MAY 2005 SOA EXAM C/CAS 4 SOLUTIONS

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1 MAY 2005 SOA EXAM C/CAS 4 SOLUTIONS Prepared by Sam Broverman, to appear in ACTEX Exam C/4 Study Guide 2brove@rogerscom sam@utstattorontoedu 1 The distribution function is JÐBÑ ' B 0Ð>Ñ> ' B %!! > Ð>Ñ& Ð>Ñ% The Kolmogorov-Smirnov statistic is Q+BÖlJ ÐB Ñ J ÐB Ñl ß lj ÐB Ñ J ÐB Ñl where the maximum is taken over all data points JÐB ÑJÐB Ñ J 8 is the empirical distribution function B3 J ÐB3Ñ J8ÐB3 Ñß J8ÐB3Ñ lj ÐB3Ñ J8ÐB3 Ñl lj ÐB3Ñ J8ÐB3Ñl Þ Þ&((! ß Þ Þ&(( Þ(( Þ( Þ! Þ ß Þ% Þ'! Þ%! Þ* Þ* Þ% ß Þ' Þ& Þ Þ Þ*%' Þ' ß Þ Þ%' Þ%' Þ Þ*'% Þ ß Þ! Þ'% Þ!&( The K-S statistic is 6803 With a sample of size 5, the critical values are Level of Significance Critical Value The K-S statistic is Þ''* so L! is rejected at the 025 significance level, but the statistic is Þ(*!, so L is not rejected at the 01 significance level Answer: D! 2 We are using the full credibility standard based on the number of claims needed for full credibility of aggregate losses of a compound distribution W This standard is Þ*' Ð Þ Ñ Z+<ÒWÓ ÐIÒWÓÑ IÒRÓ, where W is aggregate losses and R is the frequency distribution (negative binomial in this case The factor 196 comes from the 95% probability requirement (196 is the 975-percentile of the standard normal With a severity random variable ], the mean of the compound distribution W is IÒWÓ IÒRÓ IÒ] Ó and the variance is Z+<ÒWÓIÒR ] Z+<Ò]ÓZ+<ÒRÓ ÐIÒ]ÓÑ For this problem, IÒRÓ< < and Z+<ÒRÓ< Ð Ñ<, IÒ] Ó ÐÑÐÞ%Ñ Ð!ÑÐÞ%Ñ Ð!!ÑÐÞÑ %Þ%, IÒ] Ó ÐÑ ÐÞ%Ñ Ð!Ñ ÐÞ%Ñ Ð!!Ñ ÐÞÑ!%!Þ%, and Z +<Ò] Ó!%!Þ% Ð%Þ%Ñ %%&Þ!% Then, IÒWÓ Ð<ÑÐ%Þ%Ñ (Þ< and Z +<ÒWÓ Ð<ÑÐ%%&Þ!%Ñ Ð<ÑÐ%Þ%Ñ ß %(*Þ%%< ß%(*Þ%%< The standard for full credibility becomes Ð%Þ'Ñ Ð(Þ<Ñ Ð<Ñ %'*Þ The expected number of claims needed is at least 2470 Answer: E 2brove@rogerscom sam@utstattorontoedu

2 3 There is no censoring and there is exactly one death at each time of death, and therefore this is a complete data set If there are 8 individuals alive at time!, the number at risk at the first death point is 8 The number at risk at the second death point is 8, etc The product-limit estimate 85 of survival to the 5-th time of death is Ð 8 ÑÐ 8 ÑâÐ 85 Ñ 8, so that 8* WÐ>Ñ 8 * 8 If we knew the value of 8, we could find WÐ>Ñ 8 * The Nelson-Aalen estimate of LÐ> Ñ is LÐ> s = = * Ñ < < 8 8! From inspection it can be seen that 8! (alternatively, 8 8 is approximately equal to *, and if we solve 8 8!, we get 8 *Þ**, but 8 must be an integer!* Then, W8 Ð>ÑW *! Ð>Ñ *! Þ&& Answer: A 4 The empirical distribution is a 2-point random variable with :ÐÑ and :Ð%Ñ The mean of the empirical distribution is ÐÑÐ Ñ Ð%ÑÐ Ñ Þ The third central moment of the empirical distribution is ÐÑ Ð ÑÐ%Ñ Ð Ñ The estimator is s 1Ð\ ß\ ß\ Ñ D Ð\ 3 \Ñ for a sample \ ß\ ß\ Þ There are ( possible bootstrap samples (of size 3 from the original sample For each sample we calculate the estimate s based on the values for that sample These 27 samples can be described as follows: - all 1's, 8 samples, \ß s!ß Ðs Ñ Ð!Ñ %à - two 1's and one %, 12 samples, \ßs ÒÐÑ ÐÑ Ð%Ñ Óß and Ðs Ñ ÐÑ! à - one 1 and two %'s, 6 samples, \ßs ÒÐÑ Ð%Ñ Ð%Ñ Ó ß and Ðs Ñ ÐÑ ' à - all %'s, 1 sample, \ % ß s! ß Ðs Ñ Ð!Ñ % The bootstrap estimate to the MSE of the estimator is ' Ð ( ÑÐ%Ñ Ð ( ÑÐ!Ñ Ð ( ÑÐ'Ñ Ð ( ÑÐ%Ñ ( %Þ* An example of the count of the number of samples given above is the number of samples with one 1 and two 4's: Ðß4ß4Ñß Ðß4ß4Ñß Ð4ßß4Ñß Ð4ßß4Ñß Ð4ß4ßÑß Ð4ß4 ßÑ The reason that there are two samples of the form Ðß4ß4 Ñ is that there are two sample points equal to 1 If we label them 1a and 1b, then Ð+ß 4ß 4 Ñ and Ð,ß 4ß 4 should be regarded as separate samples The same applies to the other samples Answer: E 2brove@rogerscom sam@utstattorontoedu

