
ry=ffix34.66
and this is a decreasing function
years. Based on this
0=
-p
,-o.o6"ue
f
-r=t=oj# 0.02
so that
P=
=L .rb'?
az 0.0086.
o.s2 I
(c) We have
0=l-PZ* = P- o,| =u €x .
since with zero force of intere st e8t
=t and f (aV) = E(T\ = Z*.
Exercise 6.5 We have
Fru
v' . 1p,p"(t) dt
gI =
_
.
Ivt ' ,p" dt
0
This shows that
FQe)
is weighted averige
of p,(t) for
r>
0.
As all values
of p,(t)
exceed
p(x), we conclude that F(A,)> p(x). Exercise 6.6
If the force is constant, then
Z, =\n*+6)t . pdt = 0
#g,
and
')-
U .,, - p+26.
Copynght O ACTEX Publications 2007
Actuarial Mathematics: Solutions Manual for Exercises
Chapter
Tberefore
'.q,-Z:' _'Z-_ei
(6a*)'
6. lll
equals
(l-A,)'
pp2
Vop-
(p3
+2p26+p621-2p26 62 (p+26)
_p
p+2
_
'L.
Erercise 6.7
Ifd=0,thenyt=land @
I t p,p,(t) dt
Ff+l=s-
I tP, dt
=*. €a
0
Erercise 6.8 We would like to show that
var(vu) < Var(vr -F.dT), where F
=FU),
or, equivalently, that
var(v?) But this follows directly from the fact that
. (t.5)'
.var(vr).
(t. #) > 1, as both F and d are posifive.
Exercise 6.9 We have d-a* dn
"= (u@)+6)a,-t
and
dZ, dx
-,(p(*)*6\4-pG).
(r +
(p(x)
Based on this
(,. *)uG,)
-
*
=
+a)a - r)F1 a) - ((pr.1 + d) a - p@)
= (t + (p(x) + a)a, -r)-f; -@@) + d)7" + p(x) = (p(x)+ a)a -(p(xy+ a)A + pG) = P(x).
Copyright @ ACTEX Publications 2007
Actuarial Mathematics: Solutions Manual for Exercises
Exercise 6.10
calculations in the first row are based on the following identities: z:s"iol
= ,,io *n:r^*toErs l('hs-rc'Ers)+
rcEts'
i(,4ts- rc Ets' /as) + 5'
Asid = (4s-rcEx'&s)+
__ 5s:iol : D
rcEzs,
#(4s-rcE$'&s)+ d' =mt;, =ffi l-(4s-rcEts'4s
P(zrr,rr)
rcEE
rcEgs
+ roErs) '
: ilr*roi E*n - T:4m Ats.n-=i
ns'&s) + vro' to pts) . 4Ar_vto.topts.,4cr)+rto.rofrs)' d'((,hs-v\o'
ro
Calculations in the other rows follows analogous formulas.
Exercise 6.11 We have
,o4u-P.1a
=HH=W=#=
zoP(zolrc4).
Exercise 6.12 We calculate
4
@@6
= iv**t . *lq, = iro*t1t- r)rk = v(l-r)l(vr)& = v(l-r) k=0
t=0
&=0
#
:- l-r l+i-r'
'd.=+=('-ff|) +=#+=#, #=
P* = ax Furthermore
,'A"
\-.r,. l+,
is calculated the same way as A, ,butthat it is based on doubled force
interest, resulting in
Based on these values
Copyright @ ACTEX Publications 2007
of
24=& 24-L2 _ (l-r)r (1-A,)2 l+2i+i2 -r Actuarial Mathematics: Solutions Manual for Exercises
l16 o CtrapterT
Exercise 7.5
In example 6.I.1 elq, : .2 for k : 0, 1,2,3,4, futrs, f is fte uniform discrete distribution on 0,1, ..., 4. Assuming the UDD, Iis rmiformly dishibuted on [0, 5]. Thus this problem is a repeat of example 6.1.1 to the situation where the variables are now continuous instead of discrete. By the IDD a here is 6.1.1, etc.. Exponential reseryes are not worth ino ^exaryle the effort of calculation.
k:
Exercise 7.6
& : vu - PZ,,;)a6ifor0 ( U < n - t,andvn-t - F.a;4,U ) n Nowrz : vu -P.uq: vu-P(#) : ,u[r + #] - f
:
rhen
var(,e
since
var(vu)
from (4.2.10),
Note:
. f]'uuruur: (4)'.
