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

ASM Study Manual for Exam P, Second Edition By Dr. Krzysztof M. Ostaszewski, FSA, CFA, MAAA (krzysio@krzysio.net) Errata 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: http://smarturl.it/errata 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 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= 3 4 4 =. 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.5.6.5 + x 3.5 dx.5.6.5 + 5.6 =.5 dx + x dx =.5 x 3.5.6 x=.5.6.5 = x.6.5.5.6.5.5.6 = + +.6.5.93473..5x.5 x.5.5.5.5.5 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 [,] [,], should be rectangle [,] [,], This expression appears twice in the solution. 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

Y =, when there is no damage,with probability.94, 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 should be: instead of = 4 + 3 4 + E X = E X E X = 3. Second moment of T = 4 + 3 4 ( +) = 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 should be: 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 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 should be 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 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 April 6, 9 In the text of Problem 4 in Practice Examination, the sentence: We put that chip aside and pick a second chip from the same contained. should be:

We put that chip aside and pick a second chip from the same container. Posted April 6, 9 In the text of Problem 7 in Practice Examination 3, the word whwther should be whether. Posted March 5, 9 The first sentence of Problem 6 in Practice Examination 6 should end with t <, instead of t >. Posted March, 9 The properties of the cumulant moment-generating function should be: The cumulant generating function has the following properties: ψ X ( ) =, E e tx d dt ψ ( t) X = d E Xe tx dt d 3 dt 3ψ X ( t) but for k > 3, d k dt ψ k X ( t) = E X (( ) ) 3, = E X E X ( k) =ψ X ( ) < E X E X d ln E( etx ) dt ( e tx )E e tx (( ) ) k. E Xe tx E( e tx ) = E X = E XetX E e tx E Xe tx Also, if X and Y are independent (we will discuss this concept later), ( t) =ψ X ( at) + bt, and ψ X+Y ( t) =ψ X ( t) +ψ Y ( t). ψ ax+b, = Var( X), 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). Posted July 3, 8 In the solution of Problem 5 in Practice Examination 3 the expression Var( X ) = 4, should be 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. should be Of the five numbers, neither nor 5 can ever be the median. 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 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. 6 should be.636, not.66. It is the correct answer. Posted May 8, 7 Problem No. 4 in Practice Examination No. 5 should be (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,... 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 ( x) =..9 x for x =,,, 3,. Answer A. Posted May 5, 7 Problem No. in Practice Examination No. 3 should be (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. 44.4 C. 66.7 D.. E. 466.7 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 ( h) = f W ( w) = 4 Furthermore, x F X This implies that s X for h 4, and w 4. Thus E ( H ) = E( W ) =. = = Pr( X x) = Pr max( H,W ) x ( { }) = Pr H x = Pr { H x} W x ( x) = x 6 Pr W x = x 4 x 4 = x 6.

for x 4, and 4 E( X) = x 6 dx = 4 43 48 = 3 4 3 = 8 3. 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 over the region where the wife lives longer and one over the region where the husband lives longer, as illustrated below: w 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 w=h = h 4 h=4 w=4 4 dw dh + hw 4 4 dw dh = h= w= h= 4 h w=h w 4 hw = w=4 4 h dh + 6 3 4 dh = w= 3 w=h 6 dh 6h h3 + dh = 3 4 w=h = h + 3 h3 dh = 4 h + 8 h4 h=4 Cov( X, H ) = E( XH ) E( X)E( H ) = 6 8 3 = 66 3. h= = 4 4 + 8 44 = 6. 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 dt = d + e t dt 3et 3 + e t 8 + e t = 3e t 3 + 8e t 3 8 7 = e t =.

instead of = d M t E X dt = d + e t dt 3et 3 8 + e t 8 + e t = 3e t 3 + 8e t 3 et = =. Posted April 7, 7 In the solution of Problem No. 9 in Practice Examination 3, the expression max( X,) should be min( X,). Posted April, 7 Problem No. 6 in Practice Examination No. should be: 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 be the joint density of them. Furthermore, let f X be the density of X and let F X hence f U,V below, we have: v ( u,v) = e u v e = e u e v for < u <, < v <. Also, noting the graph u = vx or v = u/x u

