Posted June 26, 2013 In the solution of Problem 4 in Practice Examination 8, the fourth item in the list of six events was mistyped as { X 1

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1 ASM Stud Manual for Exam P, Sixth Edition B Dr. Krzsztof M. Ostaszewski, FSA, CFA, MAAA Web site: krzsio@krzsio.net Errata Effective Jul 5, 3, onl the latest edition of this manual will have its errata updated. You can find the errata for all latest editions of m books at: Posted June 6, 3 In the solution of Problem 4 in Practice Examination 8, the fourth item in the list of six events was mistped as { X < X < X 3 }, while it should have been { X < X 3 < X }. Posted April 5, 3 The second to last formula in the solution of Problem in Practice Examination 5 should be: x 3 Pr( X > ) 4 x ddx 3 4 x( x)dx 3 4 x 4 x3 3 x The calculation shown was correct but there was a tpo in limits of integral calculation. x Posted April, 3 The last formula in the solution of Problem in Practice Examination 4 should be.5.6 E( Y ).5 x x 3.5 dx x 3.5 dx dx + x dx.5 x x x x.5 x x.6 x The calculation shown was correct, but a factor in the numerator of the last fraction was mistped as.5, when it should have been..6 Posted March 6, 3 In the solution of Problem 6 in Practice Examination, the expression rectangle [, ] [,3], should be rectangle [,] [,], This expression appears twice in the solution.

2 Posted June 6, In the solution of Problem 3, Practice Examination, the second method of solving the problem should be: You can also ask ourself these two questions: What is the total number of possible outcomes? It is n. What is the total number of favorable outcomes? It is equal to the number of possible was to get the third head on the n-th toss, which is the number of was to put two heads n among the first n tosses, i.e.,. This gives the probabilit sought as n n+ ( n ) ( n ) n. Posted April 3, The fourth formula in the solution of Problem 4 in Practice Examination 7 should be: intead of M ± M ± ± ± , , Posted April, The first formula in the last line of the solution of Problem 9 in Practice Examination 7 should be Pr( X 8) 4 instead of Pr ( X 8 ) 4. Posted March, The second sentence of the solution of Problem in Practice Examination should be: As the polic has a deductible of (thousand), the claim pament is, when there is no damage,with probabilit.94, Y max(, X ), when < X < 5, with probabilit.4, 4, in the case of total loss,with probabilit..

3 Posted Januar 5, In Problem 6 in Practice Examination 6, the calculation of the second moment of X should be: instead of E X E X E X 3. Second moment of T ( +) 3. Second moment of T Posted Jul 4, In the solution of Problem in Practice Examination 6, the statement under the first expression on the right-hand side of the third to last formula should be: number of was to pick ordered samples of size from population of size n instead of number of was to pick ordered samples of size n from population of size n Posted Jul 3, In the solution of Problem, Practice Examination 5, at the end of the first part of the fourth sentence of the solution, 5/6 is a tpo, it should be 5/36, as used in the formula for Pr (Y 6). Posted June 9, 99% 5% Knows Baes In the alternative solution of Problem 8, Practice Examination, the probabilit tree diagram should be: % 5% Actuarial Knows no Baes 5% 98.5% 3% Knows Baes 3% Random student Mathematics.5% 3% Knows no Baes 45% Statistics 98% 45% Knows Baes % 45% Knows no Baes Some numbers in the diagram were mistped.