(r7,*,,;4
-
var(vu).
7,*,nn1r),
2Z "+t"n-d then Var(rz) - "x+tin-i --/ ' ar,;)z -2 16
Z- tlr>t: U, the future lifetime of (x*l).
Exercise 7.7
f'-t _aA uPx+t ltx+t(u) du * d;4 n-tp*+t J, vu - uPxtt
El'Ll :
Var(1L) Note:
:
[t
upx*rurrl,'o-'
* Io' ' vu upx+tdu *
u) : $ v"'t, .fi|r7,*,,v
:
d;4n-tpx*t
:
dx+t:Frr
- V,*,,42f
?- tlr>,: U, the future lifetime of (x*/).
Exercise 7.8 1a) [email protected])
:
V+s,x:1
-
2sP(A35,,s1) z+s,i6i
(b) There are no future premiums, so 5 Z+1,O
:
Vlost
Exercise 7.9
: -lnCP(A,Xd + P(,aS1ti from Equati on 7.2.5. o) 6 : ln(1.06),P(Vd : .020266 =+ uo : T\35) - 20 :23.25 years. (a)
uo
t.
Chapter 7
. ll7
Exercise 7.10
(35J0 :65 t.-Setting the minimum loss to 100 The minimum loss occurs when zero is the same as setting 65 t equal to -ln(P(As)/(6 + P(135))/6 23.25 years. Thus
U:
t:
47.75 years.
-
-
-
:
Erercise 7.11 Analogous to the development of (7.2.9).
Exercise 7.12 Same comment as problem 11. Nothing is done in the text vtdth these densities except to €xhibit another formula.
Erercise 7.14 fl-v(.1'aq1:
(a) Prospective: Zso - 2sF(Vai att,Tdi (b) Retrospective: zoP@+i.s+o,i6i - ,oEoo (c) From the prospective formula, we have
It_^rce,,r$]z*:lr-fu*urn^ln* (d) Alternatively, from the prospective,
l*
- roP(tdfdso,6i :
I roF(Zso)
-
roF('n^1] aro3q
Exercise 7.15
,o-tr(A*.6):
-
(a)
Prospective: Vso,Tot
O)
Rehospective: [email protected]+0,I0
(c)
Anarogous to
(d)
Analogous to
4(c): I
t
J
-
P(A+o.ro)]aso,O
: | - 6 Z56.iq,
- p@ao,ro1) + a]aro.t
From (e), since F(7*.61)
,\ (g)
.From(rr,
air,t-z1y-
*6
dn;tg
---v4ofrl
sinceZ
roEoo
'
4(d): F(7ro,ioj)
(0
,^
-
t -'+#4lr,,t t'"so:lol/
L
(e) From (a), since Vro5ol we have
P1Ao*5paro,6i
| -V : ---T-'
:
wehave
#^,
Ato,Tol
Aoo,^l
- - l=z;{,
-
t - 3t'g u4otzol
Exercise 7.16 Retrospectively, there have been no benefits, so the rreserve is just the accumulation of past soF($l zrsFssfrl.
premiums: tz|V(tola34
:
Exercise 7.17 Begrn with the retrospective resewe formula:
Multiply AyP,,*t : ! P,,;t' and
-/-
\
^r(,,^*;1) : Vlr,A
ffi.
:
^T @,.-*,t)
: FF,*-)
s,.6
-
^8,.
n"rproduces \
-/P\A,,^*,1)
-
--r P(ai,a),since,E,'s".4
:
d,d
^E*, ^8,
This establishes (a). It is interpreted as seeing the premium
PF,,^ ) in two pieces: one
which provides the coverage for those m years,n@ia),and the other which provides for the reserve after m years, if alive. Thus the reserve is in the nature of a pure endowment benefit, so the premium for it is a P.E. premium. Now multiply equation (a) by Fr"1, and subtract ,E* fromboth sides. This yields
totally parallel to that for (a).
