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Unformatted text preview: SECTION 22 PRACTICE EXAMINATION NUMBER 18
SOLUTIONS 1. Study Note P0908, Problem No. 126
Under an insurance policy, a maximum of ﬁve claims may be ﬁled per year by a policyholder.
Let p,l be the probability that a policyholder files n claims during a given year, where n = 0,1,2,3,4,5. An actuary makes the following observations:
(i) p" 2 p”+1 forn = 0.1.2.3,4.
(ii) The difference between p“ and p"+1 is the same for n = 0,1,2,3,4. (iii) Exactly 40% of policyholders ﬁle fewer than two claims during a given year.
Calculate the probability that a random policyholder will ﬁle more than three claims during a
given year. A. 0.14 B. 0.16 C. 0.27 D. 0.29 E. 0.33
Solution.
We know that p0 + p1 =% and po+P1+P2+P3+P4+Ps=L
Let pn—pn+l=c forall nS4. Then pn=ponc for ISnS5.Thus po+(po—c)+(po—2c)+...+(po5c)=6po_15c=1_ Also
2
P0+P1=Po+(Po_c)=2Poc=g
This results in the system of equations
6po—15c=1,
2
2 — =—,
P0 C 5
from which we et ci and —2—5 Based on this
g _60 p° 120‘
17 15 32
+ = —4 + —5 =—+—=——=0.2667.
1’“ p5 0’” c) (p° c) 120 120 120
AnswerC. 2. November 2003 Society of Actuaries Course 3 Examination, Problem No. 34
You are given: (i) Losses follow an exponential distribution with the same mean in all years. (ii) The loss elimination ratio this year is 70%. (iii) The ordinary deductible for the coming year is 4/3 of the current deductible. ASM Study Manual for Course Pll Actuarial Examination. © Copyright 20042008 by Krzysztof Ostaszewski  614  PRACTICE EXAM 18 (w, Compute the loss elimination ratio for the coming year.
A. 70% B. 75% C. 80% D. 85% E. 90%
Solution. Let us write d for a deductible, and X for the loss covered by the policy, assumed to be
exponential with unknown mean [1. The loss elimination ratio equals :1 d _5
. s xdx e"dx xx=d
__E(mm<Xd» I X" J 1H7 ]...:.
x=0
" 10 0
In the current year, this equals 0.7, so that l— e7 = 0.7, or d = plug. In the coming year, the 1500 — E (X) it u
. . 4 10 . . . . .
deductible IS Eulng, so that the loss elimination ratio IS gumn
3
_._]n_ 4
3 4 10 '
l—e .. =1—e3 3:1[§)3=o.7992. Answer C. 3. November 1986 Course 110 Examination, Problem 36
Let X,, X2, and X3 be a random sample with replacement from a discrete distribution with probability function 0.4 forx = —3,
Pr(X=x)= 0.6 forx=6,
0, otherwise. Let Z = %(XI + X2 + X3) be the sample mean. What is the probability function for X ? 0.064 forx = —3,
0.4 forx = —3, 0.360 forx = 0,
A. Pr(X=x)= 0.6 forx=6, B. Pr(X=x)= 0.360 forx=3,
0, otherwise. 0.216 forx = 6, 0, otherwise. ASM Study Manual for Course Pll Actuarial Examination. © Copyright 20042008 by Knysztof Ostaszewski  615  SECTION 22 0.25 forx = —3, 0.064 forx = 3,
0.25 forx = 0, 0.096 forx = 0,
C. Pr(X=x)= 0.25 forx=3, D. Pr(X=x)= 0.144 forx=3,
0.25 forx=6, 0.216 forx=6,
0, otherwise. 0, otherwise.
0.064 forx = 3,
0.288 forx = 0,
E. Pr(X=x)= 0.432 forx= 3,
0.216 forx = 6,
0, otherwise.
Solution.
We have X _ —3 with probability 0.4,
’ 6 with probability 0.6. Therefore,
9 with probability 0.43 = 0.064,
3
0 with probability [2} 0.42 0.6 = 0.288,
X1 + X2 + X3 = 3
9 with probability [1}04062 = 0.432,
18 with probability 0.63 = 0.216,
and
—3 with probability 0.43 = 0.064,
3
0 with probability 0.42 .06 = 0.288,
— XI «1 X2 + X3 2
X = _ =
3 3
3 with probability 1 0.4  0.62 = 0.432,
6 with probability 0.63 = 0.216,
Answer E. 4. November 1986 Course 110 Examination, Problem 38
Let XI and X2 be independent random variables, each with density function 2—2x for0<x<1, x = .
fx ( ) {0 otherw15e.