4 Posted Januar 5, In the description of the gamma distribution in Section, the condition for the range of its MGF should be t < β, not < t < β. Posted June 8, 9 The last formula in the solution of Problem 9 of Practice Examination 5 should be Pr( X ) 4 6 instead of Pr ( X ) < 4 6. Posted June 7, 9 In the statement of Problem 9 in Practice Examination 7, Pr( X > 8) should be Pr( X 8), and in the solution, all inequalities should be changed accordingl. Posted April 6, 9 In the text of Problem 7 in Practice Examination 3, the word whwther should be whether. Posted March 9, 9 The solution of Problem 3 in Practice Examination 8 should be: Let E be the event that a new insured is accident-free during the second polic ear, and F be the event that a new insured is accident-free during the first polic ear, and let G be the event that this new insured was accident-free the last ear, before the polic was issued. Note that for an ear onl the previous ear affects a given ear, but not the ear before that. Therefore Pr( E) Pr ( E F G) ( E F G C ) ( E F C G) ( E F G C ) + Pr E F G C + Pr( E F C G) + Pr( E F C G C ) + Pr( G C ) Pr( F G C ) Pr( E F G C ) + + Pr( G C ) Pr F C G C Pr E F G Pr( G) Pr( F G) Pr E F G + Pr( G) Pr( F C G) Pr E F C G Pr E F C G C (.8) (.6) Answer E. Posted March 5, 9 The first sentence of Problem 6 in Practice Examination 6 should end with t <, instead of t >.

5 Posted March, 9 The properties of the cumulant moment-generating function should be: The cumulant generating function has the following properties: ψ X, d ln E( etx ) dt t E X t E XetX E e tx, E e tx d dt ψ ( t) X d E Xe tx dt t d 3 dt 3ψ X ( t) but for k > 3, d k dt ψ k X ( t) t t t E X (( ) ) 3, E X E X ( k) ψ X ( e tx )E e tx (( ) ) k. < E X E X E Xe tx E( e tx ) E Xe tx t Also, if X and Y are independent (we will discuss this concept later), ( t) ψ X ( at) + bt, andψ X+Y ( t) ψ X ( t) +ψ Y ( t). ψ ax+b Var( X), Posted Januar 3, 9 In Practice Examination 8, Problem 4, the calculation of the expected value had a tpo, an extra, unnecessar p in the second line, and it instead should be: k Pr( X k) E X Pr X k + k Pr( X k) k p + ( k +) Pr ( X k +) p + k Pr( X k +) + Pr( X k +) k p + k ( p) Pr( X k) + Pr( X k +) k k Pr( X +) Pr( Xk+) k this sum is equal to p + ( p) k Pr( X k) + Pr X k p + ( p) E X ( ) + ( p) + p E( X). k Posted September, 8 In the solution of Problem 6 in Practice Examination, the left-hand side of the second formula should be E( X ), not E( X).

6 Posted August 5, 8 In the solution of Problem in Practice Examination 8, the phrase Then the total number of claims among less than $,5 follows the binomial distribution with probabilit of success p.843 should be Then the total number of claims among less than $,5 follows the binomial distribution with probabilit of success p.5793 Posted Jul 4, 8 In the solution of Problem 6 of Practice Examination 7, the phrase Therefore, recalling that for a discrete random variable, whose onl possible values are possible integers, should be Therefore, recalling that for a discrete random variable, whose onl possible values are positive integers, Posted Jul 3, 8 In the solution of Problem 5 in Practice Examination 3 the expression Var( X ) 4, should be Var( X ) 5,. Posted Jul, 8 The formula at end of the third sentence of the solution of Problem 8 in Practice Examination should be % instead of %. Posted Februar 7, 8 The discussion of the lack of memor propert of the geometric distribution should have the formula Pr X n + k X n Pr( X k) corrected to: Pr( X n + k X > n) Pr( X k). Posted November 3, 7 In the solution of Problem No. 6 of Practice Examination No. 5, the sentence: Of the five numbers, can never be the median. should be Of the five numbers, neither nor 5 can ever be the median.

7 Posted October 5, 7 In the solution of Problem No. in Practice Examination No. 3, the random variable Y should refer to class B. Posted September 3, 7 The beginning of the third sentence in the solution of Problem No. in Practice Examination No. should be The mean error of the 48 rounded ages, instead of The mean of the 48 rounded ages. Posted August, 7 In the solution of Problem No. 3, Practice Examination No., the sentence But the median is given as should be But the mode is given as Posted August 9, 7 In the solution of Problem No. 4, Practice Examination No. 9, the formula should be Pr( Y ( 5) ) ( { }) Pr Y ( 5) > Pr Y ( 5) > Pr { Y } { Y } { Y ( 3) } { Y ( 4) } Y ( 5) Pr { X } { X } { X 3 } { X 4 } { Y 5 } 5 ( e ) F X Pr( Y ( 5) ) ( { }) Pr { Y } { Y } { Y ( 3) } { Y ( 4) } Y ( 5) Pr { X } { X } { X 3 } { X 4 } { X 5 } 5 ( e ) F X Posted August 5, 7 In the solution of Problem No., Practice Examination 8, the formula

