April 2018 Short-Term Actuarial Mathematics Sample . - SOA

Question # 47Key: CThe maximum likelihood estimate for the Poisson distribution is the sample mean:λˆ x 50(0) 122(1) 101(2) 92(3) 1.6438.365The table for the chi-square test is:Number of days0123 Probabilitye 1.6438 0.193241.6438e 1.6438 0.317651.64382 e 1.6438 re5.980.320.3483.230.92*365x(Probability) **obtained by subtracting the other probabilities from 1The sum of the last column is the test statistic of 7.56. Using 2 degrees of freedom (4 rowsless 1 estimated parameter less 1) the model is rejected at the 2.5% significance level but notat the 1% significance level.Question # 48Key: D0.4(0) 0.1(1) 0.1(2)0.1(0) 0.2(1) 0.1(2)0.5, µ (1) 1 0.60.4µ 0.6(0.5) 0.4(1) 0.7µ (0) 20.06a 0.6(0.52 ) 0.4(12 ) 0.7 0.4(0) 0.1(1) 0.1(4)0.1(0) 0.2(2) 0.1(4) 2 0.52 7 /12, 1 0.5v(0)v(1)0.60.4v 0.6(7 /12) 0.4(0.5) 11/ 20 10 k v / a (11/ 20) / 0.06 11/1.2 55 / 6, Z 60 /115 12 / 2310 55 / 612 10 11Bühlmann credibility premium (0.7) 0.8565. .23 10 23STAM-09-18- 13 -

Question # 49 - DELETEDQuestion # 50Key: CThe four classes have means 0.1, 0.2, 0.5, and 0.9 respectively and variances 0.09, 0.16, 0.25,and 0.09 respectively.Then, µ 0.25(0.1 0.2 0.5 0.9) 0.425v 0.25(0.09 0.16 0.25 0.09) 0.1475a 0.25(0.010.04 0.25 0.81) 0.4252 0.096875 k 0.1475/ 0.096875 1.52258 Z 4 0.72434 1.52258The estimate is 5[0.7243(2/4) 0.2757(0.425)] 2.40.Question # 51 - DELETEDQuestion # 52 - DELETEDQuestion # 53Key: BFirst obtain the distribution of aggregate ) 1/5(1/5)(2/3)(2/3) 4/45(3/5)(2/3) 2/5(1/5)(2)(2/3)(1/3) 4/45(1/5)(1/3)(1/3) 1/45µ (1/ 5)(0) (1/ 5)(25) (4 / 45)(100) (2 / 5)(150) (4 / 45)(250) (1/ 45)(400) 105σ 2 (1/ 5)(02 ) (1/ 5)(25) (4 / 45)(1002 ) (2 / 5)(1502 ) (4 / 45)(2502 ) (1/ 45)(4002 ) 1052 8100Question # 54 - DELETEDSTAM-09-18- 14 -

Question # 55Key: B(1/ 2)(1/ 3) (1/ 2)(1/ 3) (1/ 3)(1/ 6) (1/ 6)(0)(1/ 3)(1/ 6)Pr(class 2 1) (1/ 2)(1/ 3) (1/ 3)(1/ 6) (1/ 6)(0)(1/ 6)(0)Pr(class3 1) (1/ 2)(1/ 3) (1/ 3)(1/ 6) (1/ 6)(0)Pr(class1 1)34140because the prior probabilities for the three classes are 1/2, 1/3, and 1/6 respectively.The class means areµ (1) (1/ 3)(0) (1/ 3)(1) (1/ 3)(2) 1µ (2) (1/ 6)(1) (2 / 3)(2) (1/ 6)(3) 2The expectation isE ( X 2 1) (3 / 4)(1) (1/ 4)(2) 1.25.Question # 56Key: EThe first, second, third, and sixth payments were observed at their actual value and eachcontributes f(x) to the likelihood function. The fourth and fifth payments were paid at thepolicy limit and each contributes 1 – F(x) to the likelihood function. This is answer (E).Question #57 - DELETEDSTAM-09-18- 15 -