Exercise 7.18 The given equation relates to formula (7.3.3). This equation states that the reserve at the beginning of the interval (at time 10, interval length 5) is the a.p.v. of benefits payable during the interval plus the a.p.v. of a P.E. for the amount of the reserve at the end of the interval, less the a.p.v. of net premiums to be received during the interval. If we rea:range the equation to read
--jt
6r-i t= ilV@ro) +' z"nP(7n)d1,6.r: ZloA + 2OTtr-7 t
sE+o.V(An),
we show that the a.p.v. of all resources available to the insurer at the beginning of the interval is equal to a.p.v. of the uses of those resources.
Exercise 7.19 This is totally analogous to Question 14. Exercise 7.20 This is totally analogous to Question 15. Exercise 7.21 This is totally analogous to Question 17.
Irrlrei,*7.22
.
Since
pZ,;
Since
drA
Ttu"
m
*
: , - d.ffi- : *,** W
5
6'
tt_
, vzl?-kn-2H : 2.ii**p,4,A* +ffir_O nxlkn-lcl ti: )'-g : t' FinallY' kY*asT: . wr*?-kn-2A qx*kn-kl
iira2la@
l--4
4 -r - r-5
1
5'
kdn723 FtI$r Continuous:
: |#^:
(a) ro-K4ss-i)
dA:
dt
-f
'l752905usingz'';
nPxdx+n and the values
fud4,ontinuous: (b) rcY6s): i(rovzs;
: a(x)ii,q-
of d" in the table
: itr - ffi): .08566
FuIIy-Disoete:
(c)
rc%lssT
pt
:
Arts.^1l-
_-;3s'301 - 3s301 atssTi
Eisf4:
dos
-
0(m),
Pt
r.o.1aor.m1:
Ats ii3s
-
-
.03273
usingl|r.,
:
A4s
-
f0
zopesAos
flgon$Aes: .oo4gl5
v"u soPssdes
- fo zoPtsiies: ll'575
hdrc724
:
: i n'.a + A,,k # *n.a : i.e*r - i ' Px' iix+k {tCL) : : I |,l.*r - P,. d*+kl : t. rr, (c) Yes. {(1,;1) - V.i*a=A ' e(/.;1).dnp.4 : t' A|**a - t P'",a' d,+*i4 (a) (b)
7',a * A,,h 7,+t - P(A,).a**e
No. Recall thatV*6 Yes.
: tlni.-^-
pta.ii,*r.frf
:
t.ov);t
:
.08846
bdc&7 Xcptacinghbyh+1,(8.3.9)becomes nV * nn Then6'1Y'v'Pr+n : (nV * ri - bn+r 'v'1x+n
*+tv:
@#P
-bn+t
Thc interpretation is most easily seen GV
+
nn\Q
i) :
+
:
bn+r'v'Qx+h
*
n+rV'v'Px+h.
ffi
if we write it
bn+r ' Qx+t
*
as
r+t Y'P*+t
Now the old reserve plus the premium, with interest to the end of the year, is sufficient to povide b6.1 if the policy dies (with probability Qx+n), or to provide the new reserve if the policy lives (with probability P,+n).
kercise
8.8
(a)
k-l
Dr'*' ,lq*
fi'w:
At :i,'APr-#:kv*
-+re,
(retrospective form). The reserve is just the accumulated value of all of the premium income, less the accumulated value of all death benefits paid out, taking account of the benefit of survivorship in the accumulations.
(b)
Since (r,Z' +
t :
PrXl +
P, - r.e,+h(I-n+rVr) k-l
Dt"' h4
n+rv,
(l - n+tV') * l+r l/*,then v.n+rV, - tVr. Thusthegivensummationbecomes
e*+r,
:
- nv,l(o*l*-n.
a telescoping series which is easily seen to reduce to kl/x. Interpretationz rY, is the accumulated value of past premiums without benefit of survivorship, less the accumulated value of past benefits without benefit of survivorship, such benefits being only the excess of the
This is
insurance amount over the reserve.
Exercise 8.9 From (8.3.14),
rn-r :
IJercrl: fr,andfuwehave
Tr
:
@n
1rV,sO
v. pV - 1,-1V,or 1'-rY *zr)(1
tV : zr(l+t) 2V : [zr(1 +i) + Tlrl:.s
pV
-
- nV) v Qx+h-r *
n.
6
n
n](1
+;) :
v ' 1V
* i) :
r'i1,etc.
-
n-tY.
