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45 NOVEMBER 2005 EXAM NOVEMBER 2005 CAS COURSE 3 EXAM SOLUTIONS 1. Maximum likelihood estimation: The log of the density is 68 0ÐBÑ œ 68 ) Ð) "Ñ 68 B. The loglikelihood function is & j œ 68 0ÐB Ñ œ & 68 ) Ð) "Ñ 68 B œ" œ & 68 ) Ð) "ÑÒ 68 Þ#" 68 Þ%$ 68 Þ&' 68 Þ'( 68 Þ(# Ó œ & 68 ) Ð) "ÑÐ $Þ("$%Ñ. ` & `) ) Then setting j œ! results in $Þ("$% œ!, so that the mle of ) is & $Þ("$% R œ œ "Þ$%'. " "!! ) ) ) ) " Moment estimation: The first moment of \ is IÒ\Óœ' B0ÐBÑ.Bœ' B.Bœ. According to the method of moments, this is set equal to the empirical estimate of the mean, which is the sample mean ) B œ Þ&"). Solving for ) in the equation œ Þ&") results in the moment estimate for ) of S œ "Þ!(&. ) " Then R S œ Þ#(", Answer: A 2. In a sample of size 5, the order statistic ] & is the largest of the 5 sample values. We write the probability as TÒ] "Óœ" TÒ] Ÿ"Ó. & Then, TÒ] & Ÿ"ÓœTÒall sample values are Ÿ1ÓœTÒÐ\ " Ÿ"Ñ â Ð\ & Ÿ"ÑÓ œtò\ " Ÿ"ÓâTÒ\ & Ÿ"Ó(this follows from independence of the random sample values \" ß ÞÞÞß \&). Since each has an exponential distribution with mean 1, we have \ 3 " T Ò\ 3 Ÿ "Ó œ " / for each 3 œ "ß #ß $ß %ß &. " & Then, T Ò] Ÿ "Ó œ Ð" / Ñ œ Þ"!", and T Ò] "Ó œ " Þ"!" œ Þ))*. & Alternatively, from the theory of order statistics, the cdf of ] 8, the largest ( 8th) order statistic 8 from a sample of size 8 (\ ß ÞÞÞß \ ), is J Ð>Ñ œ ÒJ Ð>ÑÓ. " 8 ] 8 \ & & ] \ Therefore, T Ò] Ÿ "Ó œ J Ð"Ñ œ ÒJ Ð"ÑÓ œ ÒT Ð\ Ÿ "ÑÓ. Answer: E & & & & CAS Exam 3 Study Guide S. Broverman N05-1

46 NOVEMBER 2005 EXAM 3. The power of the test is the probability of rejecting the null hypothesis when the alternative hypothesis is true. Since the alternative hypothesis in this test has a larger value of - than the null hypothesis, rejection of the null will occur if the single sample value B is "large", where "large" is translated into the inequality, B -rejected if B -( -is the critical value). Since the level of significance of the test is 10%, we This means that the null hypothesis will be want a probability of 10% of rejecting L when L is true. This means that we want!! TÒB -l- œ"óœþ"!. Since \ has a Poisson (integer) distribution, this is the same as saying TÒBŸ-l- œ"óœþ*. Therefore we want the critical point -to satisfy the requirement T Ò\ œ!ß "ß ÞÞÞß -l- œ "Ó œ Þ*. From the Poisson probability function, " " T Ò\ œ!l- œ "Ó œ / œ Þ$'(* ß T Ò\ œ "l- œ "Ó œ / œ Þ$'(* ß " / T Ò\ œ #l- œ "Ó œ # œ Þ")$*. We see that T Ò\ œ!ß "ß #l- œ "Ó œ Þ*#!. The critical value is -œ#. The reason for this is that TÒ\ #l- œ"óœþ!)! Þ", but TÒ\ "l- œ"óœþ#'% (we want a Type I error probability of at most 10%, that is significance level). Now that we have determine the critical value -œ#, the power of the test is # # # TÒ\ #l- œ#óœ" TÒ\œ!ß"ß#l- œ#óœ" Ò/ #/ #/ ÓœÞ$#$. Answer: B 4. Suppose that : is the probability of winning an individual bet. This is a binomial success probability. When the number of trials of binomial distribution is given, say 7, the maximum likelihood estimate of the binomial parameter : is the same as the moment estimator of :, which is the proportion of successes in the 7trials. In this example 7 œ '!, since Mr. Jones places 60 bets in total (20 visits to the race track, 3 bets on each visit). Mr. Jones wins 13 of the bets (he won one bet on 7 visits, and 2 bets on 3 visits, for a total of (Ð"Ñ $Ð#Ñ œ "$). "$ '! The mle of : is. Answer: A 5. Suppose that \ has a normal distribution with unknown mean. and unknown variance 5 #. Ð\ \Ñ If \ ß ÞÞÞß \ is a random sample of size 8 from this distribution, then D 3 " 8 5# has a chisquared distribution with 8 " degrees of freedom. In this example, 8œ"!, and # # # # (# DÐ\ 3 \Ñ œ D\ 3 8\ œ "ß #)# "!Ð"" Ñ œ (#, so that 5 # has a chi-squared distribution with 9 degrees of freedom. We can construct a confidence interval for 5 # using any interval from the chi-squared distribution which has 95% probability. # N CAS Exam 3 Study Guide S. Broverman 2006

