Errata for the ASM Study Manual for Exam P, Fourth Edition By Dr. Krzysztof M. Ostaszewski, FSA, CFA, MAAA

Transcription

1 Errata for the ASM Study Manual for Exam P, Fourth Edition By Dr. Krzysztof M. Ostaszewski, FSA, CFA, MAAA Effective July 5, 3, only the latest edition of this manual will have its errata updated. You can find the errata for all latest editions of my 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 mistyped 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 : x 3 Pr( X > ) 4 x dydx 3 4 x( x)dx 3 4 x 4 x3 3 x The calculation shown was correct but there was a typo in limits of integral calculation. x Posted April, 3 The last formula in the solution of Problem in Practice Examination 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 mistyped 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], rectangle [,] [,], This expression appears twice in the solution. Posted April 3,

2 The fourth formula in the solution of Problem 4 in Practice Examination 7 should be: M ± ± , intead of M ± ± , Posted April, The first formula in the last line of the solution of Problem 9 in Practice Examination 7 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 policy has a deductible of (thousand), the claim payment is, when there is no damage,with probability.94, Y max(, X ), when < X < 5, with probability.4, 4, in the case of total loss,with probability.. Posted January 5, In Problem 6 in Practice Examination 6, the calculation of the second moment of X : instead of E X E X E X 3. Second moment of T ( +) 3. Second moment of T Posted July 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 : number of ways to pick ordered samples of size from population of size n instead of number of ways to pick ordered samples of size n from population of size n

3 Posted July 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 typo, it 5/36, as used in the formula for Pr (Y 6). Posted January 5, In the description of the gamma distribution in Section, the condition for the range of its MGF t < β, not < t < β. Posted June 8, 9 The last formula in the solution of Problem 9 of Practice Examination 5 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) Pr( X 8), and in the solution, all inequalities changed accordingly. Posted April 6, 9 In the text of Problem 7 in Practice Examination 3, the word whwther whether. Posted March 9, 9 The solution of Problem 3 in Practice Examination 8 : Let E be the event that a new insured is accident-free during the second policy year, and F be the event that a new insured is accident-free during the first policy year, and let G be the event that this new insured was accident-free the last year, before the policy was issued. Note that for any year only the previous year affects a given year, but not the year 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.

4 Posted March 5, 9 The first sentence of Problem 6 in Practice Examination 6 should end with t <, instead of t >. March, 9 The properties of the cumulant moment-generating function : 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 X E X ( e tx )E e tx (( ) ) k. E Xe tx E( e tx ) E Xe tx t Also, if X and Y are independent (we will discuss this concept later), ψ ax+b ( t) ψ X ( at) + bt, and ψ X+Y ( t) ψ X ( t) +ψ Y ( t). Var( X), Posted January 3, 9 In Practice Examination 8, Problem 4, the calculation of the expected value had a typo, an extra, unnecessary p in the second line, and it instead : 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

5 Posted September, 8 In the solution of Problem 6 in Practice Examination, the left-hand side of the second formula E( X ), not E( X). 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 probability of success p.843 Then the total number of claims among less than $,5 follows the binomial distribution with probability of success p Posted July 4, 8 In the solution of Problem 6 of Practice Examination 7, the phrase Therefore, recalling that for a discrete random variable, whose only possible values are possible integers, Therefore, recalling that for a discrete random variable, whose only possible values are positive integers, Posted July 3, 8 In the solution of Problem 5 in Practice Examination 3 the expression Var( X ) 4, Var( X ) 5,. Posted February 7, 8 The discussion of the lack of memory property 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. Of the five numbers, neither nor 5 can ever be the median.