3 5 The sample is of size 9 For the :: - plot, we assign smoothed percentiles to the sample points &! &! & **!! % & ' ( * Smoothed!!!!!!!!! Percentile J8ÐBÑ We then find JÐB3Ñfor the fitted cdf J and the :-: plot has the smoothed percentile as the horizontal coordinate and the fitted cdf value JÐB Ñ 3 as the vertical coordinate We can determine which is the correct fitted model by elimination For Model B, J ÐÑ Î, but in the plot, we have J ÐÑ! (this is the vertical coordinate of the first plotted point This eliminates B B For Model C, the uniform on Òß!!Ó, J ÐBÑ ** ß so J Ð!!Ñ, but in the plot we have J Ð!!Ñ Þ( (approx(this is the vertical coordinate of the last plotted point This eliminates C BÎ! For Model D, the cdf is J ÐBÑ /, so that J ÐÑ Þ!*&, but in the plot, we have JÐÑ! This eliminates D B%! For Model E, J ÐBÑ FÐ %! Ñ, so that J Ð!Ñ FÐ Þ&Ñ Þ&, but in the plot we have J Ð!Ñ Þ& This eliminates Model E Þ& Model A can also be seen to fit by considering JÐB Ñ B for each B : &! &! & **!! % & ' ( * Smoothed!!!!!!!!! Percentile J8ÐBÑ B 3Þ&! Þ' Þ% Þ%* Þ&( Þ' Þ' Þ' Þ' These are the vertical coordinates in the plot Answer: A 6 With 8& observations, the Buhlmann credibility factor is ^ & + We are told that \l is Poisson with mean Therefore, the hypothetical mean Ð Ñ IÒ\l@ and the process variance Ñ Z+<Ò\l@ From the form of the cdf we see has a Pareto distribution with Þ' and (not the of the prior - same letter, different variable Therefore + Z +<Ò@ Ð ÑÓ Z +<Ò@ Ó ÐÞ'ÑÐÞ'Ñ Ð Þ' Ñ Þ'*(, IÒZ +<Ò\l@ ÓÓ IÒ@ Ó Þ' Þ'& Þ & Finally, ^ Þ* Answer: E & Þ'& Þ'*( 2brove@rogerscom sam@utstattorontoedu

4 7 Using the notation associated with the Kaplan-Meier approximation for large data sets, we have -!!! ß - &!! ß -!!! ß - &!!! ß -%!ß!!!! %!! (there are 400 observations with deductible of 300, && deducible of 525,! % (observations with (there are no observations with deducible higher than 5000 B &!, B & ß B!! ß B!! ß?! ß? %! Þ! 4 4 With and!, the number at risk at - is < ÐB?Ñ, and we use the estimator JÐ- s B! B4 4 Ñ Š < â Š < Then with! 4 - &!!!, we have <!! %!! ß ! 3! < Ð Ñ ÐB? Ñ Ð%!! &&Ñ Ð&!!Ñ (& ß!!! < Ð ÑÐB B?? ÑÐ%!!&&!ÑÐ&!&!!Ñ(&! Then Js B! B B Ð&!!!Ñ Š < Š < Š <! &! &!! Š %!! Š (& Š (&! Þ&& Answer: E!!! 8 The two knots are ÐB! ß C! Ñ Ð!ß!Ñ and ÐB ß C Ñ Ðß!Ñ The cubic spline on the two knots can be written in the form 0ÐBÑ+!,B-B!! B!,!ŸBŸÞ The spline must satisfy the relationships 0ÐB! Ñ C!, from which we get +!!, and 0ÐB Ñ C, from which we get,! %-!!! (A w w!!!!!! Also, 0 Ð!Ñ,, and 0 ÐÑ, - ÐÑ ÐÑ %- (B Equations A and B become %-!! % and %-!! % Solving this system results in -! and!!, along with,! and +!! The spline function is 0ÐBÑ BB!! The squared norm measure is ' ww ' Ò0 ÐBÑÓ B B Answer: D 9 If \ has an exponential distribution with mean, and \ and \ forms a random sample of size two, then \ \ has a gamma distribution with parameters and the same as in the exponential distribution If ] sample mean of \ and \, then >/ >Î T Ò]!Ó T Ò\ \!Ó '! > This integral can be found by integration by parts once is known Since the original random variable is exponential, and we have a sample of size two, the maximum likelihood estimate of is the sample mean, which is 6 The maximum likelihood estimate of the probability above is >Î' >Î' >!Î' ' >/ >/ >Î'!/!Î'! ' > ' / ¹! Ò >! ' / Ó Þ& +> +> We have used the integration by parts relationship ' +> >/ / >/ > + + ; in this case + ' Answer: D 2brove@rogerscom sam@utstattorontoedu