8 should be f x, f X x Y f X f ( x, ) f ( x, ) f Y ( x Y ) f x, f ( x, ) f ( x, ) f Y dx dx x x dx x x dx x x x dx x dx x 5 x.5 x Posted August 5, 7 On page 7, the expression s Y should be s Y Pr( Y > ) Pr min( X,a) > Pr( min( X,a) > ), < a,, a, Pr( { X > } { X a} ) + Pr( { a > } { X > a} ), < a,, a, Pr( < X a) + Pr( X > a), < 3,, 3, Pr( Y > ) Pr min( X,a) > s X, < a,, a. Pr( min( X,a) > ), < a,, a, Pr( { X > } { X a} ) + Pr( { a > } { X > a} ), < a,, a, Pr( < X a) + Pr( X > a), < a,, a, s X, < a,, a. Posted Ma 6, 7 In the solution of Problem No., Practice Examination, the expression: Therefore, using Poohsticks, f Y( 3) f X + f X + f X3 f Y f Y( 3) e + e + e 3e 3 3e + 6e 3e 3 6e 6e 3. should be: Therefore, using Poohsticks,

9 f Y f X + f X + f X3 f Y f Y( 3) e + e + e 3e 3 3e + 6e 3e 3 6e 6e 3. There was a tpo in the subscript on the left-hand side of the formula. Posted Ma 5, 7 While the solution of Problem No. 6, Practice Examination No., is correct, I decided to post a better explanation of it, and here is the whole problem: Ma 98 Course Examination, Problem No. 8 Two balls are dropped in such a wa that each ball is equall likel to fall into an one of four holes. Both balls ma fall into the same hole. Let X denote the number of unoccupied holes at the end of the experiment. What is the moment generating function of X? A. 7 4 t if t or 3, otherwise B. 9 4 t + 8 t C. ( 4 3et + e 3t ) D. 4 et + 3e 3t E. 4 e 3t t e Solution. Each ball had four possible holes, so that the total number of was for the balls to fall is There can be onl two or three holes unoccupied. If three holes are unoccupied, there are four was to put two balls in one occupied hole. Hence, Therefore, Pr( X 3) 4 6 4, Pr( X ) Pr( X 3) 3 4. M X Answer C. ( t) 3 4 et + 4 e3t ( 4 3et + e 3t ). Posted Ma 5, 7 In the solution of Problem No., Practice Examination No. 7, both events were mistakenl labeled E. In order to correct that, the expression Define the following events: E { X } { X }, E { X } { X }. should be replaced b: Define the following events: E { X } { X },

10 E { X } { X }. Posted Ma, 7 The solution of Problem No. in Practice Examination No. is correct, but the correct answer choice is B (not C). Posted Ma, 7 The first displaed formula of the solution of Problem No. in Practice Examination No. should be: F T ( t) Pr( T t) Pr( max( T,T ) t) Pr( { T t} { T t} ) Pr( T t) Pr( T t) ( e 4t ) e 3t instead of F T ( t) Pr( T t) Pr( max( T,T ) t) Pr( { T t} { T t} ) Pr( T t) Pr( T t) ( e 4t ) e 3t e 3t e 4t + e 7t. e 3t e 4t + e 7t. Posted Ma 9, 7 The first sentence of Problem No. 5 in Practice Examination should be: A fair coin is tossed until a head appears. instead of: A fair coin is tossed until a head appears until a head appears. Posted Ma 9, 7 Problem No. in Practice Examination No. 9 should be as below (some of the letters τ were mistped as r): Casualt Actuarial Societ November 5 Course 3 Examination, Problem No. 9 Claim size X follows a two-parameter Pareto distribution with parameters α and θ, for which the densit and the cumulative distribution function are given below: αθ f X ( x) α x +θ F X ( x) α +, θ α ( x +θ ). α A transformed distribution Y is created b Y X τ. Which of the following is the probabilit densit function of Y? A. f Y ( +θ ) τ + τθτ

