SOCIETY OF ACTUARIES EXAM STAM SHORT-TERM ACTUARIAL MODELS EXAM STAM SAMPLE SOLUTIONS

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1 SOCIETY OF ACTUARIES EXAM STAM SHORT-TERM ACTUARIAL MODELS EXAM STAM SAMPLE SOLUTIONS Questions -37 have been taken from the previous set of Exam C sample questions. Questions no longer relevant to the syllabus have been deleted. Some of the questions in this study note are taken from past examinations. The weight of topics in these sample questions is not representative of the weight of topics on the exam. The syllabus indicates the exam weights by topic. Copyright 8 by the Society of Actuaries PRINTED IN U.S.A. STAM

2 Question # - DELETED Question # Key: E.645 Var( X ) The standard for full credibility is + where X is the claim size variable.. EX ( ) (.5) For the Pareto variable, EX ( ) =.5 / 5 =. and Var( X ) = (.) =.5. Then the 5(4) standard is + = 6,93 claims... Question #3 - DELETED Question #4 Key: A The distribution function is ( ) x α α x α = α = = F x t dt t x L= f(3) f(6) f(4)[ F(5)] α α α α = α3 α6 α4 (5 ) 3 α α [3(6)(4)(65)]. Taking logs, differentiating, setting equal to zero, and solving: ln L = 3lnα αln57,5 plus a constant dln L/ dα = 3α ln57,5 = ˆ α = 3 / ln57,5 =.57.. The likelihood function is Question #5 π( q,) p( q) p( q) π( q) = q( q) q( q)4 q q ( q) 5 5 q ( q) dq= /68, π ( q,) = 68 q ( q). 3 5 The expected number of claims in a year is EX ( q) = q and so the Bayesian estimate is 5 (,) = (68) ( ) = 4 / 3. E q q q q dq The answer can be obtained without integrals by recognizing that the posterior distribution of q is beta with a = 6 and b = 3. The posterior mean is Eq (,) = a/( a+ b) = 6/9= /3 The posterior mean of q is then 4/3. STAM

3 Question #6 - DELETED Question #7 - DELETED Question #8 Let N be the Poisson claim count variable, let X be the claim size variable, and let S be the aggregate loss variable. µθ ( ) = ES ( θ) = EN ( θ) EX ( θ) = θθ= θ 3 v( θ) = Var( S θ) = E( N θ) E( X θ) = θθ = θ 6 d θ θ θ θ 6 θ θ θ θ µ = E( θ ) = θ (5 θ ) θ = 5 / 3 EPV = E( ) = (5 ) d = 5 VHM = Var( ) = ( ) (5 ) d (5 / 3) =. k = 5 /. =.5. Question #9- DELETED Question # - DELETED Question # Key: D f (5 θ = ) Pr( θ = ) Pr( θ = X = 5) = f(5 θ = ) Pr( θ = ) + f(5 θ = 3) Pr( θ = 3) (/ 36)(/ ) = = 6 / 43 (/ 36)(/ ) + (3 / 64)(/ ) Pr( X > 8 X = 5) = Pr( X > 8 θ = ) Pr( θ = X = 5) + Pr( X > 8 θ = 3) Pr( θ = 3 X = 5) = (/ 9)(6 / 43) + (3 /)(7 / 43) =.6. For the last line, Pr( X > 8 θ) = θ( x + θ) dx = θ(8 + θ) is used. 8 STAM

4 Question # The sample mean for X is 7 and for Y is 67. The mean of all 8 observations is 695. vˆ = [(73 7) + + (7 7) + (655 67) + + (75 67) ] / [(4 )] = 3475 aˆ = [(7 695) + (67 695) ] / ( ) 3475 / 4 = 38.5 kˆ = 3475 / 38.5 = 9.48 Zˆ = 4 / ( ) =.35 P =.35(67) +.695(695) = C Question #3 There are 43 observations. The expected counts are 43(.744) = 7.99, 43(.35) = 5., and 43(.3744) = The test statistic is ( 7.99) (8 5.) ( ) + + = Question #4 From the information, the asymptotic variance of ˆ θ is /4n. Then Var( ˆ θ) = 4 Var( ˆ θ) = 4(/4 n) = / n Question #5 Key: A π( p,,,,,,,) Pr(,,,,,,, p) π( p) p 8 8 p p π ( p,,,,,,,) = = = 9(.5 ) p p dp.5 / p9(.5 ) p dp 9(.5 )(.5 ) / = Pr( X =,,,,,,,) = Pr( X = p) π ( p,,,,,,,) dp = = Question #6 - DELETED Question #7 - DELETED STAM

5 Question #8 Key: D The means are.5(5) +.3(,5) +.(6,) =,875 and.7(5) +.(,5) +.(6,) = 6,675 for risks and respectively. The variances are.5(5) +.3(,5) +.(6,),875 = 556,4,65 and.7(5) +.(,5) +.(6,) 6,675 = 36,738,5 respectively. The overall mean is (/3)(,875) + (/3)(6,675) =,88.33 and so EPV = (/3)(556,4,65) + (/3)(36,738,5) = 476,339,79 and VHM = (/3)(,875) + (/3)(6,675),88.33 = 8,54,. Then, k = 476,339,79/8,54, = and Z = /( ) =.767. The credibility estimate is.767(5) (,88.33) =,6. Question #9 - DELETED Question # - DELETED Question # From the Poisson distribution, µλ ( ) = λand v( λ) = λ. Then, µ = E( λ) = 6 / =.6, EPV = E( λ) =.6, VHM = Var( λ) = 6 / =.6 where the various moments are evaluated from the gamma distribution. Then, k =.6 /.6 = and Z = 45/(45 + ) = 9/ where 45 is the total number of insureds contributing experience. The credibility estimate of the expected number of claims for one insured in month 4 is (9/)(5/45) + (/)(.6) = For 3 insureds the expected number of claims is 3(.56364) = 6.9. Question # The likelihood function is α αθ L( αθ, ) = j α and its logarithm is = ( x + θ ) + j j= l( αθ, ) = ln( α) + αln( θ) ( α+ ) ln( x + θ). When evaluated at the hypothesized values of.5 and 7.8, the loglikelhood is The test statistic is ( ) = 7.7. With two degrees of freedom ( free parameters in the null hypothesis versus in the alternative), the test statistic falls between the 97.5 th percentile (7.38) and the 99 th percentile (9.). j STAM

6 Question #3 Key: E Assume that θ > 5. Then the expected counts for the three intervals are 5( / θ) = 3 / θ,5(3 / θ) = 45 / θ, and 5( θ 5) / θ = 5 75 / θ respectively. The quantity to minimize is [(3 5) (45 5) (5 75 5) θ + θ + θ ]. 5 Differentiating (and ignoring the coefficient of /5) gives the equation 3 (3θ 5)3θ (45θ 5)45θ + ( 75 θ )75θ =. Multiplying through by θ and dividing by reduces the equation to (3 5 θ)3 (45 5 θ)45 + (θ 75)75 = θ = for a solution of ˆ θ = 855 /5 = 7.6. Question #4 Key: E.5.5 πθ ( ) θ(.5 θ ) θ. The required constant is the reciprocal of πθ ( ).5θ.5 =. The requested probability is Pr( θ >.6 ) =.5 θ dθ = θ =.6 = θ dθ =.4 and so Question #5 Key: A k knk / nk Positive slope implies that the negative binomial distribution is a good choice. Alternatively, the sample mean and variance are.6 and.93 respectively. With the variance substantially exceeding the mean, the negative binomial model is again supported. STAM

