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Unformatted text preview: SECTION 18 PRACTICE EXAMINATION NUMBER 14
SOLUTIONS 1. The number of typos per chapter of an actuarial examination study manual follows a binomial distribution with n = 5 and p = 0.1 . Given that there are m typos in a given chapter, the number
of calculation errors in the same chapter is 0 with probability 0.60, m with probability 0.30, and m + l with probability 0.10. Calculate the expected number of typos in a chapter given that there
are 2 calculation errors in that chapter. A.1 3.3 c.3 6.3 E2
5 5 5 7 7 Solution.
Let us write X for the random number of typos in a chapter, and onr the random number of calculation errors in the same chapter. We are given that 5
Pr(X = m) =( ]0.1"' 095‘“
m form=0, 1,2, 3,4,5,and
0 with probability 0.6,
(YX = m) = m with probability 0.3,
m +1 with probability 0.1.
Note that Y: 2 has positive probability only forX = 1 or X = 2. Therefore,
Pr(Y=2)= iPr({Y=2}n{X=m})= iPr(Y=2X=m)Pr(X= m): m=0 m=0
=Pr(Y=2X=1)~Pr(X=l)+Pr(Y=2X=2)Pr(X=2)=
= 0.1 . 5 01 0.9“ +0.3 . 10.0.12 093 = 0.054675.
The quantity we are looking for is E (X [Y = 2). It equals
1Pr(X=lY= 2)+2Pr(X=2Y= 2): _Pr({x=1}n{y=2}) ,Pr({x=2}n{r=2}) =1 +2 = Pr(Y=2) Pr(Y=2)
=1_Pr(Y=2X=l).Pr(X=l)+2'Pr(Y=2X=2)Pr(X=2) =
Pr(Y=2) Pr(Y =2)
=1.0.150.1‘0.9“ +2.0.3100.110.9’ =7
0.054675 0.054675 5 The whole calculation can be done slightly more efﬁciently by noticing that ASM Study Manual for Course Pll Actuarial Examination. © Copyright 20042008 by Knysztof Ostaszewski  500  PRACTICE EXAM 14
Pr({x=1}n{r=2}) PI'(X=1Y=2)=—Pr'zy—=2)——=
___Pr_({X=_l}o_{_E})___
' Pr({X=l}n{Y=2})+Pr({X=2}n{Y=2})_
_ 0.1.5o.1o.9‘ _3
‘ 0.1.5o.1o.9‘ +0.3.1o.o.120.93 ' 5’
Pr(X=2Y=2)=1—3=3.
5 5
and
E(XY=2)=1.Pr(x=1[Y=2)+2.Pr(x=2Y=2)=1%+2%=%.
AnswerA. 2. YI is a lognormal random variable such that lnYl is normally distributed with mean 16 and
standard deviation 1.5. Similarly, Y2 is a lognormal random variable such that In Y2 is normally
distributed with mean 15 and standard deviation 2. Y1 and Y2 are independent. Calculate the probability that min(l"l ,Yz) > e“. A. 0.1250 B. 0.1333 C. 0.1474 D. 0.1543 E. 0.1667 Solution. Note that a minimum of two numbers if greater than a speciﬁc value if, and only if, both of them
are greater than that number. Therefore, if we write (I) for the cumulative distribution function of
the standard normal distribution, and use the data from the table, we have Pr(min(Yl,Y2) > e“): Pr({Y, > e'°}n{Y2 > e'6})=
= My, >e“‘)Pr(r2 > e'°)= Pr(lnY, > l6)Pr(lnY2 >16):
=Pr(lnY, 16 > 16115)},{111r'2 —15 >16—15]= 15 15 2 2
=(1<b(0))(1<1>(0.5))=(1—o.5)(10.6915)=o.15425.
Answer D. 3. Let X be a continuous random variable with probability density function fx (x) = xe“ for x > 0 and fx (x) = 0 otherwise. Let Y be the greatest integer less than or equal to X. Find the
expected value of Y. A. 1.3333 B. 1.4545 C. 1.5027 D. 1.6173 E. 2.0000 ASM Study Manual for Course Pll Actuarial Examination. © Copyright 20042008 by Knysztof Ostaszewski  501  SECTION 18
Solution. It is interesting to observe that X actually has the gamma distribution, obtained as the sum of two
independent identically distributed exponential variables with mean 1. Therefore, the mean of X
is 2, and since we can reasonably expect the mean of Yto be smaller than the mean of X, answer E is impossible. Note also that Y = IX]] is a discrete random variable, which is nonnegative almost surely. Therefore E(Y) = gm)! > n): gnu? 2n). Furthermore, for a positive integer n u=x v=—e”‘ du = dx dv= e‘xdx b—ﬁ—d
INTEGRATION BY PARTS Pr(Y2n)=Pr([[X12n)=Pr(X2n)=Txe"dx= +08
_ J'(—e")dx = M" + e'" = (n + l)e"'. This results in E(Y) = 2(12 + l)e"‘ = 2e‘l + 3e’2 + 4e”3 + ...= n=l =2(e'lte'2te'3+e'4 +...)+(e'2 +e"+e"4 +...)+(e'3+e" +...)+...= 1 2 3 1 1
= 28_ + e _ + e _ +...= e _ + e _ (l+e“+e'2+e‘3+...)=
lel l—el l—e1 l—el lel
" ‘1 1 1 2 —
= e ,+ e _,' 1_,=——+—i= e 1,z1.5027.
l—e l—e l—e e—l e—l el (e—1)r Answer C. 4. Four European actuaries, a Dutchman, a German, a Pole, and a Frenchman, are independently
hired to appraise the value of an American insurance company that a Finnish conglomerate wants to buy. The true value of the company is 9 million euros, and each actuary’s estimate is uniformly distributed between 9 1 million euros and 9 + 5 million euros. Find the probability that the actual value of 0 lies between the lowest and the highest estimate. .m C.0.46 13.0.52 E. w A.0.28 B
6+1 6+1 Solution.