11 B. f Y C. f Y D. f Y E. f Y Solution. αθ α τ τ ( τ +θ ) α + θα θ ( +θ ) θ+ ατ θ + θ αθ α ( τ +θ ) α + τ τ α + We have X Y τ. Therefore, dx d τ τ. We conclude that f Y Answer B. f X ( x ) dx d αθ α ( τ +θ ) τ α + τ αθ α τ τ τ +θ α +. Posted Ma 9, 7 The first sentence of the solution of Problem No. in Practice Examination No. 9 should be: The densit of the exponential distribution with mean is e x for x >, while the densit of the exponential distribution with mean is e x for x >. instead of The densit of the exponential distribution with mean is e x densit of the exponential distribution with mean is e x for x >. for x >, while the Posted Ma 9, 7 In the solution of Problem No. 5, Practice Examination No. 3, the derivative is missing a minus, and it should dx dy ( 3 8 Y ) 3. The rest of the solution is unaffected.

12 Posted Ma 9, 7 Answer C in Problem No. 7 in Practice Examination No. 6 should be.636, not.66. It is the correct answer. Posted Ma 9, 7 In Problem No., Practice Examination No. 7, in the first part of the Practice Examination (not in the Solutions part), the expression E X µ X. E X µ Y should be Posted Ma 8, 7 Problem No. 4 in Practice Examination No. 5 should be (it had a tpo in the list of possible values of x) Ma 99 Course Examination, Problem No. 35 Ten percent of all new businesses fail within the first ear. The records of new businesses are examined until a business that failed within the first ear is found. Let X be the total number of businesses examined prior to finding a business that failed within the first ear. What is the probabilit function for X? A...9 x, for x,,,3,... B..9. x, for x,,,3,... C..x.9 x, for x,,3,... D..9x. x, for x,,3,... E.. ( x ).9 x, for x,3,4,... Solution. Let a failure of a business be a success in a Bernoulli Trial, and a success of a business be a failure in the same Bernoulli Trial. Then X has the geometric distribution with p., and therefore f X Answer A. ( x)..9 x for x,,, 3,. Posted Ma 5, 7 Problem No. in Practice Examination No. 3 should be (it had a tpo in one of the integrals and I decided to reproduce the whole problem to provide a better explanation): Sample Course Examination, Problem No. 35 Suppose the remaining lifetimes of a husband and a wife are independent and uniforml distributed on the interval (, 4). An insurance compan offers two products to married couples: One which pas when the husband dies; and

13 One which pas when both the husband and wife have died. Calculate the covariance of the two pament times. A.. B C D.. E Solution. Let H be the random time to death of the husband, W be the time to death of the wife, and X be the time to the second death of the two. Clearl, X max( H,W ). We have f H E H Furthermore, x F X This implies that ( h) f W ( w) 4 E( W ). Pr( X x) Pr max( H,W ) x Pr H x s X ( x) x 6 for x 4, and for h 4, and w 4. Thus ({ } { W x} ) Pr( H x) Pr( W x) x 4 x 4 x 6. 4 E( X) x 6 dx In order to find covariance, we also need to find E( XH ) E( H max( H,W )). We separate the double integral into two parts: one based on the region where the wife lives longer and one based on the region where the husband lives longer, as illustrated in the graph below w 4 max h,w w here max( h,w) h here 4 h