47 NOVEMBER 2005 EXAM 5. continued For the chi-squared distribution with 9 degrees of freedom, the interval Ð! ß "'Þ*#Ñ has 95% probability (16.92 is the 95-th percentile from the chi-squared table with 9 degrees of freedom). (# Since 5 # has a chi-squared distribution with 9 degrees of freedom, it follows that (# (# (# T Ð! 5 # "'Þ*#Ñ œ Þ*&, which is the same as T Ð "'Þ*# 5 #! Ñ. Therefore, Ð%Þ$ ß Ñ is a 95% confidence interval for 5 # (# (# (since œ %Þ$ and œ ). "'Þ*# 0 In a similar way, Ð$Þ$$ ß Ñ is a 95% interval for the chi-squared with 9 df (since 3.33 is the 5th (# percentile), and therefore $Þ$$ 5 # is a 95% confidence interval. This translates into (# # (# #! œ 5 $Þ$$ œ #"Þ' (or Ð! ß #"Þ'Ñ ) as a 95% confidence interval for 5. Finally, Ð#Þ(! ß "*Þ!#Ñ is a 95% interval for the chi-squared with 9 df, since 2.70 is the 2.4 percentile and is the 07.5 percentile. This translates into a 95% confidence interval for 5 # of Ð$Þ) ß #'Þ(Ñ. All three intervals are valid 95% confidence intervals for 5 #. Answer: E # 6. The mean square error is IÒÐs) "!!Ñ Ó. This is equal to # IÒs #!! s # # ) ) "!! Ó œ IÒs) Ó #!!IÒs) Ó "!ß!!!. * Since the pdf of s "!C ) œ] is 0ÐCÑœ, we have IÒs "!! * "!C "! "!! ) Óœ'! C.Cœ Þ s) #!"!! # * "!C "!!"! "!!"! "" # "! "!! "!!"! "# "! "!! "! "!! "# "" Answer: E and IÒ Ó œ ' C.C œ Þ # The mean square error is #!! "!ß!!! œ "&"Þ& Þ 7. Since the alternative hypothesis specifies a larger value of. than the null, the critical region (region for rejection of L ) is of the form \ -. Type I error is the probability of rejecting the! null hypothesis when it is true. Since the probability of Type I error is.05, we have \ "! TÒ\ -llóœþ!&. But if L is true, then has a standard normal distribution, and \ "!!! "ÎÈ8 T Ò "Þ'%&lL Ó œ Þ!& - œ "! "ÎÈ8! true, so that È. 8 \ "" The probability of Type II error is TÒ\ Ÿ -ll true Ó. If L is true, then has a standard normal distribution, so that T Ò\ Ÿ "! ll true Ó œ T Ò Ÿ È8 "Þ'%&lL trueó "Þ'%& " " "ÎÈ8 "Þ'%& \ "" È8 " "ÎÈ8 " is the probability of Type II error. We want this to be no more than This means that È 8 "Þ'%& must be Ÿ "Þ&&& (the 6-th percentile). Therefore, È8 "Þ'%& Ÿ "Þ&&& p È # 8 $Þ# p 8 $Þ# œ "!Þ#% p 8 "" ( 8 must be an integer). Answer: B CAS Exam 3 Study Guide S. Broverman N05-3

48 NOVEMBER 2005 EXAM DÐ\ \Ñ 3 8. Since the sample is from a normal distribution with. unknown, 5# has a chi-square distribution with 5 degrees of freedom (since there are 6 data points in the sample). If the null hypothesis ( 5 # # DÐ\ 3 \Ñ œ&! ) is true, then &! has a chi-square distribution with 5 df. # # # # Based on the given data set, DÐ\ 3 \Ñ œ D\ 3 '\ œ "'!*Þ*' 'Ð"$ Ñ œ &*&Þ*', # DÐ\ 3 \Ñ so that &! œ ""Þ*#. The : -value is the probability that a chi-square random variable with 5 df is ""Þ*#. From the table, this is between.025 and.05, since the 95-th percentile is and the 97.5 percentile is What this means is the probability of is less than.05 (the prob. of 11.07), but it is larger than.025 (the prob. of "#Þ)$). Answer: D # 9. We apply least squares estimation to the log-transformed model: w ^ œ 68 ] œ 68 α "\ œ α "\. The data set becomes \ " # $ % & ^ œ 68 ] 'Þ*#) (Þ!#" (Þ!$! (Þ!*) (Þ"&& D\^ 8\^ D\ 8\ "!'Þ##( &Ð$ÑÐ(Þ!%'%Ñ && &Ð$ Ñ 3 3 The least squares estimates are "s œ œ œ Þ!&$", 3 # # # w and α s œ^ " s \œ'þ))(. The estimated relationship is ^s œ 'Þ))( Þ!