6 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. The mean error of the 48 rounded ages, instead of The mean of the 48 rounded ages. Posted August 9, 7 In the solution of Problem No. 4, Practice Examination No. 9, the formula 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 f X ( x Y y) f ( x, y) f Y ( y) f ( x, y) xy xy y y x 5y f x, y dx xy dx y x dx f X ( x Y y) f x, y y f ( x, y) f ( x, y) f Y dx xy y xy dx xy y y x dx x.5y x y Posted May 5, 7

7 In the solution of Problem No., Practice Examination No. 7, both events were mistakenly labeled E. In order to correct that, the expression Define the following events: E { X } { X }, E { X } { X }. replaced by: Define the following events: E { X } { X }, E { X } { X }. Posted May 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. Posted May 9, 7 Answer C in Problem No. 7 in Practice Examination No , not.66. It is the correct answer. Posted May 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 Posted May 8, 7 Problem No. 4 in Practice Examination No. 5 (it had a typo in the list of possible values of x) May 99 Course Examination, Problem No. 35 Ten percent of all new businesses fail within the first year. The records of new businesses are examined until a business that failed within the first year is found. Let X be the total number of businesses examined prior to finding a business that failed within the first year. What is the probability 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,...

8 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 May 5, 7 Problem No. in Practice Examination No. 3 (it had a typo 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 uniformly distributed on the interval (, 4). An insurance company offers two products to married couples: One which pays when the husband dies; and One which pays when both the husband and wife have died. Calculate the covariance of the two payment 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. Clearly, 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 E( X) x 6 dx In order to find covariance, we also need to find E( XH ) E( H max( H,W )). 4 x 4 x 6.

9 We separate the double integral into two parts: one based on the region where the wife lives longer w and one based on the region where the husband lives longer, as illustrated in the graph below 4 max( h,w) w here max( h,w) h here 4 h Finally, 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 9, 7 The second formula to the last in the solution of Problem No. 6 in Practice Examination No. 3 : instead of d M t E X dt t d + e t dt 3et 3 + e t 8 + e t 3e t 3 + 8e t t e t t.

10 d M t E X dt t d + e t dt 3et 3 8 t + e t 8 + e t 3e t 3 + 8e t 3 et t. Posted April 7, 7 In the solution of Problem No. 9 in Practice Examination 3, the expression max( X,) min( X,). Posted April, 7 Problem No. 6 in Practice Examination No. : May Course Examination, Problem No. 6, also Study Note P-9-5, Problem No. 9 A company offers earthquake insurance. Annual premiums are modeled by an exponential random variable with mean. Annual claims are modeled by an exponential random variable with mean. Premiums and claims are independent. Let X denote the ratio of claims to premiums. What is the density 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 density of them. Furthermore, let f X be the density 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

11 we have: 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 + +. Finally, 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 ( u,v) e u e v for u >, v >. It follows that f X,Y ( ) ( u,v) ( x, y) e xy e ( x, y) f U,V u( x, y),v x, y for xy > and y >, or just x > and y >. Therefore, f X ( x) ye xy e y dy y x + x+ ye y dy e x+ y y y y y x det ye xy e w y z x + e x+ y dw x+ y dy dz e dy INTEGRATION BY PARTS x + e x+ y dy x+ + x + e y dy x + x + e x+ y y y y ( x + ).

12 Answer B, again. The second approach is probably more complicated, but it is a good exercise in the use of multivariate transformations. Posted January, 7 Exercise.8 should read as follows: November Course Examination, Problem No. A company takes out an insurance policy to cover accidents that occur at its manufacturing plant. The probability that one or more accidents will occur during any given month is 3. The number of accidents that occur in any given month is independent 5 of the number of accidents that occur in all other months. Calculate the probability 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 probability 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). But Answer D. Pr( X ) Pr( X ) Pr( X ) Pr( X 3) Pr X Posted December 3, 6 In Section, the general definition of a percentile 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 mistyped 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:

13 is also defined as the sent that consists of all elementary events that belong to any one of them. is also defined as the set that consists of all elementary events that belong to any one of them. The word set was mistyped as sent. Posted December, 6 In Practice Examination 6, Problem, had inconsistencies in its assumptions, and it replaced by the following:. A random variable X has the log-normal distribution with the density: 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 density 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 By 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,