1V. Then with
sV:0,
we find
Exercise 8.10
: Du^i h=t
(a) v'ii*,4: PYB
n-rlq,
: hDO-r-\l h=l
It(,/ ,-,1a"- r',-te,) if, - dii*d #-ii"fr: Thus
(b)
uo
n:
;:'a
ux:nl
-f
nE*
: hViu - ,"
rqrf
+ tr ,Pr)
dit-d*;t
.
PW - d;:A - d"apffi. a-**.4) - n'ii"*p.4.
From part (a) we see that, attime ft, the
n .d,*1,,;4.Thus
n-ilq*
1z : (o^ -
Clearly the PYP is
Exercise 8.11 (8.4.3) sals pasZ
:
bk+t
rr-s
t-sQx+k+"
*
r+t
V'vr-t
l-,rPr+/c+s'
Multiplying bY ,P*+k,we obtain spx+k.
k+sv :
:
sPx+k'*+sV
*
vr-t
b*+t vr-t r[-"qr+,t bk+t vr-t (qr+*
,qt+kbk+t :
*
t+r
V'vl-t pr+t
- sQx+k) *
t+r
V'r'-'
p**r
vr-t(b*t' Qx+* * *tY'P*+t) ,t-sQ,V + zrrXl + t)
: : (t+i)GVqri
Interpretation: The old reserve plus premium, \nith interest to time s, will provide the reserve at time s if (x * &) has suruived to that time, or provide for the then present value of death benefrt(bwr to be paid a?year-end) if (x * &) has died.
Exercise 8.12
Interpretation for both (a) and (b): The reserve is sought at a duration between two consecutive premium-payment points. This reserve is approximated by interpolating linearly between the two adjacent policy year terminal reserves, and adding the unearned premium for the current premium period. The interpolation coefficients on the two ierminal reserves are easily obtained. Since r is the fraction of the year beyond the last premium payment point, then (j - r) is the fraction of the year remaining to the next premium payment point, so that is the appropriate fraction of annual premium unearned. Note that this fraction multiplies the annual premium, not the fractional premium actually paid at each premium payment point.
ffit2t (e) ElT
-t375(l) +
(b) Yt(4 :
375(3)l (362.12)
6,450,962
as
+
t2s0(1)
+
2s0(3)l(561.08):t,104,260
before
c:l.&5l@+l,lo4,260:l,108,483,whichis1.00378timesthereserve. (c)
vu(z) = [37s(l) + 37s(9)X1187.14) + [250(l) + 2s0(9)](343.84) : ct :
(d)
:
\ffi
1.645
s,3rt,37s
as
before
37gl.l4,which is .00343 times the reverse'
44 : 110,426,000 Ya{Q : 645,096'250 c : tto,426,ooo+ 1.645\@250 : 110,467,780, which is 1.000378 times the aggregate reserve
Yar(Z) : c1 :
531,137,500 1.645
:
\@
37,gll,whichis.o0o34timesthereverse.
Erercise 8.29 We seek 10,000
rcrpY{tl(4') 10,000
ll.rcv{t}(Vto)
s,000 [ toV (Ail
*
+ }.rrv{r}(Ato) +
t.p{t}(7ro)]
n-vGzi + P{t}(7ro)].
Exercise 8.30
SinceP,a
: *, - 4
Since12".1
then d",Tt"
: t-W:
:
: ffi
f,**d,+r't:
T-iE
.78ii*.4:
(a) d*,Tt : l*uprd,*1a,soQx:1-(1'?'*q83) : (b)
dx+ril1
:
L
(c) It'r, : f
*
(b^*t
vPx+r,
-
so4r+t
r,+rY)
p"+h'
:
1
-
(1'2X'625)
(d)
2.083.
t.625 .2
:'25
Qx+h' hPx
As : (#)' G-.66)2(.2X.8) : .6084 A1 : (i=)' (3 - l.s6)2 (.8x.7s)(.2s) : * I Ar :
:
.216
+ .69M4 (.216) :
Var (62)
:
Var(12)
: X : # : is no risk in the final year of an annual premium endowment.)
Ao
(Note that there
.6084 -27
.7584
186
.
Chapter 15
Exercise 15.5 Since a select-and-ultimate table is used the life insured is [a0]. Let G be the expense-loaded premium. Then this premium pays for the following items Qisted in terms of their actuarial present value at issue):
Commissions Premium tax
of
0.40G + 0.0sG\",l,sl
0.35G + 0.O5Gajao1iol.
of 0.02G\+o1lil.