5 ! >/ >Î 10 From 9 above we have J] Ð!Ñ T Ò] Ÿ!Ó T Ò\ \ Ÿ!Ó '! > Using the mle s! >Îs, the estimated probability is ' >/ > 1Ðs! Ñ s Using the delta method, the estimated variance of 1ÐsÑ is Z +<Ð1Ðs w ÑÑ Ò1 ÐIÐs ÑÑÓ Z +<Òs Ó Since s is the mle of the mean of the exponential distribution, we know that s B, which is the sample mean of the first two observed values Therefore, Z+<ÒsÓZ+<ÒBÓ Z+<Ò\Ó (the variance of the exponential distribution is the square of the mean Using the antiderivative formula mentioned in the solution to 9 above, we have >! >Îs >Îs!Îs! ' >/ >/ >Îs!/!Îs! > s s / ¹ s /! >!!Îs!/!Îs / 1Ðs s Ñ! s!îs!!îs / Ð Ñ!/ w Then, 1Ðs s!îs! Ñ / Ð Ñ s s Using the estimate value s ', we get!î'!!î'!ð'ñ/ Ð Ñ!/ w '!Î' 1 Ð'Ñ! ' / Ð ' Ñ Þ!'' Then Z+<Ð1Ðs w ÑÑis approximately equal to Ò1Ðs ÑÓ Z+<Òs w ÓÒ1Ðs ÑÓ, which is w ' estimated to be Ò1 Ð'ÑÓ ÐÞ!''Ñ ÐÑ Þ!(* Answer: A 11 Credibility is being applied to aggregate losses W, which has a compound distribution The frequency R depends on the parameter -, and the severity ] depends on the parameter, so W depends on both parameters In general, for a compound distribution, the mean is IÒWÓIÒRÓ IÒ]Ó and the variance is Z+<ÒWÓIÒRÓ Z+<Ò]ÓZ+<ÒRÓ ÐIÒ]ÓÑ The hypothetical mean in this example is IÒWl- ß Ó IÒRl- Ó IÒ] l Ó - and the process variance is Z +<ÒWl- ß Ó IÒRl- Ó Z +<Ò] l Ó Z +<ÒRl- Ó ÐIÒ] l ÓÑ The variance of the hypothetical mean is +Z+<ÒIÒWl ß-ÓÓZ+<Ò- ÓIÒÐ- Ñ ÓÐIÒ- ÓÑ Since - and are independent, we have IÒ- Ó IÒ- Ó IÒ Ó, and IÒÐ- Ñ Ó IÒ- Ó IÒ- Ó IÒ Ó ÐÑÐÑ % (since - has an exponential distribution with mean 1, the second moment of - is Ðmean Ñ, and since has a Poisson distribution with mean 1, Z +<Ò Ó IÒ Ó ÐIÒ ÓÑ IÒ Ó ÐÑ, so that IÒ Ó Therefore, +% The expected process variance IÒZ +<ÒWl- ß ÓÓ IÒ - Ó IÒ- Ó IÒ Ó ÐÑÐÑ % Then 5 + Answer: B 2brove@rogerscom sam@utstattorontoedu

6 12 The number of deaths R, from 100 people has a binomial distribution with 7!! and ;!Þ! The inversion method requires knowing the distribution function The probability!! 5!!5 function for the binomial is T ÒR 5Ó Š 5 ÐÞ!Ñ ÐÞ*(Ñ 5! ÞÞÞ T ÒR 5Ó Þ!%(&& Þ%(!( Þ&& J Ð5Ñ Þ!%(&& Þ*%' Þ%*(( From? Þ!, the simulated value of R is 8, since Þ*%' Ÿ Þ Þ%*(( Þ From? Þ!, the simulated value of R is 8!, since Þ! Þ!%(&& Þ From? Þ!*, the simulated value of R is 8, since Þ!%(&& Ÿ Þ!* Þ*%' Þ! The average of the simulated values is Answer: B 13 If ] is a mixture of \ and \ with mixing weights + and +, we can define the Öß, with T Ò@ Ó + ß T Ò@ Ó + Then IÒ]ÓIÒIÒ]l@ ÓÓ IÒ]l@ Ó TÒ@ ÓIÒ]l@ Ó TÒ@ Ó IÒ\ Ó +IÒ\ Ó Ð+Ñ (this is the usual way the mean of a finite mixture is formulated Z+<Ò]ÓIÒZ+<Ò]l@ ÓÓZ+<ÒIÒ]l@ ÓÓ, where IÒZ+<Ò]l@ ÓÓZ+<Ò\ Ó +Z+<Ò\ Ó Ð+Ñ and Z +<ÒIÒ] l@ ÓÓ ÐIÒ\ Ó IÒ\ ÓÑ +Ð +Ñ The last equality follows from the fact that if ^ is a two-point random variable? prob : ^š, then Z+<Ò^ÓÐ?@Ñ :Ð:Ñ; IÒ]l@ prob : prob a two-point random variable IÒ] l Ó š Ó + IÒ\ Ó prob + Therefore Z +<Ò] Ó Z +<Ò\ Ó + Z +<Ò\ Ó Ð +Ñ ÐIÒ\ Ó IÒ\ ÓÑ +Ð +Ñ B If \ß\ are Poisson random variables then IÒ\ÓZ+<Ò\Ó and IÒ\ÓZ+<Ò\Ó, so that IÒ]ÓIÒ\ Ó +IÒ\ Ó Ð+ÑZ+<Ò\ Ó +Z+<Ò\ Ó Ð+Ñ It follows that Z +<Ò] Ó Z +<Ò\ Ó + Z +<Ò\ Ó Ð +Ñ ÐIÒ\ Ó IÒ\ ÓÑ +Ð +Ñ IÒ] Ó C For a negative binomial random variable with parameters < and, the mean is < and the variance is < Ð Ñ, so the variance is larger than the mean If \ and \ have negative binomial distributions, the IÒ\ Ó Z +<Ò\ Ó and IÒ\ Ó Z +<Ò\ Ó 2brove@rogerscom sam@utstattorontoedu