Question #58Key: BBecause the Bayes and Bühlmann results must be identical, this problem can be solved eitherway. For the Bühlmann approach, µ (λ ) v( λ ) λ . Then, noting that the prior distributionis a gamma distribution with parameters 50 and 1/500, we have:µ v E (λ ) 50(1/ 500) 0.1(λ ) 50(1/ 500) 2 0.0002a Var / a 500k v 500) 0.75Z 1500 / (1500 x (75 210) / (600 900) 0.19The credibility estimate is 0.75(0.19) 0.25(0.1) 0.1675. For 1100 policies, the expectednumber of claims is 1100(0.1675) 184.25.For the Bayes approach, the posterior density is proportional to (because in a given year thenumber of claims has a Poisson distribution with parameter λ times the number of policies)e 600 λ (600λ )75 e 900 λ (900λ ) 210 (500λ )50 e 500 λ λ 105e 2000 λ which is a gamma density withλΓ(50)75!210!parameters 335 and 1/2000. The expected number of claims per policy is 335/2000 0.1675and the expected number of claims in the next year is 184.25.uestion #59Key: EThe q-q plot takes the ordered values and plots the jth point at j/(n 1) on the horizontal axisand at F ( x j ;θ ) on the vertical axis. For small values, the model assigns more probability tobeing below that value than occurred in the sample. This indicates that the model has aheavier left tail than the data. For large values, the model again assigns more probability tobeing below that value (and so less probability to being above that value). This indicates thatthe model has a lighter right tail than the data. Of the five answer choices, only E isconsistent with these observations. In addition, note that as you go from 0.4 to 0.6 on thehorizontal axis (thus looking at the middle 20% of the data), the q-q plot increases from about0.3 to 0.4 indicating that the model puts only about 10% of the probability in this range, thusconfirming answer E.STAM-09-18- 16 -

Question #60Key: CThe posterior probability of having one of the coins with a 50% probability of heads isproportional to (0.5)(0.5)(0.5)(0.5)(4/6) 0.04167. This is obtained by multiplying theprobabilities of making the successive observations 1, 1, 0, and 1 with the 50% coin times theprior probability of 4/6 of selecting this coin. The posterior probability for the 25% coin isproportional to (0.25)(0.25)(0.75)(0.25)(1/6) 0.00195 and the posterior probability for the75% coin is proportional to (0.75)(0.75)(0.25)(0.75)(1/6) 0.01758. These three numberstotal 0.06120. Dividing by this sum gives the actual posterior probabilities of 0.68088,0.03186, and 0.28726. The expected value for the fifth toss is then (.68088)(0.5) (.03186)(0.25) (.28726)(0.75) 0.56385.Question #61Key: ABecause the exponential distribution is memoryless, the excess over the deductible is alsoexponential with the same parameter. So subtracting 100 from each observation yields datafrom an exponential distribution and noting that the maximum likelihood estimate is thesample mean gives the answer of 73.Working from first principles,f ( x1 ) f ( x2 ) f ( x3 ) f ( x4 ) f ( x5 ) θ 1e 125/θ θ 1e 150/θ θ 1e 165/θ θ 1e 175/θ θ 1e 250/θL(θ ) [1 F (100)]5(e 100/θ )5 θ 5e 365/θ .Taking logarithms and then a derivative givesl (θ ) 5ln(θ ) 365 / θ , l ′(θ ) 5 / θ 365 / θ 2 0 θˆ 365 / 5 73.STAM-09-18- 17 -

Question #62Key: DThe number of claims for each insured has a binomial distribution with n 1 and q unknown.We have) q, v(θ ) q (1 q )µ (q (q ) 0.1 (see iv) µ E a Var (q ) E (q 2 ) 0.12 0.01 (see v) E (q 2 ) 0.02v E[q (1 q )] E (q ) E (q 2 ) 0.1 0.02 0.08k v / a 8, Z 10 / (10 8) 5 / 9Then the expected number of claims in the next one year is (5/9)(0) (4/9)(0.1) 2/45 andthe expected number of claims in the next five years is 5(2/45) 2/9 0.22.Question #63 - DELETEDQuestion #64Key: EThe model distribution is f ( x θ ) 1/ θ , 0 x θ . Then the posterior distribution isproportional to1 1 500π (θ 400, 600) θ 4 , θ 600.2θθ θIt is important to note the range. Being a product, the posterior density function is non-zeroonly when all three terms are non-zero. Because one of the observations was equal to 600, thevalue of the parameter must be greater than 600 for the density function at 600 to be positive.Or, by general reasoning, posterior probability can only be assigned to possible parametervalues. Having observed the value 600 we know that parameter values less than or equal to600 are not possible.The constant is obtained from 600θ 4 dθ 1and thus the exact posterior density is3(600)3π ( θ 400, 600) 3(600)3θ 4 , θ 600. The posterior probability of an observation exceeding550 isθ 5503(600)3θ 4 dθ 0.3125.θwhere the first term in the integrand is the probability of exceeding 550 from the uniformdistribution. 600600Pr( X 3 550 400, 600) Pr( X 3 550 θ )π (θ 400, 600)dθ STAM-09-18- 18 -