Maintenance expenses Death benefit
=
of 12.50+
4o4oj = 8.50 + 4\+01,A.
of 1000{+01F.
Therefore,
GAt*lul=0.35G+0.05G41+ol,iol +0.02Giip1.;.1+8.50+4\ao1a+1000{+ol,zs,t,
or c(o'laa1+01.- - 0.05d1+01,a - 0.35) = 8.50 + 4a1*1,Tit+ 1000{+o;,rsl, and hence
G-
8.50+
a\*l4+1000{+o[H
0.98i1+o16
1
-0.05&1+olrol-0.35
ooo{+o1B + 4d1+o16i+ 8'50
0'93d1+ol,E +0.05'roE1+01'A[+o]+ro:i3
-0.35
Exercise 15.6 This policy has a single premium
fI
that pays for all benefits and expanses, so that premium is calculated as the actuarial present value at issue of all benefits and expenses. The premium pays for the following items (listed in terms of their actuarial present value at issue):
-
Taxes
of 0.025II.
Commissions
of 0.04n"
Other expenses
of
5
+ 2.50a*--t
=
2.50 +2.50ii".;1.
Benefits of 10001,;1.
Therefore,
fI
=
0.025fI + 0.04fI + 2.50 + 2.5Oii,.A+ I 0001- .r,
so that
0.935n =
2.50 +
2.50ii,s+
10001-
.r,
and
2.50+2.50A xint 1+10007x.:nt0.93s
Copyright @ ACTEX Publications 2007
Actuarial Mathematics: Solutions Manual for Exercises
Chapter 15 o 187
Exercise 15.7
ThelevelannualconfractpremiumG=aPr+cpaysforthefollowing(listedintermsoftheir actuarial present value at issue):
-
An initial expense of eo. Annual expenses of (e, + ezPr)iir.
of q.A,.
The cost of claim settlement
Benefit of
A*
Therefore,
(aPr+c)ii" = Recallthat ^(, '
=#,h"n""(recalltlrat 4x'
("+ * ")r, aA* +
cii* =
€o
€0
1
=
* (q+qPr)ii,
+ er. A* + Ar.
= .4r+diir)
"o
*(",. rt)r"
+
q. A, + A,,
* erii, + erA, + qA, + A, = Ar(l+eo+er+er)
+ iir(er+deo),
and we conclude that
a = l+eo+e2+%, c = e1+dq. Exercise 15.8 We have, for
Z>
0,
t(r 61, a) We assume that
(a)
"
T(x) andB
= Bvr
+ a B a7 +
0
aa + p (B n + f)a7 - (B n + f)a7.
are independent.
under the conditional equivalence principle ,
n(t(r{u),8)"lB - b) 0. Therefore,
o = n(r(r6y,a)"ln=t)
t(atf + aBaa + laa+ p(ar + f)an -(ao + y)aola = t) = s(t ' + abaa+ ilaa+ p(tn + f)afl-(tr + f)dn) =
=
b-4 + abd* + 0d* + p(bn+
Copyright @ ACTEX Publications 2007
f)a. - (br+ f)a,.
Actuarial Mathematics: Solutions Manual for Exercises
Chapter 9 Exercise 9.1
Totaty anarogous to Exampre g.2.r. Aseries of tedious integration tricks:
(i)
(ii)
/-#
*= *irnlr
fi(s):
#isadensiryon[0,oo)
qirrdtds : r';ri6,=T-' irn > 2 I[isajointdensityfor0(s,t(oo
(iii) From(i),Et(r+s)'l=
,r-tll c#
+ fs,r6r)= @, ,t\(n,-=?) ' (l+s+Dr
vffiirm<.n : T5 simitarly =
*l.rhus
sl: FL: I +EF] =+ E[^SJ H : EKt +,S)2J : | + 2Etsl+ E[^r2J can be used with E[^iJ to find E[^s2]
EU
(iv)
=+
+
Exercise 9.2
/,* /,*'a', l,*r#
dvdx
-n *
F
Exercise 9.3 Analogous to Example 9.2.3. Exercise 9.4
(a) Pr(T > n) = npry : ,px . npy,by independence. (b) Pr[T(x) ) n and W) S ,,.?, \y) ) n and Kx) < n] : i l
(c)
np*(l-np) * ,py(l'-,pr)' :
Pr [at least one survives]
npx
: I - prfneither survivesl : I - pr{max[t\r),W)]
*
npy
: l-ngl,_:nPfi: ,p, 1rp, (d) Pr[T
_ 2.np".npy.
npx.npy
npx.npy.