&$"\, which becomes ] s 'Þ))( Þ!&$"\ œ /, when reversing the log transformation. If \ œ ', the value of ] s is "ß $%'. Answer: D 10. The survival model is DeMoivre's Law with = œ%!. Therefore, = B in general, and %! #& / œ / œ œ (Þ& in this case. Answer: A B # #& # 11. With constant force of mortality., an 8 -year survival probability is : œ / (for any age B ). We are given ; œþ", so that : œþ*œ/. #! B #! B #!. If the force is doubled to #., the survival probability for 20 years becomes #!Ð#. Ñ %!. #!. # # / œ / œ Ð/ Ñ œ ÐÞ*Ñ œ Þ)". The 20-year mortality probability based on force #. is " Þ)" œ Þ"*. Answer: C 8 B 8. N CAS Exam 3 Study Guide S. Broverman 2006

49 NOVEMBER 2005 EXAM 12. Since a force of mortality is given, it is convenient to use the exponential form for survival probability. "!l;'! œ "!:'! "":'!. "! '.> "! "! Þ$& : œ /!. '! >, where '.> œ ' # >..> œ Þ$&, so that : œ /. "! '!! '! >! #!! "! '! "" '.> Þ%"#& In a similar way, "!:'! œ /!. '! > œ /. Þ$& Þ%"#& Then, "!l;'! œ / / œ Þ!%#*. Answer: C 13. The survival model is DeMovire's Law with = œ "!!. Note that this automatically implies that UDD is true. > ; B Þ#& ; %& We can use the UDD rule ; œ for! Ÿ = > Ÿ ", so that ; œ. > B = " = ; Þ#& %&Þ(& B =Ð%'Ñ :%& œ =Ð%&Ñ œ Þ*)") ;%& œ Þ!")# " " " B = B %& "!! %& && Þ#& ÐÞ!")#Ñ " Þ(& ÐÞ!")#Ñ " Þ(& ; From the survival function, we have and, or alternatively, from DeMoivre's Law we have ; œ, so that ; œ œ œ Þ!")#. Then, ; œ œ Þ!!%'. Answer: C Þ#& %&Þ(& %& 14. The force of mortality for DeMoivre's Law with upper age = is. B " = B, so the survival > model is DeMoivre's law with = œ "!!. Then ; œ. & %&À&! & %&À&! & %& & &! > B & "!! B & : œ " ; œ " Ð ; ÑÐ ; Ñ œ " Ð "!! %& ÑÐ "!! &! Ñ œ Þ**!*. Answer: E 15. : œ " Þ!#> is the same as =Ð>Ñ œ " > (survival from birth). Mortality (battery failure) >! &! follows DeMoivres law with = œ&!. The device fails on the first battery failure. This is a jointlife status for the two batteries. The probability of the device not failing within 15 years is the joint-life status probability "&:!À! œ "&:! "&:! œ Ð" Þ$ÑÐ" Þ$Ñ œ Þ%*, so the probability that the device fails within the 15 year warranty period is.51. Answer: D 16. The expected time to the first death is #! #! #! / œ ' :.>œ' : :.>œ' #! > # #! > # Ð ÑÐ Ñ.> BB! > BB! > B > B! " œ "'ß!!! ' #! %! Ð#! >Ñ.> œ %. #! #! The expected time to the second death is / BB œ / B / B / BB œ #/ B %. #! #! #! / B œ ' > : B.> œ ' #! > # "!! Ð Ñ.> œ ' # #! %!!! Ð#! >Ñ.> œ 'Þ'(. Then / BB œ #Ð'Þ'(Ñ % œ *Þ$%. The difference is *Þ$% % œ &Þ$%. Answer: B CAS Exam 3 Study Guide S. Broverman N05-5

50 NOVEMBER 2005 EXAM 17. We are asked to calculate the actuarial present value of a 2 year term insurance of $1000 for someone at "age" 2 (2 years in the facility). At the start of the second year there are j# œ %) individuals still in the facility. The APV of the insurance will be # " $ " " # "l # "Þ!& %) Ð"Þ!&Ñ # %)..# $. $ " ;# œ j œ %) "l;# œ j œ # # %) ; Ó œ "!!!Ò Ó œ ()Þ%# We are calculating the probabilities from, and. Answer: B 18. If 3œ!, the value of an 8-year term insurance of 1 is ' :..> œ ' 8 :..> œ ; in the continuous insurance case, and >! > B B >! > B B > 8 B it is also ; 8 B in the discrete term insurance case. This problem involves a multiple decrement model, but since 3œ!, insurance values will still be probabilities. #! ' > Ð 7 Ñ Ð"Ñ Ð"Ñ! > B. B > #! B #! For the 20-year term insurance for other causes of, the APV is ' > Ð Ñ Ð#Ñ > : 7 B. B >.> œ #! ; B. Ð"Ñ > Ð#Ñ > Ð7Ñ 3> We now note that. B > œ #! ß. B > œ "! ß and. B > œ #!. Ð"Ñ " Ð7Ñ Ð#Ñ # Ð7Ñ Therefore,. B > œ $. B >, and. B > œ $. B >, and it follows that Ð"Ñ " Ð7Ñ Ð#Ñ # Ð7Ñ 8; B œ $ 8; B and 8; B œ $ 8; B. From the combined force of decrement, we have #! B Ð 7 Ñ '.