1. The sample is of size 15. We assign smoothed percentiles to the sample points:
"*& Þ Þ Þ #)! $&! Þ Þ Þ %&! %*! Þ Þ Þ
" % & "! ""
"' Þ Þ Þ "' "' Þ Þ Þ "' "' Þ Þ Þ
Þ!'#& Þ Þ Þ Þ#&! Þ$! Þ$"#& Þ Þ Þ Þ'#& Þ'&! Þ')(& Þ Þ Þ
‚ "' % %Þ) & "! "!Þ% ""
Since 4.8 is 80% of the way from 4 to 5, the smoothed empirical estimate of the 30th percentile is 80% of the way from 280 to 350, which is 336. Since 10.5 is 40% of the way from 10 to 11, the smoothed empirical estimate of the 65th percentile is 40% of the way from 450 to 490, which is 466.
The cdf of the Burr distribution with α œ # is J ÐBÑ œ "
Ò
"ÐBÎ Ñ" ) #Ó
.#
Applying the percentile matching method, we use the estimated percentiles in the cdf to get J Ð$$'Ñ œ "
Ò
"Ð$$'Î Ñ" ) #Ó
œ Þ$! J Ð%''Ñ œ " Ò
"Ð%''Î Ñ" ) #Ó
œ Þ'&# #
and .
After a little algebraic juggling, these equations become
Ð$$'Î Ñ œ) # " " œ Þ"*&##* Ð%''Î Ñ œ) # " " œ Þ'*!$!*
We find the posterior probabilities as follows, and note that if R is Poisson with mean , then
-3œ"
NOVEMBER 2006 SOA EXAM C/CAS EXAM 4 SOLUTIONS
The empirical estimate of Z +<Ò\Ó is
" " The cdf of the Weibull distribution is
J ÐBÑ œ " /\ ÐBÎ Ñ)7 œ " /ÐBÎ#!!Ñ# œ " /B Î%!ß!!!# .
Given uniform Ð!ß "Ñ number , the simulated value of is the solution of the equation? B
? œ " /B Î%!ß!!!# , or equivalently, B œ Ò %!ß !!! 68Ð" ?ÑÓ"Î# . The simulated values of the loss amounts are:
from ? œ Þ#($), the simulated is B ""$Þ"$ , claim amount is 0 after deductible of 150 ;
NOVEMBER 2006 SOA EXAM C/CAS EXAM 4 SOLUTIONS
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5. From the table of distributions, for the inverse exponential distribution with parameter , the) mode is . The pdf of the inverse exponential is )# 0 ÐBÑ œ )/B ÎB#) , and the log of the pdf is
The derivative of the loglikelihood is
. " " " " " " " (since we are told that the prior distribution of has mean 600).)
The Buhlmann credibility premium based on Years 1, 2 and 3 is ^\ Ð" ^Ñ ,
From the Buhlmann credibility premium based on Years 1 and 2, we have
")!! œ ^\ Ð" ^Ñ \ œ œ "'&! œ #%!! truncation since there is no indication that there is any truncation. The first payment amount is C œ %" , and there are < œ "!" at risk and = œ #" losses of that amount. The second payment amount is C œ )# , and there are < œ &# at risk (only count losses above the last censoring point), and there is = œ "# loss of that amount. The product-limit estimate is
WÐ)Ñ œ Ð" s # ÑÐ" Ñ œ" "' WÐ)Ñs
"! & #& . The Greenwood approximation to the variance of is
ÒWÐ)ÑÓ † Òs # = = Ó œ Ð Ñ † Ò# Ó œ Þ!$!(#
NOVEMBER 2006 SOA EXAM C/CAS EXAM 4 SOLUTIONS
8. The interpolating polynomial can be found by the Lagrange method.
With points ÐB ß C Ñ ß ÐB ß C Ñ ß ÐB ß C Ñ! ! " " # # , the 2nd degree interpolating polynomial is The model distribution is Poisson with a mean of .
-There are a total of 8 œ "!! observations (B ß B ß ÞÞÞß B" # "!!) in Year 1, and the total number of claims is DB œ (Ð"Ñ #Ð#Ñ "Ð$Ñ œ "%3 .
The gamma-prior/Poisson-model combination results in a posterior distribution which is also gamma, with updated parameters, αw œα B œ % "% œ ")D 3 and
)w œ 8 ")) œ "!!ÐÞ!#Ñ"Þ!# œ Þ!#$ .
The Bayesian premium for one risk in Year 2 is
IÒ\"!"lB ß B ß ÞÞÞß B" # "!!Ó œ ! IÒ\"!"l Ó † Ð lB ß B ß ÞÞÞß B" # "!!Ñ .
NOVEMBER 2006 SOA EXAM C/CAS EXAM 4 SOLUTIONS
©S. Broverman, 2006 www.sambroverman.com 11. We want T Òl\ IÐ\Ñl Þ!&IÐ\ÑÓ Þ*& .
