( ) is proportional to ( 10 + x)!2. Calculate the

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1 PRACTICE EXAMINATION NUMBER 6. An insurance company eamines its pool of auto insurance customers and gathers the following information: i) All customers insure at least one car. ii) 64 of the customers insure more than one car. iii) of the customers insure a sports car. iv) Of those customers who insure more than one car, 5 insure a sports car. Calculate the probability that a randomly selected customer insures eactly one car and that car is not a sports car. A..6 B..9 C..6 D..9 E..3. The lifetime of a machine part has a continuous distribution on the interval, 4) with probability density function f X, where f X probability that the lifetime of the machine part is less than 5. ) is proportional to + ). Calculate the A..3 B..3 C..4 D..58 E An insurer s annual weather-related loss, X, is a random variable with density function.5.5, for >, f X ) 3.5, otherwise. Calculate the difference between the 5-th and 75-th percentiles of X. A. 4 B. 48 C. 67 D. 4 E A device runs until either of two components fails, at which point the device stops running. The joint density function of the lifetimes of the two components, both measured in hours, is, y) + y f X,Y for < < and < y <. What is the probability that the device fails during 8 its first hour of operation? A..5 B..4 C..39 D..65 E Let X ), X ),, X 6 ) be the order statistics from a random sample of size 6 from a distribution with density function ASM Study Manual for Course P/ Actuarial Eamination. Copyright 4- by Krzysztof Ostaszewski

2 f X ) What is E X 6) )?, for < <,, otherwise. PRACTICE EXAMINATION NO. 6 A. B. 3 C. 5 6 D. 6 7 E Let X be a normal random variable with mean and variance a >. Calculate Pr X < a). A..34 B..4 C..68 D..84 E An urn contains lottery tickets. There is one ticket that wins 5, three tickets that win 5, si tickets that win, and fifteen tickets that win 3. The remaining tickets win nothing. Two tickets are chosen at random, with each ticket having the same probability of being chosen. Let X be the amount won by the one of the two tickets that gives the smaller amount won if both tickets win the same amount, then X is equal to that amount). Find the epected value of X. A..348 B..44 C..636 D..79 E. Does not eist 8. X, X, X 3 ) is a random vector with a multivariate distribution with the epected value,,) and the variance/covariance matri: If a random variable W is defined by the equation X ax + bx 3 + W and it is uncorrelated with the variables X and X 3 then the coefficient a must equal: A. B. 4 3 C. 5 3 D. E A random variable X has the eponential distribution with mean. Let be the greatest integer function, denoting the greatest integer among those not eceeding. Which of the following is the correct epression for the epected value of N X? ASM Study Manual for Course P/ Actuarial Eamination. Copyright 4- by Krzysztof Ostaszewski

3 SECTION A. B. C. D. e e E. e. X and Y are independent and both distributed uniformly from to. Find the probability density function of Z 5X Y. A. f Z z).5 where non-zero B. Stepwise formula: f Z z) C. f Z z) ) e z, z <, 3, z < 3, 5 + z, 3 z 5. where non-zero D. f Z z z, for z >, zero otherwise E. Stepwise formula: f Z z) + z, z <, 5, z < 3, 5 z, 3 z 5.. Let X ), X ),, X 8) be the order statistics from a random sample X, X,, X 8 of size 8 from a continuous probability distribution. What is the probability that the median of the distribution under consideration lies in the interval X ), X 7)? A. 8 B. 8 9 C. 8 D. 4 8 E. Cannot be determined. In a block of car insurance business you are considering, there is a 5 chance that a claim will be made during the upcoming year. Once a claim is submitted, the claim size has the Pareto ASM Study Manual for Course P/ Actuarial Eamination. Copyright 4- by Krzysztof Ostaszewski

4 PRACTICE EXAMINATION NO. 6 distribution with parameters 3 and, for which the mean is given by the formula, and the second moment is given by the formula. Only one claim will ) ) happen during the year. Determine the variance of the unconditional distribution of the claim size. A. 65 B C. 5 D. 75 E. 3. A random vector X,Y ) has the bivariate normal distribution with mean,) and the variance-covariance matri at the origin.. Find the probability that X,Y ) is in the unit circle centered A.. B C D..63 E You are given that X and Y both have the same uniform distribution on [, ], and are X independent. U X + Y and V. Find the joint probability density function of U, V) X + Y evaluated at the point,. A. B. 4 C. 3 D. E. 5. Three fair dice are rolled and X is the smallest number of the three values resulting if more than one value is the smallest one, we still use that value). Find Pr X 3 ). A B C D E The moment-generating function of a random variable X is M X t) Calculate the ecess of Pr X > Chebyshev s Inequality., for t >. t over its best upper bound given by the two-sided) ASM Study Manual for Course P/ Actuarial Eamination. Copyright 4- by Krzysztof Ostaszewski