ASM Study Manual for Course P/l Actuarial Examination. © Copyright 20042008 by Knysztof Osmszewski  616  PRACTICE EXAM 13 What is the probability that exactly 1 of the 2 variables exceeds %?
l 6 7 8 9 A — B — C. — D — E —
16 16 16 16 16 Solution. First, 34%): H(X>% =) J:'(2— —2x)dx= (2x— 1:): 2 :‘2=(2—1)[1%]=:12. 1
Let Y be a Bernoulli trial with success deﬁned as X (with PDF fx) exceeding 2. We perform two such Bernoulli trials and want the probability that the total number of successes is one. That 2
probability is [1]'_'_=_' Answer B. 5. November 1986 Course 110 Examination, Problem 40
Let X be a continuous random variable with density function e_ l
e2’+2e" for0<x<oo, fx(x)= 0 otherwise.
Let Y = e—zx . Then for 0 < y <1 the density function for l: is given by f, (y) = 2y+J§ 2+5 3 1 1 (—32y+\/§)
A. 2 B. 422 C. 2y +y .—+— E. D'24J‘ 5 Solution. 1 ’2‘ is onetoone and its inverse is x= —§ln y. The derivative of the The transformation y = e dx 1 inverse is F=2—.Note that for O<x<oo we have 0<y=e'2"<1.Therefore
y y
dx 42:») 1[§my] 1 ( 1y)1 1
+—e ——= + — —+—.
Hymn WW 2 2yy2‘/'2y=24‘/—.
AnswerD. ASM Study Manual for Course P/l Actuarial Examination. © Copyright 2004—2008 by Krzysztof Ostaszewski  617  SECTION 22
6. November 1986 Course 110 Examination, Problem 41 Let E (XIY = y) = 3y, Var(XY = y) = 2 and let Y have density function
_ e” fory > 0,
My) _ {0 otherwise.
What is Var(X)?
A.3 3.5 C.9 D.ll E.20 Solution.
Note that Y is exponential with mean 1, so that its variance is the square of 1, i.e., also 1. We
have Var(X) = Var(E(XY)) +E(Var(XY)) = Var(3Y)+ 13(2) = 9Var(Y) + 2 = 9 + 2 =11.
Answer D. 7. November 1986 Course 110 Examination, Problem 42
Let X and Y be random variables with joint density function f ( ) 24xy forx>0,y>0,andx+y<1, x, =
x“? y 0 otherwise.
What is the conditional probability Pr[X <% Y = i)?
A. —l B. i C. l D. 3 E. 3 4 9 2 3 4
Solution.
The condition x + y < 1 is equivalent to x < l — y, and for y = i we have 11 2
1 1 4 1 ‘ 2 x=2 27
3(2) = aujxf“, [x.2)dx= i 24»de = £62m 3x ”3 _ E“ Therefore 21(le
fx[x‘y=l]=  4 _6x_162x_32 for 0 < x <: and zero otherwise. Finally ASM Study Manual for Course Pll Actuarial Examination. © Copyright 20042008 by Krzysztof Ostaszewski  618  PRACTICE EXAM 18 Answer B. 8. November 1986 Course 110 Examination, Problem 43 LetSand The independent events, Pr(SnT)=%, and Pr(SnTc)=%. m((5ur)°)=
11.1 13.2 c.1 13.1 13.1
10 3o 15 15 10
Solution.
Notethat
Pr(s)=Pr(SnT)+Pr(SnT‘)=l+l=3.
10 5 10 and due to independence 1
PIS T _ l
”(T)=%=%=§ 10 Therefore
Pr((SuT)C)=l—Pr(SUT)=1—(Pr(S)+Pr(T)‘Pr(5nT))'' 3 1 1 1 l 15 3 5 7
=1— —+——— =1———=———————_—,
5 3 15 15 15 15 AnswerC. 9. November 1986 Course 110 Examination, Problem 44
Let (X ,Y ) have joint density function ( )_ 2 for0<x<y<l,
f“ x,y _ 0 otherwise.
ForO <x< 1,what is Var(YX=x)?
_ 2 _ 3
Al B. (I x) C.l+—x D. 1 x
18 12 2 3(lx) E. Cannot be determined from the given information E Solution.