7 Question #6 /( θ) /( θ) 3/( θ) 5/(3 θ) 8/ θ e e e e e The likelihood function is =. The loglikelihood function is 4 θ θ θ 3θ 4θ ln(4) 4ln( θ) 8 / θ. Differentiating with respect to θ and setting the result equal to yields 4 8 θ + θ = which produces ˆ θ =. Question #7 Key: E The absolute difference of the credibility estimate from its expected value is to be less than or equal to kµ (with probability P). That is, [ ZX + ( ZM ) ] [ Zµ + ( ZM ) ] kµ partial kµ ZX partial Zµ kµ. Adding µ to all three sides produces answer choice (E). Question #8 In general, 5 E( X ) E[( X 5) ] = x f ( x) dx x f ( x) dx 5 f ( x) dx ( x 5 ) f ( x) dx. 5 = 5 Assuming a uniform distribution, the density function over the interval from to is 6/74 (the probability of 6/74 assigned to the interval divided by the width of the interval). The answer is x 6 ( x 5 ) dx = 5 x = STAM

8 Question #9 The probabilities are from a binomial distribution with 6 trials. Three successes were observed Pr(3 I) = (.) (.9) = Pr(3 II) = (.) (.8) = Pr(3 III) = (.4) (.6) = The probability of observing three successes is.7(.458) +.(.89) +.(.7648) = The three posterior probabilities are:.7(.458) Pr( I 3) = = (.89) Pr( II 3) = = (.7648) Pr( III 3) = = The posterior probability of a claim is then.(.887) +.(.38) +.4(.5975) =.833. Question #3 - DELETED Question # 3 - DELETED Question # 3 Key: D N is distributed Poisson( λ ) µ = E( λ) = αθ = (.) =. v = E( λ) =., a = Var( λ) = αθ = (.) = k = =, Z = = / 6 7 Thus, the estimate for Year 3 is (.5) + 5 (.) = Note that a Bayesian approach produces the same answer. Question # 33 - DELETED STAM

9 Question # 34 The likelihood is: n x j n rr ( + ) ( r+ xj ) β xj L = β ( β) r+ xj + x!( + β ) j= j j= The loglikelihood is: r xj. n l = [ x ln β ( r+ x )ln( + β)] j= n xj r+ xj l = = j= β + β n = [ x ( + β) ( r + x ) β] = x rnβ = nx rnβ j j j j= j= ˆ β = x / r. j j n Question # 35 The Bühlmann credibility estimate is Zx + ( Z ) µ where x is the first observation. The Bühlmann estimate is the least squares approximation to the Bayesian estimate. Therefore, Z and µ must be selected to minimize + µ + + µ + + µ [ Z ( Z).5] [ Z ( Z).5] [3 Z ( Z) 3]. Setting partial derivatives equal to zero will give the values. However, it should be clear that µ is the average of the Bayesian estimates, that is, µ = ( ) =. 3 The derivative with respect to Z is (deleting the coefficients of /3): ( Z +.5)( ) + (.5)() + ( Z )() = 4Z 3 =, Z =.75. The answer is.75() +.5() =.5. Question # 36 - DELETED STAM

10 Question # 37 The likelihood is: α α α 3 3α α5 α5 α5 α 5 L = = (5 + 5) (5 + 55) (5 + 95) [(375)(675)()] The loglikelihood is: α+ α+ α+ α+ l = 3lnα + 3αln5 ( α + ) ln[(375)(675)()] l = α + = α ˆ α = 3 / 4.48 = ln5 ln[(375)(675)()] Question # 38 Key: D For this problem, r = 4 and n = 7. Then, vˆ= =.4, aˆ = =.9. 4(7 ) 4 7 Then, k = =, Z = = = (4 / 9) 77 Question # 39 X is the random sum Y+ Y + + Y N. N has a negative binomial distribution with r = a =.5 and β = θ =.. E( N) = rβ =.3, Var( N) = rβ( + β) =.36 E( Y ) = 5,, Var( Y ) = 5,, EX ( ) =.3(5,) =,5 Var( X ) =.3(5,,) +.36(5,,) = 6,5, Number of exposures (insureds) required for full credibility n = (.645 /.5) (6,5, /,5 ) = 7, FULL Number of expected claims required for full credibility ENn ( ) FULL =.3(7,937.67) =,38. STAM

11 Question # 4 Key: E X Fn ( x ) Fn ( x ) Fn ( x) F ( x) Fn ( x ) F ( x) where: ˆ / θ = x and F ( ). x = e x The maximum value from the last two columns is.73. Question # 4 Key: E µ = E λ = v = E σ = a = Var λ = ( ), ( ).5, ( ) / k = v/ a= 5, Z = / ( + 5) = /6. Thus, the estimate for Year is (/6)() + (5/6)() = Question # 4 - DELETED Question # 43 Key: E The posterior density, given an observation of 3 is: θ 3 3 f (3 θπθ ) ( ) (3 + θ) θ (3 + θ) 3 ( 3) = = = = 3(3 + ), >. 3 πθ θ θ f(3 θπθ ) ( ) dθ (3 + θ) dθ (3 + θ ) Then, Pr( ) 3(3 ) 3 6(3 ) Θ> = + θ dθ = + θ = 6 / 5 =.64. STAM

12 Question # 44 L= F() [ F() F()] [ F()] = ( e ) ( e e ) ( e ) / θ 7 / θ / θ 6 / θ 7 = ( p) ( p p ) ( p ) = p ( p) where p e / θ = The maximum occurs at p = /33 and so ˆ ln( / 33) θ = = Question # 45 Key: A EX ( θ) = θ / θ 6 EX ( 4,6) EX ( θ) f( θ 4,6) dθ 3 dθ 3 3 = = θ 3 3 3(6 ) θ 3(6 )(6 ) = = = Question # 46 - DELETED STAM

13 Question # 47 The maximum likelihood estimate for the Poisson distribution is the sample mean: ˆ 5() + () + () + 9(3) λ = x = = The table for the chi-square test is: Number of days Probability Expected* Chi-square.6438 e = e = e = ** *365x(Probability) **obtained by subtracting the other probabilities from The sum of the last column is the test statistic of Using degrees of freedom (4 rows less estimated parameter less ) the model is rejected at the.5% significance level but not at the % significance level. Question # 48 Key: D.4() +.() +.().() +.() +.() µ () = =.5, µ () = =.6.4 µ =.6(.5) +.4() =.7 a =.6(.5 ) +.4( ).7 =.6.4() +.() +.(4).() +.() +.(4) v = = v = =.6.4 v =.6(7 /) +.4(.5) = / k = v/ a = (/ ) /.6 = /. = 55 / 6, Z = = 6 /5 = / / 6 ().5 7 /, ().5 Bühlmann credibility premium = + (.7) = STAM

14 Question # 49 - DELETED Question # 5 The four classes have means.,.,.5, and.9 respectively and variances.9,.6,.5, and.9 respectively. Then, µ =.5( ) =.45 v =.5( ) =.475 a =.5( ).45 = k =.475 / = Z = = The estimate is 5[.743(/4) +.757(.45)] =.4. Question # 5 - DELETED Question # 5 - DELETED Question # 53 First obtain the distribution of aggregate losses: Value Probability /5 5 (3/5)(/3) = /5 (/5)(/3)(/3) = 4/45 5 (3/5)(/3) = /5 5 (/5)()(/3)(/3) = 4/45 4 (/5)(/3)(/3) = /45 µ = (/ 5)() + (/ 5)(5) + (4 / 45)() + ( / 5)(5) + (4 / 45)(5) + (/ 45)(4) = 5 σ = (/ 5)( ) + (/ 5)(5) + (4 / 45)( ) + ( / 5)(5 ) + (4 / 45)(5 ) + (/ 45)(4 ) 5 = 8 Question # 54 - DELETED STAM

15 Question # 55 ( / )( / 3) 3 Pr( class ) = = ( / )( / 3) + ( / 3)( / 6) + ( / 6)() 4 ( / 3)( / 6) Pr( class ) = = ( / )( / 3) + ( / 3)( / 6) + ( / 6)() 4 (/ 6)() Pr( class3 ) = = ( / )( / 3) + ( / 3)( / 6) + ( / 6)() because the prior probabilities for the three classes are /, /3, and /6 respectively. The class means are µ () = ( / 3)() + ( / 3)() + ( / 3)() = µ () = (/ 6)() + ( / 3)() + (/ 6)(3) = The expectation is EX ( ) = (3 / 4)() + (/ 4)() =.5. Question # 56 Key: E The first, second, third, and sixth payments were observed at their actual value and each contributes f(x) to the likelihood function. The fourth and fifth payments were paid at the policy limit and each contributes F(x) to the likelihood function. This is answer (E). Question #57 - DELETED STAM