Question #65Key: Cβ 0.4E ( N ) r Var ( N ) r β (1 β ) 0.48E (Y ) θ / (α 1) 500 Var (Y )θ 2α 750, 000(α 1) 2 (α 2)Therefore, E ( X ) 0.4(500) 200Var ( X ) 0.4(750, 000) 0.48(500) 2 420, 000.2 1.645 420, 000The full credibility standardis n 11,365, Z 2 0.05 200Question #66 - DELETEDQuestion #67Key: E µ (r ) E ( X 1) 100r r) E(N )E(Y ) r βθ / (α v(r ) Var ( X r ) Var ( N ) E (Y ) 2 E ( N )Var (Y )r β (1 β )θ 2r βαθ 2 210, 000r(α 1) 2(α 1) 2 (α 2) v E (210,000r ) 210,000(2) 420, 0002 a Var(100r ) (100)(4) 40, 000 k v Z 100 / (100 10.5) 0.905/ a 10.5, Question #68 - DELETEDSTAM-09-18- 19 -2,500 /11,365 0.47.

Question #69Key: BFor an exponential distribution the maximum likelihood estimate of the mean is the samplemean. We have( X ) E ( X ) θ , Var ( X ) Var ( X )/n θ2 /nE (θ / n ) / θ 1/ n 1 5 0.447.cv SD( X ) / E ( X ) If the maximum likelihood estimator is not known, it can be derived:0 θˆ X .L(θ ) θ n e Σx /θ , l (θ ) n ln θ nX / θ , l ′(θ ) nθ 1 nX θ 2 Question #70Key: DBecause the total expected claims for business use is 1.8, it must be that 20% of businessusers are rural and 80% are urban. Thus the unconditional probabilities of being businessrural and business-urban are 0.1 and 0.4 respectively. Similarly the probabilities of beingpleasure-rural and pleasure-urban are also 0.1 and 0.4 respectively. Then,µ 0.1(1.0) 0.4(2.0) 0.1(1.5) 0.4(2.5) 2.05v 0.1(0.5) 0.4(1.0) 0.1(0.8) 0.4(1.0) 0.932a 0.1(1.0) 2 0.4(2.0) 2 0.1(1.5) 2 0.4(2.5) 2 2.05 0.2225k v / a 4.18, Z 1/ (1 4.18) 0.193.Question #71Key: ANo. claims123456 22Chi-square15(15)/250 0.9015(15)/350 0.6410(10)/240 0.421(1)/110 0.017(7)/40 1.2312(12)/10 14.40The last column sums to the test statistic of 17.60 with 5 degrees of freedom (there were noestimated parameters), so from the table reject at the 0.005 significance level.STAM-09-18- 20 -

Question #72Key: CIn part (ii) you are given that µ 20 . In part (iii) you are given that a 40. In part (iv) youare given that v 8,000. Therefore, k v/a 200. Then,800(15) 600(10) 400(5) 100X 180091800Z 0.9 1800 2000.9(100 / 9) 0.1(20)P 12.CQuestion #73 - DELETEDQuestion #74 - DELETEDQuestion #75 - DELETEDQuestion #76Key: DThe posterior density is proportional to the product of the probability of the observed valueand the prior density. Thus, π (θ N 0) Pr( N 0 θ )π (θ ) (1 e θ )θ e θ .The constant of proportionality is obtained from 0θ e θ θ e 2θ dθ θ N 0) (1/ 0.75)(θ e θ θ e 2θ ).The posterior density is π ( 1 1 0.75.12 22Then,Pr( N 2 0 N1 0) 0Pr( N 2 0 θ )π (θ N1 0)dθ 0(1 e θ )(4 / 3)(θ e θ θ e 2θ )dθ4 θ4 1 2 1 θ e 2θ e 2θ θ e 3θ dθ 2 2 2 0.8148. 3 03 1 2 3 Question #77 - DELETEDQuestion #78Key: BFrom item (ii), µ 1000 and a 50. From item (i), v 500. Therefore, k v/a 10 andZ 3/(3 10) 3/13. Also, X (750 1075 2000) / 3 1275. ThenPc (3/13)(1275) (10 /13)(1000) 1063.46.STAM-09-18- 21 -

Question #79Key: C1 x /1001ee x /10,000 (1 p)10010, 000L(100, 200) f (100) f (2000)f ( x) p pe 1 (1 p)e 0.01 pe 20 (1 p)e 0.2 10, 000 10010, 000 100Question #80 - DELETEDQuestion # 81 - DELETEDQuestion #82 - DELETEDQuestion #83 - DELETEDQuestion #84Key: A c(400 x) x 400B x 4000 100 E ( B) c 400 cE ( X 400)300 c 400 c300 1 300 400 4 c 400 300 7 100 c 0.44228.6STAM-09-18- 22 -