7 Therefore, IÒ]ÓIÒ\ Ó +IÒ\ Ó Ð+ÑZ+<Ò\ Ó +Z+<Ò\ Ó Ð+Ñ, and Z+<Ò\ Ó +Z+<Ò\ Ó Ð+Ñ Z +<Ò\ Ó + Z +<Ò\ Ó Ð +Ñ ÐIÒ\ Ó IÒ\ ÓÑ +Ð +Ñ Z +<Ò] Ó ; therefore IÒ] Ó Z +<Ò] Ó A For a binomial random variable with parameters 7 and ;, the mean is 7; and the variance is 7;Ð ;Ñ, which is smaller than the mean Therefore IÒ]ÓIÒ\ Ó +IÒ\ Ó Ð+ÑZ+<Ò\ Ó +Z+<Ò\ Ó Ð+Ñ, and it is possible that when we add ÐIÒ\ Ó IÒ\ ÓÑ +Ð +Ñ to the right side, we get approximate equality So it is possible that IÒ] Ó Z +<Ò] Ó Answer: A for a mixture of binomials 14 We wish to find IÒ\ l\!ó This can be formulated as '! IÒ\ l-ó 1Ð-l\!Ñ -, where 1Ð-l\!Ñ is the posterior density of - given 0Ð!ß-Ñ \! The posterior density is 1- Ð l\!ñ 0 Ð!Ñ, where the numerator is the joint density of \ and -, and the denominator is the marginal probability that \! / The joint density is 0Ð!ß -Ñ 0Ð!l-Ñ 1Ð-Ñ -! - -Î' -Î!x ÒÐ!Þ%Ñ ' / Ð!Þ'Ñ / Ó Þ%! ( -Î' Þ'! -Î This can be written as 0Ð!ß -Ñ '!x - /!x - / The marginal probability 0\ Ð!Ñ is 0\ Ð!Ñ ' 0Ð!ß Ñ ' Þ%!! Ò! / - Î' Þ' ! / - '!x!x Î Ó - Þ%!x Þ'!x Þ% ' Þ' '!x Ð(Î'Ñ!x ÐÎÑ ' Ð ( Ñ Ð Ñ We have used the relationship '! >/ 5 +> > 5x for integer 5! The posterior density is Þ%! ( -Î' Þ'! - Î Þ% ' Þ' 1- Ð l\!ñ Ò '!x - /!x - / Ó Ò ' Ð ( Ñ Ð Ñ Ó The Bayesian premium IÒ\ l\!ó is '! - 1Ð-l\!Ñ - Using the integral relationship mentioned above, we have ' Þ%! ( -Î' Þ'! - Î! - Ò '!x - /!x - / Ó - Þ% x Þ' x ÐÞ%ÑÐÑ ' ÐÞ'ÑÐÑ '!x Ð(Î'Ñ!x ÐÎÑ ' Ð ( Ñ Ð Ñ ÐÞ%ÑÐÑ ' ÐÞ'ÑÐÑ ' Ð ( Ñ Ð Ñ Finally, IÒ\ l\!ó Þ% ' Þ' *Þ Ð Ñ Ð Ñ ' ( + 5 \ 2brove@rogerscom sam@utstattorontoedu

8 14 continued Another approach to solving this problem follows from a careful look at the algebraic form of the Þ%! ( -Î' Þ'! -Î joint density 0Ð!ß -Ñ '!x - /!x - / We know that the posterior density is proportional to this (as far as the parameter - is concerned, so the posterior must be a mixture of two gamma distributions The first gamma in the posterior has ß ' (, and the second gamma has ß Therefore, the posterior density will be of the form - +! ( - Î' - Ð+Ñ! - / / Î ' Ð Since the posterior density is proportional to the joint ( Ñ!x Ð Ñ!x density, the ratio of the factors Ò Þ' Ó Ò Þ% Ó!x Î '!x % in the joint density, must be the same as + + ' Ð+Ñ the ratio of the factors Ò Ó Ò Ó Ð Ñ Î '!x Ð Ñ!x ( + ( ' Therefore, ( Ð + Ñ %, from which we get + Þ( ß + Þ'* Then IÒ\ l\!ó is '! - 1Ð-l\!Ñ - is the mean of the posterior distribution, which is the mixture of the two means + Ð +Ñ *Þ Answer: D 15 The interval is LÐ%Þ&Ñ s Þ*' ÉZ s+<òlð%þ&ñó s There is no indication of any censoring or truncation (monitoring is from starting date of policy to time of first claim, so the number at risk is reduced only by the numbers of claims at the various claim time points LÐ%Þ&Ñ s! * ( Þ((%' (the most recent claim time before 45 is time 4 The Aalen estimate of the variance of LÐ%Þ&Ñ s is Z s+<òlð%þ&ñó s! * ( Þ!*% The interval is Þ((%' Þ*' ÈÞ!*% ÐÞ* ß Þ'Ñ Answer: A 16 For any estimator s we have the relationship s QWIs Ð Ñ Z+<Ò ÓÒ,3+= s Ð ÑÓ From (iv it follows that Z+<Òs ÓÒ,3+= s Ð ÑÓ 5 Since s 5 \, it follows that,3+= Ð Ñ IÒs s Ó IÒ 5 \Ó 5 IÒ\Ó Also, Z+<Òs ÓZ+<Ò 5 \ÓÐ 5 Ñ Z+<Ò\ÓÐ 5 Ñ & Þ Therefore, from Z+<Òs ÓÒ,3+= s Ð ÑÓ, we get 5 Ð Ñ 5 & Ð 5 Ñ, from which 5 &, and 5& Answer: D 2brove@rogerscom sam@utstattorontoedu