" ' ASM Study Manual for Course P/l Actuarial Examination. 9 Copyright 20042008 by Krzysztof Osmszewski  619  SECTION 22
For a given value of x we have 0 < x < y < 1. Therefore, fx(x)= Ifx.y(x.y)dy=j2dy=2(1x) any x
for0 <x<1.This gives
fr (YIX = x)= fxr (x,y) ____ 2 = l fx(x) 2(1—x) l—x for x < y < 1. This, of course, is constant with respect to y, and thus this distribution is uniform
on the interval [x,l]. Of course, we could have seen this from the very beginning: the joint
density is constant, thus the joint distribution is uniform, and so the conditional distributions
must be uniform as well. You can see that for a ﬁxed x, the values of the joint density are
positive only for y in the interval [x,l], just by looking at the joint density formula. Therefore,
the variance sought is given by the wellknown formula for the variance of the uniform
distribution Var(YX = x): (11:): . Answer B . 10. November 1986 Course 110 Examination, Problem 47 Let (X ,Y) be the coordinates of a point randomly chosen on the xyplane, and let R = \l X 2 + Y 2
be the distance from (X ,Y ) to the origin. If X and Y are independent, normally distributed random variables, each with mean 0 and variance 0'2, what is the value of r such that the
probability that R exceeds r is 0.95? A. 0.10'2 B. 03166 C. 1.710' D. 2.450' E. 5.9962 Solution.
We have 0.95 =Pr(R>r)=Pr(\/X2 +Y2 > r =
_ 2 _ 2 2
p.(xz.yz>.2)pr[[2r_°) +[y_0] _] 0' 0'
[X—OJZ [if—0T
+ _
0' 0'
is chisquare with two degrees of freedom, which is the same distribution as exponential with
mean 2. Therefore, But ASM' Study Manual for Course Pll Actuarial Examination. © Copyﬁgh120042008 by Krzysztof Ostaszewski  620  PRACTICE EXAM 18 _ 2 _ 2 2 _i
0.95=Pr[[x 0) +[Y 0) >r—2]=e 26’.
O' 0' O' This means that r2 = 2 1110.950“. From this, we obtain r = owl2 In 0.95 = 0.32030". Answer B. 11. November 1986 Course 110 Examination, Problem 48
The random variable X has density function Ic+l "
cx 1x for0<x<1,
fx(x)= ( ’ .
0 otherWISe,
where c>0 and 1< k<2. What is the mode ofX? A. k“ B.k+l c.—k D.—k+l E. “2
2k+l k+l 2k 2k+3 Solution.
The mode is the point where the density reaches its highest value. Here, the density outside of the interval (0,1) is zero, so we look for the mode only inside that interval. Inside that interval, we have f; (x) = fixox“l (l  x)‘ = C(k +1)x"(l— x)k  clock“ (1  x)"l . In order to ﬁnd the maximum, we set the derivative equal to 0 and obtain
C(k + 1)x" (l  x)k  ckx"+1 (1  x)"1 = 0, or
(k+1)(1—x)—kx=0,
01'
k+1
x= .
2k+1 Note that since c > 0 and 1<k<2, the density is zero at 0 and at 1, while positive and differentiable in between. This means that the point at which derivative is zero gives the
maximum of the function.
Answer A. 12. November 1986 Course 110 Examination, Problem 50
Let X have the density function 32 f (x) = {e forx < 2, 0 otherwise. What is the 75th percentile of X?
ASM Study Manual for Course Pll Actuarial Examination. (9 Copyright 20042008 by Knysztof Ostaszewski  621  SECTION 22 A.2+1n2 112—1113 C.1n[1+3e2] D.—1n[1—3e2) 1=..2+1nl
4 4 4 4 4 Solution.
The 75th percentile of X, let us write x03, for it, must satisfy the condition
*0 ’on .75
0.75 = Fx (xm) = J' fx (x)dx = J‘ exZax = e"2 —oo 3:30.75 .—
= €10.75 1 _
x—)—n Therefore xt,7s = 2+1n0.75 = 2+1n%. Answer A. 13. May 1988 Course 110 Examination, Problem No. 3
A fair coin is tossed. If a head occurs, 1 die is rolled; if a tail occurs, 2 dice are rolled. If Y is the total on the die or dice, then Pr(Y = 6) = A. l B. i C. 1—1 D. l E. 1—1
9 36 72 6 36
Solution.