14 Finall, Answer C. hmax h,w E H max( H,W ) f H ( h) f W ( w) dwdh h 4 wh h 4 h 4 w 4 4 dw dh + hw 4 4 dw dh h w h 4 h wh w 4 hw w 4 4 h dh dh w 3 wh 6 dh 6h h3 + dh 3 4 wh h + 3 h3 dh 4 h + 8 h4 h 4 h Cov( X, H ) E( XH ) E( X)E( H ) Posted April 3, 7 Problem No. 3 in Practice Examination No. should be: An insurance polic has a deductible of. Losses follow a probabilit distribution with densit f X ( x) xe x for x > and f X ( x) otherwise. Find the expected pament. A. e B. e C. e D. e E. e Solution. Let X be the random variable describing losses, and let Y be the amount of pament. Then, X, Y X, X >. We can calculate the expected value of Y directl x E Y xe x dx x e x dx xe x dx u x v e x du xdx dv e x dx ( x e x ) x x + Integration b parts for the first integral xe x dx xe x dx e 8xe x dx x u 8x v e du 8dx dv e x dx Integration b parts for the first integral e + ( 8xe x x ) 8 x e x dx Survival function of exponential distribution with hazard rate evaluated at x e 8e 8e e. Answer D. We could also do this problem b appling the Darth Vader Rule. We have

15 s Y Pr( Y > ) Integration b parts Pr( X > ), if, Pr( X > ), if >, Pr X > + x u x v e xe x x dx dx dv e x x+ + e x dx + Survival function of exponential with hazard rate evaluated at + xe x dx, ( + )e + e ( + )e for an nonnegative. As the pament random variable is non-negative almost surel, the expected pament is Answer D. ( + )e d e e d This equals one because e for > is a densit of an exponential random variable with hazard rate + e e d This equals one because the integrand is the densit in this problem. + e. Posted April 9, 7 The second formula to the last in the solution of Problem No. 6 in Practice Examination No. 3 should be: d M t E X instead of E X dt t d M t dt t d + e t dt 3et 3 d + e t dt 3et t t + e t 3e t 3 + e t 3e t e t + 8e t 3 7 e t + e t + 8e t 3 et t t.. Posted April 7, 7 In the solution of Problem No. 9 in Practice Examination No. 3, the expression max( X,) should be min( X,). Posted April, 7 Practice Examinations: An Introduction, the sentence Practice examinations 6- are meant to be more challenging. should read Practice examinations 6- are meant to be more challenging.

16 Posted April, 7 Problem No. 6 in Practice Examination No. should be: Ma Course Examination, Problem No. 6, also Stud Note P-9-5, Problem No. 9 A compan offers earthquake insurance. Annual premiums are modeled b an exponential random variable with mean. Annual claims are modeled b an exponential random variable with mean. Premiums and claims are independent. Let X denote the ratio of claims to premiums. What is the densit function of X? A. x + B. ( x + ) C. e x D. e x E. xe x Solution. Let U be the annual claims and let V be the annual premiums (also random), and let f U,V u,v be its cumulative distribution function. We are given that U and V are independent, and hence be the joint densit of them. Furthermore, let f X be the densit of X and let F X f U,V ( u,v) e u v e e u e v for < u <, < v <. Also, noting the graph below, v u vx or v u/x u we have:

17 F X ( x) Pr X x vx Pr U V x Pr U Vx vx f U,V ( u,v) u vx dudv e u e v dudv e u e v dv e vx e v + v e dv u x+ v x+ e v + e dv x + e v v e x + +. Finall, f X ( x) F X ( x) ( x + ). Answer B. This problem can also be done with the use of bivariate transformation. We will now give an alternative solution using that approach. Consider the following transformation X U, Y V. Then the inverse transformation is U XY, V Y. We V know that f U,V f X,Y ( u,v) e u e v for u >, v >. It follows that ( ) ( u,v) ( x, ) e x e ( x, ) f U,V u( x, ),v x, for x > and >, or just x > and >. Therefore, f X ( x) e x e d x + x+ e d e x+ x det e x e w z x + e x+ dw x+ d dz e d INTEGRATION BY PARTS x + e x+ d x+ + x + e d x + x + e x+ ( x + ). Answer B, again. The second approach is probabl more complicated, but it is a good exercise in the use of multivariate transformations. Posted April 8, 7