5 SECTION A..35 B.. C. D..5 E.. 7. The time to failure X of an MP3 player follows a Weibull distribution, whose survival function is s X ) e ) ) e for >. It is known that Pr X > 3 Find the probability that this MP3 player is still functional after 4 years. A..498 B..8 C..353 D..69 E..3, and that Pr X > 6 ) e You are given that the hazard rate for a random variable X is X ) for >, and zero otherwise. Find the mean of X. A. B. C..5 D. 3 E Let P be the probability that an MP3 player produced in a certain factory is defective, with P assumed a priori to have the uniform distribution on [, ]. In a sample of one hundred MP3 players, is found to be defective. Based on this eperience, determine the posterior epected value of P. A. B. C. 99 D. 5 E. 5. You are given that Pr A) 5, Pr A B) 3 5, Pr B A ) 4, Pr C B ) 3, and Pr C A B). Find Pr A B C). A. 3 B. 5 C. 3 D. E. 4. Let X ),..., X n) be the order statistics from the uniform distribution on [, ]. Find the correlation coefficient of X ) and X n). A. n B. n + C. D. n + E. n ASM Study Manual for Course P/ Actuarial Eamination. Copyright 4- by Krzysztof Ostaszewski - 6 -

6 PRACTICE EXAMINATION NO. 6. A random variable X has the log-normal distribution with density f X ) e for >, and otherwise, where µ is a constant. You are given that Pr X E X). ).4. Find A. 4.5 B C D. 5. E. Cannot be determined ln µ ) 3. You are given a continuous random variable with the density f X ) + for, and otherwise. Find the density of Y X, for all points where that density is nonzero. A. y B. y C. 3 y D. 4 3 y E. yln 4. An insurer has independent one-year term life insurance policies. The face amount of each policy is. The probability of a claim occurring in the year under consideration is.. Find the probability that the insurer will pay more than the total epected claim for the year. A.. B.. C..6 D..6 E An insurance policy is being issued for a loss with the following discrete distribution:, with probability.4, X, with probability.6. Your job as the actuary is to set up a deductible d for this policy so that the epected payment by the insurer is 6. Find the deductible. A. B. 5 C. 7 D. E For a Poisson random variable N with mean find lim E N N ). A. B. C. D. e E. Cannot be determined 7. X is a normal random variable with mean zero and variance and Y is distributed ASM Study Manual for Course P/ Actuarial Eamination. Copyright 4- by Krzysztof Ostaszewski - 6 -

7 SECTION eponentially with mean. X and Y are independent. Find the probability Pr Y > X ). A. e B. e C. D. E. 8. X ), X ),, X 4) are order statistics from a continuous probability distribution with a finite mean, median m and variance. Let be the cumulative distribution function of the standard ) normal distribution. Which of the following is the best approimation of Pr X ) m Central Limit Theorem? ) B..49 C..53 D..56 E. A..5 using the 9. N is a Poisson random variable such that Pr N ) Pr N ). Find the variance of N. A..5 B.. C..68 D. 3.5 E There are two bowls with play chips. The chips in the first bowl are numbered,, 3,,, while the chips in the second bowl are numbered 6, 7, 8,, 5. One chip is chosen randomly from each bowl, and the numbers on the two chips so obtained are compared. What is the probability that the two numbers are equal? A. B. 5 C. D. 4 E. 5 ASM Study Manual for Course P/ Actuarial Eamination. Copyright 4- by Krzysztof Ostaszewski - 6 -

8 PRACTICE EXAMINATION NUMBER 6 SOLUTIONS PRACTICE EXAMINATION NO. 6. P Sample Eam Questions, Problem No., also Dr. Ostaszewski s online eercise posted September, 7 An insurance company eamines its pool of auto insurance customers and gathers the following information: i) All customers insure at least one car. ii) 64 of the customers insure more than one car. iii) of the customers insure a sports car. iv) Of those customers who insure more than one car, 5 insure a sports car. Calculate the probability that a randomly selected customer insures eactly one car and that car is not a sports car. A..6 B..9 C..6 D..9 E..3 Always start by labeling the events. Let C be the event of insuring a sports car not S, because we reserve this for the entire probability space), and M be the event of insuring multiple cars. Note that M C is the event of insuring eactly one car, as all customers insure at least one car. We are given that: Pr M ).64, Pr C)., and Pr C M ).5. We need to find Pr M C C C ). We recall De Morgan s Law and obtain Answer C. ) Pr M C Pr M C C C Pr M ) C ) Pr M C) Pr M ) Pr C) + Pr M C) ) Pr C) + Pr C M ) )Pr M. P Sample Eam Questions, Problem No. 35, also Dr. Ostaszewski s online eercise posted May 3, 8 The lifetime of a machine part has a continuous distribution on the interval, 4) with probability density function f X, where f X probability that the lifetime of the machine part is less than 5. ) is proportional to + ). Calculate the A..3 B..3 C..4 D..58 E..97 We know the density has the form C + ) for < < 4 and is zero otherwise), where C is a certain constant. We determine the constant C from the standard condition f X d : ASM Study Manual for Course P/ Actuarial Eamination. Copyright 4- by Krzysztof Ostaszewski )