Let us write X for the random outcome of the coin toss. Then
Head with probability 1 ,
X _ 2
_ 1
Tail with probability 2'
Also Pr(Y = 61X = Head) =% and Pr(Y = 6X = Tail) = 35—6. Based on the law of total
probability,
Pr(Y = 6) = Pr(X = Head)  Pr(Y = 6X = Head)+ Pr(X = Tail)  Pr(Y = 6X = Tail) =
_ 1.1 + l _5_ _ E
2 6 2 36 72 '
Answer C. 14. May 1988 Course 110 Examination, Problem No. 44
Let X“ X2, X3, X4, and X5 be independent normally distributed random variables each with mean 2 and standard deviation 3. Which of the following has a chi—square distribution? ASM Study Manual for Course P/l Actuarial Examination. © Copyright 20042008 by Krzysztof Ostaszewski  622  PRACTICE EXAM 18
5 14.25le 3%in C.%i(X,—2)2 D.%Z(X,—2)2 E.%i(Xi—2)2 1:! i=1 i=1 i=1 in] Solution.
Chisquare distribution is obtained as a sum of independent identically distributed standard normal random variables, so we need to standardize these variables, add those standardized variables, and hope we get one of the answers. In fact
5 5 x. —2]2 1 2 —‘— = — X — 2
g 3 9g, )
is exactly the expression in D.
Answer D. 15. May 1990 Course 110 Examination, Problem No. 1
Let R, S, and T are independent, equally likely events with common probability %. What is Pr(RuSuT)?
A —1 B. 3 C. 2 D. 2—6 E. l
27 3 27 27
Solution.
We have
Pr(RUSUT)= =Pr(R)+Pr(S)+Pr(T)—Pr(RnS)Pr(SnT)—Pr(RnT)+Pr(RnSnT)= = Pr(R)+ Pr(S)+ Pr(T)— Pr(R)  Pr(S) Pr(S)  Pr(T) Pr(R).Pr(T) +
+Pr(R).Pr(s)Pr(T)= 111111111 11_27—9+1_Q + ————————————— + ————— =l——+_ 1 l
=—+ —— .
3 3 3 3 3 3 3 3 3 3 3 3 27 27 27 Answer C. 16. May 1990 Course 110 Examination, Problem No. 2
Let X and Y be continuous random variables with joint density function 0.25 forOSxSZandx—ZSny, fx.y(x.y)={0 otherwise.
What is E(X2Y)?
A. 2 B. i C. 2 D4 E. 344
3 3 5 ASM Study Manual for Course P/l Actuarial Examination. © Copyright 20042008 by Krzysztof Ostaszewski  623  SECTION 22
Solution. We have Answer A. 17. May 2000 Course 3 Examination, Problem No. 30
X is a random variable for a loss. Losses in the year 2000 have a distribution such that: E(XAd)= 0.025d2 +1.475d—2.25 for d = 10, 11, 12, ..., 26, where X A d = min(X,d). Losses are uniformly 10% higher in 2001. An insurance policy reimburses 100% of losses subject to a deductible of 11 up to a maximum
reimbursement of 11. Calculate the ratio of expected reimbursements in 2001 over expected
reimbursements in the year 2000. A. 110.0% B. 110.5% C. 111.0% D. 111.5% E. 112.0% Solution.
We will conﬁne the notation X to the 2000 loss. As the policy pays all losses subject to a
deductible of 11 up to a maximum reimbursement of 11, the amount paid by the policy in 2000 is Y = (X A 22)— (X A 1 1) = min(X,22)— min(X,11).
In 2000, expected value of that amount paid was E(Y) = E(XA 22) E(XA11)= = (—0.025  222 +1.475  22 2.25)—(—0.025 112 +1.475 11 2.25) = 7.15. In 2001, the loss is 1.1X, where X is the 2000 loss. The amount paid by the policy in 2001 is W = (1.1x A’22)(1.1XA11)= min(1.1X,22)— min(1.1X,11)= = l.1min(X,20)1.1min(X,10)= 1.1(min(X,20)min(x,10)). Therefore E(W) = E(1.1(min(x,2o) min(X,10)))=1.1(E(min(X,20)) E(min(x,10))) = = 1.1((41025 2o2 +1.475 . 20 — 2.25)(—o.025 .102 + 1.475 10 — 2.25)) = 7.975.
The ratio sought is ASM Study Manual for Course Pll Actuarial Examination. © Copyright 20042008 by Krzysztof Ostaszewski  67A  PRACTICE EXAM 18 & E(W)=_7_9_7_§=1.1154=11154%.