I
16 o Chanter 9
Exercise 9.5 We seek nPx' n-tPy, which is Px' n-tPx+t ' n-rPy, 91 P*_n-lPxlty
Alternatively, npy-r
: Py-r'
n-rPytso that n-rPy
:
#,producing
nPx:y-r/py-t'
Exercise 9.6
Intuitively, tpo PoQ) is the p.d.f' of the R.V. T : T\xl). Thus the integral is PdT I n) - ,qo. Aninteresting algebraic approach is to note that p*(t) :2pr(t),and po: tPx.lpr. Then the integral becomes
P
,lo" *-(,p'p,'(t)) dt
/ 1 ^l'r since : ,(-rdl|r), P,F,Q) :
Thenwehave 1 - nP? :
- rPu :
|
-ft,P,'
nQn.
Exercise 9.7
T: IW),FrU): 1- S4,yr6,y(t, t): L - # (a) fr4): F[(t):
fromproblem2 above
tf
ffi
O) S1(t):l-F(r):dF (c) Elr@y)l: Io* tpxydt : lr* &
:
Ir*
(s
:2t, ds :
,61s" : th
2dt)
(see solution to problem
l)
Exercise 9.8 Analogus to the given example and equation (9.3.8).
Exercise 9.9. tP-ry
- 1- tQx.tQy - 1 - (1 -rpJ(t -,P) : I : tP* * p, - P*'Py : tP, * tp, - ZtP''tPv I tP'v : tp, (l - ,py) * ,P, (l - tP) * tP"y. = 1-tQ7
(1
-tP,-tPy*tP"'tPy)
Reasoningly, the event of at least one out of x and y surviving t years is obtained survives andy does not, or ify survives and x does not, or ifboth survive.
if
x
Erercise 9.10
l) : I pr[xdiesintr+
Pr [at least onedies in (n +
: I - Jt -
pr[neither dies in (n 1) +
r1]'{t _ rrgai"rirl"i
rX}
: I - (1 -npx*,+e)(l-,py*,*ipr) : I - (l _n.lq,_,1q, I ,lq,.,lqr) : nlq, + nlq, - nlq,.,lq, t
y l ea is the probability that the second death out of
the
same event as above.
Algebraically,
:
nlqy
nlq,
x
andy occurs in (z
+ ,lqy - ,lq,y. Clearly ,lq,y *
*
l), which is not
,lq,.nlqy.
Exercise 9.11
(a) FrwG): tQo: tQxtQy: F4aQ)F49Q). fromproblem 1. NowfslG): F[oot!)
o)
calculate
(c)
u4Q)
Etw)l
: ##'
as
E[(-r)] + EVU)I
-
Now plug
in
FaaQ)
- 1-
nv@v)1. use problem s #7
and,#t.
from (a)
use results
Exercise 9.12 We seek tsp+0. We note tsPzs' 35p40
:
that
zspzs:so
:
zspzs
soP2s,so that 35pas
. zspso
: #,
:
5ep25. Furthennore,
: 3 :
?.
Exercise 9.13 We
will
need
p, :
e-[ip'6)as
-
eXp
: ex'
[-
(a)
:
:
-, - "l-' a"l
lr,t*-'-",
: TO#-. and,pastta(t) : # and, 75str5s!) : ,,10.'
We will also make use of 1p*1t,(r)
Ir'(1000
Then rpqo
:
lr]
:
f _ f6
;
rcpq.rcpso
(t
2
3
t
T0o=.
and tPso
- *B)(t _ *B) : eg # : (b) roPa*o: toPco* ropso- ropto:so:;8+f8-?:H rcpqo:so
-
: 1- #,
I
(c)
T:
The p.d.f. of
Z(40:50) is
:
tp+o:so ttq,so(t)
tpqo
.
tpso?-tq(t)
+ pn?))
