Practice_Exam_14-Solutions

Practice_Exam_14-Solutions - SECTION 18 PRACTICE...

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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, (Y|X = 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=2|X=m)-Pr(X= m): m=0 m=0 =Pr(Y=2|X=1)~Pr(X=l)+Pr(Y=2|X=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 1-Pr(X=l|Y= 2)+2-Pr(X=2|Y= 2): _Pr({x=1}n{y=2}) ,Pr({x=2}n{r=2}) =1 +2 = Pr(Y=2) Pr(Y=2) =1_Pr(Y=2|X=l).Pr(X=l)+2'Pr(Y=2|X=2)-Pr(X=2) = Pr(Y=2) Pr(Y =2) =1.0.1-5-0.1‘-0.9“ +2.0.3-10-0.11-0.9’ =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 2004-2008 by Knysztof Ostaszewski - 500 - PRACTICE EXAM 14 Pr({x=1}n{r=2}) PI'(X=1|Y=2)=-—P-r'zy—=2)——= ___Pr_({X=_l}o_{_E})___ ' Pr({X=l}n{Y=2})+Pr({X=2}n{Y=2})_ _ 0.1.5-o.1-o.9‘ _3 ‘ 0.1.5-o.1-o.9‘ +0.3.1o.o.12-0.93 ' 5’ Pr(X=2|Y=2)=1—3=3. 5 5 and E(X|Y=2)=1.Pr(x=1[Y=2)+2.Pr(x=2|Y=2)=1-%+2%=%. AnswerA. 2. YI is a log-normal random variable such that lnYl is normally distributed with mean 16 and standard deviation 1.5. Similarly, Y2 is a log-normal 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 > 16-115)},{111r'2 —15 >16—15]= 15 15 2 2 =(1-<b(0))-(1-<1>(0.5))=(1—o.5)-(1-0.6915)=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 2004-2008 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 non-negative 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([[X1|2n)=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'l-t-e'2-t-e'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+...)= l-el l—el l—e1 l—el l-el " ‘1 1 1 2 — = e -,+ e _,' 1_,=——+—--i-= e 1,z1.5027. l—e l—e l—e e—l e—l e-l (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 2004-2008 by Knysztof Ostaszewski -502- PRACTICE EXAM 14 0, x<0-l, 0, x<9-l, Fx‘(x)= %(X-(9-1)), 0—ISxSG+5, = P69“, e-1sxse+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 Manama-Pr {n.izewrse} =1-Pr(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 two-parameter Pareto distribution with 6 = 50 and a = 3. The cumulative distribution function of a two-parameter 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 — 20|T 2 20). A. 5 B. 15 C. 25 D. 35 E. 45 Solution. Let us write U = (T - 20|T 2 20). We have sU(u)=Pr(T-20>u|T220)=Pr(T>u+20|T220)= _ 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 2004-2008 by Knysztof Ostaszewski - 503 - SECTION 18 E(T—20|T 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 — 20|T 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—10-d9]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,(05|e=9) _ _ l-e'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 2004-2008 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 two-day 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 two-year 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 non-smokers, Z “Mm” , with the weight of Its expected value is the weighted average of the expected values for smokers and non-smokers, i.e., = l . E(ZSmokets)+2 .E(ZNon-Smokcrs). 3 3 We will calculate the expected values for smokers and non-smokers 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 non-smokers 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) = 2x|y| 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 y-axis, 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=j2x1y|dy=2leyldy=2x-[lt-y)dy+jydy]=2x for 0 < x < 1, and fx (x) = 0 otherwise, while ASM Study Manual for Course Pll Actuarial Examination. © Copyright 2004-2008 by Knysztof Ostaszewski - 507 - - 8m - Ema...“ng 355 3 “838” 23.38 © eozéaaxm 3333. 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A "Em mon ma .ommgofio o u A3 xx can 4 v A v 7 8m N o a . o _A_um._a_~u§h._a_~u§_a_i "fiaszl "Be _ _ _ 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>X|Y<-;-)= 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 75-th 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 (Y|X=x)~N[py +poy-x;#x ,(1-p2)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, 500-1000 p = px, = 0.7. Therefore ASM Study Manual for Course P/l Actuarial Examination. © Copyright 2004-2008 by Krzysztof Ostaszewski - 509 - SECTION 18 (YIX = 1200) ~ N[20000+0.7-1000.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 75-th 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 (100000-13): D 100000 200000 mp 100000 2 2 2 =50000— D -D+ D =50000-D+ 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 2004-2008 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)=l-4l= 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 20-th 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 2004-2008 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(X-20<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 one-to-one, 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(‘y5X-es)’)=Pr(e-ySXSe+y)= — e.¢(¢-y) _ e—e(e+y), ife _ y > 0’ e-G’ e‘y _ e'e’ e-‘y’ < e, l—e"(‘+y), ife-ySO, 1—e"‘e"’. if yZe- Note that F, (e) = l - e'z‘: , regardless of which of the two formulas we use. To the left-hand side of e, the derivative of the CDF is i( "2e" - e"1e"’) = e“: -e-e"' + e": ~e-e"" = e“‘1-(e"’ + i"), dy and to the right-hand 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 left-hand side derivative formula, we get e + eH‘z. If we substitute y = e in the right-hand 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 2004-2008 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;] dt26-5t+t2 _-12(6-—5t+t2)2—6(5—2r)-2(6—5t+t2)(-5+2t) _19 (6—5t+t2)4 [=0 18’ _ 2- 2-2-2112_§_£ 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 2004-2008 by szysztof Ostaszewski - 514 - PRACTICE EXAM l4 5 l 5 2 Pr(X=1000)=Pr(N =1)=[ )021-03‘ =0.8‘, 0.22 -0.83 = 0.4-0.83, 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 2004-2008 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(Y|X = 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 (Y|X=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(Y|X = 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. 1-2e'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)=1-e‘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 2004-2008 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 2004-2008 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(2XSl|X+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 2004-2008 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 '1-T’f-r'I-1-T- 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 Ink-r) 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+0.3-l=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 2004-2008 by Knysztof Ostaszewski - 521 - SECTION 1 8 28. Study Note P-09-08, 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))=1oo-i+200-i+300-3+350-3+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 2004-2008 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 3.4-52421960 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 2004-2008 by Knysztof Ostaszewski - 523 - ...
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