Let us write X1, X2, X3, and X4 for the four appraisals. We have ASM Study Manual for Course P/l Actuarial Examination. © Copyright 20042008 by Knysztof Ostaszewski 502 PRACTICE EXAM 14 0, x<0l, 0, x<9l,
Fx‘(x)= %(X(91)), 0—ISxSG+5, = P69“, e1sxse+5,
1, 9+5<x, 1, 9+5<x, for i = 1, 2, 3, 4. Let us write 11,),1/(2),Y(3),Y(4) for the order statistics of the random sample
X,,X2,X3,X4. Then, because 9—] S 9 S 0 + 5, we have ManamaPr {n.izewrse} =1Pr(n.lze)Pr(nse>=
Manually cxclusiveevenu
4 =1—Pr[fS]{X, 29}]Pr(ﬂ{x, 9}): i=1 i=1 =1—ﬁ(1—FX‘ (OD—115‘ (e)=1[1 Q—EHI—[O‘SHT = i=1 4 4
6 6 1296 1296 1296 1296 648 Answer D. 5. Fall 2004 Society of Actuaries Course 3 Examination, Problem No. 24
The future lifetime of a newborn follows a twoparameter Pareto distribution with 6 = 50 and
a = 3. The cumulative distribution function of a twoparameter Pareto random variable T is given by the formula F, (t) = l — for t 2 0, F, (t) = 0 for t < 0. Calculate
E(T — 20T 2 20).
A. 5 B. 15 C. 25 D. 35 E. 45 Solution.
Let us write U = (T  20T 2 20). We have sU(u)=Pr(T20>uT220)=Pr(T>u+20T220)=
_ Pr({T>u+2o}n{T220}) Pr(T>u+20) s,(u+2o) _——=— Pr({T 2 20}) Pr(T 2 20) s, (20)
Therefore ASM Study Manual for Course P/l Actuarial Examination. (9 Copyright 20042008 by Knysztof Ostaszewski  503  SECTION 18 E(T—20T 2 20): E(U)= Isy(u)du =T:P1:—I:((i‘%uO—))du u=t+20 u=20=>t=0 _TPr(T>t+20)
du=dt u—)°o=>t—>°°_ Pr(T.>.20) \—__d
Integration by substitution dt. Based on the CDF F, (t) = l we get s, (t) = 1— FT (t) = for t > 0. Hence 6 a
E(T — 20T 220): E(U) = lama = ﬂ—E‘ﬁeiﬂjaJ—dt = 20+9 +oo a +ee 3 3
=I[ 20+9 J dt: 70 )dt: 70 .(70+t)3+l
o o f—>+°o 20+t+9 70+t 3+l i=0 Answer D. 6. May 2000 Course 3 Examination, Problem No. 17 The future lifetimes of a certain population can be modeled as follows: (i) Each individual’s future lifetime is exponentially distributed with constant hazard rate 9.
(ii) Over the population, 6 is uniformly distributed over (1, 11). Calculate the probability of surviving to time 0.5 for an individual randomly selected at time 0. A.0.05 3.0.06 C.0.09 D. 0.11 E.0.12 Solution.
We use the Fubini Theorem for the appropriate double integral: Pr(T >0.5)= IfT(t)dt=T BTlfT(tI(3=9)«fe(0)d9 dt= T[9T19e’9’1—10d9]dt= _ W—I
05 Gal airﬂow) 05
\__—_v—_J
=frl‘) 9:11 t—m 9:11 [9:11 =0.1 j J' Ge'a‘dt d9=0.l. j e45°d0=0.1. j ewgdG—Te'ojode]:
9=l 9=0 9:0 8=l t=05
s,(05e=9)
_ _ le's's l—e'o"
=0.10(al—“50%aﬁ50%)=0.10[ 0.5 — 0.5 240.1205. Answer E. ASM Study Manual for Course P/l Actuarial Examination. © Copyright 20042008 by Kizysztof Ostaszewski  504  = 35. PRACTICE EXAM 14 B a
7. You are given a random variable X with the moment generating function Mx (t) = ,
where a and ,6 and positive parameters and t < B. Y is the sum of a random sample of size 2
taken from the distribution of X. Determine the coefﬁcient of variation of Y. 1 l 1 ﬁ 1
A. — B. — C. — D. — E. —
J20: J5 J32 J20: 2
Solution.