: k k(#=7*#) :T5#,0<'
:
B+o,so
(d) 8ao;m :
EtTf
50, not
: r#',/"nrs-t)dt: r+00
+ 8ro -
9qo
is
55).
: lf r-+rlr'] ,r.ou
8ro,ro
: fo* (t - .6) * (t - #) : (uo - fS}3) + (so - qf}l) - : o,
dt
t8.06
fo'o
18.06
(e) Err2t:
tbo/" rlss-t1 at :
Then Var(Z)
(D Elr2f :
lo*
E[72]
,rtaoo,
fft'-inl,'] :
: {ttn}' :"hf (486.1r; -
*
loto
-
(g)
(h)
:
1547.2222
: : '" :
cov lr1+o:so), (40s0)]
r4+oso;,44613n-q
-
:
pe.S+72
8* 8ro louo
-
Cov[(40:50),
:
160.11.
-
:
ts47.22222.
n2.66.
goo,ro
dnlo,
gae,
. {r- s6l dt - (t8.06)(36.s4) fo'o
(30x2s)-(18.06x36.94)
var[(40:50)J.
Note:
{,
1ra.Oo;2
+se.rrrri
- +se.rrrri
f no at
: tf$ * Gr+# - 486.1r1n : Then Var(Z)
36.e4
I(40:50)] :
va444651]
:
82.86.
82.86
y'(t6o.llxl82.66)
_ jcj<
These answers differ slightly from the text answers since we have rounded prior answers for use in later calculations.
140 o Chanter9
Exercise 9.17
ForZ:
(1,1), we seekPr(2 < T < 4). Now
:
- expl- I,' r(tf) *]
,pl
($)-' :
(s),,
- eXpl- Ir' (10-x-"1-'*] : t - #. we seek 2pr,r - 4pt:t: ef (8) - (.e)4(;) : ,%t - frtrE : ,?$fo :531/2000 and
,p{
Exercise 9.18
(a)
:
FrwQ)
Fq,rft)
+
FrelQ)
-
Fao{t)
:t*t- (2,-[n[t.ss])
:[rnlt.$$] O)
Differentiate the answer above.
Exercise 9.19
F4*1ror(5,s1:[hlt.W] Exercise 9.20
(a)
As
o
sQ4
:
a:
o)
If
(c)
If a
:
+ 0 (x) sqx.
and7|O)) are independent so
:
sQy: (.05X.03)
3, fromproblem 19
-3, fromproblem
.0015
sQd:
19
.000266
srv:
.004232
Exercise 9.21
In general, dfr
:
au
*
du
-
auv,so
o@A
:
ctry
* ai -
a"y.nl
:
a4
* rlar.
This aruruity will pay until the first failure out of x andy, or until time n, whichever is later. Thus, it pays for nyearc for certain, and beyond that as long as the joint status (ry) survives.
Exercise 9.22
will
This insurance
:
V*r=,
:
7* +V;t -Vr,;1, where ZA
u n.
(vn Tln
tetZ- J trt T) n
Altematively,
7.: x:nl
Then
pay atthe death ofx, or at time z, whichever is later.
EI4 :
vn nQ,
*
nl7,
- 9"'a-v'nP') * *Ar-A*4*vn Exercise 9.23
Cov [v?@), ,r@Df
: : :
: :
: ," -
yn -
Zt,A
*
- np")
Vn npx
[email protected])]
- n[vw]n[vw1
ElvT\n . urrDl
z[vzwtlnlvxvtl
-
v, (l
nfvx,tlafvnt)] _ ZfvrootlnlrW)l due to independence 7: 7r-- Z7'VZ ,: 7*'7,r - (7, *V, -7r)7, 17, -71117, -Vr)
Exercise 9.24
For 0
20, the annuity will p.ay.if either is alive, since both are under age the annuitv wili pav onrv ir (25) i;
.ii*. iy tr,. .uo.nt p"y-r*
':n;r'";:' aPv
f2o
:
J,
:
l"
:
vt ,pE,Ta O,
r, pzs dt
Azs,4
*
*
dzo'fr1
*
r25
Jro
I'o
-
vt
p25 dr
r: pso dt _
I'o
r, *2ssodt
dzs,to,q
Exercise 9.25 kr this case, the annyltf will pay for k 26,... if either is aiive.
and for
:
/r
:
21,...,25 onlyif (30) is alive,
25
Thus apv
: tru b2l
kho
+ iro *pxso tol;
oo
: Dur kPn * f ,o kpzs t'r F2t n:ie
:
oo
lF26
zolazo
*
zslazs
-
zslazs,to
oorr,ro
50.