9 17 There is 8 sample value B, so BB, and the Buhlmann credibility estimate for the number of claims in Year 2 is ^B Ð ^Ñ, where + The model distribution \l has a geometric distribution and the parameter has a Pareto distribution with parameters and The hypothetical mean is IÒ\l Ó and the process variance is Z +<Ò\l Ó Ð Ñ Then IÒIÒ\l ÓÓIÒ Ó, +Z+<ÒIÒ\l ÓZ+<Ò ÓIÒ ÓÐIÒ ÓÑ ÐÑÐÑ Ò Ó ÐÑ ÐÑ ÐÑ ÓÓ IÒ Ð ÑÓIÒ ÓIÒ Ó ÐÑÐÑ ÐÑÐÑ ÐÑÐÑ Then + Ò ÐÑÐÑ Ó Ò ÐÑ ÐÑ Óß so that Þ + The Buhlmann credibility premium is B ^B Ð ^Ñ B Ð ÑÐ Ñ Answer: D 18 Survival in the context of this problem means not having had an accident The baseline survival distribution is parametric with a constant hazard rate A constant hazard rate corresponds to an exponential distribution, therefore baseline survival is exponential with a >Î mean of The pdf for baseline time of accident is 0Ð>Ñ! /, and the baseline survival >Î probability is WÐ>Ñ/ The hazard rate for individual 4with factors ^ and ^ is ^ ^! 2Ð>Ñ/ 4 2Ð>Ñ! The survival probability for that individual 4is / W4 Ð>Ñ ÒW! Ð>ÑÓ ^ ^, and the pdf for time of accident of that individual is 04 Ð>Ñ W4 w Ð>Ñ Ð/ ^ ^ ÑÒW! Ð>ÑÓ / ^ ^ 0! Ð>Ñ There are four individuals considered Individual 1 has ^!ß ^! and the accident is observed to occur at time 3 Since this is an observed accident time, in the likelihood function, we would include the density ^ ^ / ^ ^ 0 ÐÑ Ð/ ÑÒW Ð>ÑÓ 0 Ð>Ñ!! ÐÞÑÐ!ÑÐÞ!ÑÐ!Ñ ÐÞÑÐ!ÑÐÞ!ÑÐ!Ñ Î / Î Ð/ ÑÒ/ Ó / ; the log of that density is ÐÞÑÐ!ÑÐÞ!ÑÐ!Ñ ÐÞÑÐ!Ñ ÐÞ!ÑÐ!Ñ Ð/ ÑÐ Ñ 68ÐÑ Þ*% Individual 2 has ^!ß ^! and the accident occurs after time 6, so this is a right censored observation In the likelihood function we will include the survival probability WÐ'Ñ, and the log of that is ^ ^ ÐÞÑÐ!ÑÐÞ!ÑÐ!Ñ 68ÒW Ð'ÑÓ Ð/ Ñ 68ÒW! Ð'ÑÓ Ð/ ÑÐ Ñ Þ%&! ' 2brove@rogerscom sam@utstattorontoedu

10 18 continued Individual 3 has ^ ß ^! and the accident is observed to occur at time 7 Since this is an observed accident time, in the likelihood function, we would include the density ÐÞÑÐÑÐÞ!ÑÐ!Ñ ÐÞÑÐÑÐÞ!ÑÐ!Ñ (Î / (Î 0Ð(ÑÐ/ ÑÒ/ Ó / ; the log of that density is ÐÞÑÐÑÐÞ!ÑÐ!Ñ ( ( ÐÞÑÐÑ ÐÞ!ÑÐ!Ñ Ð/ ÑÐ Ñ 68ÐÑ Þ!(!& Individual 4 has ^ ß ^ %! and the accident occurs after time 8, so this is a right censored observation In the likelihood function we will include the survival probability WÐÑ %, and the log of that is ^ ^ ÐÞÑÐÑÐÞ!ÑÐ%!Ñ 68ÒW% ÐÑÓ Ð/ Ñ 68ÒW! ÐÑÓ Ð/ ÑÐ Ñ Þ( Þ The total loglikelihood function value is Þ*% Þ%&! Þ!( Þ( (Þ& Answer: C 19 A False See the bottom of page 427 of the Loss Models book B False The K-S test is applied to individual data C False See page 430 of the Loss Models text D False The critical value depends on the number of cells, the number of estimated parameters, and the level of significance Answer: E 20 The Buhlmann estimate is ^\ Ð ^Ñ _ In this problem we are given a single value of \(2 claims in year 1, so 8 and \ The hypothetical mean is IÒ\l Ó Ð!ÑÐ Ñ ÐÑÐ Ñ ÐÑÐ Ñ & The process variance is Z+<Ò\l ÓIÒ\ l ÓÐIÒ\l ÓÑ ÒÐ! ÑÐ Ñ Ð ÑÐ Ñ Ð ÑÐ ÑÓ Ð & Ñ * & Then IÒ IÒ\l Ó Ó IÒ & Ó &IÒ Ó &ÒÐÞ!&ÑÐÞÑ ÐÞÑÐÞÑÓ Þ&, + Z +<Ò IÒ\l Ó Ó Z +<Ò & Ó &Z +<Ò Ó &ÐÞ!& ÞÑ ÐÞÑÐÞÑ Þ&, ÓÓIÒ* & Ó*IÒ Ó&IÒ Ó *ÐÞÑ &ÒÐÞ!&Ñ ÐÞÑ ÐÞÑ ÐÞÑÓ Þ% Then ^ Þ%', and the Buhlmann credibility estimate is Þ% Þ& ÐÞ%'ÑÐÑ Ð Þ%'ÑÐÞ&Ñ Þ'* Answer: B 2brove@rogerscom sam@utstattorontoedu