> Þ"&>.> $! : œ / #! Ð7Ñ. '! œ / #! B >! œ /. Ð"Ñ " $! Ð#Ñ # $! Therefore, 20; B œ $ Ð" / Ñ and 20; B œ $ (" / Ñ. " $! # $! % $! The term insurance single premium is #Ò $ Ð" / ÑÓ $ (" / Ñœ $ (" / Ñ. For the 20-year term insurance for accidental death, the APV is #@ :.> œ # ;. Answer: E 19. We use the following transformation rule. If \ has pdf 0\ ÐBÑ and if ] œ 2Ð\Ñ, then we w find the inverse transformation \ œ 5Ð] Ñ. The pdf of ] is 0 œ 0 Ð5ÐCÑÑ l5 ÐCÑl. In this example, 0 ÐBÑ œ \ α α) ÐB ) Ñα " (Pareto). ] ÐCÑ \ "Î7 7 The transformation is ]œ2ð\ñœ\. The inverse transformation is \œ] œ5ð]ñ. The pdf of ] is 0 ÐCÑ œ 0 ÐC Ñ 7C œ 7C œ. Answer: B ] \ α α 7 " 7 7 " α) 7 " α) 7C ÐC7 ) Ñα " ÐC7 ) Ñα " N CAS Exam 3 Study Guide S. Broverman 2006

51 NOVEMBER 2005 EXAM 20. With deductible., the expected payment per loss is IÒÐ\.Ñ Ó œ IÒ\Ó IÒ\.Ó. For the exponential with parameter ), from the table of distributions, we have IÒ\Ó œ ) and.î).î) IÒ\.Ó œ ) Ð" / Ñ, so that IÒÐ\.Ñ Ó œ )/. "!!Î) We are given that IÒÐ\ "!!Ñ Ó œ #ß!!!, and therefore, )/ œ #ß!!!. We are asked to find IÒÐ\ &!!Ñ Ó œ /. IÒÐ\ &!!Ñ Ó &!!Î) )/ %!!Î) We see that œ œ /, and therefore IÒÐ\ "!!Ñ Ó )/ "!!Î) &!!Î) ) %!!Î) %!!Î) IÒÐ\ &!!Ñ Ó œ IÒÐ\ "!!Ñ Ó / œ #ß!!!/. Answer: C 21. Let \ denote the claim random variable for this year. \ has a Pareto distribution with α œ # and ) œ %!!ß!!!. The expected proportion of claims that exceed 750,000 is the same as T Ò\ (&!ß!!!Ó, which is the probability that a claim exceeds 750,000. This probability is %!!ß!!! (&!ß!!! %!!ß!!! # (from the distribution table) " J Ð(&!ß!!!Ñ œ Ð Ñ œ Þ"#!*)$. Next year after inflation of 10%, the claim random variable will be ] œ "Þ"\. The Pareto distribution is a "scale distribution" with scale parameter ). This means that if \ is Pareto with parameters α and ), and if ]œ-\(-!), then ] also has a Pareto distribution with the same α, and with ) w œ -). In this case, ] œ "Þ"\ will have a Pareto distribution with %%!ß!!! (&!ß!!! %%!ß!!! # α œ # and ) œ %%!ß!!!, and T Ò] (&!ß!!!Ó œ Ð Ñ œ Þ"$'("%. Þ"$'("% The ratio is Þ"#!*)$ œ"þ"$. Answer: D " $ 22. The bonus payment is F œ Ð'&!ß!!! PÑ if P '&!ß!!! and 0 otherwise. This is the same as F œ " IÒPÓ œ $ Ò'&!ß!!! IÐP '&!ß!!!ÑÓ. For the given Pareto random variable P, " $ Ò'&!ß!!! ÐP '&!ß!!!ÑÓ, and the expected bonus is '!!ß!!! '!!ß!!! $ " '&!ß!!! '!!ß!!! $ ". " $ Ò'&!ß!!! #&!ß ))!Ó œ "$*ß (!( IÐP '&!ß!!!Ñ œ Ò" Ð Ñ Ó œ #$!ß ))! The expected bonus is. Answer: C CAS Exam 3 Study Guide S. Broverman N05-7

52 NOVEMBER 2005 EXAM Ð#ß$Ñ 23. The transition from Standard to Substandard at time 2 has probability.2. This is U # in the matrix notation U # U " (note that the use of the notation in the transition models study note would use the to describe transition probabilities from time 1 to time 2, so the notation used in this problem is not the same as that in the study note). But this is the probability that someone classified as Standard at time 1 will transition to Substandard at time 2. In order for a new driver to make that transition, the new driver must transition from whatever the initial classification is to Standard at time 1. Let \ 8 denote the classification state at time 8. The probability of being Standard at time 1 is T Ò\ " œ #Ó Þ We can write this as TÒ\ " œ#l\! œ"ó TÒ\! œ"ó TÒ\ " œ#l\! œ#ó TÒ\! œ#ó T Ò\ œ #l\ œ $Ó T Ò\ œ $Ó "!! Ð"ß#Ñ Ð#ß#Ñ Ð$ß#Ñ œ U " ÐÞ"Ñ U " ÐÞ$Ñ U " ÐÞ'Ñ œ ÐÞ%ÑÐÞ"Ñ ÐÞ'ÑÐÞ$Ñ ÐÞ$ÑÐÞ'Ñ œ Þ%. What we have done is looked at the three possible classifications of a new driver, and found the combined probability of being reclassified as Standard at time 1. Then the overall probability of a new driver making the transition from Standard to Substandard from time 1 to time 2 is TÐbeing classified as Standard at time 1Ñ TÐtransition from Standard to Substandard from time 1 to time 2 ÑœÐÞ%ÑÐÞ#ÑœÞ!). Answer: A 24. the claim count R is Poisson with - œ. Þ!" before the deductible is applied. When the deductible of 500 is applied, the probability of a loss being above the deductible is w T Ò\ &!!