16 Question #58 Because the Bayes and Bühlmann results must be identical, this problem can be solved either way. For the Bühlmann approach, µλ ( ) = v( λ) = λ. Then, noting that the prior distribution is a gamma distribution with parameters 5 and /5, we have: µ = v= E( λ) = 5(/ 5) =. a = Var λ = = k = v/ a= 5 Z = 5 / (5 + 5) =.75 x = (75 + ) / (6 + 9) =.9 ( ) 5(/ 5). The credibility estimate is.75(.9) +.5(.) =.675. For policies, the expected number of claims is (.675) = For the Bayes approach, the posterior density is proportional to (because in a given year the number of claims has a Poisson distribution with parameter λ times the number of policies) 6λ 75 9λ 5 5λ e (6 λ) e (9 λ) (5 λ) e 5 λ λ e which is a gamma density with 75!! λγ (5) parameters 335 and /. The expected number of claims per policy is 335/ =.675 and the expected number of claims in the next year is uestion #59 Key: E The q-q plot takes the ordered values and plots the jth point at j/(n+) on the horizontal axis and at F( xj; θ ) on the vertical axis. For small values, the model assigns more probability to being below that value than occurred in the sample. This indicates that the model has a heavier left tail than the data. For large values, the model again assigns more probability to being below that value (and so less probability to being above that value). This indicates that the model has a lighter right tail than the data. Of the five answer choices, only E is consistent with these observations. In addition, note that as you go from.4 to.6 on the horizontal axis (thus looking at the middle % of the data), the q-q plot increases from about.3 to.4 indicating that the model puts only about % of the probability in this range, thus confirming answer E. STAM

17 Question #6 The posterior probability of having one of the coins with a 5% probability of heads is proportional to (.5)(.5)(.5)(.5)(4/6) =.467. This is obtained by multiplying the probabilities of making the successive observations,,, and with the 5% coin times the prior probability of 4/6 of selecting this coin. The posterior probability for the 5% coin is proportional to (.5)(.5)(.75)(.5)(/6) =.95 and the posterior probability for the 75% coin is proportional to (.75)(.75)(.5)(.75)(/6) =.758. These three numbers total.6. Dividing by this sum gives the actual posterior probabilities of.6888,.386, and.876. The expected value for the fifth toss is then (.6888)(.5) + (.386)(.5) + (.876)(.75) = Question #6 Key: A Because the exponential distribution is memoryless, the excess over the deductible is also exponential with the same parameter. So subtracting from each observation yields data from an exponential distribution and noting that the maximum likelihood estimate is the sample mean gives the answer of 73. Working from first principles, f( x) f( x) f( x3) f( x4) f( x5) θ e θ e θ e θ e θ e L( θ ) = = 5 / θ 5 [ F()] ( e ) 5 365/ θ = θ e. Taking logarithms and then a derivative gives l( θ) = 5ln( θ) 365 / θ, l ( θ) = 5 / θ / θ = ˆ θ = 365 / 5 = 73. 5/ θ 5/ θ 65/ θ 75/ θ 5/ θ STAM

18 Question #6 Key: D The number of claims for each insured has a binomial distribution with n = and q unknown. We have µ ( q) = qv, ( θ) = q( q) µ = Eq ( ) =. (see iv) a= Varq = Eq = Eq = v= Eq q = Eq Eq = = k = v/ a= 8, Z = / ( + 8) = 5 / 9 ( ) ( ).. (see v) ( ). [ ( )] ( ) ( )...8 Then the expected number of claims in the next one year is (5/9)() + (4/9)(.) = /45 and the expected number of claims in the next five years is 5(/45) = /9 =.. Question #63 - DELETED Question #64 Key: E The model distribution is f( x θ) = / θ,< x< θ. Then the posterior distribution is proportional to 5 4 πθ ( 4,6) θ, θ> 6. θθ θ It is important to note the range. Being a product, the posterior density function is non-zero only when all three terms are non-zero. Because one of the observations was equal to 6, the value of the parameter must be greater than 6 for the density function at 6 to be positive. Or, by general reasoning, posterior probability can only be assigned to possible parameter values. Having observed the value 6 we know that parameter values less than or equal to 6 are not possible. 4 The constant is obtained from θ dθ = and thus the exact posterior density is 6 3 3(6) 3 4 πθ ( 4, 6) = 3(6) θ, θ> 6. The posterior probability of an observation exceeding 55 is θ Pr( X3 > 55 4, 6) = Pr( X 6 3 > 55 θπθ ) ( 4, 6) dθ= 3(6) θ dθ= θ where the first term in the integrand is the probability of exceeding 55 from the uniform distribution. STAM

19 Question #65 EN ( ) = rβ =.4 Var( N) = rβ( + β) =.48 EY ( ) = θ / ( α ) = 5 θα Var( Y ) = = 75, ( α ) ( α ) Therefore, EX ( ) =.4(5) = Var X ( ) =.4(75, ) +.48(5) = 4,. The full credibility standard is n.645 4, = =,365, Z =,5 /,365 = Question #66 - DELETED Question #67 Key: E µ ( r) = E( X r) = E( N) E( Y) = rβθ / ( α ) = r v( r) = Var( X r) = Var( N) E( Y ) + E( N) Var( Y ) rβ ( + β ) θ rβαθ = + =, r ( α ) ( α ) ( α ) v= E(, r) =, () = 4, a = Var( r) = () (4) = 4, k = v/ a=.5, Z = / ( +.5) =.95 Question #68 - DELETED STAM

20 Question #69 For an exponential distribution the maximum likelihood estimate of the mean is the sample mean. We have E( X ) = E( X ) = θ, Var( X ) = Var( X )/ n = θ / n cv = SD( X ) / E( X ) = ( θ / n) / θ = / n = 5 =.447. If the maximum likelihood estimator is not known, it can be derived: n Σx/ θ L( θ) = θ e, l( θ) = n ln θ nx / θ, l ( θ) = nθ + nxθ = ˆ θ = X. Question #7 Key: D Because the total expected claims for business use is.8, it must be that % of business users are rural and 8% are urban. Thus the unconditional probabilities of being businessrural and business-urban are. and.4 respectively. Similarly the probabilities of being pleasure-rural and pleasure-urban are also. and.4 respectively. Then, µ =.(.) +.4(.) +.(.5) +.4(.5) =.5 v =.(.5) +.4(.) +.(.8) +.4(.) =.93 a = = k = v/ a= 4.8, Z = / ( + 4.8) =.93..(.).4(.).(.5).4(.5).5.5 Question #7 Key: A No. claims Hypothesized Observed Chi-square (5)/5 = (5)/35 = ()/4 =.4 4 ()/ = (7)/4 =.3 6+ ()/ = 4.4 The last column sums to the test statistic of 7.6 with 5 degrees of freedom (there were no estimated parameters), so from the table reject at the.5 significance level. STAM

21 Question #7 In part (ii) you are given that µ =. In part (iii) you are given that a = 4. In part (iv) you are given that v = 8,. Therefore, k = v/a =. Then, 8(5) + 6() + 4(5) X = = Z = = P =.9( / 9) +.() =. C Question #73 - DELETED Question #74 - DELETED Question #75 - DELETED Question #76 Key: D The posterior density is proportional to the product of the probability of the observed value and the prior density. Thus, ( N ) Pr( N ) ( ) ( e θ θ πθ > > θπθ = ) θe. θ θ The constant of proportionality is obtained from θe θe dθ = =.75. The posterior density is ( N ) (/.75)( e θ θ πθ θ > = θe ). Then, θ θ θ Pr( > > ) = Pr( > ) ( > ) = ( )(4 / 3)( ) N N N θπθ N dθ e θe θe dθ 4 4 = + = + + = θ θ 3θ θe θe θe dθ Question #77 - DELETED Question #78 From item (ii), µ = and a = 5. From item (i), v = 5. Therefore, k = v/a = and Z = 3/(3+) = 3/3. Also, X = ( ) / 3 = 75. Then P = (3/3)(75) + ( /3)() = c STAM