E(Y) 7.15
Answer D. 18. November 2000 Course 3 Examination, Problem No. 27 Total hospital claims for a health plan were previously modeled by a Pareto distribution with
a = 2 and 9 = 500. The health plan begins to provide ﬁnancial incentives by paying a bonus of
50% of the amount by which total hospital claims are less than 590. No bonus is paid if total
claims exceed 500. Total hospital claims for the health plan are now modeled by a new Pareto
distribution with a = 2 and 6: K. The expected claims plus the expected bonus under the
revised model equals expected claims under the previous model. Calculate K. A. 250 B. 300 C. 350 D. 400 E. 450 Solution.
9 a  l
0 = 500, this gives the mean of 500. With the new parameters of a = 2 and 9 = K the expected
9 a—l S (X) = 6“ = [ 6 J“
x (x + 9)“ x + 9 '
Let X be the amount of claims under the new model. The bonus equals %(§00 — nun(§1590));
Recall this version of the Darth Vader Rule: 7 The expected value of a Pareto distribution with parameters a and 0 is , and for a = 2, claims are = K. Recall that the survival function of the Pareto distribution is E(min(X,a)) = E(X A a) = jsx (x)dx.
0
Using it, we obtain the expected value of the bonus under the new model £64500 — min(x,500))] = 5500— E(min(X,500))) = 250 %5}0[ K )2 = x=500 2 2
=250— _K_+K_ =
2(500+K) 2K =250—[—%K2 ~(x+K)"] x=0 K2 K 5002
= 250+——— = —.
2(500 + K) 2 2(500+ K)
This means that 5002 500=K+——,
2(500+K) w 01' ASM Study Manual for Course Pll Actuarial Examination. © Copyright 20042008 by Krzysztof Ostaszewski  625  SECTION 22 5002 + 1000K =1000K + 216,
resulting in =§® = 353 5.534 J5 Answer C. 19. May 2001 Course 3 Examination, Problem No. 25
For a discrete probability distribution, you are given the recursion relation fx(k)=k2 "fo‘ 1)
fork: 1,2,3, ....Determine fK(4). A.0.07 B.0.08 C. 0.09 D.0.10 E. 0.11 Solution.
Note that the random variable K considered here only assumes values of 0, l, 2, ..., i.e., non
negative integers. We have 2 2 fx(k)=fx(k1)=ZﬁfAkZF~=;'k—_T~ffx(0)=%fx(0) Therefore gfo‘): 2%: fx(0)= fx(0) 2—=fx (0) ~2.e But the sum of all values of the probabihty function must be one, so that fx (0)e2 =1 and therefore fK (0)oe2 =1. Based on this we see immediately that K has Poisson distribution with 4
mean2so that fK(4 )= e'z i—z0.().902 Answer C. 20. November 2001 Course 3 Examination, Problem No. 28
The unlimited severity distribution for claim amounts under an auto liability insurance policy is given by the cumulative distribution Fx (x) = 1— 0.85”“ — 0.2e'om" for x 2 0. The insurance policy pays amount up to a limit 1000 per claim. Calculate the expected payment under this
policy for one claim. A. 57 B. 108 C. 166 D. 205 E. 240 Solution.
We have sx (x) = 1— Fx (x) = 0.8  e'om" + 0.2  e'oml".
ASM Study Manual for Course Pll Actuarial Examination. © Copyright 20042008 by Knysztof Ostaszewski  626  PRACTICE EXAM 18
Because of the policy limit, the payment will be min(X,1000). The expected value we want to calculate is E(min(X,1000)) and it equals 1000 l000 1000
E(min(X,1000))= j sx(x)dx=0.8 j e'm’dx+0.2 j e‘°”°“dx=
0 0 0
= 123.. (—e*’°2*) "'°°° + £ . (—e'm“) "'°°° z 166.4241.
0.02 x=° 0.001 x=°
Answer C.
21. November 2001 Course 3 Examination, Problem No. 35
The random variable for a loss, X, has the following characteristics:
Calculate the mean excess loss for a deductible of 100.
A. 250 B. 300 C. 350 D. 400 E. 450
Solution.
The mean excess loss equals
I sx (x)dx l
——‘°° = E X—min X,100 E=
3x000) sx (100) ( ( ))
=5(x) becausel’x (1000)::
r—*—
_ E(min(X,]000)) —E(min(X,100)) _ 331 — 91 _ 300 l—FX(100) 1—0.2
Answer B. 22. November 2001 Course 3 Examination, Problem No. 36 WidgetsRUs owns two factories. It buys insurance to protect itself against major repair costs.