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 V, Y = V. Then the inverse transformation is U = XY, V = Y. We know that f U,V ( u,v) = e u e v for u >, v >. It follows that = e xy e f X,Y ( x, y) = f U,V ( u( x, y),v( x, y) ) u,v y det x, y for xy > and y >, or just x > and y >. Therefore, y x = ye xy e y

f X ( x) = + ye xy e y dy = = y x + + x+ ye y dy = e x+ y y + y= 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= = ( x +). 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 4) = Pr( X = ) Pr( X = ) Pr( X = ) Pr( X = 3) = = 3.8979. 3 5 4 4 3 5 4 5 5 3 5 4 5 6 3 3 5 4 3 5 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 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: is also defined as the sent that consists of all elementary events that belong to any one of them. should be 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 No. 6, Problem No. had inconsistencies in its assumptions, and it should be 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 Find E( X). =.4. A. 4.5 B. 4.66 C. 4.75 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.5 =.66. By linear interpolation, =.5987 and is standard normal, Φ.6.6.5987 z.6.5 + (.6.5 ).66.5987 =.5 +..533. 3 This gives µ = ln + z.6.946485. The mean of the log-normal distribution is E( X) = e µ+ σ, so that in this case E( X) = e.946485+ 4.48369. Answer A. 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 ) =.3. =.6. Previously it had a typo: Pr( A E 99 ) = Pr( A E 99 ) Pr( E 99 ) =.3. =.36. Posted August 3, 6 In the solution of Problem No. 4 in Practice Examination No., the formula f Y ( y) = k y, should be y f Y = k. Posted July 3, 6 The relationship between the survival function and the hazard rate, stated just after the definition of the hazard rate, should be x s X ( x) = e instead of λ X ( u)du

= e x λ X ( u)du s X x. 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, should be x f X for every x instead of f X ( x) for every x Posted May 6, 6 The solution of Problem No. 3 in Practice Examination No. 3 should be (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) = 3.93. 3 + 3.93. Answer E..8654 +.3433684 =.564337. 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= j=i+ 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 should be: n M X ( t) = t. (its argument was mistakenly written as x instead of t). Posted February, 6 The solution of Problem No. 8 of the Practice Examination No. 4 should be: 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 occurs, is uniform on [,5]. This means that we can calculate all probabilities by 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] in the figure below to the area of the rectangle i.e., Y [,5] [,5], 3 4 = 6. 5

However, this is conditional on the occurrence of the loss of the employee whose loss is Y (but no longer 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% =.4. Therefore the probability sought is.4 6 = 5 6 = 5. Answer B. Posted January 3, 6 In the solution of Problem No. in Practice Examination No., May Course Examination, Problem No. 9, the fourth sentence should begin The mean of the 48 rounded ages, E = 48 48 E i i= Posted January 6, 6 In the solution of Problem No. 5 in Practice Examination No., May Course Examination, Problem No. 7, the survival function should be written as: ( x) = ( + x) 3. s X Posted January 8, 6 In the description of the geometric distribution: M X ( t) = pet. Some texts use a different version of this distribution, which only t qe counts failures until the first success. That s just too pessimistic for this author. The two

distributions differ by one. For this other form of random variable (similar in its design to the negative binomial distribution, just below) Y = X, and E( Y ) = q p, Var( X) = q p, p M X ( t) = qe. t should be M X ( t) = pet. Some texts use a different version of this distribution, which only t qe counts failures until the first success. That s just too pessimistic for this author. The two distributions differ by one. For this other form of random variable (similar in its design to the negative binomial distribution, just below) Y = X, and E( Y ) = q p, Var( Y ) = q p, M Y ( t) = p qe t. Posted September 8, 5 In the solution of Exercise., the first sentence should be: It is important to realize that there is a point-mass at -- this can be seen by analyzing the limit of CDF at from the left (equal to ) and the right-hand side limit of CDF at (equal to.5). Posted August, 5 In the solution of Problem No. in Practice Examination No. 6 these words is the probability that either none of eight elements of the random Similarly, Pr X ( 7) < m sample X, X,, X 8 are greater than or equal to m should be replaced by is the probability that either none or one of eight elements of the Similarly, Pr X ( 7) < m random sample X, X,, X 8 are greater than or equal to m Posted August, 5 In Exercise 3., the second to last formula should be: = f X It was missing y. y 3 dy = 4 y = 4. Posted August, 5