Ó œ " J Ð&!!Ñ œ Þ(&. The number of claims R that are above the deductible also has a Poisson distribution, but with mean - w œ Þ(& - œ Þ!!(&. The probability of paying at least one claim after the deductible is implemented is w w Þ!!(& T ÒR "Ó œ " T ÒR œ!ó œ " / œ Þ!!(%(. We are using the following property of the Poisson distribution. Suppose R has a Poisson distribution with mean - and is a count of the number of events occurring in a unit time period. Suppose that the events can be classified as Type 1 and Type 2 events, and each time an event occurs the probability of being Type 1 is :, and Type 2 is " :. If we define R w to be the number of Type 1 events occurring in the unit time period, then R w also has a Poisson distribution, with mean - w œ -:. In this question, claims can be described as Type 1 (over 500) and Type 2 ( Ÿ 500). Then : œ T Ð\ &!!Ñ œ Þ(&, and the number of Type 1 claims is Poisson with mean -: œ ÐÞ!"ÑÐÞ(&Ñ œ Þ!!(&. Answer: B N CAS Exam 3 Study Guide S. Broverman 2006

53 NOVEMBER 2005 EXAM 25. Assuming that policy sales and hailstorms are independent Poisson processes, if we combine the two processes we get a Poisson process with a rate of 210 events per year. Each event is either a policy sale (Type 1 event) or a hailstorm (Type 2 event). When an event occurs, the probability that the event is a policy sale is :œ #! #" (the proportion of annual events that are Type 1 on average). Since each event has the same probability of being Type 1, the probability of 80 Type 1 events (80 policy sales) before the first Type 2 event (hailstorm) is )! #! )! : œ Ð #" Ñ œ Þ!#!. Answer: B 26. The number of reindeer injured in the last hour of the day, say Q has a Poisson distribution with mean ' #% ' #$ -Ð>Ñ.> œ #% "Î# #$ Ð>Î"#Ñ.> œ "Þ% œ IÒQÓ. The probability that no reindeer will be injured during the last hour of the day is "Þ% TÒQœ!Óœ/ œþ#%(. Answer: A 27. The number of accidents reported by noon has a Poisson distribution with a mean of 12 (3 accidents per hour, on average, for 4 hours from 8AM to noon). This is the frequency R. The severity \ for an accident is the number of claimants associated with the accident. \ has a negative binomial distribution with <œ$ and " œþ(&. The total number of claimants before noon is W œ \" \ # â \ R, a compound distribution (actually, a compound Poisson distribution). The mean and the variance of Rare IÒRÓœZ+<ÒRÓœ"#, and the mean and variance of \ are IÒ\Ó œ < " œ #Þ#& and Z +<Ò\Ó œ < " Ð" " Ñ œ $Þ*$(&. # Then Z+<ÒWÓœIÒRÓ Z+<Ò\Ó Z+<ÒRÓ ÐIÒ\ÓÑ œ"!). Answer: D 28. The number of customer being served by the tellers follows a Poisson process with a rate of 18 per hour. The number of customers being served in a 10 minute period, say Q, has a Poisson distribution with a mean of 3 (10 minutes is " ' of an hour). The assumption of "independent increments" for a Poisson process implies that when a new customer arrives, what has happened prior to his arrival is irrelevant, and the number of customers being served follows the same Poisson process that we can regard as "starting over". If a new customer arrives and finds two customers waiting in line, this customer will get to a server served when 3 customers are served (one already at a teller plus the two in line). For this new customer who finds 2 in line when he arrives, the probability of waiting more than 10 minutes for the next available teller is TÒQ Ÿ #Ó(if 2 are served in the next 10 minutes, he is next in line, if 0 or 1 are served, he has 2 or 1 in front of him in line). Q has a Poisson distribution with a mean of 3, so # $ $ $ $/ TÒQŸ#Óœ/ $/ œþ%#$. Answer: C # CAS Exam 3 Study Guide S. Broverman N05-9