For technique of
148 o
CamF l0
Exercise 10.4
,pg
p{l' p9' p[)
: :
ll
-
(.02
+.os)l[1
:
(.93X.91X.89)
: ,qtl : et#--e
,k[']
:
l-*)ll,-osl
f'-nsl ,p{l
.qtl
-
(.03 +.06)][1
.7s3207
(.7s3207)(.0s)
:
-
(.04
+.07)]
.0376603s
: if# :
Jrso4,usingthetableof Example 10.3.1
Exercise 10.5
(a)
:
Probability of graduation ir op[")
.3024.
Then the number of graduates, G, is a binomial R.IZ., with
n
:
p : .3024. ElGl:nP:302.4 1000,
V(G)
:
np(I
- p) :
210.95424
(b) Similarly, number of failures,4 is binomial with n : 1000, and p: 4qt): .15 * (.60X.10) + (.60X.70)(.05) : .231 ThenE[F] : np:231, V(F): np(l-p):177.639 Exercise 10.6
k
_a_
0
1000
I 2
600 420
3
336
al'
:
"60
d?
4)ql])
250
150
120
2l
63 33.6
0
res
BT
(a) fie)
: aE#rl* :
fi(z) :
.231
total others withdrawals
T
fi(3): q#:
: $ .q?)
W:
:
.4666
so24
: 2) : Pr(termination at k : 2 iamodeT) Thenf(l lk : 2) : # : .25 andfl(Zlk : 2) : # : .75 Of coursef (3lk:2) : 0.
(b) fi(i lk -
2) . Pr(k
Excrcin f0.? (a) From (10.4.1),
*) :'
* -l-[
p'U)
* rrot
ar]
-''*[-1,t, * &)dvf :;.*[-'+ k+]: (b) 4D : @)
8)
fo' tJ,r:r<,1o,
: I'c{J,p!r)@at:
:
lo'
: "*-, at "-x - "-x-l
fo'%lL:ry
: - (a - x - t) "--,l,to -
:'-'
Irt
e-*-t dt
I "-x-' .:-; :"; -? -:':l-"':'"-x
Erercise 10.8 Aga^,
*)
roooexp[-1,'"* 1000exp
Erercise 10.9
(a) *,n9'
&*]
[-* * n (' - *) - rn | :
1000
e-d
-t
: -4& F) dx
:-
(b)
t,u9
:
*l,lry,ri', ,o,r.ff') :
:
Lry clx ltr)
- r<,t1gf
- 0.,
bt,pP
(c) #,nP
og,[ug,a>
*)pa(,) *
p?al
*,ryff'l* (c-\*,)l,tatl
* ,oyt ro@) - pa(r)
: #W:
fi[4],rgatl:
,pg)pga>
(a-x1s-,
I
{? r
Chanter
l0
Exercise 10.17
Firstwe
findpf)
rhus,pf]
pg p{} rlter_q]|/2
qt)
: : : :
[r
- n't"1 [t -
Similarly, qt]
qr)
:
.23952
.85027,
h\ : qi;
.t4973
.821,1,5,
q{} :
.17885
:
qzl :
q?:
.02578;
o']
: - .87478166 tnp'S
.01767; qg)
.02054;
qg
-f
.76048,
: ffi'mP', :
o'o'] [t
.02665 qg) :
.Igszo
:
.0s726
.031e3; q?l
.03705; q?]:.11603
Exercise 10.18 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.
Exercise 10.19
(a) *',a on9) is justified by the constant force assumption. (b) Accepting that d.a x m9, *"n i d"r?:i6 = ; f :O. if r-l'lx r-2'Yx
decrements are
uniformly distributed in both the single decrement tables and the multiple decrement model.
(c)
Clearly this leads to
(d) l,'1, d,alr
-
L nl'l:
*
n9'
+
t
4f) = oglt
t'
- i'qf)
- L n:,'1, which in turn implies
og)
:
q9, or d,a
Exercise 10.20
: 4* m'a:4 *9
,orto
v.e.,-tl:
cfu
:
#
-q
:
'02073'etc'