Because you studied his manual thoroughly, you see immediately that X has gamma distribution
with E(X) =% and Var(X) = g. This implies that E(Y) =2?a and Var(Y)= 2—? The coefﬁcient of variation of Y is therefore Answer A. 8. You failed your ﬁrst actuarial examination taken while in college and, as a result, upon
graduation did not obtain an actuarial job, and had to settle for working at an unpleasant place,
where your boss gets mad at you often, and the number of daily rages of your boss follows the
Poisson distribution with the mean of 3. The number of rages in a given day is independent of
the number of rages in any other day. Unfortunately, you have to work every day, including
weekends. Find the probability that during a particular weekend (i.e., on a Saturday and a
Sunday) you will suffer fewer than 4 rages of your boss. A. 0.1252 B. 0.1512 C. 0.1675 D. 0.1822 E. 0.2021 Solution. Since the number of rages on a given day is independent of the number of rages in any other day,
the total number of rages in a twoday period follows a Poisson distribution with mean 6. Let us
write onr the random total number of rages on a Saturday and the Sunday that follows it. We
have Pr(Y<4)=Pr(Y=0)+Pr(Y=l)+Pr(Y=2)+Pr(Y=3):
4 6 4 <52 4 63 =e'6+e .—+e —+e 1! 2! 3! Answer B. ASM Study Manual for Course P/l Actuarial Examination. © Copyright 20044008 by Krzysztof Ostaszewski  505  SECTION 18
9. May 2000 Course 3 Examination, Problem No. 8 For a twoyear term insurance on a randomly chosen member of a population:
(i) 1/3 of the population are smokers and 2/3 are nonsmokers.
(ii) The future lifetimes follow a Weibull distribution with r = 2 and 0 = 1.5 for smokers, and
1' = 2 and 6 = 2.0 for nonsmokers.
(iii) The death beneﬁt is 100,000 payable at the end of the year of death.
(iv) The interest rate is i = 0.05.
The survival function for a Weibull random variable T with parameters 1' and 9 is
f a
s,. (t) = e_[°‘) . Calculate the expected present value of the death beneﬁt of this insurance. A. 64,100 B. 64,300 C. 64,600 D. 64,900 E. 65,100 Solution.
The random present value of the beneﬁt is: 100,000 _
Z= 1.05"+" k_0’l’
o, k=2,3,..., where k is the number of whole years lived by an insured. We treat this random variable as a mixture of the random present value of the beneﬁt for smokers, Z 5mm” , with the weight of 31;, and the random present value of the beneﬁt for nonsmokers, Z “Mm” , with the weight of Its expected value is the weighted average of the expected values for smokers and nonsmokers,
i.e., = l . E(ZSmokets)+2 .E(ZNonSmokcrs).
3 3
We will calculate the expected values for smokers and nonsmokers separately. Let us start with
smokers. We use the information about the Weibull distribution given E(ZS'*‘°“°“) = 100’000 Pr(Death in year 1) + 1:06:20 Pr(Death in year 2) =
_ a ’  i1 — i 2
= 100,000. 1_ e its] +M. e (15]  e (15) z 77,000.2417.
1.05 1.05
Now we turn to the calculation for nonsmokers
E(ZN°“'sm°“°”) = 10:3;00 Pr(Death in year 1) + 12%;)? 'Pr(Death in year 2) =
_ 1 ’  l ’ — 3 1
= 100,000. 1_ e [2) + 100,020. e [2] _ e (2] z 58,338.3691.
1,05 1.05 Substituting these values we get ASM Study Manual for Course P/l Actuarial Examination. © Copyright 20042008 by Krzysztof Ostaszewski  506 ~ PRACTICE EXAM 14 5(2) = 77,000.2417 + g  58,338.3691 = 64,558.9933. Answer C. 10. In a given student population, 70% major in actuarial science, and 60% of students majoring
in actuarial science have an insurance minor. 15% of the student population major in insurance.
Of the students who major in something else than actuarial science or insurance, 20% have
insurance minor. What is the probability that a randomly selected student from this population is
either an insurance major or an insurance minor? Assume that every student must have a major,
but not all students must have a minor, that a student cannot have more than one major, and that some students major in something else than actuarial science or insurance.