54 NOVEMBER 2005 EXAM 29. When time between events is exponentially distributed with mean ), and successive event times are independent, the claim number process is a Poisson process with average rate - œ " ). In this case, the number of claims follows a Poisson process with an average rate of 2 per year. The number of claims in the next 2 years, say Q, has a Poisson distribution with a mean of 4. We can separate claims into two types, Type 1 claims have a payment, and Type 2 claims have no payment. We are told that the probability of a claim being Type 1 is.7. If we define O to be the number of Type 1 claims in 2 years, then O has a Poisson distribution with mean Ð%ÑÐÞ(Ñ œ #Þ). In order for no payment to be made in the next two years, O must be 0. #Þ) This probability is TÒOœ!Óœ/ œþ!'. Answer: A 30. For a single policy, the number of hospitalizations in the year, R, is a Bernoulli random variable with TÒRœ!ÓœÞ*& and TÒRœ"ÓœÞ!&. When a hospitalization takes place, the cost is ] œ "!!\, where \ has a lognormal distribution with. œ "Þ!$* and 5 œ Þ)$$. The total cost ßWßin 2005 for a single policy is a compound distribution based on frequency R and severity ]. The mean and variance of R are IÒRÓ œ Þ!& ß Z +<ÒRÓ œ ÐÞ!&ÑÐÞ*&Ñ œ Þ!%(&. The mean and second moment of ] are " # " #. 5 "Þ!$* ÐÞ)$$ Ñ IÒ] Ó œ "!!IÒ\Ó œ "!!/ # œ "!!/ # œ %!! # # # #. # 5 IÒ] Ó œ "!! IÒ\ Ó œ "!ß!!!!/ # #Þ!() #ÐÞ)$$ Ñ œ "!ß!!!/ # œ $#!ß!"$. # # The variance of ] is IÒ] Ó ÐIÒ] ÓÑ œ "'!ß!"$. The mean and variance of W are IÒWÓ œ IÒRÓ IÒ] Ó œ #! and # Z+<ÒWÓœIÒRÓ Z+<Ò]Ó Z+<ÒRÓ ÐIÒ]ÓÑ œ"&ß'!". The aggregate cost for 10,000 policies is [ œ W" W# â W"!ß!!!. The mean and variance of [ are IÒ[ Ó œ "!ß!!!IÒWÓ œ "!ß!!!Ð#!Ñ and Z +<Ò[ Ó œ "!ß!!!Z +<ÒWÓ œ "!ß!!!Ð"&ß '!"Ñ (we assume that the policies are independent, so the variance of the sum does not involve any covariances). We apply the normal approximation to find the fund size J which satisfies the probability [ IÒ[ Ó J "!ß!!!Ð#!Ñ ÈZ +<Ò[ Ó È"!ß!!!Ð"&ß'!"Ñ TÒ[ ŸJÓœÞ*. The probability can be written as TÒ Ÿ ÓœÞ*. Then J "!ß!!!Ð#!Ñ È "!ß!!!Ð"&ß'!"Ñ is the 90th percentile of the standard normal distribution, which, from the table, is "Þ#)#. Solving for J results in J œ #"'ß!"$. #"'ß!"$ "!ß!!! This is the premium for 10,000 policies. The premium per policy is œ #"Þ'!. This solution interprets the question as meaning that the policy will pay $100 for each day for a hospitalization that occurs in 2005 (so that if the hospitalization occurs on Dec. 31, 2005, all days are still covered, even though the stay may carry over to 2006). Answer: C N CAS Exam 3 Study Guide S. Broverman 2006