Question #85Key: CLet N number of computers in departmentLet X cost of a maintenance callLet S aggregate cost2Var(X ) [Standard Deviation(X )]2 20040, 000 ( X 2 ) Var(X ) [ E ( X )]2E 246, 400 40, 000 80 E ( S ) N (3)(80) 240 Nλ E ( X ) N Var(S ) N λ E( X 2 ) N (3)(46, 400) 139, 200 NWe want0.1 Pr ( S 1.2 E ( S ) ) S E (S )0.2 E ( S ) Pr 139, 200 N 139, 200 N 0.2(240) N 1.282 Φ (0.9) 373.1N 2 1.282(373.1) 99.3N 48 Question #86Key: DThe modified severity, X*, represents the conditional payment amount given that a paymentoccurs. Given that a payment is required (X d), the payment must be uniformly distributedbetween 0 and c(b – d).The modified frequency, N*, represents the number of losses that result in a payment. Theb dbdeductible eliminates payments for losses below d, so only 1 FX (d ) of losses willrequire payments. Therefore, the Poisson parameter for the modified frequency distribution isλb d. (Reimbursing c% after the deductible affects only the payment amount and not thebfrequency of payments).STAM-09-18- 23 -

Question #87Key: E f ( x) 0.01, 0 x 80 0.01 0.00025( x 80) 0.03 0.00025 x, 80 x 120E ( x) 80 00.01x dx 12080(0.03 x 0.00025 x 2 )dx0.01x 2 80 0.03 x 2 120 0.00025 x3 800223 32 120 101.33 50.66667 E ( X 20) E ( X ) 2001208020x f ( x)dx 20 1 f ( x) dx 0 0.01x 2 20 20 1 0.01x02 50.6667 2 20(0.8) 32.6667Loss Elimination Ratio 1 32.6667 0.355350.6667 50.6667 STAM-09-18(- 24 -200)

Question #88Key: BFirst restate the table to be CAC’s cost, after the 10% payment by the auto owner:Towing Cost, x7290144p(x)50%40%10%Then E ( X ) 0.5(72) 0.4(90) 0.1(144) 86.4 .E ( X 2 ) 0.5(722 ) 0.4(902 ) 0.1(1442 ) 7905.6Var( X ) 7905.6 86.42 440.64Because Poisson,( N ) 1000E ( N ) Var ( X ) E ( N ) 86.4(1000)E ( S ) E 86, 400Var( S ) E ( N )Var( X ) E ( X ) 2 Var( N ) 1000(440.64) 86.42 (1000) 7,905, 600 S E ( S ) 90,000 86,400 Pr( S 90,000) Pr Pr( Z 1.28) 1 Φ (1.28) 0.10 Var( S )7,905,600 Since the frequency is Poisson, you could also have used 7,905,600 .Var(S ) λ E ( X 2 ) 1000(7905.6)That way, you would not need to have calculated Var(X).Question #89Key: CLER E ( X d ) θ (1 e d /θ ) 1 e d /θθE( X )Last year0.70 1 e d /θ d θ log(0.30)Next year: d new θ log(1 LER new )43Hence θ log(1 LER new ) d new θ log(0.30)log (1 LER new ) 1.6053(1 LER new ) e 1.6053 0.20LER new 0.80STAM-09-18- 25 -

Question # 90Key: EThe distribution of claims (a gamma mixture of Poissons) is negative binomial.E(N ) EΛ [ E ( N Λ )] EΛ (Λ ) 3Var ( N ) EΛ [Var ( N Λ )] VarΛ [ E ( N Λ )] EΛ (Λ ) VarΛ (Λ ) 6rβ 3r β (1 β ) 6(1 β ) 6 / 3 2; β 1r r 3β 3; p0 (1 β ) r 0.125rβ 0.1875(1 β ) r 1Pr(at most 1) p0 p1 0.3125. p1Question # 91Key: A 121,110E ( S ) E ( N ) E ( X ) 110(1,101)Var ( S ) E ( N )Var ( X ) E ( X ) 2 Var ( N ) 110(702 ) 11012 (750) 909, 689, 750StdDev( S ) 30,161100, 000 121,110 0.70 Pr( S 100, 000) Pr Z 0.24230,161 where Z has standard normal distribution.STAM-09-18- 26 -