A. 45% B. 50% C. 55% D. 60% E. 70%
Solution. Actuarial science majors who are insurance minors constitute 0.60  0.70 = 0.42 = 42% of the
student population. Insurance majors are 15% of the student population. Majors other than
actuarial science or insurance constitute 100%  70% — 15% = 15% of the population, and 20%
of them, i.e, 3% of the student population, also have an insurance minor. This gives us a total of all insurance majors and minors as
5% + 42% + 5% lnmnce Acnminl science Majors other than
majors mjors who are also atomic! science or
insurance minors insurance who minor
in insurance = 60%. Answer D. 11. The joint probability density function of X and Y is f” (x, y) = 2xy for
0 < x <1,1< y <1, and f” (x,y) = 0 otherwise. Find E(X’Y‘). A. i B. l c. l o. 3 E. l
15 15 5 5 3
Solution. Because the joint density f“, (x, y) is positive only on a rectangle with sided parallel to the x
axis and the yaxis, and on that rectangle f“, (x, y) is a product of a function of x and a function of y, X and Y are independent. This implies that E(X3Y" ) = E(X3) . E(Y‘ ). Furthermore n (x)=jammy=j2x1ydy=2leyldy=2x[lty)dy+jydy]=2x for 0 < x < 1, and fx (x) = 0 otherwise, while ASM Study Manual for Course Pll Actuarial Examination. © Copyright 20042008 by Knysztof Ostaszewski  507   8m  Ema...“ng 355 3 “838” 23.38 © eozéaaxm 3333. E 02.50 a. :35: >an 22
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_ _ _ 3 2020mm PRACTICE EXAM 14 The region where the event {Y < happens is marked in the graph with vertical dotted lines, and the region where the event {Y > X} n {Y < happens is marked with dotted lines parallel to the line given by the condition y = 1:, also shown and marked in the graph. As the joint
distribution is uniform, probabilities equal ratios of areas of regions to the area of the entire unit
square. But the area of the unit square is 1, so that the probability we are looking for is Prunmlal Pr[{Y<%}J _ Pr[Y>XY<;)= Area ofthe region where {Y > X} n{Y < happens 1
Area of the region where {Y < happens I 6 Answer A. 13. A health insurance policy covers the cost of anesthesia, denoted by X , and the cost of
surgery, denoted by Y. The joint distribution of X and Y is modeled as bivariate normal, with E(X)= ,ux = 1000, a, =./Vm(x) = 500, E(Y) = p, = 20,000, a, = Var(Y) = 1000, and
E (XY ) = 20350000. The policy will only pay a “reasonable expense” for the cost of surgery, and the reasonable expense is deﬁned as no more than the 75th percentile of the conditional
distribution of Y given the value of X. For a given patient, the cost of anesthesia is 1200. Find the
maximum amount paid for the surgery, assuming the bivariate normal joint distribution of X and
Y, with parameters given above. A. 20280 B. 20525 C. 20762 D. 20994 E. 362525 Solution.
Recall that for bivariate normal distribution (YX=x)~N[py +poyx;#x ,(1p2)0',’,). X In this case, we have
Cov(X,Y) = E(XY)  E(X) o E(Y) _ 20350000  1000 c 20000 _
ox ~a, ox .0, 5001000 p = px, = 0.7. Therefore ASM Study Manual for Course P/l Actuarial Examination. © Copyright 20042008 by Krzysztof Ostaszewski  509  SECTION 18 (YIX = 1200) ~ N[20000+0.71000.m—‘mg9 ,(1 072110002]:
= N(20280,510000).
In the table of the standard normal distribution we ﬁnd (11(0.67)= 0.7486 and <D(0.68) = 0.7517. Using linear interpolation, we obtain the 75~th percentile of the standard normal distribution to be
2075 = 0.67 + 0.75  0.7486
' 0.7517  0.7486 The maximum amount paid is the 75th percentile of the distribution of (Y X = 1200), and that
is calculated as 20280 + 20.7, \/5 10000 = 20280 + 0.6745  J510000 z 20761.6893.
Answer C. .(0.68 —O.67) = 0.6745. 14. A homeowner insurance coverage is provided with a uniform loss distribution over the
interval from 0 to 100,000. Calculate the deductible amount if the desired expected amount paid
by the insurance company is 40,500. A. 8000 B. 10,000 C. 12,000 D. 14,000 E. 16,000 Solution.
Let us write D for the deductible. Then the expected amount paid is
100000 _ 2 ﬁlm
I x D (ix: x — D (10000013):
D 100000 200000 mp 100000
2 2 2
=50000— D D+ D =50000D+ D .
200000 100000 200000
This results in the following equation
2
50000  D + D = 40500,
200000
or
DZ
 D + 9500 = 0,
200000 and this gives these two solutions of the quadratic equation: D=—dm)=100000(11J0.81)={ ’ ’ 2 10,000. 200000
But a deductible of 190,000 is not a feasible solution, as it would produce a zero insurance payment for all claims. Thus D = 10,000.
Answer B.
ASM Study Manual for Course Pl] Actuarial Examination. (9 Copyright 20042008 by Krzysztof Ostasuwski  510  PRACTICE EXAM 14 15. X is a continuous random variable with probability density function fx (x) = éxze" for x > 0 and fx (x) = 0 otherwise. Find the second moment of X. A B. l C. 4 D. 12 E. 24 l
' 2 Solution. This is the PDF of the gamma distribution obtained by adding three independent identically
distributed exponential distributions with hazard rate 1. The mean of such a gamma distribution
is l + l + l = 3, and the variance of it is also 1 + l + l = 3. Therefore, its second moment is E(X2) = Var(X)+ (E(X))2 = 3+ 32 =12.