9 SECTION 4 ) C + d C + ) 4 C C 5 5 C, and therefore C 5.5. The probability that X < 5 is found as the integral of the density over the interval, 5): 5 ).5 + d.5 + Answer C. ) ) P Sample Eam Questions, Problem No. 6, and Dr. Ostaszewski s online eercise posted December 5, 7 An insurer s annual weather-related loss, X, is a random variable with density function.5.5, for >, f X ) 3.5, otherwise. Calculate the difference between the 5-th and 75-th percentiles of X. A. 4 B. 48 C. 67 D. 4 E. 98 The cumulative distribution function of X is given by F X ).5.5 dt.5 t 3.5 t for >. Therefore, the p-th percentile p of X is given by.p F p ).5.5 p p 5, or.p) 5, resulting in p p.p) 5 Answer A It follows that 4. P Sample Eam Questions, Problem No. 77, also Dr. Ostaszewski s online eercise posted March 4, 9 A device runs until either of two components fails, at which point the devise stops running. The ASM Study Manual for Course P/ Actuarial Eamination. Copyright 4- by Krzysztof Ostaszewski

10 PRACTICE EXAMINATION NO. 6 joint density function of the lifetimes of the two components, both measured in hours, is, y) + y f X,Y for < < and < y <. What is the probability that the device fails during 8 its first hour of operation? A..5 B..4 C..39 D..65 E..875 Probability that the device fails within the first hour is calculated by integrating the joint density function over the shaded region shown below. This is best done by integrating over the un-shaded region and then subtracting the result from : { }) Pr { X < } Y < + y d dy + y 8 dy y y)dy y) dy 3y + y ) 6 Answer D May 983 Course Eamination, Problem No. 5, and Dr. Ostaszewski s online eercise posted October, Let X ), X ),, X 6) be the order statistics from a random sample of size 6 from a distribution with density function f X ) What is E X 6) )?, for < <,, otherwise. ASM Study Manual for Course P/ Actuarial Eamination. Copyright 4- by Krzysztof Ostaszewski

11 SECTION A. B. 3 C. 5 6 D. 6 7 E. 3 The CDF of the original distribution is F X ) t dt t t t for < <. We also have F X ) for, and F X ) for. Hence, F X 6) ) Pr X 6) ) ) for and F X 6) ) for >. Therefore, for < <, ), and s X 6 ) for. We conclude that < <. Also, F X 6) s X 6) Answer E. E X 6 ) ) ) )d ) F X ) 6 for 6. February 996 Course Eamination, Problem No., and Dr. Ostaszewski s online eercise posted October 9, Let X be a normal random variable with mean and variance a >. Calculate Pr X < a). A..34 B..4 C..68 D..84 E..9 Since a >, and a is the variance, a is the standard deviation of X. Therefore, if we denote a standard normal random variable by Z, and the standard normal CDF by, we get Answer C. ) Pr Pr X < a) Pr a < X < a ) ) ) ) a < X a a < a a ) Pr < Z < ) ) ) Dr. Ostaszewski s online eercise posted March, 5 An urn contains lottery tickets. There is one ticket that wins 5, three tickets that win 5, si tickets that win, and fifteen tickets that win 3. The remaining tickets win nothing. Two tickets are chosen at random, with each ticket having the same probability of being chosen. Let X be the amount won by the one of the two tickets that gives the smaller amount won if both tickets win the same amount, then X is equal to that amount). Find the epected value of X. A..348 B..44 C..636 D..79 E. Does not eist ASM Study Manual for Course P/ Actuarial Eamination. Copyright 4- by Krzysztof Ostaszewski

12 PRACTICE EXAMINATION NO. 6 Note that you cannot have X 5, because there is only one 5 ticket. Thus, the possible values of X are: 5,, 3, and. Furthermore, X 5 when we choose one 5 ticket and one 5 ticket, are and there are 3 3 ways to do that, or when we choose two 5 tickets, and there ways to do that. Since there are ways to choose tickets out of, Pr X 5) Furthermore, X for one ticket and one higher amount ticket, or two tickets, so that Pr X ) , while X 3 for one 3 ticket and one higher amount ticket, or two 3 tickets, so that, Pr X 3) , and finally, X, one ticket and one other ticket, or two tickets, so that Pr X ) Thus E X) Answer C. 8. Dr. Ostaszewski s online eercise posted March 9, 5 X, X, X 3 and the variance/covariance matri: ) is a random vector with a multivariate distribution with the epected value,,) ASM Study Manual for Course P/ Actuarial Eamination. Copyright 4- by Krzysztof Ostaszewski

13 SECTION If a random variable W is defined by the equation X ax + bx 3 + W and it is uncorrelated with the variables X and X 3 then the coefficient a must equal: A. B. 4 3 C. 5 3 D. E. 7 3 We have W X ax bx 3, and therefore, ) Cov X ax bx 3, X ) Cov X, X ) avar X ) bcov X 3, X ) Cov W, X Cov W, X 3.5 a.5b, ) Cov X ax bx 3, X 3 ) Cov X, X 3 ) acov X, X 3 ) bvar X 3 ).5a b, Hence, 3 a.5a, so that.5a, and a 4 3. Answer B. 9. Dr. Ostaszewski s online eercise posted October 6, A random variable X has the eponential distribution with mean. Let be the greatest integer function, denoting the greatest integer among those not eceeding. Which of the following is the correct epression for the epected value of N X? A. B. C. D. e e E. e Note that N is a discrete non-negative random variable, so that its epected value can be calculated as note that e <, because > ): ) Pr N n) E N Answer E Pr X n) Pr X n) e n e e e. n n n n. Dr. Ostaszewski s online eercise posted May 4, 5 X and Y are independent and both distributed uniformly from to. Find the probability ASM Study Manual for Course P/ Actuarial Eamination. Copyright 4- by Krzysztof Ostaszewski