55 NOVEMBER 2005 EXAM 31. We define a Type 1 shot to be a shot that results in a goal. The number of Type 1 shots in a game has a Poisson distribution with a mean of $!ÐÞ"Ñ œ $. The variance of the Poisson distribution is the same as the mean, so the variance of the number of goals scored in a game is 3, and the standard deviation is È$ œ "Þ($. Answer: C 32. Let \ denote the normal claim random variable with mean 4,000 and variance 1,000,000, and let ] denote the normal claim random variable with mean 5,000 and variance 1,000,000. A randomly selected claim will have claim size ^ that is a mixture of \ and ] with mixing weight.6 for \ and.4 for ]. Then T Ò^ 'ß!!!Ó œ ÐÞ'ÑT Ò\ 'ß!!!Ó ÐÞ%ÑT Ò] 'ß!!!Ó. \ &ß!!! 'ß!!! &ß!!! "ß!!!ß!!! "ß!!!ß!!! ] %ß!!! 'ß!!! %ß!!! "ß!!!ß!!! "ß!!!ß!!! T Ò\ 'ß!!!Ó œ T Ò È È Ó œ " F Ð"Ñ œ Þ"&)(. T Ò] 'ß!!!Ó œ T Ò È È Ó œ " F Ð#Ñ œ Þ!##). Then T Ò^ 'ß!!!Ó œ ÐÞ'ÑÐÞ"&)(Ñ ÐÞ%ÑÐÞ!##)Ñ œ Þ"!%. Answer: B 33. The claim amount random variable in 2005, say \, has a Pareto distribution with α œ $ and ) œ )!!. The claim amount random variable in 2006 is ] œ "Þ!)\. Since the Pareto is a scale w distribution, ] will also have a Pareto distribution, but with α œ $ and ) œ "Þ!)) œ )'%. Expected cost per loss after deductible Expected cost per loss without deductible w ) )'% α " $ " The loss elimination ratio is ". In 2006, the denominator is IÒ] Ó œ œ œ %$# Þ The numerator is the expected cost per loss with a franchise deductible of 300. For a franchise deductible., if a loss is above., then the insurer pays the full loss ] (note ].). The expected cost per loss with a franchise deductible. is equal to IÒÐ].Ñ Ó. Ò" J] Ð.ÑÓ. Note that it is always true that IÒÐ].Ñ ÓœIÒ]Ó IÒ].Ó. Since ] is Pareto with α œ $ and ) w œ )'%, from the table of distributions, we get )'% )'% IÒ] $!!Ó œ # Ò" Ð $!! )'$ Ñ # Ó œ "*%. The expected cost per loss with the franchise deductible of 300 is IÒÐ] $!!Ñ Ó $!! Ò" J] Ð$!!ÑÓ œ IÒ] Ó IÒ] $!!Ò $!! Ò" J] Ð$!!ÑÓ )'% œ %$# "*% $!!Ð Ñ œ $'" $!! )'$ $. $'" %$# The loss elimination ratio is " œ Þ"'%. Answer: B CAS Exam 3 Study Guide S. Broverman N05-11

56 NOVEMBER 2005 EXAM 34. The mean and variance of the frequency Rare IÒRÓœZ+<ÒRÓœ"!. The mean and variance of the severity \ are IÒ\Óœ#ß!!! and Z+<Ò\Óœ#ß!!! #. The mean of the compound distribution W is IÒWÓ œ IÒRÓ IÒ\Ó œ #!ß!!!, and the variance # of W is Z +<ÒWÓ œ IÒRÓ Z +<Ò\Ó Z +<ÒRÓ ÐIÒ\ÓÑ œ )!ß!!!ß!!!. Using the lognormal approximation to W and the method of moments, we have, W has lognormal. 5 parameters. and 5 that satisfy the relationships IÒWÓ œ / œ #!ß!!! and # #. # 5 IÒW Ó œ / # # # œ %)!ß!!!ß!!! (note that IÒW Ó œ Z +<ÒWÓ ÐIÒWÓÑ ). " # # # Then,. 5 œ 68Ð#!ß!!!Ñ œ *Þ*!$& and #. # 5 œ 68Ð%)!ß!!!ß!!!Ñ œ "*Þ*)*. We can solve for. and 5 from these equations:. œ *Þ)"# and 5 œ Þ")# (and 5 œ Þ%#'' ). Then, using the estimated lognormal distribution, T ÒW "Þ!&IÐWÑÓ œ T ÒW #"ß!!!Ó œ " T ÒW Ÿ #"ß!!!Ó (we now use the cdf of the lognormal given in the table of distributions) 68 #"ß!!! *Þ)"# œ " FŠ œ " FÐÞ$#*Ñ œ " Þ'#* œ Þ$(" " # # # Þ%#''. To find FÐÞ$#*Ñ we have done a linear interpolation in the normal table between FÐÞ$Ñ œ Þ'"(* and FÐÞ%Ñ œ Þ'&&%. Answer: D # 35. For a fully continuous whole life policy of face amount 1, with annual premium at rate U, the loss random variable is Pœ^ U], where ^is the present value random variable for a continuous whole life insurance of 1, and ] is the PVRV for a continuous whole life annuity of " ^ $ " ^ U U $ $ $. U # U $ $ 1 per year. We use the relationship ]œ to write Pin the form Pœ^ U] œ^ UÐ ÑœÐ" Ñ^ The variance of Pis Z+<ÒPÓœÐ" Ñ Z+<Ò^Ó (the constant disappears when the variance is found). The wording of the question implies that Uis the benefit premium TÐEBÑ.. # With constant force of mortality. and constant force of interest $, we have EB œ $. œ $ #. " EB $ E and EB œ # $. œ #. We also have TÐEBÑ œ + œ œ œ Þ". " E B. B B # # " # # " Þ" # " Then Z +<Ò^Ó œ E ÐE Ñ œ Ð Ñ œ, and Z +<ÒPÓ œ Ð" Ñ Ð Ñ œ Þ&. Answer: D B B # $ ") Þ!& ") N CAS Exam 3 Study Guide S. Broverman 2006