11 21 The prior distribution is gamma with & and The model distribution is Poisson given - The combination of gamma prior distribution and Poisson model distribution results in a posterior distribution that is also gamma If there are 8 observed values of B, say B ß B ß ÞÞÞß B8, the posterior distribution has parameters w w D B3 and 8 We are given 8 observed values, B & and B w w Î Therefore, the posterior distribution has parameters &ß ÐÐÎÑÑ % The mean of the posterior distribution is w w ÐÑÐ % ÑÞ& Answer: B 22 The graph below illustrates the three kernel functions The first curve on the left is the kernel density function for the sample point C ; 5 ÐBÑ ÐB Ñ! Ÿ B Ÿ È 1 The middle curve is the kernel density function for the sample point C ; 5ÐBÑ È 1 ÐBÑ ŸBŸ% The curve on the right is the kernel density function for the sample point C & ; 5ÐBÑ & È 1 ÐB&Ñ %ŸBŸ' The empirical probabilities are :ÐÑ Þ& ß :ÐÑ Þ& (2 sample values at 3, and :Ð&Ñ Þ& The kernel density estimator is :ÐC Ñ5 ÐBÑ ÐÞ&Ñ5 ÐBÑ ÐÞ&Ñ5 ÐBÑ ÐÞ&Ñ5 ÐBÑÞ 4 4 C & 4 We see that the middle curve has double the coefficient as the curves on the left and right, so the middle curve is doubled, with resulting graph Answer: D 2brove@rogerscom sam@utstattorontoedu

12 23 A False See the middle paragraph on page 486 of the Loss Models book B False Cubic splines pass through all data points C False The stated requirement is for the natural spline D True See page 495 of the Loss Models book E False The inequality should be reversed See Theorem 155 on page 501 of the Loss Models book Answer: D 24 The first and second moments of the Pareto distribution are and ÐÑÐÑ The sample estimate of the first and second moments are!!&! & &!!! &! 2 and & 701,4016 ÐÑÐÑ Dividing then second expression by the square of the first expression results in 2 ÐÑ (!ß%!Þ' ÐÑÐÑ Š Ð&!Ñ Þ( Then solving for results in %Þ(' Solving for results in &!Ð Ñ *Þ The limited expected value with limit 500 is Þ(' *Þ *Þ IÒ\ &!!Ó Š Ò Š &!! Ó Š Þ(' Ò Š &!!*Þ Ó *' Answer: C 2 25 The credibility factor is ^s 7, where 7! and s+ is given as Þ 7@s +s 83 s@ Ð8 ÑÐ8 Ñ 734Ð\ 34 \Ñ 3 Then ÐÑÐÑ 3 4 Ò&!Ð!! (Þ(Ñ '!Ð&! (Þ(Ñ!!Ð'! (Þ*&Ñ *!Ð!! (Þ*&Ñ Ó (ß *&Þ' ^s! Þ%** Þ Answer: B! (ß*&Þ' '&Þ! 2brove@rogerscom sam@utstattorontoedu

13 26 The ogive is the cdf of the data based on linear interpolation between interval endpoints From the data set, 16 of 100 claims are below 1,000, so the empirical cdf at 1,000 is J!! Ðß!!!Ñ Þ' Similarly, J!! Ðß!!!Ñ Þ ß J!! Ð&ß!!!Ñ Þ' ß J!! Ð!ß!!!Ñ Þß etc We wish to find T Òß!!! \ Ÿ 'ß!!!Ó We implicitly assume that losses are uniformly distributed within each interval (this implies the linearity of J ÐBÑ within each interval, so that the estimated probability is!! J Ð'ß!!!Ñ J Ðß!!!Ñ Þ''' Þ(! Þ*'!!!! Answer: B 27 The pdf for the Pareto with!ß!!! is 0ÐBÑ!ß!!! ÐB!ß!!!Ñ!ß!!!!ß!!! and the cdf is JÐBÑ Š B!ß!!!, so that JÐBÑ Š B!ß!!! Þ The log of the density is 68 0ÐBÑ 68 68!ß!!! Ð Ñ 68ÐB!ß!!!Ñ,!ß!!! and the log of J ÐBÑ is 68Ð J ÐBÑÑ 68Š B!ß!!! The derivative with respect to of 68 0ÐBÑ is 68 0ÐBÑ 68!ß!!! 68ÐB!ß!!!Ñ,!ß!!! and the derivative with respect to of 68Ð J ÐBÑÑ is 68Š B!ß!!! For each of the 3 losses of amount 750 with deductible 200, the likelihood function has a factor of the conditional density 0Ð(&!l\!!Ñ 0Ð(&!Ñ J Ð!!Ñ, and the derivative with respect to of the log of the conditional density is!ß!!! Ò68 0Ð(&!Ñ 68Ð J Ð!!ÑÑÓ 68!ß!!! 68Ð!ß (&!Ñ 68Š!ß!! 68!ß!! 68Ð!ß (&!Ñ For each of the 3 losses of amount 200 with no deductible (and policy limit 10,000, the likelihood function has a factor of 0Ð!!Ñ, and the derivative of the log is 68!ß!!! 68Ð!ß!!Ñ For each of the 4 losses of amount 300 with no deductible (and policy limit 20,000, the likelihood function has a factor of 0Ð!!Ñ, and the derivative of the log is 68!ß!!! 68Ð!ß!!Ñ For each of the 6 losses above 10,000 with non deductible and policy limit 10,000, the likelihood!ß!!! factor is J Ð!ß!!!Ñ, and the derivative of the log of the likelihood factor is 68Š!ß!!! For each of the 4 losses of amount 400 with deductible 300, the likelihood function has a factor of the conditional density 0Ð%!!l\!!Ñ 0Ð%!!Ñ J Ð!!Ñ, and the derivative with respect to of the log of the conditional density is!ß!!! Ò68 0Ð%!!Ñ 68Ð J Ð!!ÑÑÓ 68!ß!!! 68Ð!ß %!!Ñ 68Š!ß!! 68!ß!! 68Ð!ß %!!Ñ 2brove@rogerscom sam@utstattorontoedu