18 On page 3, the sentence An infinite intersection of a sequence of sets of all elementar events that belong to an one of them. should read An infinite intersection of a sequence of sets of all elementar events that belong to all of them. A n is also defined as the set that consists n A n is also defined as the set that consists n Posted April 7, 7 On page 4-5, this sentence Also, for the points x where the jump occurs, in the weighted average of the two PDF s, f X ( x) w f X ( x) + w f X ( x), with X being the discrete one and X being the continuous one, and w + w, w, w, we actuall must have w f X ( x), and f X ( x) w f X ( x) for an x >. for which f X x should replace Also, for the points x where the jump occurs, in the weighted average of the two PDF s, f X ( x) w f X ( x) + w f X ( x), with X being the discrete one and X being the continuous one, and w + w, w, w, we actuall must have w, and f X ( x) w f X ( x) for an x for which f X ( x) >. The tpo was in the statement w instead of w f X ( x). Posted Februar, 7 Problem 3 (and its solution) in Practice Examination, should be: A new super-size passenger airplane has 4 seats in coach class and seats in the business class, with these being the onl two classes available. The weight of the luggage (all luggage pieces combined) of an individual coach passenger follows a continuous probabilit distribution with mean of pounds and the standard deviation of pounds, while the weight of the luggage (all luggage pieces combined) of a business class passenger follows a continuous probabilit distribution with mean of 8 pounds and the standard deviation of 5 pounds. The weights of individual passengers luggages are independent. You are given that the 95-th percentile of the standard normal distribution is Approximate the smallest possible number w such that the probabilit that the total weight of the entire luggage on the plane is less than w with probabilit.95, assuming that the airplane is full. A B. 55 C. 554 D E

19 Solution. Let us write X for the total weight of all of coach passengers luggage. Then X is the sum of weights of individual passengers luggages, and since these are independent and identicall distributed, and there are 4 coach passengers, X is approximatel normal. We also know that E( X) 4 4 and Var( X) 4 4. Let us also write Y for the total weight of all of business passengers luggage. Then Y is the sum of weights of individual passengers luggages, and since these are independent and identicall distributed, and there are business passengers, Y is approximatel normal. We know that E Y 8 8 and Var( Y ) 5 5. The total weight of all luggage is the sum of X and Y. Furthermore, X and Y are independent, and since a sum of two independent normal random variables is normal, X + Y is approximatel normal, with E X + Y and Var( X + Y ) If we write Φ for the CDF of the standard normal distibution, then from the table given for Course P, Φ(.644).9495, while Φ(.645).955. Using linear interpolation, we approximate the 95-th percentile of the standard normal distribution as The 95-th percentile of X + Y is therefore approximatel equal to Answer A. Posted Februar 9, 7 In Problem No., Practice Examination No., answer choice is D, instead of E as stated now (tpo). Posted Februar 6, 7 The last formula in the solution of Problem No. 8 in Practice Examination No. should be: kx x kx dxd dx x k4 d k5 x x k. It contained a tpo: x in place of, which did not materiall affect the solution or the answer, but was confusing. Posted Februar 5, 7 Problem No. in Practice Examination No. should be: X, X, X 3 is a random sample from the exponential distribution with mean. Let Y, Y, and Y ( 3) be the corresponding order statistics. Find the variance of the second order statistic.

20 A B C. 5 6 D E. 5 4 Solution. We have s Y Pr Y ( > ) Pr( X > ) Pr( X > ) Pr X 3 > so that Y is exponential with hazard rate 3, and f Y F Y( 3) Pr Y ( 3) Pr X so that f Y( 3) e Pr X 3 3e + 3e e 3, F Y( 3) Therefore, using Poohsticks, f Y( 3) f X 3e 6e + 3e 3. e 3, 3e 3. On the other hand, Pr X 3 + f X + f X3 f Y f Y 3 e + e + e 3e 3 3e + 6e 3e 3 6e 6e 3. You could also use the general formula for the PDF of the order statistics. Let us write X for a general exponential random variable with mean. Using that general formula Therefore, f Y E Y 3! ( )! ( 3 )! ( F X ) s X 6( e )( e ) e 6( e e 3 ). 6 e e 3 d 3 f X 6 e d 3e 3 d mean of EXP mean of EXP with hazard rate with hazard rate 3 Similarl, the second moment can be calculated as E( Y ) 6( e e 3 ) d 6 e d 3e 3 d 3 second moment of EXP second moment of EXP with hazard rate with hazard rate 3 Finall