Proﬁt equals revenues less the sum of insurance premiums, retained major repair costs, and all
other expenses. WidgetsRUs will pay a dividend equal to the proﬁt, if it is positive. You are
given: (i) Combined revenue for the two factories is 3. (ii) Major repair costs at the factories are independent. (iii) The distribution of major repair costs for each factory is @ ASM Study Manual for Course P/l Actuarial Examination. (9 Copyright 20042008 by Knysztof Ostaszewski  627  SECTION 22 (iv) At each factory, the insurance policy pays the major repair costs in excess of that factory’s
ordinary deductible of l. The insurance premium is 110% of the expected claims. (v) All other expenses are 15% of revenues.
Calculate the expected dividend. A. 0.43 B. 0.47 C. 0.51 D. 0.55 E. 0.59 Solution.
The insurance benefit for each factory is 2 — 1 = 1 with probability 0.2 and 3 — 1 = 2 with probability 0.1 , and the expected value of that beneﬁt is 0.2 1+ 0.1  2 = 0.4. The total for both factories is 0.8. The insurance premium per factory is 110% of that, i.e., 0.88. Also, the losses
paid by WidgetsRUs, either under the deductible, or equal to the deductible, are 0 with probability 0.4 ~ 0.4 = 0.16,
X = l with probability 2 0.4 06 = 0.48,
2 with probability 0.6 . 0.6 = 0.36. The proﬁt of WidgetsRUs is
' 1.67 with probability 0.16, 3 0.88 — 0.15  3 X = 1.67 — X = 0.67 with probability 0.48,
—0.33 with probability 0.36. The dividend will not be paid when the proﬁt is negative. Therefore, the expected value of the
dividend is 0.16 1.67 +0.48 0.67 = 0.5888.
Answer E. 23. Study Note P0908, Problem No. 127
Automobile losses reported to an insurance company are independent and uniformly distributed
between 0 and 20,000. The company covers each such loss subject to a deductible of 5,000. Calculate the probability that the total payout on 200 reported losses is between 1,000,000 and
1 ,200 ,000. A. 0.0803 B. 0.1051 C. 0.1799 D. 0.8201 E. 0.8575 Solution.
Because the number of payouts (including payouts of zero when the loss is below the deductible) is large, and losses are independent and identically distributed, we can apply the Central Limit
Theorem and assume the total payout S is normal. For a given loss there is no payout with
probability 0.25, and otherwise, with probability 0.75, the payout is uniformly distributed ASM Study Manual for Course Pll Actuarial Examination. © Copyright 20042008 by Krzysztof Ostaszewski  628  PRACTICE EXAM 18
between 0 and 15000. Therefore, if we write X for the payout, E(X) = 0.25 0 + 0.75  7500 = 5625,
and
150002 E=(X’) 0 .25. o2 + o 75 (75002 + J: 56,250,000.
Therefore, the variance of a single claim payout is Var(X) = E(X2)— E (X )2 = 24,609,375. Applying the Central Limit Theorem, and writing Z for a standard normal random variable and
(I) for the CDF of Z, we conclude that Pr(1000000 < S < 1200000) = _ Pr[1000000— —200 5625< s 200 5625 <1200000 200. .5625)
Jzoo 24609375 <J200 24609375 Jzoo 24609375 = Pr(—1 .7817 < z < 1.0690) = <1>(1.069o)— <D(1.7817)= = <1)(1.0690)+ <1>(1.7817)—1== 0.8577 + 0.9625 — 1 = 0.8202.
Answer D. 24. November 2002 Course 3 Examination, Problem No. 16 Prescription drug losses, S, are modeled assuming the number of claims has a geometric
distribution with mean 4, and the amount of each prescription is 40. Calculate the expected value
of the excess of S over 100. Note: Assume the geometric distribution counts the number of
failures until the ﬁrst success in a series of independent identically distributed Bernoulli Trials. A. 60 B. 82 C. 92 D. 114 H.146 Solution. Let N be a geometric random variable with mean 4, as deﬁned in this problem. Since the
distribution counts the number of failures (as opposed to the number of attempts), l—_p = 4, and
P the probability of success in the underlying Bernoulli Trial is p = 0.20. The total claim amount is
S = 40N, and its expected value is 160. The excess of S over 100 is S—min(S,100), and it equals exactly the amount paid in a policy with a deductible of 100. S will exceed 100 if N = 3, 4, 5, etc., and O with probability Pr(N—  0)—  0.20, min(S,100)= 40 with probability Pr(N: l)= 0.800.20 = 0.16, ' 80 with probability Pr(N ==2) 0 .802 0.20 = 0.128,
100 with probability Pr(N > 2): 1— 020— 016—0128 = 0512. This gives E(min(S,100)) = 0.200 +0.1640 +0.128  80 +0512  100 = 67.84.