14 density function of Z 5X Y. PRACTICE EXAMINATION NO. 6 A. f Z z).5 where non-zero B. Stepwise formula: f Z z) C. f Z z) ) e z, z <, 3, z < 3, 5 + z, 3 z 5. where non-zero D. f Z z z, for z >, zero otherwise E. Stepwise formula: + z, z <, f Z z) 5, z < 3, 5 z, 3 z 5. This problem can be solved in a simplified way, but we will discuss three possible solutions, in a drawn-out fashion, in order to fully eplain possible approaches such problems involving sums or differences of random variables. To begin with, note the following: f X ) for < <, and zero otherwise, as well as f Y f X,Y, y) y) for < y <, and zero otherwise, and finally for < < and < y <, and zero otherwise. There are actually 4 three possible approaches to solving this: the multivariate transformation approach, the convolution approach, and the CDF approach. You should know all of them for Eam P/, so all three will be presented here. Multivariate transformation Consider a transformation W Y, Z 5X Y, i.e., W,Z ) X,Y ) Y,5X Y ), whose inverse is W,Z 5,W. You might wonder: how do I know that I am supposed to pick W Y? You are not supposed to pick anything. Your objective is to find a ) W + Z ASM Study Manual for Course P/ Actuarial Eamination. Copyright 4- by Krzysztof Ostaszewski

15 SECTION second function of X and Y such that you will be able to find the inverse of the transformation so obtained. There is no unique answer. In this case, W Y would do the job, so would W 5X + Y, and so would inifinitely many other choices. The key point is that you must be able to find and then find the determinant of its derivative the Jacobian). Let us find that derivative now we switch to lower case variables because this is what we will use in the density): Therefore, This gives ) w,z) ) ) det, y w,z f W,Z w y w w y w z y z z y z * * * * ) w,z) f X,Y w,z), y w,z) ), y w,z * *. * * ) ) ) 4 5. We also have to figure out the ranges for w and z. As w y and < y <, we have < w <. Also, as < < and z + w z + w, so that < <, and this means that 5 5 w < z < 5 w or equivalently z < w < 5 z. Graphically, z w z 5 w z 3 z w z w Given that, we can now figure out the marginal density of Z: ASM Study Manual for Course P/ Actuarial Eamination. Copyright 4- by Krzysztof Ostaszewski - 7 -

16 f Z z) f W,Z w,z) dw all values of w 5z PRACTICE EXAMINATION NO. 6 dw, z <, + z z, z <, dw, z < 3, ) 5, z < 3, 5 z dw, 3 z 5., 3 z 5. * Answer E. The convolution method Recall that if X and Y have a continuous joint distribution and are continuous, then the density of + X + Y is f X+Y s) f X,Y,s )d. If X and Y are independent, then f X+Y + s) f X ) f Y s )d. In this case, we are adding U 5X, which has the uniform distribution on, 5), and is independent of V Y, which has the uniform distribution on, ). The density of U is 5, where non-zero, and the density of V is, where non-zero. Thus f U+V + s) f U u) f V s u)du 5 du 5 du u5 and su u5 and sus+ s+ * du, s <, + s, s <, s+ du, s < 3, + s 5, s < 3, 5 5 s du, 3 s 5., 3 s 5. s ) ), Answer E, again. The CDF method We have Z 5X Y. Let us find the CDF of it directly. We have: F Z z) Pr Z z) Pr 5X Y z). This probability can be obtained by simply taking the integral of the joint density of X and Y over the region where 5X Y z. The figure below shows the region where the joint density is ASM Study Manual for Course P/ Actuarial Eamination. Copyright 4- by Krzysztof Ostaszewski - 7 -

17 SECTION allowed to play, and the shaded area is where 5X Y z. y y.5.z z The problem is that the position of the line y.5.z can vary as z varies, and we get different results in different cases. The line crosses the point,) when z 5. If z 5 then the line does not go through the twenty by twenty square at all and the probability of being above the line is. Since the slope of the line is more than, the net point where a change occurs is when the line crosses the point,), which occurs for z 3. The value of the CDF of Z for any z between 3 and 5 is the area shown in this figure: y y.5.z z The point where the line crosses the ais is at, y) ),5 z z 5, and the point where it crosses the line is at, y. The area of the bottom-right triangle left out of the calculation of probability is therefore z 5 5 z, and the corresponding ASM Study Manual for Course P/ Actuarial Eamination. Copyright 4- by Krzysztof Ostaszewski - 7 -