22 Question #79 x/ x/, f( x) = p e + ( p) e, L(, ) = f() f() pe ( p) e pe ( p) e = + +,,.. Question #8 - DELETED Question # 8 - DELETED Question #8 - DELETED Question #83 - DELETED Question #84 Key: A c(4 x) x< 4 B = x 4 = E( B) = c4 ce( X 4) 3 = c4 c = c c = = STAM

23 Question #85 Let N = number of computers in department Let X = cost of a maintenance call Let S = aggregate cost Var( X) = [Standard Deviation( X)] = = 4, EX ( ) = Var( X) + [ EX ( )] = 4, + 8 = 46, 4 ES ( ) = NλEX ( ) = N(3)(8) = 4N Var( S) = Nλ EX ( ) = N(3)(46, 4) = 39, N We want. Pr S>. ES ( ) ( ) S ES ( ). ES ( ).(4) N Pr >.8 =Φ(.9) 39, N 39, N 373. N.8(373.) N = Question #86 Key: D The modified severity, X*, represents the conditional payment amount given that a payment occurs. Given that a payment is required (X > d), the payment must be uniformly distributed between and c(b d). The modified frequency, N*, represents the number of losses that result in a payment. The b d deductible eliminates payments for losses below d, so only FX ( d) = of losses will b require payments. Therefore, the Poisson parameter for the modified frequency distribution is b d λ. (Reimbursing c% after the deductible affects only the payment amount and not the b frequency of payments). STAM

24 Question #87 Key: E f( x) =., x 8 =..5( x 8) =.3.5 x, 8 < x x 8 x x E( x) =. x dx + (.3x.5 x ) dx..3.5 = + = = E( X ) + = E( X ) x f ( x) dx f ( x) dx.x ( = x ) = (.8) = Loss Elimination Ratio = = STAM

25 Question #88 First restate the table to be CAC s cost, after the % payment by the auto owner: Towing Cost, x p(x) 7 5% 9 4% 44 % Then EX ( ) =.5(7) +.4(9) +.(44) = EX ( ) =.5(7 ) +.4(9 ) +.(44 ) = Var( X ) = = Because Poisson, E( N) = Var( N) = ES ( ) = EXEN ( ) ( ) = 86.4() = 86,4 Var( S) = EN ( )Var( X) + EX ( ) Var( N) = (44.64) () = 7,95, 6 S ES ( ) 9, 86,4 Pr( S > 9,) + Pr > = Pr( Z >.8) = Φ(.8) =. Var( S) 7,95,6 Since the frequency is Poisson, you could also have used Var( S) = λex ( ) = (795.6) = 7,95,6. That way, you would not need to have calculated Var(X). Question #89 d / θ EX ( d) θ ( e ) LER = = = e EX ( ) θ d / θ d / θ Last year.7 = e d = θ log(.3) Next year: d = θ log( LER ) new new log( LER ) = d 4 = θ log(.3) 3 Hence θ new new ( new ) =.653 ( ) e log LER.653 LER = =. new LER =.8 new STAM

26 Question # 9 Key: E The distribution of claims (a gamma mixture of Poissons) is negative binomial. EN ( ) = EΛ[ EN ( Λ )] = EΛ( Λ ) = 3 Var( N) = EΛ[ Var( N Λ )] + VarΛ[ E( N Λ )] = EΛ( Λ ) + VarΛ( Λ ) = 6 rβ = 3 rβ( + β) = 6 ( + β) = 6 / 3 = ; β = rβ = 3; r = 3 p r = ( + β ) =.5 rβ p = =.875 r+ ( + β ) Pr(at most ) = p + p =.35. Question # 9 Key: A ES ( ) = ENEX ( ) ( ) = (,) =, Var S = E N Var X + E X Var N = + = StdDev( S) = 3,6 ( ) ( ) ( ) ( ) ( ) (7 ) (75) 99,689,75,, Pr( S <, ) = Pr Z < =.7 =.4 3,6 where Z has standard normal distribution. STAM

27 Question # 9 Let N = number of prescriptions then n fn ( n ) FN ( n ) F ( n) N EN ( ) = 4 = [ F( j)] j= E[( S 8) ] = 4 E[( N ) ] = 4 [ F( j)] + + j= = 4 [ F( j)] [ F( j)] j= j= = 4(4.44) =.4 E[( S ) + ] = 4 E[( N 3) + ] = 4 [ F( j)] [ F( j)] = 4(4.95) = 8.9 j= j= Because no values of S between 8 and are possible, ( ) E[( S 8) + ] + ( 8) E[( S ) + ] E[( S ) + ] = = Alternatively, E[( S ) ] = (4 j ) f ( j) + f () + 6 f () + f () + j= N N N N (The correction terms are needed because (4j ) would be negative for j =,, ; we need to add back the amount those terms would be negative) = 4 jf ( j) f ( j) + (.) + 6(.6) + (.8) N j= j= N = 4 EN ( ) = = 9.6 STAM

28 Question #93 Key: E Method : In each round, N = result of first roll, to see how many dice you will roll X = result of for one of the N dice you roll S = sum of X for the N dice EX ( ) = EN ( ) = 3.5 Var( X ) = Var( N) =.967 ES ( ) = ENEX ( ) ( ) =.5 Var S = E N Var X + Var N E X = + = ( ) ( ) ( ) ( ) ( ) 3.5(.967).967(3.5) Let S the sum of the winnings after rounds ES ( ) = (.5) =, 5 SD( S ) = (45.938) = 4.33 After rounds, you have your initial 5,, less payments of,5, plus winnings for a total of,5 + S Since actual possible outcomes are discrete, the solution tests for continuous outcomes greater than In this problem, that continuity correction has negligible impact., 499.5, 5 Pr(,5 + S > 4,999.5) = Pr( S >, 499.5) Pr Z > =.7 = Question #94 b pk = a+ pk k.5 = ( a+ b).5 a = b b.875 = a+ (.5).875 = (.5 b)(.5) b=.5, a=.5.5 p3 =.5 + (.875) =.5 3 STAM

29 Question #95 Key: E k k β = mean = 4, = β / ( + β ) + p k n P(N = n) x f () ( x ) f () ( x ) f (3) ( x ) fs() =., fs() =.6(.5) =.4 fs () =.6(.5) +.8(.65) =.49 fs (3) =.6(.5) +.8(.5) +.4(.56) =.576 F (3) = =.346 S Question #96 Key: E Let L = incurred losses; P = earned premium = 8, L Bonus.5.6 P =.5(.6 P L ) =.5(48, L ) if positive. P This can be written.5[48, ( L 48,)]. Then, E(Bonus) =.5[48, EL ( 48, )] From Appendix A..3. E(Bonus)=.5{48, [5, ( (5, / (48,+5,))]} = 35,65 Question # 97 Key: D Severity after increase Severity after increase and deductible Expected payment per loss =.5() +.5() +.5(8) +.5() = 75 Expected payments = Expected number of losses x Expected payment per loss = 3(75) =,5 STAM