21 Answer B. E( Y ) E( Y ) Var Y Posted Februar 5, 7 Problem No. in Practice Examination No. should be: Five different bidders are submitting independent bids for a race horse. The all believe the value of the horse to be between $, and $,, and as a result of that common belief, their bids, while independent, are all be uniforml distributed between $, and $,. Find the expected value of the difference between the winning (i.e., the highest) bid and the lowest bid. A. 6,667 B. 33,333 C. 5, D. 66,666 E. 83,333 Solution. Let us denote the five bids b X, X, X 3, X 4, and X 5. This is a random sample from the uniform distribution on, [ ] with PDF, x, f X ( x), otherwise, and CDF, x <, x F X ( x), x,, < x. Let Y ( 5) max( X, X, X 3, X 4, X 5 ) be the winning bid and Y min X, X, X 3, X 4, X 5 the lowest bid. Then F Y( 5) Pr( Y ( 5) ) Pr( max( X, X, X 3, X 4, X 5 ) ) Pr { X } { X } { X 3 } { X 4 } { X 5 } for, while F Y 5 >. Therefore 5 be for < and F Y 5 for

22 + s Y 5 E Y 5 5 d ( )d , On the other hand, s Y Pr( Y > ) Pr( min( X, X, X 3, X 4, X 5 ) > ) Pr { X > } { X > } { X 3 > } { X 4 > } X 5 > Therefore, E Y The answer is ( { }) + E Y ( 5) Y Answer D. 5 d + 6 ( ) E Y ( 5) , E( Y ) 83, , , Posted Februar 5, 7 The second sentence of Problem No. 4, Practice Examination No. should be: The amount of the doctor s bill X is a continuous random variable with standard deviation of 5, and the amount of the hospital bill Y is a continuous random variable with standard deviation of. Posted Februar 5, 7 The second answer choice in Practice Examination No., Problem No. 9 should be: B et It was mistakenl written as the same as A.

23 Posted Januar, 7 Exercise.8 should read: November Course Examination, Problem No. A compan takes out an insurance polic to cover accidents that occur at its manufacturing plant. The probabilit that one or more accidents will occur during an given month is 3. The number of accidents that occur in an given month is independent 5 of the number of accidents that occur in all other months. Calculate the probabilit that there will be at least four months in which no accidents occur before the fourth month in which at least one accident occurs. A.. B.. C..3 D..9 E..4 Solution. Consider a Bernoulli Trial with success defined as a month with an accident, and a month with no accident being a failure. Then the probabilit of success is 3. Now consider a 5 negative binomial random variable, which counts the number of failures (months with no accident) until 4 successes (months with accidents), call it X. The problem asks us to find Pr( X 4). We calculate Pr X 4 Pr( X ) Pr( X ) Pr( X ) Pr( X 3) Answer D Posted December 3, 6 In Section, the general definition of a percentile should be the -p-th percentile of the distribution of X is the number x p which satisfies both of p and Pr( X x p ) p. the following inequalities: Pr X x p It was mistped as the -p-th percentile of the distribution of X is the number x p which satisfies both of the following inequalities: Pr X x p p and Pr( X x p ) p. Posted December 8, 6 On page 3 in the bottom paragraph, the words: is also defined as the sent that consists of all elementar events that belong to an one of them. should be

24 is also defined as the set that consists of all elementar events that belong to an one of them. The word set was mistped as sent. Posted December, 6 In Practice Examination 6, Problem, had inconsistencies in its assumptions, and it should be replaced b the following: A random variable X has the log-normal distribution with the densit: f X ( x) ( ln x µ x π e ) for x >, 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 Solution. X is log-normal with parameters µ and σ if W ln X N µ,σ. The PDF of the lognormal distribution with parameters µ and σ is ( ln x µ ) f X ( x) xσ π e σ for x >, and otherwise. From the form of the densit function in this problem we see that σ. 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 random variable. Then Pr( Z z.6 ).4. But Z ln X µ thus ln µ z.6, and µ ln + z.6. From the table, Φ B linear interpolation, Φ.6 z This gives µ ln + z and (.6.5 ) The mean of the log-normal distribution is E( X) e µ+ σ, so that in this case E( X) e Answer A. is standard normal, Posted December, 6