You could also obtain this answer immediately, if you remember the deﬁnition of the gamma
function, since 5(x2) =1}. Answer D. x’e“dx=  Ix‘e"dx=l'l‘(5)=l4l= 12.
o 2 2 l l
2 2 16. Jack and Meg are meeting for lunch, but they are very busy at work, and as a result their time
of arrival at the restaurant where they meet is uncertain, modeled here as a random variable
uniformly distributed over 60 minutes between noon and 1 pm. Because Jack and Meg are very
busy at work, if any one of them has to wait for 20 minutes or more, they will miss each other,
because the one waiting this long will leave at the end of the 20th minute of waiting. Find the
probability that they will actually meet each other. A .1
9 B C. 3 D. 2 E.
3 9 belt 1
' 2 Solution. Let X be the difference between the time of Jack’s arrival and noon, in minutes, and Y be the
difference between the time of Meg’s arrival and noon, in minutes. Then X and Y are
independent identically distributed, following uniform distribution on the interval [0, 60]. The question asks for the probability Pr(Y  XI < 20). Note that the joint distribution of X and Y is
uniform on [0,60]2 , and
Pr(Y—X <20)=Pr(—20 <Y—X<20)=Pr(X—20 <Y <X+20). Since we are working with the bivariate uniform distribution, we calculate probabilities by
comparing areas ASM Study Manual for Course P/l Actuarial Examination. © Copyright 20042008 by Knysztof Ostaszewski  511  SECTION 18 20 60 The region marked by dotted lines (let’s call it R), between the lines given by the equations
y = x— 20 and y = x + 20, corresponds to the event {X — 20 < Y < X + 20}, for which we are trying to ﬁnd its probability. The square [0,60] x [0,60] , whose area is 3600, corresponds to the entire probability space. The ratio of the area of R to 3600 is the probability we are looking for.
But the area of R is 3600 minus the sum of the areas of two triangles not covered with dotted
lines, and the two triangles combined form a square with side of length 40, so that 3600—40«40__5_ Pr(Y—X<20)=Pr(X20<Y<X+20)= 3600 9 Answer D. 17. You are given that a random variable X is exponentially distributed with hazard rate of e.
Deﬁne a new random variable Y as Y = X — e. Find the probability density function of Y. 0 y<0. 0
_ . 0 y< ,
A, e1¢z(e'y+e'3) O<y<e, B,fy(y)= I”: 0
2 e e y20.
e"‘ e"’ yZe.
0 y<0.
O y<e, (L
C fy()’)= ,2 D. f,(y)= e '(e"+e”) 0<y<e,
e" ~e"’ yZe. 1
e'" e"’ y>e ASM Study Manual for Course [’11 Actuarial Examination. © Copyright 20042008 by Knysztof Ostaszewski  512  PRACTICE EXAM l4 0 y<0,
E' f”(y)={ e"'y y>0 Solution.
This is a problem about transformation of a random variable into another one. But it is unusual, because the transformation is not onetoone, so we cannot get the density from the standard
formula for the density of a result of a transformation. But we know two approaches in handling
transformation: memorize the PDF formula, or use the CDF technique. When one does not work,
or is inconvenient, we try the other one. So we use the CDF technique. Y assumes only positive values and for an arbitrary y> 0
500:1“),Sy)=Pr(IX‘eI5Y)=Pr(‘y5Xes)’)=Pr(eySXSe+y)= — e.¢(¢y) _ e—e(e+y), ife _ y > 0’ eG’ e‘y _ e'e’ e‘y’ < e,
l—e"(‘+y), ifeySO, 1—e"‘e"’. if yZe Note that F, (e) = l  e'z‘: , regardless of which of the two formulas we use. To the lefthand side of e, the derivative of the CDF is i( "2e"  e"1e"’) = e“: ee"' + e": ~ee"" = e“‘1(e"’ + i"),
dy and to the righthand side of e, the derivative of the CDF is d _(1 (“16'”): '8": ' (e) ' 6"” = e”: e"’. dy If we substitute y = e in the lefthand side derivative formula, we get e + eH‘z. If we substitute
y = e in the righthand side derivative formula, we obtain eH‘z. This means that the derivative
of P} at e does not exist, and the PDF value at that point cannot be established uniquely. But changing one value of the PDF of a continuous distribution does not change any probabilities,
because for a continuous distribution probabilities are calculated as integrals. Therefore, we can 0 y<0,
fy(y)= e"? (e"+e") 0<y<e,
e”: e’” yZe 18. You are given a random variable X whose moment generating function is 6
M x (t) = —— for t < 2. Find the variance of this random variable.
6— 5t +t2 ASM Study Manual for Course P/l Actuarial Examination. © Copyright 20042008 by Knysztof Osmszcwski  513  SECTION 18 A. 3 B. 1 C. l D. l E. E
2 3 36
Solution.
You can immediately notice that
6 2 3
M.(r)= — —— so that this MGF is a product of the MGF of an exponential distribution with the hazard rate of 2
and an exponential distribution with the hazard rate of 3, and X can be represented as a sum of
two independent exponential random variables with those parameters, so that
1 l l 1 9 + 4 13
Var(X)=—2+—2=—+—=—=—.
2 3 4 9 36 36 Alternatively, you can calculate the ﬁrst and second moment from the MGF and get the variance
from them: E(x)=M;.(o)=[——) _ = 16—5—91.
,30 d’ (6—5t+t2)2 E(X2)=Msz(0)=[d—z;] dt265t+t2 _12(6—5t+t2)2—6(5—2r)2(6—5t+t2)(5+2t) _19
(6—5t+t2)4 [=0 18’
_ 2 222112_§_£
var(X)'E(X) (Em) '18 [6] _18 36—36' Answer E. 19. A mutual life insurance company issued a participating life insurance policy with death
beneﬁt of 1000. The number of claims in a given year follows the binomial distribution with
parameters p = 0.2 and n = 5 . The company pays a dividend to policyholders given by the following formula: Y = max(4000 — X ,0) , where X is the total amount of death beneﬁts paid in
the year. Find the standard deviation of Y. A. 769 B. 823 C. 846 D. 859 E. 893 Solution.