30 Question # 98 Key: A E(S) = E(N)E(X) = 5() =, Var( S) = E( N) Var( X ) + E( X ) Var( N) = 5(4) + () = 4,, 8,, Pr( S < 8, ) Pr Z < =.998 =.6 4,, Question #99 Let S denote aggregate loss before deductible. E(S) = () = 4, since mean severity is. e fs () = =.353, since must have losses to get aggregate loss =.! e fs () = =.9, since must have loss whose size is to get aggregate loss =.! 3 ES ( ) = fs() + fs() + [ fs() fs()] = (.353) + (.9) + (.353.9) =.639 E[( S ) ] = ES ( ) ES ( ) = = Question # Limited expected value =.x.x.x.x [ F( x)] dx =.8e +.e dx = 4e e = = Question # Mean excess loss = EX ( ) EX ( ) 33 9 = = 3 F().8 Note that EX ( ) = EX ( ) because F() =. STAM

31 Question # Key: E Expected insurance benefits per factory = E[( X ) + ] =.() +.() =.4 Insurance premium = (.)( factories (.4 per factory) =.88. Let R = retained major repair costs, then R() =.4 =.6, R() = (.4)(.6) =.48, S() =.6 =.36 f f f Dividend = 3.88 R.5(3) =.67 R, if positive. Expected Dividend =.6(.67 ) + (.48)(.67 ) +.36() =.5888 Question #3 - DELETED Question: #4 - DELETED Question: #5 Key: A Using the conditional mean and variance formulas and that N has a conditional Poisson distribution:. = EN ( ) = EEN [ ( Λ )] = E( Λ).4 = Var( N) = E[ Var( N Λ )] + Var[ E( N Λ )] = E( Λ ) + Var( Λ ) =. + Var( Λ) Var( Λ ) =.4. =. STAM

32 Question: #6 N = number of salmon in t hours X = eggs from one salmon S = total eggs. E(N) = t Var(N) = 9t ES ( ) = EN ( ) E(X) = ( t)(5) = 5 t = + Var N = t + t = t Var(S) E(N) Var(X) E(X) ( ) ( )(5) (5 )(9 ) 3,, 5t, 5t.95 < Pr( S >, ) = Pr Z >.645 3, t = 3, t, 5t =.645(5.66) t = t 5t t, = ± ( 49.48) 4(5)(, ) t = = 4.73 (5) t =.6 Round up to 3 Question: #7 X = losses on one life E(X) =.3() +.() +.(3) = S = total losses E(S) = 3E(X) = 3() = 3 3 E[( S ) ] = ES ( ) [ F()] = 3 [ f()] = 3 (.4 ) =.64 + STAM S S Alternatively, the expected retained payment is f () + [ f ()] =.936 and the stop-loss premium is =.64. Question: #8 pk ( ) = pk ( ) = + pk ( ) k k Thus an (a, b, ) distribution with a =, b =. Thus Poisson with λ =. 4 e p(4) = =.9 4! S S

33 Question: #9 By the memoryless property, the distribution of amounts paid in excess of is still exponential with mean. With the deductible, the probability that the amount paid is is F = =. / () e.393 Thus the average amount paid per loss is (.393)() + (.67)() =.4 The expected number of losses is ()(.8) = 6. The expected amount paid is (6)(.4) = 94. Question: # Key: E EN ( ) =.8() +.() =. EN ( ) =.8() +.(4) =.6 Var( N) =.6. =.6 EX ( ) =.() +.7() +.() = 7 EX ( ) =.() +.7(, ) +.(,, ) = 7, Var( X ) = 7, 7 = 78, ES ( ) = ENEX ( ) ( ) =.(7) = 4 Var( S) = E( N) Var( X ) + E( X ) Var( N) =.(78,) + 7 (.6) = 98,344 SD( S) = 98,344 = 33.6 So Budget = = 58 STAM

34 Question: # Key: E For a compound Poisson distribution, Var( S) = E( N) E( X ). EX ( ) = (/ )() + (/ 3)(4) + (/ 6)(9) = / 3 Var( S) = ( / 3) = 4 Alternatively, the total is the sum N+ N + 3N3 where the Ns are independent Poisson variables with means 6, 4, and. The variance is 6 + 4(4) + 9() = 4. Question: # Key: A N = number of physicians E(N) = 3 Var (N) = X = visits per physician E(X) = 3 Var (X) = 3 S = total visits ES ( ) = ENEX ( ) ( ) = 3(3) = 9 Var S = E N Var X + E X Var N = + = ( ) ( ) ( ) ( ) ( ) 3(3) 9() 89 SD( S) = 89 = Pr( S > 9.5) = Pr Z > =.68 = Φ(.68) 43.5 Question: #3 Key: E EN ( ) =.7() +.() +.(3) =.7 Var( N) =.7() +.(4) +.(9).7 =. EX ( ) =.8() +.() = Var( X ) =.8() +.() = 6 ES ( ) = ENEX ( ) ( ) =.7() =.4 Var S = E N Var X + E X Var N = + = ( ) ( ) ( ) ( ) ( ).7(6) 4(.) 6.4 SD( S) = 6.4 = 4 3 Pr( S >.4 + (4) = 9.4) = Pr( S = ) =.7.(.8).(.8) =. The last line follows because there are no possible values for S between and. A value of can be obtained three ways: no claims, two claims both for, three claims all for. STAM

35 Question: #4 Key: A 5 5 λ λ 5 5 λ 5 5 P() = e d = ( e ) = ( e ) = λ λ λ 5 λ λ λ P() = e d = ( e e ) = ( 6 e ) =.99 PN ( ) = =.694 Question: #5 Key: D Let X be the occurrence amount and Y = max(x, ) be the amount paid. / E( X ) =, Var( X ) =, Pr( X > ) = e = The distribution of Y given that X >, is also exponential with mean, (memoryless property). So Y is with probability =.9563 and is exponential with mean with probability EY ( ) =.9563() () = EY ( ) =.9563() ()( ) =,89, 675 Var( Y ) =,89, = 99,944 Question: #6 Expected claims under current distribution = 5/( ) = 5 K= parameter of new distribution X = claims EX ( ) = K/ ( ) = K K E(claims + bonus) = E{ X +.5[5 ( X 5)]} = K K 5 5 K = + ( K K)(5 + K) +.5K = 5(5 + K).5K + 5K + 5, +.5K = 5, + 5K K = 5, K = 5, = 354 STAM

36 Question: #7 - DELETED Question: #8 Key: D EN ( ) = 5 Var( N) = 5 EX ( ) = (5 + 95) / = 5 Var( X ) = (95 5) / = 675 ES ( ) = ENEX ( ) ( ) = 5(5) = 5 Var S = E N Var X + E X Var N = + = ( ) ( ) ( ) ( ) ( ) 5(675) 5 (5) 79,375 SD( S) = 79,375 = PS ( > ) = P Z> =.66 = Φ(.66) 8.74 Question: #9 - DELETED Question: # Key: E For : EX ( ) = / ( ) = EX ( 3) = = 3 + So the fraction of the losses expected to be covered by the reinsurance is ( )/ =.4. The expected ceded losses are 4,, and the ceded premium is 4,4,. For : Inflation changes the scale parameter, here to (.5) =. The revised calculations are EX ( ) = / ( ) = EX ( 3) = = The revised premium is.(,5,)( 35)/ = 4,757,5. The ratio is 4,757,5/4,4, =.8. Question # - DELETED STAM

37 Question # - DELETED Question #3 α θ θ x E( X x) = = = α x θ x + + x+ x EX ( x) [ ].75[ EX ( 5) EX ( 5)] +.95 EX ( ) EX ( 5).75(59 ) +.95( 437) = 6.6 The 5 breakpoint was determined by when the insured s share reaches 36: 36 = (5 5) + (5 5) Question #4 - DELETED Question #5 Key: A NI, N II denote the random variables for # of claims for Types I and II in one year XI, X II denote the claim amount random variables for Types I and II SI, S II denote the total claim amount random variables for Types I and II S = SI + SII E( NI ) = Var( NI ) =, E( NII ) = Var( NII ) = 4 E( X I) = ( + ) / = /, Var( X I) = ( ) / = / E( X II ) = ( + 5) / = 5 /, Var( X II ) = (5 ) / = 5 / ES ( ) = EN ( I ) EX ( I ) + EN ( II ) EX ( II ) = (/ ) + 4(5 / ) = 6 Var( S) = E( N ) Var( X ) + E( X ) Var( N ) + E( N ) Var( X ) + E( X ) Var( N ) I I I I II II II II = (/) + (/ ) () + 4(5 /) + (5 / ) (4) = Pr( S > 8) = Pr Z > =.37 = Φ (.37) = STAM