14 27 continued The derivative of the log likelihood is the sum of all the derivatives listed above, and it is set equal to 0 to solve for the mle of : 68!ß!! 68!ß (&! 68!ß!!! 68Ð!ß!!Ñ % 68!ß!!! 68Ð!ß!!Ñ ' 68!ß!!! 68Ð!ß!!!Ñ % 68!ß!! 68Ð!ß %!!Ñ % 68!ß!!! 68!ß (&! ' 68!ß!!! % 68!ß %!!! Solving for results in Þ!* Answer: C 28 Since the information provided is the number of claims per policyholder over a 2-year period, we will use \ to denote the number of claims in a two year period Since the number of claims per year for an individual has a Poisson distribution, the number of claims in a 2-year period will also have a Poisson distribution Each policyholder has a mean, say, for the expected number of claims in 2 years, where varies from policyholder to another The 100 observations vary over different values of Using the semiparametric nonempirical Bayes approach, we have &!Ð!Ñ!ÐÑ&ÐÑ%ÐÑÐ%Ñ s!! Þ(' and s+!! Ò D\ 3!!\ Ó \ ** Ò%!!ÐÞ('Ñ Ó Þ(' Þ!*!* For a policyholder who has had 1 claim over the 2-year period, we have 1 observation (since the random variable \ refers to the number of claims in a 2-year period, we have information for one 2-year period Therefore the credibility factor for this policy holder is ^ Þ!' Þ(' The estimate for the expected number of claims in the next 2-year period Þ!*!* for a policyholder who had 1 in claim in the first 2-year period is ÐÞ!'ÑÐÑ Ð Þ!'ÑÐÞ('Ñ Þ(' The estimate for the expected number of claims in the Year 3 (the first half of the next 2-year period is ÐÞ('Ñ Þ* Þ Answer: C 29 The baseline hazard rate is parametric so we can formulate a full (no partial likelihood > function We find LÐ>Ñ! ' 2ÐBÑB >!! ÞThen for baseline survival, the survival LÐ>Ñ! >Î w > >Î probability is W! Ð>Ñ / /, and the density is 0! Ð>Ñ W! Ð>Ñ / Baseline is loss amount distribution for Class A For Class B, B/ > / > / Î the hazard rate is 2F ÐBÑ / 2! ÐBÑ, and LFÐ>Ñ ß WFÐ>Ñ /, >/ and 0 Ð>Ñ > / / Î F The likelihood function for the given data, with Class A losses of 1 and 3 and Class B losses of 2 and 4 is 2brove@rogerscom sam@utstattorontoedu

15 29 continued ' %/ / P 0! ÐÑ 0! ÐÑ 0FÐÑ 0FÐ%Ñ / / / / % / Þ % Ð!!/ ÑÎ Î *Î %/ Î '/ Î!!/ The loglikelihood is 68 P 68 % %68 Taking partial derivatives and setting equal to 0, we get ` %!!/ `!/ ` 68 P! ` 68 P and! From the second equation we have!/, and then substituting this into the first equation, %!!/ we have!!/!!/ Solving for / results in /, so that 68 Þ'* Þ Answer: A 30 Notice that at the three knots we have 2ÐÑ ß2Ð!Ñ!ß2ÐÑ, so the three data points actually lie in a straight line The natural cubic spline is the spline whose 2nd derivative is 0 at the end knots But the straight line through the three points has a 2nd derivative of 0 (any linear function is of the form C7B5 and has 2nd derivative 0 Therefore, the natural spline is simply the straight line through the knots The 2nd derivative of this spline is 0 at all points We can also find the spline using the algebraic approach The natural spline on the interval Òß!Ó is of the form ww 0! ÐBÑ+!,! ÐBÑ-! ÐBÑ! ÐBÑ, and 0! ÐBÑ-! '! ÐBÑ For a natural spline one 3 knots, we have 7! 7! and CC CC!!!ÐÑ Ð2! 2 Ñ7 ' Š 2 2, so that Ð Ñ7 ' Š! Solving for 7 results in 7! We use the relationships -4 and 4 '24 to get -!! and!! Therefore, 0! ww Ð!Þ&Ñ! Answer: C Þ ÐBÎ Ñ Þ 31 The pdf of this Weibull with 7 Þ is 0ÐBÑ ÞB / Þ, its log is B 68 0ÐBÑ 68 Þ Þ 68 B Þ 68 Ð Ñ Þ, and the derivative of 68 0ÐBÑ with respect to is Þ ÞB 68 0ÐBÑ Þ Þ Þ B Þ ÞB The derivative of 68Ð J ÐBÑÑ is 68Ð J ÐBÑÑ Ò Ð Ñ Ó Þ For the given data points, the derivative of the loglikelihood function is the sum of the derivatives of 68 0ÐBÑ for the four given claim amounts plus 2 the derivative of 68Ò J Ð!!!ÑÓ 2brove@rogerscom sam@utstattorontoedu