Let us write N for the number of death claims in the year under consideration. Note that X =1000N and N is binomial with parameters p = 0.2 and n = 5. Therefore Pr(X =0) =Pr(N = o) = [302° .035 = 0.35, ASM Study Manual for Course Pll Actuarial Examination. © Copyright 20042008 by szysztof Ostaszewski  514  PRACTICE EXAM l4
5 l 5
2 Pr(X=1000)=Pr(N =1)=[ )02103‘ =0.8‘, 0.22 0.83 = 0.40.83, and U! 0.23 0.82 =0.08 »O.82. Pr(X= 2000)=Pr(N = 2)=[
Pr(X= 3000)=Pr(N= 3)=[ 3
We have the following distribution for Y 4000, ifN = o, with probability 0.85,
3000, if N =1, with probability 0.84 , Y = max(4000 — x,o) = 2000, ifN = 2, with probability 0.4 0.83,
1000, if N = 3, with probability 0.08 . 0.82,
0, otherwise. Therefore,
E(Y) = 4000 0.85 + 3000  0.84 + 2000 »0.4 .083 + 1000 0.08 0.82 = 3000.32, E(YZ) = 40002 08’ + 30002 0.84 + 200020.401;3 +10002 0.08 082 = 9799680,
Var(Y) = E(Y‘) — (E(Y))2 = 9799680  3000.322 == 797759.8976, and ﬁnally
0', = ,iVar(Y) = J797759.8976 = 893.17406.
Answer E. 20. Dwizeel and Satellite Component are musicians who travel to concerts around the world by
ﬂying a private jet. They decided to purchase a life insurance policy that will pay 1,000,000 upon
the ﬁrst death of the two of them. The future lifetime (in years) of either Dwizeel or Satellite
Component is exponential with hazard rate 0.0001 , if death by means other than plane crash is
considered. But the time until the crash of their plane follows the exponential distribution with
mean 25 years, and is independent of any other causes of death of either Dwizeel or Satellite
Component. Assume that, unfortunately, nobody survives their private plane crash if one
happens. Let Wrepresent the time of the ﬁrst death. You are given that the present value of the beneﬁt paid on this policy is 1000000e'ww . Find the expected value of the death beneﬁt
payment on this policy, i.e., the expected value of W. A. 275,500 B.389,750 C.445,675 D.495,075 E.525,100 Solution. Let us define the following random variables: X : time until death od Dwizeel by causes other than their private plane crash, ASM Study Manual for Course Pll Actuarial Examination. © Copyright 2004.2008 by Krzysztof Ostaszewski  515  SECTION 18
Y : time until death od Satellite Component by causes other than their private plane crash,
Z 2 time until death od Dwizeel and Satellite Component as a result of their private plane crash. Then the time until the first death is W = min(X,Y,Z). Furthermore
SW (w): Pr(W > W): Pr({X > w}n{Y > w}n{Z > w})= e [ﬁle = 3x (W) ' Sr 52 (W) = 3.0mm” ‘ (1
Therefore, fw (w) = —d—wsw (w) = 0.0402e'°m" and the expected value of the death beneﬁt 0.000lw . 0.0402w e =8 payment on this policy is 5(1000000e4’05‘”) = 1000000 I e'°°‘" 0.0402e'°”4°2“’dw =
0 4a
=40200 j (099°2de = 40200 =445,676.2749.
a 0.0902
Mm
Mthhawdmewgggmoz Answer C. 21. May 1985 Course 110 Examination, Problem No. 5
Let X and Y be random variables with variances 2 and 3, respectively, and covariance —1.
Which of the following random variables has the smallest variance? A. 2X+Y B. 2X—Y C. 3X—Y D.4X E.3Y Solution.
We have Var(2X + Y) = 4Var(X)+Var(Y)+4Cov(X,Y) = 8 + 3 4 = 7, Var(2X Y) = 4Var(X)+Var(Y) 4Cov(X,Y) = 8 + 3+ 4 =15, Var(3X — Y) = 9Var(X)+Var(Y) 6Cov(X,Y) = 18 + 3+ 6 = 27,
Var(4X) = l6Var(X) = 32, and Var(3Y) = 9Var(Y) = 27. Note that when you add two random
variables with negative covariance, you reduce the variance of the sum in relation to the sum of
variances, while subtracting them increases variance. This eliminates answers B and C. Similar
consideration eliminates D, as variance of Ydoes not overcome reduction from negative
covariance. E is eliminated if we observe that Y, Y, and Y (summands of 31’) are perfectly correlated, while X, X, and Y (summands of 2X + Y) are not, and the variance of X is smaller than the variance of Y.
Answer A. 22. May 1985 Course 110 Examination, Problem No. 7
Let X and Y have a bivariate normal distribution. Which of the following statements must be true?