38 Question #6 Let X be the loss random variable. Then, ( X 5) + is the claim random variable. EX ( ) = = EX ( 5) = = E[( X 5) ] = EX ( ) EX ( 5) = = Expected aggregate claims = ENE ( ) [( X 5) + ] = 5(3.69) = 8.5. Question #7 A Pareto ( α =, θ = 5) distribution with % inflation becomes Pareto with α =, θ = 5(.) = 6. In 4 6 EX ( ) = = EX ( ) = = E[( X ) + ] + = EX ( ) EX ( ) = E[( X ) + ].5 LER = = =.65 EX ( ) 6 Question #8 - DELETED Question #9 - DELETED STAM

39 Question #3 Key: E Begin with ( ) ( N N EW = E ) = EE [ ( Λ )]. The inner expectation is the probability generating ( ) function of Poisson distribution evaluated at, () Λ Λ PN = e = e. Then, Ee ( Λ ) λ (.5).5 λ = e dλ = e =.5( e ) = 3.4. If the pgf is not recognized, the inner expectation can be derived as Λ n Λ n Λ n N n e Λ e ( Λ) Λ e ( Λ) Λ E( Λ ) = = = e = e n= n! n= n! n= n! probabilities for a Poisson distribution with mean Λ and so must be. noting that the sum is over the Question #3 - DELETED Question #3 - DELETED Question #33 E( X q) = 3 q, Var( X q) = 3 q( q) µ µ = E(3 q) = 3qqdq = q = v = E[3 q( q)] = 3 q( q)qdq = q.5q =.5 a = Var(3 q) = E(9 q ) = 9q qdq = 4.5q 4 = =.5 k = v/ a=.5 /.5 =, Z = =.5 + The estimate is.5() +.5() =. Question #34 - DELETED Question #35 - DELETED STAM

40 Question #36 Pr(claim = 5 class ) Pr(class ) Pr(class claim = 5) = Pr(claim = 5 class ) Pr(class ) + Pr(claim = 5 class ) Pr(class ).5( / 3) = =.5( / 3) +.7(/ 3) 7 E(claim class ) =.5(5) +.3(,5) +.(6, ) =,875 E(claim class ) =.7(5) +.(,5) +.(6, ) = 6, 675 E(claim 5) = ( /7)(,875) + (7 /7)(6, 675) =,3 Question #37 Key: D L( p) = f(.74) f(.8) f(.95) = ( p+ ).74 ( p+ ).8 ( p+ ).95 = p + 3 p ( ) (.56943) l( p) = ln L( p) = 3ln( p+ ) + pln(.56943) 3 l'( p) =.5639 = p + 3 p+ = = , p = p p p Question #38 - DELETED STAM

41 Question #39 Let X be the number of claims. EX ( I) =.9() +.() =. E( X II) =.8() +.() +.() =.3 E( X III) =.7() +.() +.() =.4 Var( X I) =.9() +.(4). =.36 Var( X II) =.8() +.() +.(4).3 =.4 Var( X III) =.7() +.() +.(4).4 =. 44. µ = (/ )( ) =.3 v = (/ 3)( ) = a = (/ 3)( ).3 =.6667 k =.43333/.6667 = Z = = For one insured the estimate is.4549(7/5) (.3) =.38. For 35 insureds the estimate is 35(.38) =.3. Question #4 Key: A For the given intervals, based on the model probabilities, the expected counts are 4.8, 3.3, 8.4, 7.8,.7,.5, and.5. To get the totals above 5, group the first two intervals and the last three. The table is Interval Observed Expected Chi-square infinity Total Question #4 - DELETED STAM

42 Question #4.575 = Pr( N = ) = Pr( N = θπθ ) ( ) dθ e k θ θ k k k θ e e e = e dθ = = + e ( e ) ( e ) ( e ) k e + e = = k ( e ) = (.575) =.5 k =.9. k k k k k k Question #43 - DELETED Question #44 - DELETED Question #45 The subscripts denote the three companies. 5, 5, 5, xi = = 5, xi = = 5, xii = = 3 5 5, 5, 5, xii = = 5, xiii = = 3,, xiii = =, 3 5 5, 3, 3, 7, xi = = , xii = = 375, xiii = =,5, x = = ,3 ( ) + ( ) + 5(3 375) + 3(5 375) + 5(3,,5) + 5(,,5) vˆ = = 53,888, ( ) + ( ) + ( ) 3( ) + 8( ) + (, ) 53,888, (3 ) aˆ = = 57, ,3,3 53,888, k = = , Z = = , STAM

43 Question #46 Key: D n m Let α α α j be the parameter for region j. The likelihood function is L = α + α+ i= xi i= yi α α The expected values satisfy =.5 α α and so 3α α =. Substituting this in the + α likelihood function and taking logs produces n m 3α + 4α l( α) = ln L( α) = nln α ( α+ ) ln xi + mln ln yi i= + α + α i= n m n m 6 l'( α) = ln xi + ln y. i = α α ( + α ) ( + α ) i= i= Question #47 - DELETED Question #48 STAM The mean is mq and the variance is mq( q). The mean is 34,574 and so the full credibility standard requires the confidence interval to be ± which must be.96 standard deviations. Thus, =.96 mq( q) =.96 34,574 q q=.9, q =.. Question #49 - DELETED Question #5 - DELETED Question #5 EN ( ) = 5, EN ( ) = 8(.55) = 4.4, µ =.5(5) +.5(4.4) = 4.7 Var( N ) = 5, Var( N ) = 8(.55)(.45) =.98, v =.5(5) +.5(.98) = 3.49 a =.5(5) +.5(4.4) 4.7 =.9, k = 3.49 /.9 = r Z = =.78, 4.69 = (4.7) Var( N ) = 5, Var( N ) = 8(.55)(.45) =.98, v =.5(5) +.5(.98) = 3.49 a = + = = =.5(5).5(4.4) 4.7.9, k 3.49 / r Z = =.78, 4.69 = (4.7) The solution is r = 3. Question #5

44 Key: A These observations are truncated at 5. The contribution to the likelihood function is x / θ f( x) θ e =. Then the likelihood function is 5/ θ F(5) e 6/ θ 7/ θ 9/ θ θ e θ e θ e 3 7/ θ L( θ) = = θ e 5/ θ 3 e ( ) l( θ) = ln L( θ) = 3lnθ 7θ l '( θ) = 3θ + 7θ = θ = 7 / 3 = Question #53 - DELETED Question #54 Key: E.5 For a compound Poisson distribution, S, the mean is ES ( λµσ,, ) = λex ( ) = λ e µ + σ and the variance is Var( S,, ) E( X ) e µ + λµσ = λ = λ σ. Then, µ +.5σ µ +.5σ.5σ ( ).5 ES ( ) = EES [ ( λµσ,, )] = λe σdλ dµ dσ = e σ dµ dσ = e e σdσ = ( e )( e ) =.4686 µ + σ [ (,, )] µ + σ σ.5( ) v = E Var S λµσ = λe σdλ dµ dσ = e σ dµ dσ = e e σdσ =.5( e ).5( e ) =.5( e ) = µ σ a= VarES [ ( λµσ,, )] = λe σdλ dµ dσ ES ( ) = = 3 3 = e e ES = e e ES = k = = µ + σ σ e σ dµ dσ ES ( ) ( e ) e σdσ ES ( ) ( ) ( ) ( ) ( )( ) / 6 ( ) Question #55 - DELETED STAM