In Exercise 3.5, November Course Examination, Problem No. 4, in the solution, first two displayed formulas should be: E X + Y = E( X) + E( Y ) = 5 + = 7, = Var( X) + Var( Y ) + Cov( X,Y ) = 5 + 3 + =. Var X + Y The problem with existing formulas was that Y was improperly replaced by X. Posted August 8, 5 In Problem No. 7 of Practice Examination No. 4, the five choices of answers should be: A. B. C. D. E. 6 5, 6 5, 6 5, 6 5, 6 5, 3 3 5 5 ( 5 x y) dydx 5 x 5 x y 5 x 5 x y ( 5 x y) dydx ( 5 x y) dydx ( 5 x y) dydx ( 5 x y) dydx Posted August 5, 5 The solution of Problem No. 3, Practice Examination No. 5 should be: Note the region where the density is positive and the region describing the probabilities we are calculating in the figure below y xy = w Therefore, x

F W ( w) = Pr( W w) = Pr ( XY w ) = Pr Y w X = w w y y w = 8xy dxdy + 8xydx dy = 4x x=y ( y x= ) dy + 4x x= w y y x= dy = w w Hence, f w Answer A. w = 4y 3 dy + ( w) = = w 4w dy = y 4 y w y= + 4w ln ln w F W y= y= w y= w + 4w ln y = w 4w ln w. ( w) = w 8wln w 4w w w = = w 8wln w w = 8wln w. = Posted June, 5 In Section, the statement of the Binomial Theorem should be: ( a + b) n n n = k ak b n k, k= i.e., the sum should start from k =, not k =. Posted May, 5 On the second page of Section 4: Risk and Insurance, page 77 of the manual, the formula for the variance should be: Var( X) = E X E( B) ( E X ) = q E( B ) ( q E B ) = q E( B ) q p ( ( E B ) ) + pq ( E( B) ) = q Var( B) + pq ( E( B) ). instead of pq ( E( B) ) ) = q E B ( i.e., + pq E( B) = Posted May 8, 5 In Exercise.3, the five answer choices should be: A. y.8 8 y. e B. 8y. y.8 e C. 8y. e (.y).5 D. (.y).5.5.y e.5

.5 e.y.5 E..5.y (parentheses were missing in answers D and E) Posted May 8, 5 The last line of the solution of Problem No., Practice Examination No. 3, should be: = E max( Y + Z,) X = ( is multiplied by 6 7, not by 5 7 ). 4 7 + 5 7 + 6 7 = 7 9. Posted May 8, 5 The last line of the solution of Problem No. 6, Practice Examination No. 3, should be: E( X ) = d dt M ( t) X = d dt e3t+t = d dt = + 9 =. = e 3t+t + ( 3+ t) e 3t+t (( 3+ t)e ) 3t+t (the superscript was missing in the symbol of the second derivative) = Posted May 5, 5 In Problem No. 3, Practice Examination No. 5, the figure illustrating the solution should be: y y = x x

Posted May 4, 5 In Problem No. 5, Practice Examination No. 5, in the solution, this formula Y X = Y X + Y X + + Y X, should replace Y X = X Y + X Y + + X Y. Posted May 4, 5 In Problem No. 9, Practice Examination No. 5, the solution is correct but the answer choice should be E. Posted May 4, 5 In Problem No. 4, Practice Examination No. 4, the solution is correct but the answer choice should be C. Posted May 4, 5 The density of the log-normal distribution should be: f X ( x) = xσ ( ln x µ ) π e σ.