The expected value we are looking for is ASM Study Manual for Course PI] Actuarial Examination. (9 Copyright 20042008 by Krzysztof Ostaszewski  629  SECTION 22
E(S — min(S,100)) = E(S) — E(min(S,100)) = 160 — 67.84 = 92.16.
Answer C. 25. Study Note P0908, Problem No. 128 An insurance agent offers his clients auto insurance, homeowners insurance and renters
insurance. The purchase of homeowners insurance and the purchase of renters insurance are
mutually exclusive. The proﬁle of the agent’s clients is as follows: i) 17% of the clients have none of these three products. ii) 64% of the clients have auto insurance. iii) Twice as many of the clients have homeowners insurance as have renters insurance. iv) 35% of the clients have two of these three products. v) 11% of the clients have homeowners insurance, but not auto insurance. Calculate the percentage of the agent’s clients that have both auto and renters insurance. A. 7% B. 10% C. 16% D. 25% E. 28% Solution.
Let H be the event of a client having a homeowners insurance, R be the event of a client having a
renters insurance, and A be the event of a client having an auto insurance. We are given that Pr((HuRuA)C)=0.l7, so that Pr(HuRuA)=0.83. But
0.83=Pr(HuRuA)=
=Pr(H)+Pr(R)+Pr(A)Pr(HnR)—Pt(HnA)—Pr(RnA)+Pr(HanA)=
=(Pr(H)—Pr(HnA))+(Pr(R)—Pr(RnA))+0.64=
=Pr(HnAC)+Pr(RnAC)+0.64=0.11+Pr(RnAC)+O.64=Pr(RnAC)+0.75.
This means that Pr(RnAc) = 0.08. Also, Pr(H nA)+Pr(RnA) = 0.35 while
Pr(H):Pr(HnA)+Pr(HnAc)=Pr(HnA)+0.ll=
=2Pr(R)=2(Pr(RnA)+Pt(RnA‘‘))=
=2Pr(RnA)+0.16.
This means that we simultaneously have Pr(H n A)+ Pr(R nA) = 0.35 and
Pr(HnA)=2Pr(RnA)+0.05, so that 2Pr(RnA)+0.05 +Pr(RnA) = 0.35, and Pr(R n A) = 0.10.
Answer B. 26. November 2003 Society of Actuaries Course 3 Examination, Problem No. 29
The graph of the density function for losses is: ASM Study Manual for Course P/l Actuarial Examination. © Copyright 20042008 by Krzysztof Ostaszewski  630  PRACTICE EXAM 18 0 80 120
Los amount, x Calculate the loss elimination ratio for an ordinary deductible of 29. A. 0.20 B. 0.24 C. 0.28 D. 0.32 E. 0.36
Solution.
From the graph we determine that 0.01 for 0 S x S 80, fx(x)= 0.030.00025x for80Sx5120, (WW, 0, otherwise. Based on this
80 120
E(X) = jo.01xdx+ J'(o.03— 0.00025x)xdx =
0 80 x=120 x=tzo _ 0.00025
x=80 3 x=80 Let Y will be the losses eliminated by a deductible of 20. Then Y = min(X,20) and = 0.0051:2 x3 z 50.6667. :"+o.015x2 X:
x: 20 120 20
E(Y) = Ixfx(x)dx+ I20«fx(x)dx= Io.01m+2on(2o s X 3120):
0 20 0 = 0.005}:2 :f’ + 200 Pr(X < 20)) = 2+ 200 200.01) = 13. The loss elimination ratio equals M = 18 = 0.3553.