ASM Study Manual for Course Pll Actuarial Examination. © Copyright 20042008 by Krzysztof Ostaszewski  516  PRACTICE EXAM 14
I. Any nonzero linear combination of X and Yhas a normal distribution. II. E(Y IX = x) is a linear function of x.
111. Var(YX = x) .<_ Var(Y). A. I and H only B. I and 111 only C. II and III only
D. I, II and III E. The correct answer is not given by A, B, C, or D Solution.
The fact that any nonzero linear combination of X and Y has a normal dism'bution is one of the
key properties of the bivariate normal distribution, so I is true. Another key property is that (YX=x)~ N[p, +pr, .o, .x—itill— p§,,.)o;). O'X
Therefore
3"”
E(YIX=x)=“Y+pX.Y '0'? ‘ O, x ’
x so that 11 is true, and Var(YX = x): (1 —p§l,.)o32, s a; =Var(Y),
so that III is true, as well.
Answer D. 23. May 1985 Course 110 Examination, Problem No. 9 The number of automobiles crossing a certain intersection during any time interval of length t
minutes between 3:00 pm. and 4:00 pm. has a Poisson distribution with mean I. Let Wbe the
time elapsed after 3:00 pm. before the ﬁrst automobile crosses the intersection. What is the
probability that W is less than 2 minutes? A. 12e'1— e'2 B. e'2 C. 2e'I D. l e'2 E. 2e"l + e'2 Solution.
The number of automobiles crossing the intersection considered in this problem during the time
interval starting at 3:00 pm. and ending at 3:02 pm. is Poisson with mean 2. Let us write N for 0
that Poisson random variable. Then Pr(W > 2) = Pr(N = 0) = %  e'2 = e‘z. Note that W is a continuous random variable, so that Pr(W < 2) = Pr(W S 2), and Pr(W $2)=1—Pr(W >2)=1e‘2.
AnswerD. 24. May 1985 Course 110 Examination, Problem No. 10
Let X and Y be continuous random variables with joint density function ASM Study Manual for Course Pll Actuarial Examination. © Copyright 20042008 by Krzysztof Ostaszewski  517  SECTION 18
12x, 0<y<2x<1, fx'y (x'y) = { 0, elsewhere.
What is the conditional density function of Y given X = x? A. i for 0 < x < l, and 0 elsewhere
2x 2
B. 2: for 0 <y < 2x <1, and 0 elsewhere
x
C. i for 0 < x <1, and 0 elsewhere
x
D. i for 0 < y < 2x <1, and 0 elsewhere
x
16x
E.l 2 2 for 0 <y < 2x <1, and 0 elsewhere
— y
Solution. Note the triangular region where the joint density is positive, marked‘with dotted lines below: x
The conditional density function of Y given X = x is, by deﬁnition
fx y (xx?)
f y X = x = —' .
’( ' ) fx (x) The crucial insight is that the formula for the joint density does not depend on y, and this implies
that the ratio of the joint density and the marginal density of X does not depend on y either. This means that the distribution of (YI X = x) is uniform, and since for a ﬁxed 1:, y varies between 0
and 2x, we must have ASM Study Manual for Course P/l Actuarial Examination. © Copyright 20042008 by Knysztof Ostaszewski  518  PRACTICEEXAM 14
1 1
MM”): 2x—0 =5 for 0 < y < 2x <1, and 0 elsewhere. That’s answer B, and we are done. We can show this by calculation as well (in a moment), but of course noticing the uniform distribution is a far more
efﬁcient way. This trick of a “hidden uniform distribution” has appeared on the actuarial
examinations quite regularly, so you should be one a lookout for it. In order to do the calculation, we note that the ratio is well deﬁned only forx between 0 and For any such x fx(x)
2x
fx(x)=112xdy=12x.2x=24x2_
0 Therefore,for 0 <x <%, 2x< y <1, 25. May 1985 Course 110 Examination, Problem No. 12
Let X and Y have the joint density function f ( ) x+y, OSxSIandOSySl,
x, =
x” y 0, elsewhere. What is the Pr(2XSlX+YSl)? 2 .3 a: 2 E3
48 16 16 16 4
Solution.
Bydeﬁnition
1
Pr[{Xs—}n{Ys—X+l}]
MZXSIIX+YSU=W= 2 .
Pr(X+Ys1) pr(Ys_X+1) The joint density is positive only in the unit square, and in the ﬁgure below we show that unit square, as well as the region where X S %, marked with horizontal dotted lines, and the region where Y S —X +1, marked with vertical dotted lines. ASM Study Manual for Course P/l Actuarial Examination. © Copyright 20042008 by Krzysztof Ostaszewski  519  . 8m  E3335 banana. 3 8838 zmcaoo @ .Saéaaxm 3.533 E 858 3. :55: 33m 2? 8m xv n AAJQ Sc. 335% ion 5:» moﬁatg Bow—HS 38528 on A 98 N81—
3 62 5030.5 £83:ng e: own—EU mag hag 6N .D 355‘ é:é:.Q  $33?ng _ 3&3 _ ucwiiﬁxaa agﬁivfimcmi g
N 88205. .T + N I w b C w NW 82—? M85 USN 6:3on 93 05 mo 2038385 05 .0958 .«o .ﬂ was: 33% 3285.5: can :89? Son 5?» USE:— :o_w8 2F I I I I I
'1T’fr'I1T 2 ZOEm—m PRACTICE EXAM 14
0 < x < J; <1. What is the marginal density of Y, where nonzero? A. 2y2 B. 2y c. y2 D. J3? E. 4.5
Solution.