45 Question #56 There are n/ observations of N = (given N = or ) and n/ observations of N = (given N = or ). The likelihood function is λ n/ λ n/ n/ nλ n/ e λe λ e λ L = λ λ λ λ = = λ λ n n. Taking logs, differentiating e + λe e + λe ( e + λe ) ( + λ) and solving provides the answer. l = ln L= ( n/ ) ln λ nln( + λ) n n l ' = = λ + λ n( + λ) nλ = λ =, λ =. Question #57 Key: D The posterior density function is proportional to the product of the likelihood function and prior density. That is, π( q, ) f( q) f( q) π( q) q( qq ) q q. posterior density, integrate this function over its range: q q q q q q dq = =.469 and so π ( q, ) =. Then, q q Pr(.7 < q <.8,) = dq = Question #58 - DELETED Question #59 Key: A STAM = To get the exact The sample mean is () + (6) + (3) + 3(8) + 4() = = ˆ µ = vˆ and the 3 sample variance is ( ˆ µ ) + 6( ˆ µ ) + 3( ˆ µ ) + 8(3 ˆ µ ) + (4 ˆ µ ) = Then, aˆ = =.83589, k = =.7684 and Z = = Question #6

46 Key: E 5 4 The cdf is x 4( ) ( ) x F x = + t dt = + t = + ) 4. x Observation (x) F(x) compare to: Maximum difference..37, , , , ,..64 K-S statistic is.4. Question #6 - DELETED Question #6 EX [ d X> d] is the expected payment per payment with an ordinary deductible of d It can be evaluated (for Pareto) as α θ θ θ α θ θ EX ( ) EX ( d) α α d + θ α d + θ d+ θ = = = = d + θ α Fd ( ) α θ θ α d + θ d + θ EX [ X> ] = (5 / 3) EX [ 5 X > 5] + θ = (5 / 3) ( 5 + θ) 3 + 3θ = θ, 5 = θθ, = 5 EX [ 5 X> 5] = 5 + θ = = 75 STAM

47 Question #63 Key: D Let S = score ES ( ) = EES [ ( θ)] = E( θ) = 75 Var( S) = E[ Var( S θ)] + Var[ E( S θ)] = E(8 ) + Var( θ) = = S is normally distributed (a normal mixture of normal distributions with constant variance is normal). F(9) F(65) Φ[(9 75) /)] Φ[(65 75) /] Pr(S < 9 S > 65) = = F(65) Φ[(65 75) /] Φ(.5) Φ(.).933 (.843) = =.96 Φ(.) (.843) Question #64 Losses in excess of the deductible occur at a Poisson rate of λ = [ F(3)] λ =.75() = 5 The expected payment squared per payment is E[( X 3) X> 3) = EX ( 6X + 9 X> 3) EX [ 6( X 3) 9 X> 3] EX ( ) EX ( 3) EX [ X> 3] 6 9 F(3) 7 5 = = The variance of S is the expected number of payments times the second moment, 5(45) = 67,5. Question #65 Key: A E[( S 3) + ] = ES ( ) fs() + fs() + fs() ES ( ) = [.6 + (.4)] =.8 f () = e, f () = e ()(.6) =.e S S e fs () = e ()(.4) + (.6) =.5e! E S e e e e [( 3) + ] = (. ) + (.5 ) = =.7365 * STAM

48 Question #66 pk c Write (i) as = c + pk k This is an (a, b, ) distribution with a = b = c. Which?. If Poisson, a =, so b = and thus p =.5 and p = p = =. The probabilities do not sum to and so not Poisson.. If Geometric, b =, so a =. By same reasoning as #, not Geometric. 3. If binomial, a and b have opposite signs. But here a = b, so not binomial. 4. Thus negative binomial. a β / ( + β) = = = so r =. b r β / β r Question #67 ( ) ( ) r ( ) ( ) p =.5 = + β = + β β = =.44 c= a = β / ( + β) =.9 The number of repairs for each boat type has a binomial distribution. For power boats: ES ( ) = (.3)(3) = 9,, Var( S) = (.3), ) + (.3)(.7)(3 ) =,9, For sail boats: ES ( ) = 3(.)(, ) = 3,, Var( S) = 3(.)(4, ) + 3(.)(.9)(, ) = 39,, For luxury yachts: ES ( ) = 5(.6)(5, ) = 5,, Var( S) = 5(.6)(.4)(,, ) + 5(.6)(.4)(5, ) = 36,, The sums are 89, expected and a variance of 4,9, for a standard deviation of,3. The mean plus standard deviation is 9,3. STAM

49 Question #68 SX (5) =. =.8 fx ( y+ 5)..6 f P( y) =, f P(5) = =.5, f P(5) = =.75 Y SX (5) Y.8 Y.8 P EY ( ) =.5(5) +.75(5) = 5 P Y = + = E ( ).5(5 ).75(5 ) 7,5 P Var( Y ) = 7,5 5 =,875 Question #69 Key: A 4 3 F() =.8 +. = Question #7 Let C denote child; ANS denote Adult Non-Smoker; AS denote Adult Smoker e P(3 CPC ) ( ) = (.3) =.67 3! 3 e P(3 ANS) P( ANS) = (.6) =.37 3! e P(3 AS) P( AS) = (.) =. 3!. P( AS N = 3) = = Question #7 ES ( ) = ENEX ( ) ( ) = 3() = 3 4 Var( S) = E( N) Var( X ) + E( X ) Var( N) = 3 + (3.6) = + 36 = 46 For 95 th percentile, E( S) Var( S) = = 65.8 STAM

50 Question #7 Key: D The CDF is F( x) = 4 ( + x) Observation (x) F(x) compare to: Maximum difference..58, , , , ,..64 The K-S statistic is the maximum from the last column,.68. Critical values are:.546,.68,.66, and.79 for the given levels of significance. The test statistic is between.66 (.5%) and.79 (.%) and therefore the test is rejected at.5 and not at.. Question #73 Key: E For claim severity, µ = (.4) + (.4) + (.) = 4.4, S S σ = (.4) + (.4) + (.) 4.4 =, For claim frequency, µ F = rβ = 3 r, σf = rβ( + β) = r, For aggregate losses, µ = µµ = 4.4(3 r) = 73. r, S F S F S F σ = µσ + σ µ = 4.4 ( r) +, 445.4(3 r) =, r. For the given probability and tolerance, λ = (.96 /.) = The number of observations needed is λσ / µ = 384.6(, r) / (73. r) = 83. / r. The average observation produces 3r claims and so the required number of claims is (83. / r)(3 r ) =, 469. Question #74 - DELETED Question #75 - DELETED STAM

51 Question #76 Key: A Pick one of the points, say the fifth one. The vertical coordinate is F(3) from the model and should be slightly less than.6. Inserting 3 into the five answers produces.573,.96,.93,.95, and something less than.5. Only the model in answer A is close. Question #77 Key: E The distribution of Θ is Pareto with parameters and.6. Then, v = EPV = E( Θ ) = =.65, a = VHM = Var( Θ ) =.65 =.697,.6.6(.6) 5 k = v/ a =.65 /.697 =.369, Z = = Question #78 - DELETED Question #79 Key: D For an exponential distribution, the maximum likelihood estimate of θ is the sample mean, 6. Let Y = X+ X where each X has an exponential distribution with mean 6. The sample mean is Y/ and Y has a gamma distribution with parameters and 6. Then x /6 xe Pr( Y / > ) = Pr( Y > ) = dx 36 x /6 /6 x /6 e /6 xe = e = + e = STAM