16 31 continued Þ Þ Þ Þ! Þ Þ %! Þ 68 P Þ Þ Þ Þ Þ Þ!! Þ Þ &%! Þ!!! Þ Þ Þ! (note that the factor 2 cancels from all terms Solving for results in the mle estimate ' An alternative solution makes use of the relation between the Weibull and exponential 7 distributions If \ has a Weibull distribution with 7 known and unknown, then ^ \ has an exponential distribution with mean 7 If B ßßßÞB8 are sample values from a Weibull distribution with 7 known, then D B ß ÞÞÞß D B behave like a sample from an exponential distribution with mean 7 (this applies to known B-amounts and limit amounts also Þ Þ Þ Þ Þ We are given 7 Þ, so the values! ß %! ß!! ß &%! and the two limit values of!!! are from an exponential distribution with mean Þ When a sample from an exponential distribution has uncensored values and censored (limit values, the mle of the exponential mean sum of all observed values (including limit values is number of uncensored values In this case, the mle of Þ is Þ Þ Þ Þ Þ! %!!! &%!!!! % &Þ!', so that the mle of is Ð&Þ!'Ñ & ( Answer: E 32 The Buhlmann estimate is ^\ Ð ^Ñ, where \ is the number of claims in the first year This is a linear function of \, so answer E can be eliminated, since the Buhlmann estimate % is not linear The Bayesian estimate is IÒ\ l\ 8Ó ' - 1Ð-l\ 8Ñ - The Bayesian estimate must be between 1 and 4, since that is the interval on which - is defined This eliminates answers B and D, since B has a Bayesian estimate greater than 4 and D has a Bayesian estimate less than 1 The following reasoning can be used to eliminate answer C If the Buhlmann estimate for the 2nd year is 5 (as in graph C when the number of claims in the Þ& Þ& first year is 0, then Ð ^Ñ Þ& Since Ÿ Ÿ %, it follows that % Ÿ Ÿ Þ&, and then Þ& since ^, we have Þ&Ÿ^ŸÞ(& But then, if \* in the first year, the Buhlmann estimate for the 2nd year will be *^ Ð ^Ñ %Þ&, which is inconsistent with the Buhlmann estimate in graph C when \* in the first year Answer: A 2brove@rogerscom sam@utstattorontoedu

17 33 The hypothesized probabilities for the intervals are J ÐÑ Þ!& ß J Ð&Ñ Þ! ß J Ð(Ñ Þ'! ß J ÐÑ Þ! ß J Ð Ñ Therefore, T Ò\ Ÿ Ó Þ!& ß T Ò \ Ÿ &Ó Þ!*&ß T Ò& \ Ÿ (Ó Þ&!! ß T Ò( \ Ÿ Ó Þ!! ß T Ò \Ó Þ(! For the 300 observations, the expected numbers for each interval, based on the hypothesized distribution are Interval Expected Number of Observations Actual Number of Observations Ò! ß Ó!!ÐÞ!&Ñ!Þ& & Ð ß &Ó!!ÐÞ!*&Ñ Þ& % Ð& ß (Ó!!ÐÞ&!!Ñ &! ( Ð( ß Ó!!ÐÞ!!Ñ '! '' Ð ß Ñ!!ÐÞ(!Ñ & &! The chi-square statistic is Ð&!Þ&Ñ Ð%Þ&Ñ Ð(&!Ñ Ð'''!Ñ Ð&&!Ñ U!Þ& Þ& &! '! & Þ! The degrees of freedom in this chi-square test is &%(there is no estimation done, so no degrees of freedom are lost due to estimation From the chi-square table with 4 degrees of freedom, we have ; Þ!& Ð%Ñ *Þ% (95-th percentile and ; Þ!& Ð%Ñ Þ% (975-percentile The null hypothesis that the given data comes from the known distribution is rejected at the 5% level of significance because U Þ! *Þ%, but the hypothesis is no rejected at the 25% level of significance Answer: C 68 B 34 The cdf of the lognormal distribution is JÐBÑ FŠ 5 According to the inversion 68 B method, given uniform number?, we solve for B from? JÐBÑ FŠ 5 68 B &Þ'? Þ'(* FŠ Þ(& ; from the standard normal distribution table, we see that 68 B &Þ' FÐÞÑ Þ'(* ; therefore, Þ(& Þ, so that the simulated value of B is 33866? Þ%'! Þ&* and Þ&* FÐÞÑ, so that Þ%'! FÐ ÞÑ, and then 68 B &Þ' Þ(& Þ and the simulated value of B is 25089? Þ*%& F ÐÞ'Ñ p B *(Þ&, so the policy limit of 400 is applied? Þ!! F Ð Þ'%Ñ p B *%Þ'? Þ( F ÐÞÑ p B %*Þ(&, and the policy limit of 400 is applied? Þ%!( F Ð ÞÑ p B Þ(' Þ''&!Þ*%!!*%Þ'%!!Þ(' The average payment per claim is ' ' Answer: A 2brove@rogerscom sam@utstattorontoedu

18 35 We wish to find IÒ\ l\ Ó Using the Bayesian methodology, this can be formulated as IÒ\ l Ó T Ò l\ Ó IÒ\ l &Ó T Ò &l\ Ó The mean of the geometric distribution with parameter is, so IÒ\l Ó and IÒ\l &Ó& The conditional probabilities can be found in the following way: TÐ Ñ TÐ &Ñ % & & TÒ\ l Ó ÐÑ ( TÒ\ l &Ó Ð&Ñ ' Ì Ì TÒ\ Ó TÒ\ &Ó % &! T Ò\ l Ó T Ð Ñ T Ò\ l &Ó T Ð &Ñ '% Ì % &! T Ò\ Ó T Ò\ Ó T Ò\ &Ó '% Þ'&% Ì T Ò\ Ó %Î TÒ l\ Ó T Ò\ Ó Þ'&% Þ*! and T Ò\ &Ó &!Î'% TÒ l\ &Ó T Ò\ Ó Þ'&% Þ'! Then, Answer: C IÒ\ l\ Ó ÐÑÐÞ*Ñ Ð&ÑÐÞ'Ñ Þ 2brove@rogerscom sam@utstattorontoedu