18 PRACTICE EXAMINATION NO. 6 probability is times that, so that the probability we are looking for is: 4 F Z z) 8 z 5 5 z. The corresponding density is: f Z z) F Z z) 8 5 ) 5 z ) 8 z 5 ) ) 4 z z z, for 3 z 5. This is the same answer in this range of values of z that we obtained before. The second case starts with z for which the point, ) is crossed by the line, i.e., z 3. This case is generally described by this figure: y.5.z z This case ends when the line crosses the origin, i.e., when z. Between z and z 3, the probability we want to find is just the area of the shaded region as a fraction of the area of the z by square. The bottom side of the shaded region has length from y and the equation 5 of the line) and the top side has length 8 + z from y and the equation of the line), so that 5 its area is 4 + z z. As a fraction of the entire area, this is 5 F Z z) 5 z z 5. Therefore, f Z z) F Z z for z 3. Again, this is the same answer we obtained 5 before. The final case is when the line crosses the y-ais above the origin, but below. When the line crosses the origin, z. When the line crosses the y-ais at the point, ), we have z. In this case, the figure looks as follows: ) ASM Study Manual for Course P/ Actuarial Eamination. Copyright 4- by Krzysztof Ostaszewski

19 SECTION y y.5.z z When y, we have 8 + z, so that the area of the marked triangle is z 5 + z. The by square has the area of 4, so that 8 + z 5 + z 8 + z 5 + z F Z z). 4 8 Therefore, z f Z z) F Z z) z z z 8, for z. This is again the same formula we obtained before. Answer E.. Dr. Ostaszewski s online eercise posted March 6, 5 Let X ), X ),, X 8) be the order statistics from a random sample X, X,, X 8 of size 8 from a continuous probability distribution. What is the probability that the median of the distribution under consideration lies in the interval X ), X 7)? A. 8 B. 8 9 C. 8 D. 4 8 E. Cannot be determined Let m be the median of the distribution. We are looking for Pr X ) m X 7) ). We note the following: ASM Study Manual for Course P/ Actuarial Eamination. Copyright 4- by Krzysztof Ostaszewski

20 ) ) Pr m < X ) Pr X ) m X 7 { }{ X 7) < m} ) Pr m < X PRACTICE EXAMINATION NO. 6 ) ) Pr X 7) < m). ) is the same as the probability that in a random sample Now we note that Pr m < X ) X, X,, X 8 of size 8 there are or elements less than or equal to m. As the probability of being less than or equal to m is eactly, this amounts to performing eight Bernoulli trials with the probability of success of with success being having a piece of the random sample less than the median) and getting only or successes, and that probability equals ) is the probability that either none or one of eight elements of the random Similarly, Pr X 7) < m sample X, X,, X 8 are greater than or equal to m. As the probability of being greater than or equal to m is eactly, this again amounts to performing eight Bernoulli trials with the probability of success this time success means getting a number more than the median) of and getting only or successes, i.e., the probability we are looking for, Pr X 7) < m), is again Hence: Answer C. ) ) Pr m < X ) Pr X ) m X 7 ) Pr X 7) < m) Dr. Ostaszewski s online eercise posted April, 5 In a block of car insurance business you are considering, there is a 5 chance that a claim will be made during the upcoming year. Once a claim is submitted, the claim size has the Pareto distribution with parameters 3 and, for which the mean is given by the formula, and the second moment is given by the formula. Only one claim will ) ) happen during the year. Determine the variance of the unconditional distribution of the claim size. A. 65 B C. 5 D. 75 E. ASM Study Manual for Course P/ Actuarial Eamination. Copyright 4- by Krzysztof Ostaszewski

21 SECTION Let Y be the random claim size. Its probability distribution is mied: 5 in the Pareto distribution with parameters 3 and and 5 in a point-mass at. Its mean is simply half of the Pareto distribution with parameters 3 and, i.e., half of, or 5. The second moment is also half of the second moment of that Pareto distribution, i.e., half of 5. The variance is therefore ) ), or That s Answer B. You could also do this by defining X when there is no claim, and X when there is a claim, with Pr X ) Pr X ), so that X is a Bernoulli Trial with p. Then we see that Y X ) is degenerate distribution equal to with probability, while Y X ) is Pareto with 3 and. Based on this E Y X ) so that E Y X ) 5, and Var Y X ) so that Var Y X, 5, 3 + -,.-, ) ) ) 75. Therefore E Var Y X) ) * 3 ) 3) 5 75, ) E 75,X) 75, E X )) Var 5X) 5 Var X) ,5, ) E Var Y X) ) + Var E Y X) ) 375, + 6,5 437,5. Var E Y X Var Y Answer B, again. ) 75, 375,, 3. Dr. Ostaszewski s online eercise posted April 9, 5 A random vector X,Y ) has the bivariate normal distribution with mean,) and the variancecovariance matri origin.. Find the probability that X,Y ) is in the unit circle centered at the ASM Study Manual for Course P/ Actuarial Eamination. Copyright 4- by Krzysztof Ostaszewski