E (X ) 50.6667
Answer E. 27. November 1986 Course 110 Examination, Problem 35
Let Z1, Z2, Z3 be independent random variables each with mean 0 and variance 1, and let ASM Study Manual for Course P/l Actuarial Examination. © Copyright 20042008 by Knysztof Ostaszewski  631  SECTION 22
X = 2Zl — Z3 and Y = 2Z2 + Z3. What is the correlation coefﬁcient forX and Y? A. —1 B. —1 C. —1 D.0 E. 2
3 5 5 Solution. We have E(X) = 5(221— 23) = 2E(Zl)— E(Z3) = o, E(Y) = 15(2z2 +z3) = 2E(Z2)+E(Z3)= 0, Var(X)= Var(2zl z,)= 4Var(Zl)+Var(Z3) = 5, Var(Y) = Var(2z2 + 2,) = 4Var(Z2)+ Var(Z3) = 5, E(XY) = 13((2zl — 23) . (2z2 + Z3» = 15(4z,z2 + 22,23 — zzzz3 — 2;) =
= 4E(Z,Z2)+ 2E(z,z3) — 2E(Z2Z3)— E(z§) =
=Muzak(25+23(z,)E(zs)—28(22)::(23)(Var(z3)+(E(z,))2)=
=0+0—0—(1+0)=—1. Therefore,
Cov(XY) =)E(XY— _ 10 f___1_
Y—‘IVar(X)‘IVar(Y) ‘IVar(X)E(‘Iva:YY) 55
Answer C. 28. Let X be an exponential random variable with mean 1. Deﬁne Y = min(X, m), where m is a
positive number. Find the moment generating function of Y. A. M},(t)=— for t¢l, with MY(1)=m+1 B. M,(t)=min(1—1—,e"") for t<1 and M,(t)=e"" for t21 em:
C.M t =—
,0 1,
me"
D. My(t)= tfor t<m and My(t)=oofor th
m—
l—t~e'”'(1")
E. My(t)=———1—t—for t<1and MY(t)=°°fort21
Solution.
Fort¢1, ASM Study Manual for Course P/l Actuarial Examination. © Copyright 20042008 by Krzysztof Ostaszewski  632  PRACTICE EXAM 18 M, (t) = E (e'y) = Te"m("”') e"dx = Te“ e“dx + Te“ e"dx =
0 0 m
(tl): ‘5'" e(vl)ru —l e _
+e""e “=—+e
tl t—l = Te("')’dx + e'"  Te"dx =
o I» am!) e(rl)m _ 1+ te(ll)m _ e(ll)rn — te(Il)m _ l  1_ teMUl)
t  I t l 1 t Fort=l My (1) = E(ey) = Tem(""') ~e"dx = ?e‘ e“dx + Te“ e“dx =
0 0 In =de+e"' Je"dx=m+e"'e'”'=m+l.
0 m Answer A. 29. Study Note P0908, Problem No. 129
The cumulative distribution function for health care costs experienced by a policyholder is
modeled by the function _ ﬁ
Fx(x)= 1 e forx>0,
0 otherwise. The policy has a deductible of 20. An insurer reimburses the policyholder for 100% of health
care costs between 20 and 120 less the deductible. Health care costs above 120 are reimbursed at
50%. Let G be the cumulative distribution function of reimbursements given that the
reimbursement is positive. Calculate C(115). A. 0.683 B. 0.727 C. 0.741 D. 0.757 E. 0.777 Solution.
Let Wbe the unconditional reimbursement amount, and let Y be the reimbursement, given that
the reimbursement is positive (i .e., conditional distribution). We have 0, X S 20,
W: X20 20<Xs120,
(120— 20)+05 ~(X— 120) X >120.
Also,
X 20 X S 120, given that X > 20,
={ 40 +0.5X X >120, given that X > 20.
We conclude that ASM Study Manual for Course P/l Actuarial Examination. © Copyright 20042008 by Krzysztof Ostaszewski  633  SECTION 22 (
= m({2o < x $120X> 20}U{{40+0.5X $115}n{X >120}X> 20}) =
Pr = ({20 < X S120X > 20}u{120 < X $150X > 20})= < .13_0
=W=Fx(uo)=l_e mo =0.727.
Pr(X>20) Answer B. 30. Study Note P0908, Problem No. 130 The value of a piece of factory equipment after three years of use is 100 ~ 0.5" , where X is a 1 for t< '3'. Calculate the random variable having moment generating function M x (t) = 1— 2t
expected value of this piece of equipment after three years of use.
A. 12.5 B. 25.0 C. 41.9 D. 70.7 E. 83.8
Solution.
We have
E(100~0.5x)=100E(O.5X)=100E(e("'°5)x)=100Mx(ln0.5)= & z 41 9
1— 2 ln0.5 Answer C. ASM Study Manual for Course Pll Actuarial Examination. (9 Copyright 20042008 by Kmysztof Ostaszewski  634  ...
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