57 NOVEMBER 2005 EXAM 36. The actuarial present value of payout (i) is T E œ Þ%"))T. " " " '!À"!l " # # " " # " # '!À"!l '!À"!l The variance of the payout in (i) is T Ò E ÐE Ñ Ó œ Þ!&)&T. Since (i) and (ii) are actuarially equivalent, we have T E œ T + ÞÞ. " " # '!À#!l '!À"!l ÞÞ " E'!À#!l ÞÞ We use the relationship + œ to get + œ *Þ#))$. '!À#!l Then, Þ%"))T œ *Þ#))$T p T œ Þ!%&"T.. '!À#!l " # # " " # #. # '!À#!l '!À#!l so the variance of payout (ii) is the variance of a 20 year life annuity-due of T # T# # # # # " # #.# Ò E'!À#!l ÐE'!À#!l Ñ Ó œ ÐÞ!%&"T" Ñ Þ!!$' ÒÞ#$"# ÐÞ%%#(Ñ Ó œ Þ!"**T". # # # Þ!&)&T" Þ!"**T" œ Þ!$*T" The variance of an 8 -year life annuity-due of 1 per year is Ò E ÐE Ñ Ó, The difference in the two variances is. Answer: A per year, which is E ".+ ÞÞ " *& + *& + *& + *& ÞÞ # $ *& *& # *& $ : *& â. j*' j*( j*) *& j # *& j $ *& j and 5 *& *& *& *& *& *& 37. The annual benefit premium is T œ ÞÞ œ ÞÞ œ ÞÞ.. : œ œþ*ß : œ œþ#&ß : œ œ! : œ! for 5œ%ß&ßÞÞÞ ÞÞ # + *& œ " ÐÞ*ÑÐÞ*Ñ ÐÞ*Ñ ÐÞ#&Ñ œ #Þ!"#&. The discount rate is. œ œ Þ". " T*& œ #Þ!"#& Þ" œ Þ%. Answer: A 38. Using recursive relationship for reserves, we have Ð Z " #*Þ&#ÑÐ" 3Ñ "!!!; œ : Z ", where " 3 œ œ. # &# &# $ &!À#!l Using the given values, we have &!À#!l " Þ*& " Þ*& $ " $ ". &!À#!l &!À#!l Ð"#Þ$' #*Þ&#ÑÐ Ñ "!!!ÐÞ!#'$Ñ œ ÐÞ*($(Ñ Z p Z œ ")Þ#' Answer: B 39. Let Kbe the actuarial present value of a "Gusher" at the beginning of the year, let R be the APV of a "Normal" at the beginning of the year, and Hœ! is the APV of a "Dry" at the beginning of the year. If the well is a "Gusher" at the start of the 2nd year. the APV at the start of the 2nd year of all future dividends is K, and if the well is :Normal" at the start of the 2nd year, the APV of future dividends at the start of the 2nd year is R. Therefore, KœExpected PV of future dividends at time 0 œ "!!!@ Expected PV at time 0 of future dividends after time 1. At time 1, expected pv of future dividends is either K, R or 0. CAS Exam 3 Study Guide S. Broverman N05-13

58 NOVEMBER 2005 EXAM 39. continued Therefore, ÐKßKÑ ÐKßRÑ Expected PV at time 0 of future dividends after time 1 œ@òk U R U Ó. Therefore K œ Þ$RÓ. This is an equation in two unknown quantities, K and R. We can get a second equation by considering R in a similar way. R œ since there is.5 chance that a well that is 'Normal" at the start of the first year is still "Normal" at the start of the 2nd year. We now have two equations in K and R. K œ Þ$RÓ and R œ From the 2nd equation we get R œ $$$. Then the first equation becomes K œ *!*Þ!* Þ%&%&K *". Solving for K results in K œ "ß )$$. Answer: B 40. Rather than consider the probability of surviving 2 years, we will consider the probability of not surviving 2 years, and then we find the complement. TÒfailing during the first 2 yearsóœtòfail in 1st yearó TÒfail in 2nd year Ó. Failure will occur in the 1st year if the loss is 50 (with initial surplus of 20 plus premium of 20 in the first year, only a loss of 50 will result in failure during the 1st year), so TÒfail in 1st year ÓœÞ$. In order to fail in the 2nd year, the company has to survive the first year, which means losses must be either 0 or 10 in the 1st year. If losses are 0, then surplus at the end of the first year is 40 and then with premium of 25 at the start of the 2nd year, the surplus is 65 and even the maximum 2nd year loss of 60 will not result in failure in the 2nd year. Therefore, failure will occur in the 2nd year only if losses for the 1st year are 10, and losses for the 2nd year are 60. This probability is ÐÞ'ÑÐÞ#Ñ œ Þ"# œ T Òfail in 2nd year Ó. Then T Òfailing during the first 2 years Ó œ Þ$ Þ"# œ Þ%#, and the probability of surviving at least 2 years is " Þ%# œ Þ&). Answer: E N CAS Exam 3 Study Guide S. Broverman 2006