22 A.. B C D..63 E PRACTICE EXAMINATION NO. 6 First observe that X and Y, being uncorrelated and having the bivariate normal distribution, are independent. We are trying to find Pr X + Y ). But X + Y has the chi-square distribution with two degrees of freedom, or, equivalently, the gamma distribution with parameters, and. But any gamma distribution with is an eponential distribution whose parameter is and mean is. So the problem simply asks what the probability is that an eponential distribution with mean is less than, and that s e Answer C. 4. Dr. Ostaszewski s online eercise posted April 6, 5 You are given that X and Y both have the same uniform distribution on [, ], and are X independent. U X + Y and V. Find the joint probability density function of U, V) X + Y evaluated at the point,. A. B. 4 C. 3 D. E. We have X UV, and Y U UV. Therefore, ) ) det, y u,v This implies that f U,V This density at Answer D. u y u v det y v v u v u ) ) u,v) f X,Y u,v), y u,v) ), y u,v, equals. uv u + uv u. u u. ASM Study Manual for Course P/ Actuarial Eamination. Copyright 4- by Krzysztof Ostaszewski

23 SECTION 5. Dr. Ostaszewski s online eercise posted April 3, 5 Three fair dice are rolled and X is the smallest number of the three values resulting if more than one value is the smallest one, we still use that value). Find Pr X 3). A. 36 B. 37 C. 38 D. 39 E Pr X 3) is the same as the probability that all three dice show a 3 or more, and at least one shows a 3. This means that the event of all three showing at least 3 happened, but the event of all three of them showing at least a 4 did not happen. This probability is: Answer B. 3 3 but not All three show 3,4,5 or 6 All three show 4,5, or Dr. Ostaszewski s online eercise posted April 3, 5 The moment-generating function of a random variable X is M X t) , for t >. t Calculate the ecess of Pr X > over its best upper bound given by the two-sided) Chebyshev s Inequality. A..35 B.. C. D..5 E.. Recall that the moment-generating function MGF) of a mied distribution is a weighted average of MGFs of the pieces of the miture. Also recall that the MGF of the eponential distribution u) ) with parameter is M T u, for u <, so with, for u <, M T u u. This looks familiar, but not what we want. Consider a miture of a degenerate random variable W, where Pr W ), with weight.5, and T, eponential with, with weight Note that M u) E W ) euw E e ). We see then that W with probability 4, X T with probability 3 4. ASM Study Manual for Course P/ Actuarial Eamination. Copyright 4- by Krzysztof Ostaszewski

24 Therefore E X) , Also, the second moment is: ) E X ) 3. Second moment of T PRACTICE EXAMINATION NO. 6 The variance of X is therefore , and the standard deviation is 5 4. Because this random variable is non-negative almost surely, the two-sided) Chebyshev s Inequality implies that 4 Pr X 3 4 > 5 4 Pr X 3 4 > 5 Pr X < Pr X > Pr X > Thus Pr X > is the best upper bound estimate for this probability provided by the 4 two-sided) Chebyshev s Inequality. The eact probability is Pr X > Pr T > e.5. The ecess equals approimately It is interesting to note that if we use the one-sided Chebyshev s Inequality Pr X µ k ) Pr X µ + k ) + k, with µ 3 4, 5 4, and k, we get Pr X > , and the ecess equals 5 approimately , approimately answer D. But in my opinion, the onesided version of the Chebyshev s Inequality is not covered on eam P/. Answer E. 7. Dr. Ostaszewski s online eercise posted May 7, 5 The time to failure X of an MP3 player follows a Weibull distribution, whose survival function is s X ) e ) ) e for >. It is known that Pr X > 3 probability that this MP3 player is still functional after 4 years. A..498 B..8 C..353 D..69 E..3, and that Pr X > 6 ). Find the 4 e ASM Study Manual for Course P/ Actuarial Eamination. Copyright 4- by Krzysztof Ostaszewski

25 SECTION From the information given in the problem we have s X 3 ) e and s X 6 ) e 4. Therefore, ) e 3 3 that e, as well as e conclude that s X 4 Answer D. ) 6, or 3. Furthermore, e ) e ) 4 e 4. We are looking for s X 4 e 4 3 ) 6 e 6 9 *.69. ) e ) 4. But e ) 3 e implies e ) e 4 implies that. Therefore, we 8. Dr. Ostaszewski s online eercise posted October 3, You are given that the hazard rate for a random variable X is X ) for >, and zero otherwise. Find the mean of X. A. B. C..5 D. 3 E. 3.5 t dt We calculate the survival function of X as e using the Darth Vader Rule we calculate the mean: ) e d E X + Substitute z, then zdzd + + ze z dz ze z d. Mean of EXP) You could also notice that this is the hazard rate X ) ) * distribution with and, so that E X ) + ) * 3 are not epected to memorize the Weibull distribution any more. Answer B. e t ) e + ) e for >, and then + + for a Weibull ), but of course you 9. Dr. Ostaszewski s online eercise posted October 3, Let P be the probability that an MP3 player produced in a certain factory is defective, with P assumed a priori to have the uniform distribution on [, ]. In a sample of one hundred MP3 players, is found to be defective. Based on this eperience, determine the posterior epected value of P. ASM Study Manual for Course P/ Actuarial Eamination. Copyright 4- by Krzysztof Ostaszewski - 8 -