52 Question #8 Key: A Let W = X+ X where each X has an exponential distribution with mean θ. The sample mean is Y = W/ and W has a gamma distribution with parameters and θ. Then w/ θ we g( θ ) = FY () = Pr( Y ) = Pr( W ) = dw θ w/ θ / θ w/ θ e / θ / θ we = e = e = e (+ θ ). θ θ / θ / θ / θ e θ 3 4 g ( θ) = e (+ ) + e =. θ θ θ At the maximum likelihood estimate of 6, g (6) = The maximum likelihood estimator is the sample mean. Its variance is the variance of one observation divided by the sample size. For the exponential distribution the variance is the square of the mean, so the estimated variance of the sample mean is 36/ = 8. The answer is (.6663) (8).79 =. Question #8 Key B µλθ (, ) = ES ( λθ, ) = λθ, v λθ = Var S λθ = λ θ (, ) (, ), v = EPV = E( λ θ ) = ()( + ) = 4, a = VHM = Var( λθ) = E( λ ) E( θ ) [ E( λ) E( θ)] = () = 3, k = v/ a= 4 / 3. Question #8 - DELETED Question #83 - DELETED STAM

53 Question #84 Key: D The posterior distribution can be found from λ e λ.4 λ/ 6.6 λ/ 7 λ/ 6 3 λ/ πλ ( ) e + e λ (.8e +.6 e ).! 6 The required constant is λ ( λ λ ).8e +.6e dλ =.8(!)(6 / 7) +.6(!)( /3) = (!). 7 / 6 3 / The posterior mean is 7 λ/6 3 λ/ E( λ ) = (.8.6 ) (!) λ e + e dλ.8(!)(6 / 7) +.6(!)( /3) = = (!) Question #85 - DELETED Question #86 - DELETED Question #87 Key: D For the geometric distribution µβ ( ) = βand v( β) = β( + β). The prior density is Pareto with parameters α and. Then, µ = E( β) =, α α v = EPV = E[ β( + β)] = + =, α ( α )( α ) ( α )( α ) α a = VHM = Var( β ) = =, ( α )( α ) ( α ) ( α ) ( α ) k = v/ a= α, Z = =. + k α The estimate is x + x + =. α α α α Question #88 - DELETED STAM

54 Question #89 Key: E A is false. Using sample data gives a better than expected fit and therefore a test statistic that favors the null hypothesis, thus increasing the Type II error probability. The K-S test works only on individual data and so B is false. D is false because the critical value depends on the degrees of freedom which in turn depends on the number of cells, not the sample size. Question #9 E( θ ) =.5(.8) +.3(.) =., E( θ ) =.5 (.8) +.3 (.) =., µθ ( ) = ( θ) + ( θ) + ( 3 θ) = 5 θ, v( θ) = ( θ) + ( θ) + ( 3 θ) ( 5 θ) = 9θ 5 θ, µ = E( 5 θ) = 5(.) =.5, v = EPV = E(9θ 5 θ ) = 9(.) 5(.) =.4, a = VHM = Var( 5 θ) = 5 Var( θ) = 5(.. ) =.5, 5 k = v/ a =.4 /.5 =.6, Z = =, P = +.5 = Question #9 λ λ λ e λ e λ λ e 4λ f( λ 5,3) λ e. This is a gamma distribution with parameters 3 5! 3! 4λ and.5. The expected value is 3(.5) = 3.5. Alternatively, if the Poisson-gamma relationships are known, begin with the prior parameters α = 5 and β = where β = / θ if the parameterization from Loss Models is considered. Then the posterior parameters are α ' = = 3 where the second 5 and the 3 are the observations and β ' = + = 4 where the second is the number of observations. The posterior mean is then 3/4 = 3.5. Question #9 - DELETED Question #93 - DELETED STAM

55 Question #94 5( 7.7) + 6(5 7.7) + ( ) + 9( 78.95) vˆ = + = 7, , kˆ = 7, / 65.3 =.57, ˆ Z = = Question #95 - DELETED Question # f f(75) (4) L= f() f(3) [ F(, )] F() F(3) α 3 α 3 α 4 α 6 α 4 α, α, α,, α,3 = α+ α+ α+ α α+, 75,,3,, α 4 3α 3 6α 4α 4 = α,,,3, 75,, 4 4 3α 3α 6α 4α α,, 75,, 4, ln L = 4 lnα + 3α ln(, ) 3α ln(, 75) 6α ln(, ) 4α ln(, 4) = 4 lnα α. The derivative is 4 / α and setting it equal to zero gives ˆ α = Question # vˆ = x = =.76. 5(.76) + 3(.76) + 5(.76) + 4(3.76) + (4.76) aˆ = =.999, ˆ.76 ˆ k = = 8.36, Z = =.684, P =.684() (.76) = The above analysis was based on the distribution of total claims for two years. Thus is the expected number of claims for the next two years. For the next one year the expected number is.78564/ =.398. Question #98 - DELETED STAM

56 Question #99 Key: E The density function is f( x).x θ.8. ( x / θ ) = e.. The likelihood function is L( θ ) = f(3) f(4) f(3) f(54)[ F()].(3).(4).(3).(54) = θ θ θ θ θ ( ) θ e, l( θ) =.8ln( θ) θ ( ). =.8ln( θ).55 θ,. l '( θ) =.8θ +.(.55) θ =,. θ =.9756, ˆ θ = 3, Question # Key: A (3/ θ) (4/ θ) (3/ θ) (54/ θ) (/ θ) (/ θ) e e e e e e.... Buhlmann estimates are on a straight line, which eliminates E. Bayes estimates are never outside the range of the prior distribution. Because graphs B and D include values outside the range to 4, they cannot be correct. The Buhlmann estimates are the linear least squares approximation to the Bayes estimates. In graph C the Bayes estimates are consistently higher and so the Buhlmann estimates are not the best approximation. This leaves A as the only feasible choice. Question # The expected counts are 3(.35) =.5, 3(.95) = 8.5, 3(.5) = 5, 3(.) = 6, and 3(.7) = 5 for the five groups. The test statistic is (5.5) (4 8.5) (37 5) (66 6) (5 5) = There are 5 = 4 degrees of freedom. From the table, the critical value for a 5% test is and for a.5% test is.43. The hypothesis is rejected at 5%, but not at.5%. Question # - DELETED STAM

57 Question #3 β For the geometric distribution, Pr( X = β ) = and the expected value is 3 β. ( + β ) Pr( X = β = ) Pr( β = ) Pr( β = X = ) = Pr( X = β = ) Pr( β = ) + Pr( X = β = 5) Pr( β = 5) 4 = 7 3 = The expected value is then.394() (5) = Question #4 Key: D Let X be the random variable for when the statistic is forgotten. Then FX ( x y) = e For the unconditional distribution of X, integrate with respect to y xy xy y y/ y( x+ /) FX ( x) = ( e ) e dy = ye dy Γ() y 4 = 4( x + / ) F (/ ) = =.75 4(/ + / ) There are various ways to evaluate the integral () Integration by parts. () Recognize that y( x+ /) y( x + / ) e dy is the expected value of an exponential random variable with mean x +. (3) Recognize that ( ) y( x /) ( / ) STAM Γ ( x + / ) ye + is the density function for a gamma random variable with α = and θ = ( x + / ), so it would integrate to. Question #5 Key: D State# Number Probability of needing Therapy Mean Number of visits E(X) E(N) Var(N) Var(X) E(S) Var(S) ,5 5, ,8,95 6,43 StdDev( S ) = 6, 43 = 5

58 S 95 5 Pr(S > 3) = Pr > = Φ (.) = The Var(X) column came from the formulas for mean and variance of a geometric distribution. Using the continuity correction, solving for Pr(S > 3.5) is theoretically better but does not affect the rounded answer. Question #6 Frequency is geometric with β =, so p = /3, p = /9, p = 4/7. Convolutions of f ( ) X x needed are x f(x) f * ( x ) fs() = / 3 =.333, fs(5) = ( / 9)(.) =.44, fs() = ( / 9)(.3) + (4 / 7)(.4) =.73 EX ( ) =.(5) +.3() +.5() = 4 ES ( ) = EX ( ) = 8 E[( S 5) + ] = ES ( ) 5[ F()] 5[ F(5)] 5[ F()] = 8 5(.333) 5( ) 5( ) = 8.8 Alternatively, E[( S 5) + ] = ES ( ) fs() + fs(5) + 5 fs() = (.333) + (.44) + 5(.73) = 8.8 STAM