Question # 92Key: CLet N number of prescriptions thennf N ( n)0123E ( N ) 4 0.20000.16000.12800.1024FN (n)0.20000.36000.48800.59041 FN (n)0.80000.64000.51200.4096 [1 F ( j )]j 0 E[( S 80) ] 40 E[( N 2) ] 40 [1 F ( j )]j 21 40 [1 F ( j )] [1 F ( j )] j 0 j 0 40(4 1.44) 102.402 E[( S 120) ] 40 E[( N 3) ] 40 [1 F ( j )] [1 F ( j )] 40(4 1.952) 81.92 j 0 j 0 Because no values of S between 80 and 120 are possible,E[( S 100) ](120 100) E[( S 80) ] (100 80) E[( S 120) ] 92.16120 80Alternatively,E[( S 100) ] (40 j 100) fj 0N( j ) 100 f N (0) 60 f N (1) 20 f N (2)(The correction terms are needed because (40j – 100) would be negative for j 0, 1, 2; weneed to add back the amount those terms would be negative) 40 jf ( j ) 100 f N ( j ) 100(0.2) 60(0.16) 20(0.128)N j 0 j 0 40 E ( N ) 100 20 9.6 2.56 160 67.84 92.16STAM-09-18- 27 -

Question #93Key: EMethod 1:In each round,N result of first roll, to see how many dice you will rollX result of for one of the N dice you rollS sum of X for the N diceE ( X ) E ( N ) 3.5 Var( X ) Var( N ) 2.9167 E ( S ) E ( N ) E ( X ) 12.25Var ( S ) E ( N )Var ( X ) Var ( N ) E ( X ) 2 3.5(2.9167) 2.9167(3.5) 2 45.938Let S1000 the sum of the winnings after 1000 roundsE ( S1000 ) 1000(12.25) 12, 250 SD( S1000 ) 1000(45.938) 214.33After 1000 rounds, you have your initial 15,000, less payments of 12,500, plus winnings for atotal of 2,500 S1000 Since actual possible outcomes are discrete, the solution tests forcontinuous outcomes greater than 15000-0.5. In this problem, that continuity correction hasnegligible impact.12, 499.5 12, 250 1.17 0.12.Pr(2,500 S1000 14,999.5) Pr( S1000 12, 499.5) Pr Z 214.33 Question #94Key: Bb p k a pk 1k 0.25 (a b)0.25 a 1 bb 0.1875 a (0.25) 0.1875 (1 0.5b)(0.25) b 0.5, a 0.52 0.5 p3 0.125 0.5 (0.1875) 3 STAM-09-18- 28 -

Question #95Key: E β mean 4, pk β k / (1 β ) k 1P(N n)nxf (1) ( x)f (2) ( x)f (3) ( 250.1250.0156 0.04f S (0) 0.2,f S (1) 0.16(0.25)f S (2) 0.16(0.25) 0.128(0.0625) 0.049f S (3) 0.16(0.25) 0.128(0.125) 0.1024(0.0156) 0.0576FS (3) 0.2 0.04 0.049 0.0576 0.346Question #96Key: ELet L incurred losses; P earned premium 800,000 L P L) 0.15(480, 000 L) if positive.Bonus 0.15 0.6 P 0.15(0.6 P This can be written 0.15[480, 000 ( L 480, 000)] . Then,E (Bonus) 0.15[480, 000 E ( L 480, 000)]From Appendix A.2.3.1E(Bonus) 0.15{480,000 – [500,000 (1 – (500,000 / (480,000 500,000))]} 35,265Question # 97Key: DSeverityafter increase60120180300Severityafter increase and deductible02080200Expected payment per loss 0.25(0) 0.25(20) 0.25(80) 0.25(200) 75Expected payments Expected number of losses x Expected payment per loss 300(75) 22,500STAM-09-18- 29 -

Question # 98Key: AE(S) E(N)E(X) 50(200) 10,000Var ( S ) E ( N )Var ( X ) E ( X ) 2 Var ( N ) 50(400) 2002 (100) 4, 020, 000 8, 000 10, 000Pr( S 8, 000) Pr Z 0.16 0.998 4, 020, 000 Question #99Key: BLet S denote aggregate loss before deductible.E(S) 2(2) 4, since mean severity is 2.f S (0)e 2 20 0.1353 , since must have 0 losses to get aggregate loss 0.0!e 2 21 1 f S (1) 0.0902 , since must have

STAM-09-18 - 1 - SOCIETY OF ACTUARIES . EXAM STAM SHORT-TERM ACTUARIAL MATHEMATICS . EXAM STAM SAMPLE SOLUTIONS . Questions 1- 307 have been taken from the previous set of Exam C sample questions . Questions no longer relevant to the syllabus have been deleted.Question 308 -326 are based on material newly added.