26 A. B. C. 99 D. 5 E. 5 PRACTICE EXAMINATION NO. 6 Let us write N for the number of defective items in a sample of. Then N P p is binomial with n and p being the probability of success in a single Bernoulli Trial. We have f P,N p, p ) f P,N p, f P p N Now note that ) ) f N ) f P,N p,)dp p ) p p ) p ) 99 ) 99 dp ) p p ) 99 dp u p v p du dp dv p) 99 dp Integration by parts + p ) p p f P f P ) ) f N P p p) f N P p)dp ). p ) p) 99 p ) p) 99 dp p p) This means that f P p N ) p p) 99. We then calculate ) p p p E P N ) 99 dp p p u p v p) du pdp dv p) 99 dp Integration by parts. p p ) 99 dp p p) p ) + p) dp p + p p) dp p p) dp u p v p p) dp 5. du dp dv p) dp Integration by parts If you remember the beta distribution, then the posterior probability distribution of P is beta with ASM Study Manual for Course P/ Actuarial Eamination. Copyright 4- by Krzysztof Ostaszewski - 8 -

27 SECTION parameters +, and Thus, its epected value is Answer E. +, same answer. 5. Dr. Ostaszewski s online eercise posted November 6, You are given that Pr A) 5, Pr A B) 3 5, Pr B A Find Pr A B C). ) 4, Pr C B ) 3, and Pr C A B). A. 3 B. 5 C. 3 D. E. 4 First note that 4 Pr B A Answer D. Pr A B C ) Pr A) ) Pr B C) ) Pr A B ) Pr A B C ) 5 Pr A B ), so that Pr A B Pr C A B )Pr A B Pr C B)Pr B) Pr C A B)Pr A B Pr C B) Pr A B) Pr A ) ) ) + Pr A B) ). Hence, ) *. Dr. Ostaszewski s online eercise posted November 3, Let X ),..., X n) be the order statistics from the uniform distribution on [, ]. Find the correlation coefficient of X ) and X n). A. n B. n + C. D. n + E. n If X is uniform on [, ], so is X. The order statistics for a random sample for X are the same as the order statistics for a random sample of X, but in reverse order. This means that we have E X ) ) E X n) ) and Var X ) ) Var X n) ) Var X n) ). Note that F X n) ) n, i.e., the n-th power of the CDF of uniform distribution on [, ], so that s X n) ) n, and ASM Study Manual for Course P/ Actuarial Eamination. Copyright 4- by Krzysztof Ostaszewski - 8 -

28 ) n E X n) )d n n +. This implies that ) E X n) ) n E X ) We also have f X n) ) n n, and ) ) n n n + n +. E X n d n n+ d n n +. This implies that: Var X n) ) n n + n ) n n + n + n + ) n n + ) ) n + ) PRACTICE EXAMINATION NO. 6 )). n n +) n + ) Var X The joint density of X ) and X n) is determined from observing that if X ) is in, + d) and X n) is in y, y + dy), with < y, then n pieces of the random sample must be in the interval, y), and the probability of this is Therefore, f X ),X n), y)ddy ) X n) ) E X n n ) number of ways to pick ordered samples of size from population of size n, or just note that there are n ways to pick the first number and n ways to pick the second one ) y probability of being in the interval,y) ASM Study Manual for Course P/ Actuarial Eamination. Copyright 4- by Krzysztof Ostaszewski n d dy. probability of being in the interval,+d ) y y n y n ) y )n d dy ny n ) y ) n d) dy v y ) n ) y ) n d probability of being in the interval y,y+dy) y u ny y ) n d) du d dv n dy INTEGRATION BY PARTS ny y y )n n ) ) ) dy y n+ dy n +. This implies that the covariance is n + n, and the correlation coefficient is: n + n + n n +) n n +) n + ) Answer E. ) n +) n n + ) n +) n + ) n n +) n + ) n.

29 SECTION. Dr. Ostaszewski s online eercise posted November, A random variable X has the log-normal distribution with density f X ) ln µ e ) for >, and otherwise, where µ is a constant. You are given that Pr X ).4. Find E X). A. 4.5 B C D. 5. E. Cannot be determined X is log-normal with parameters µ and if W ln X N µ, ). From the form of the density function in this problem we see that. Therefore ln X µ Pr ln X ln) Pr ln µ.4. Let z.6 be the 6-th percentile of the standard normal distribution. Let Z be a standard normal ).4. But ln X µ random variable. Then Pr Z z.6 ln X µ is standard normal, thus ln µ z.6, and µ ln + z.6. From the table,.5).5987 and.6).66. By linear interpolation, z ) This gives µ ln + z The mean of the log-normal distribution is E X) e µ+, so that in this case E X) e Answer A. 3. Dr. Ostaszewski s online eercise posted November 7, You are given a continuous random variable with the density f X ) + for, and otherwise. Find the density of Y X, for all points where that density is nonzero. A. y B. y C. 3 y D. 4 3 y E. yln This is an unusual problem because you cannot use the direct formula for the density, as the transformation Y X is not one-to-one for. So the approach using the cumulative distribution function will work better. We have ASM Study Manual for Course P/ Actuarial Eamination. Copyright 4- by Krzysztof Ostaszewski