By deﬁnition
f, (y) = I f,” (x,y)dx= j 4xdx = 2x no = 2y
all x for 0
which
Inkr)
nposmve
for 0 < y < 1.
Answer B. 27. May 1985 Course 110 Examination, Problem No. 15
If X is a random variable with density function 1.4e'2* + 0.9e'3‘, forx 2 0,
n0#{ 0, elsewhere,
then E (X) =
A“: 3.3 c.1 DE 3.2—3
20 6 126 10
Solution.
Note that for x > 0
1.4 0.9 fx (x) = 1.4e”2‘ + 0.9e'3" = 7 ~ 2e'2" + —3  3e'3’ = 0.7 e 262‘ + 0.3. 3e'3‘, so that this density is a weighted average of densities of an exponential distribution with hazard rate of 2 (and the mean of i) and an exponential distribution with hazard rate of 3 (and mean of g), and therefore the distribution is a mixture of those distributions with weights of 0.7 and 0.3, respectively, and the mean can be calculated as the same weighted average of the means of the
two pieces of the mixture E(X)=0.7ol+0.3l=0.35+0.1=0.45 =3.
2 3 20 Yes, you could do the calculation to ﬁnd the mean from the definition, but learning mixed
distributions is important, so I hereby ofﬁcially refuse to do this problem any other way.
Answer A. ASM Study Manual for Course P/l Actuarial Examination. © Copyright 20042008 by Knysztof Ostaszewski  521  SECTION 1 8
28. Study Note P0908, Problem No. 44 An insurance policy pays an individual 100 per day for up to 3 days of hospitalization and 50 per
day for each day of hospitalization thereafter. The number of days of hospitalization, X, is a
discrete random variable with probability function 6_—5
Pr(X = k) = 15 ’ 0, otherwise.
Calculate the expected payment for hospitalization under this policy. for k =1,2,3,4,5, A. 123 B. 210 C. 220 D. 270 E. 367
Solution.
Deﬁne H (X) to be hospitalization payments made by the insurance policy. Then
100 when X =1,
200 when X = 2,
H(X)= 300 whenX= 3,
350 when X = 4,
400 when X = 5,
and therefore
E(H(X))=1ooi+200i+3003+3503+400.i=
15 15 15 15 15
3300 1100
= é (500 + 800 + 900 + 700 + 400) = —15 = ———5 = 220. Answer C. 29. May 1985 Course 110 Examination, Problem No. 24 An urn contains 4 balls numbered 0 through 3. One ball is selected at random and removed from
the um and not replaced. All balls with nonzero numbers less than half of the selected ball are
also removed from the urn. Then a second ball is selected at random from those remaining in the
urn. What is the probability that a second ball selected is numbered 3? 1 1 11 13 A. 4 B. 24 C. 3 D. 24 E. 24
Solution. Each of the initial 4 balls in the urn has the same probability of being removed in the ﬁrst step,
thus each one of them has the probability of 0.25 of being removed. Let us consider now the four
cases of each of the initial four balls being removed in the ﬁrst step: ASM Study Manual for Course Pll Actuarial Examination. © Copyright 20042008 by Knysztof Ostaszewski  522  PRACTICE EXAM l4
0 If the ball numbered 0 is removed in the ﬁrst step, then half of that is 0, and no other ball is removed in the second step, resulting in probability of g of picking the ball numbered 3 in the last step.
0 If the ball numbered 1 is removed in the first step, then half of that is %, but the ball numbered 0 is not removed in the second step, thus no other ball is removed in the second step, resulting in probability of 21; of picking the ball numbered 3 in the last step. 0 If the ball numbered 2 is removed in the ﬁrst step, then half of that is l, but 0 is not removed
and l is not less than 1, so no other ball is removed in the second step, resulting in probability of g of picking the ball numbered 3 in the last step. 0 If the ball numbered 3 is removed in the ﬁrst step, then the probability of picking it later is 0.
Therefore, the probability we are looking for is l 1 l l l l 1 1 ——+—o—+—«—+—o0 =—. 4 3 4 3 4 3 4 4
Answer A. 30. May 1985 Course 110 Examination, Problem No. 25 An urn contains 4 red balls, 8 green balls, and 2 yellow balls. Five balls are randomly selected,
with replacement, from the urn. What is the probability that 1 red ball, 2 green balls, and 2
yellow balls will be selected? A; 3.3 Ci D; Be
512 7 512 143 7
Solution. This is a straightforward multinomial distribution question. Let Xl be the number of red balls in the ﬁve balls randomly selected, X2 be the number of green balls in the same selection of ﬁve
balls, and X3 be the number of yellow balls in it. Then 5! 4'19.2 22 3.452421960
Pr(X‘—1’X2'2’X"2)'manila) (E) (T?) =T'7'?'7=7' Answer E. ASM Study Manual for Course Pll Actuarial Examination. © Copyright 20042008 by Knysztof Ostaszewski  523  ...
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This note was uploaded on 10/27/2010 for the course PSTAT 172a taught by Professor Staff during the Winter '08 term at UCSB.
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