14 Posted December, 6 In Practice Examination No. 8, in Problem No. 5, answer B listed as B..465 It was listed as B. 465, an obvious typo. Posted November 5, 6 The third equation from the bottom in the solution of Problem No. in Practice Examination No. Pr( A E 99 ) Pr( A E 99 ) Pr( E 99 ) Previously it had a typo: Pr( A E 99 ) Pr( A E 99 ) Pr( E 99 ) Posted November, 6 The third sentence of Problem No. in Practice Examination No. 7 should read: Let X and Y be the times at which the first and second circuits fail, respectively, counting from the moment when a circuit is activated. Posted August 3, 6 In the solution of Problem No. 4 in Practice Examination No., the formula f Y ( y) k y, y f Y k. Posted July 7, 6 The last two lines of the solution of Problem No. 5 in Practice Examination No. 8 : Alternatively, M X Therefore, ( t) d dt E( X) M X ( ( t ) ) + d dt e tn t n! n + e tn. ( n )! ( ) + e n ( n )! 4 + e n n! n e

15 Posted July 3, 6 The relationship between the survival function and the hazard rate, stated just after the definition of the hazard rate, x s X ( x) e instead of x λ X ( u)du λ X ( u)du s X ( x) e. The second equation applies in the case when the random variable X is nonnegative almost surely. Posted July 3, 6 The condition concerning a continuous probability density function in its first definition, in Section, x f X for every x instead of f X ( x) for every x. Posted May 6, 6 In the solution of Problem No. in Practice Examination No. 8 the last formula y E( X Y y) x x dx y y y x3 dx xy y 4 x4 x y. Posted May 6, 6 The solution of Problem No. 3 in Practice Examination No. 3 (some of the exponents contained typos) 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 probability desired is: 3 Pr( X 3) + Pr( X 3) Answer E

16 Posted March, 6 The formula for the variance of a linear combination of random variables should be: Var( a X + a X + + a n X n ) a a a n a a a n n ( Cov ( X, X i j ) i, j,,,n ) a a a n ( Cov ( X, X i j ) i, j,,,n ) a i Var X i + a i a j Cov X i, X j. i n n i ji+ a a a n T The only incorrect piece was in the last double sum, where a a appeared instead of a i a j. Posted March, 6 The moment generating function of the chi-square distribution : n M X ( t) t. (its argument was mistakenly written as x instead of t). Posted February 8, 6 Problem 8 of Practice Examination 7 and its solution : The number of warranty claims in a month follows a Poisson distribution with mean 4. Each warranty claim requires a payment of $ by the manufacturer to a repair facility. Find the probability that the total payment by the manufacturer in a month is less than one standard deviation away from the expected value of that payment. A..5 B..397 C. 547 D E Solution. Let X be the random number of the warranty claims. The expected value of monthly payments is 4 $ $48, and the standard deviation is $ $4. The probability that the total payment is between $48 $4 $4 and $48 + $4 $7, not including the endpoints (because we are supposed to be less than one standard deviation away from the mean) is the sum of probabilities of making 3 payments (cost of $36), 4 payments (cost of $48), or 5 payments (cost of $6), i.e.,

17 5 5 Pr( X k) e 4 4 k k3 k! e k3 3! + e ! + e ! Answer C. Posted February 7, 6 In the solution of Problem No. 9 of Practice Examination No. 8, the formula Pr( Y ) k!! 3! + 4! + + ( 5! )n k ( n k)! k, Pr( Y k) k k!! 3! + 4! + + ( 5! )n k ( n k)! k. Posted February, 6 The solution of Problem No. 8 of Practice Examination No. 4 : Let X be the claim for employee who incurred a loss in excess of, and Y be the claim for the other employee. The joint distribution of X and Y, given that each loss [ ]. This means that we can calculate all probabilities by occurs, is uniform on,5 comparing areas. The probability that total losses exceed reimbursement, i.e., X + Y > 8, given that X > is the ratio of the area of the triangle in the right upper corner of the square,5 [,5] [,5], i.e., Y [ ] in the figure below to the area of the rectangle 5 3 X + Y 8 X 5 8

18 However, this is conditional on the occurrence of the loss of the employee whose loss is Y (and not conditional on the occurrence of the loss for the employee whose loss is X, because the event considered is already conditional on X >, so the loss has occurred). The probability of the loss occurring is 4%. Therefore the probability sought is 4% Answer B. Posted January 3, 6 In the solution of Problem No. (May Course Examination, Problem No. 9) in Practice Examination No., the fourth sentence gin The mean of the 48 rounded ages, E E i i