25 In Practice Examination No. 8, in Problem No. 5, answer B should be listed as B..465 It was listed as B. 465, an obvious tpo. Posted November 9, 6 The solution of Problem No. 5 in Practice Examination No. is not clear in the explanation of the linear interpolation. Below is a more complete explanation: Ma 98 Course Examination, Problem No. If a normal distribution with mean µ and variance σ > has 46-th percentile equal to σ, then µ is equal to A. 8.5σ B. 9.9σ C..σ D..73σ E. Cannot be determined Solution. Let X be the random variable described in the problem. Then X µ has the standard σ normal distribution. Let us write x.46 for the 46-th percentile of X. We have X µ.46 Pr( X < x.46 ) Pr < x µ.46 X µ σ σ Pr < σ µ σ σ. From the table, Φ and Φ(.) Therefore, Φ(.) The probabilit desired is.46, and that is and Φ of the distance between.456 and.46, down from.46. This means that the 46-th percentile is approximatel of the distance between. and., down from., i.e., we can get the desired percentile of the standard normal distribution via the following linear interpolation approximation: ( (.) ).. z Therefore, σ µ.5, so that µ.5σ. σ Answer C. Posted November 5, 6 The third equation from the bottom in the solution of Problem No. in Practice Examination No. should be Pr( A E 99 ) Pr( A E 99 ) Pr( E 99 ) Previousl it had a tpo:

26 Pr( A E 99 ) Pr( A E 99 ) Pr( E 99 ) Posted November 4, 6 Problem No. 5 in Practice Examination No. should be as below. You are given that the random variable X is exponential with mean, and that the random variable Y is uniforml distributed on the interval [,]. Furthermore, it is known that X and Y are independent. Find the densit of the joint distribution of U XY and V X Y. A. B. e uv v e uv v for u >, v >, and u < v for u >, v >, and u < v C. ve uv for u > and v > D. ve uv for u >, v >, and u < v E. uv e uv for u >, v >, and u < v Solution. We have f X,Y ( x, ) e x for x > and < <. In order to find the joint densit of U and V, we need to first express X and Y in terms of U and V, i.e., find the inverse function of the transformation. Note that all variables considered here: X, Y, U, V, are positive with probabilit one. We have UV XY X Y X, so that X UV. Furthermore, U V XY XY Y. This gives us Y UV. Therefore, the inverse transformation, written in terms of regular variables, is ( x, ) ( uv, uv ). This results in x x ( x, ) ( u,v) det u v uv v uv u det v. v u v uv uv uv Therefore

27 f X,Y x u,v f U,V u,v (, ( u,v) ) x, u,v e uv v for uv > and < uv <, i.e., u >, v >, and u < v. Answer B. Posted November 4, 6 In the description of the exponential distribution, the sentence The exponential distribution is related to the geometric distribution in that if we have a random variable T with exponential distribution, and we define a new random variable K as the integer portion of T (for example, if T 7.5, then K 7, K is simpl the value of the greatest integer function when applied to T), then so defined K has the geometric distribution with p e λ. should be replaced b The exponential distribution is related to the geometric distribution in that if we have a random variable T with exponential distribution, and we define a new random variable K as the integer portion of T (for example, if T 7.5, then K 7, K is simpl the value of the greatest integer function when applied to T), then so defined K has the geometric distribution with q e λ. The onl difference is that the smbol in the last equation should be q, not p. Posted Ma 6, 6 The solution of Problem No. 3 in Practice Examination No. 3 should be (some of the exponents contained tpos) Solution. Let X be the random number of passengers that show for a flight. We want to find Pr(X 3) + Pr(X 3). We can treat each passenger arrival as a Bernoulli Trial, and then it is clear that X has binomial distribution with n 3, p.9. Therefore, 3 Pr( X x) x.9x. 3 x. The probabilit desired is: 3 Pr( X 3) + Pr( X 3) Answer E