30 F X F Y ) t + dt t 4 + t t t PRACTICE EXAMINATION NO. 6 ) for, for <, and for >. Note that Y takes on only non-negative values. Therefore, y) Pr Y y) Pr X y F X y ) F X y ) Hence, f Y y) F Y y) y. Answer A. ) y +) ) ) Pr y X y 4 y + 4 y + y + y + y 4, y. 4. Dr. Ostaszewski s online eercise posted December 4, An insurer has independent one-year term life insurance policies. The face amount of each policy is. The probability of a claim occurring in the year under consideration is.. Find the probability that the insurer will pay more than the total epected claim for the year. A.. B.. C..6 D..6 E..3 The epected claim from each individual policy is.. The overall epected claim is. The total claim for the year will be more than if there is more than death. The number of claims N has the binomial distribution with n, p.. The probability sought is: Pr N > Answer D. ) ) Pr N ) + Pr N ) ) Dr. Ostaszewski s online eercise posted December, An insurance policy is being issued for a loss with the following discrete distribution:, with probability.4, X, with probability.6. Your job as the actuary is to set up a deductible d for this policy so that the epected payment by the insurer is 6. Find the deductible. A. B. 5 C. 7 D. E. 5 ASM Study Manual for Course P/ Actuarial Eamination. Copyright 4- by Krzysztof Ostaszewski

31 SECTION The large loss of will be paid after deductible as d, and this can happen with probability.6, so that the epected payment of that loss is.6 d 6 if d. With that large of a deductible, the smaller loss of must always be absorbed by the insured, so it will not affect the epected payment, and therefore with the deductible of, the epected payment is 6. Any smaller deductible will increase the epected payment, and a larger deductible will decrease it. Thus the deductible must equal to. Answer D. ).6d.This amount is equal to 6. Dr. Ostaszewski s online eercise posted December 8, For a Poisson random variable N with mean find lim E N N ). A. B. C. D. e E. Cannot be determined Let f N n) n n e for n,,, be the probability function of N. We have Answer C. ) lim lim E N N lim f N ) + n f N N n N + n n ) lim E N n f N n) lim f N ) ) + f n N n Pr N n ) ) lim e ) ) + f lim n N n f N lim de lhospital n. e 7. Dr. Ostaszewski s online eercise posted December 5, X is a normal random variable with mean zero and variance and Y is distributed eponentially with mean. X and Y are independent. Find the probability Pr Y > X ). A. e B. e C. D. E. We know that f X ) e e for < < +, and f Y y) e y for y >, so ASM Study Manual for Course P/ Actuarial Eamination. Copyright 4- by Krzysztof Ostaszewski

32 that f X,Y Answer E., y) e e y for < < + and y >. Therefore, Pr Y > X ) e e y dyd e d + * e d + PRACTICE EXAMINATION NO. 6 e s Y )d e e d + ) + + ) ) Integral of PDF of N, * + + * Dr. Ostaszewski s online eercise posted January, X ), X ),, X 4) are order statistics from a continuous probability distribution with a finite mean, median m and variance. Let be the cumulative distribution function of the standard ) normal distribution. Which of the following is the best approimation of Pr X ) m Central Limit Theorem? ) B..49 C..53 D..56 E. A..5 using the Let F X be the cumulative distribution function of the distribution under consideration. We have: 4 ) Pr Eactly j random sample elements are m Pr X ) m ) j 4 4 j F X m) ) j ) F X m) ) 4) j j j j. The last epression is the probability that a binomial random variable with parameters n 4, p.5, has at least successes. The mean of that binomial distribution is and the variance of it is. Denote that binomial variable by N. We have: Answer D. Pr N ) Pr N > 9.5) Pr continuity correction N ) * > ) ASM Study Manual for Course P/ Actuarial Eamination. Copyright 4- by Krzysztof Ostaszewski ).9744 ).56.

33 SECTION 9. Dr. Ostaszewski s online eercise posted January 8, N is a Poisson random variable such that Pr N ) Pr N ). Find the variance of N. A..5 B.. C..68 D. 3.5 E. 5. Let be the mean and the variance of N. We have Pr N ) e + e Pr N ) e Therefore, +, so that, and ± + 4 be positive, This is both the mean and the variance of N. Answer C... Because the parameter must 3. Dr. Ostaszewski s online eercise posted January 5, There are two bowls with play chips. The chips in the first bowl are numbered,, 3,,, while the chips in the second bowl are numbered 6, 7, 8,, 5. One chip is chosen randomly from each bowl, and the numbers on the two chips so obtained are compared. What is the probability that the two numbers are equal? A. B. 5 C. D. 4 E. 5 There are pairs of chips that can be picked, and of these there are 5 pairs of identical 5 numbers, so that the probability desired is 4. Answer D. ASM Study Manual for Course P/ Actuarial Eamination. Copyright 4- by Krzysztof Ostaszewski