According to the central limit theorem, has a distribution which is approximately normal.\ Also, IÐ\Ñ œ IÐ\Ñ and Z +<Ð\Ñ œ Z +<Ð\Ñ , where is the number of items in the sample.8
8
T Òl\ IÐ\Ñl Þ!&IÐ\ÑÓ
can be "standardized" to be written in the form
T
Òl
\IÐ\Ñl
Þ!&IÐ\ÑÓ œ
T Òl^l Þ!&IÐ\Ñ Ó ^The likelihood function for a discrete random variable is the product of the probability function values at all the sample points. For the binomial with 7 œ # and ; œ Þ#& , we
have 0 Ð!Ñ œ T Ð\ œ !Ñ œ ÐÞ(&Ñ œ Þ&'#&# , 0 Ð"Ñ œ T Ð\ œ "Ñ œ #ÐÞ#&ÑÐÞ(&Ñ œ Þ$(&
and 0 Ð#Ñ œ T Ð\ œ #Ñ œ ÐÞ#&Ñ œ Þ!'#&# .
The likelihood function for the 10,000 data points is ÐÞ&'#&Ñ&!!!ÐÞ$(&Ñ&!!! , since there are 5000 '0's and 5000 '1's . The loglikelihood is
&!!! 68ÐÞ&'#&Ñ &!!! 68ÐÞ$(&Ñ œ (ß ()" . Answer: B
Then, ^ œ & œ Þ#### The semiparametric empirical Bayes estimate of the claim frequency
&Þ!%Þ(
for a vehicle of Type A in Year 4 is ÐÞ####ÑÐÞ%Ñ ÐÞ((()ÑÐÞ(Ñ œ Þ'$$ . Answer: C
NOVEMBER 2006 SOA EXAM C/CAS EXAM 4 SOLUTIONS
NOVEMBER 2006 SOA EXAM C/CAS EXAM 4 SOLUTIONS
©S. Broverman, 2006 www.sambroverman.com
17. ;4ÐX Ñ is the probability that a person at age departs due to some decrement by time -4 -4". For a two decrement model ;4ÐX Ñœ " Ð" ;4wÐ"ÑÑÐ" ;4wÐ#ÑÑ .
Using these equations, we get
Þ" œ ;!ÐX Ñ œ " Ð" ;!wÐ"ÑÑÐ" ;!wÐ#ÑÑ œ " Ð" ;!wÐ"ÑÑÐ" Þ!&Ñ p " ;!wÐ"Ñœ Þ*%($')ß Þ")# œ ;"ÐX Ñ œ " Ð" ;"wÐ"ÑÑÐ" Þ!&Ñ p " ;"wÐ"Ñ œ Þ)'"!&$ .
Since Group A is affected only by decrement 1, the survival probability to age 40 for Group A is Ð" ;!wÐ"ÑÑÐ" ;"wÐ"ÑÑ œ Þ)"&( , and the expected number of survivors to age 40 from 1000 Group A individuals observed at 0 is "!!!ÐÞ)"&(Ñ œ )"' . Answer: D
18. To formulate the likelihood function we must first note that since observation of the bulbs begins at 4 hours, all data is conditional given that X %. There are 3 known burnout times, so these would be included as conditional density in the likelihood function, and the 2 bulbs that survive more hours would be included as conditional survival probabilities. The likelihood: function is P œ 0 Ð&lX %Ñ † 0 Ð*lX %Ñ † 0 Ð"$lX %Ñ † ÒT ÐX % :lX %ÑÓ# .
19. This Buhlmann-Straub situation can be considered an ordinary Buhlmann model with
8 œ &!! ^\ Ð" ^Ñ
observations. We will find . , which will be the Buhlmann credibility estimate of the expected number of claims for one insured for one month. The expected number of claims for 300 insureds for 12 months will be $!! ‚ "# times as large.
\ œ "!"""% œ Þ!(
&!! for the given data.
\ is the number of claims for one insured for one month.
The hypothetical mean is IÒ\l Ó œ- - and the process variance is Z +<Ò\l Ó œ- - . The first and second moments of the Weibull distribution are
IÒ Ó œ- ) >Ð" Ñ œ ÐÞ"Ñ Ð"Þ&Ñ œ Þ!))'#$"7 >
and the credibility estimate for the number of claims for one month for the insured is ÐÞ*#$(!'ÑÐÞ!(Ñ ÐÞ!('#*%ÑÐÞ!))'#$Ñ œ Þ!("%#" .
The estimate for the expected number of claims for the next year for 300 insureds is
$!! ‚ "# ‚ ÐÞ!("%#"Ñ œ #&( . Answer: B
NOVEMBER 2006 SOA EXAM C/CAS EXAM 4 SOLUTIONS
20. L Ð(!!!Ñ œs" $ " " " # " " œ "Þ&%&'$& .
"# * ) ( ' % $
L Ð(!!!Ñ œs# " " " " " œ Þ*&#$!#
* ) ( % $ .
Note that since the three observations at 2500 and the two at 6000 become censored, we lose "#$ and #' from L Ð(!!!Ñs" , so we lose a total of "#$ #' œ Þ&)$ .
The censoring at 7500 is irrelevant since the estimate at 7000 is based on deaths up to 7000 but not at 7500. Answer: D The estimate of the average unpaid losses for 2005 is
ÐÞ$(*&ÑÐ%&!ß"'"$$$ß!%"*$*ß(*)'))ß%&"Ñ œ ##)ß ()'
% . Answer: A
22. According to the Schwarz Bayesian criterion, we compare j <#68 8 for each estimated model, where is the maximized loglikelihood, is the number of parameters estimated and isj < 8 the number of data points. The model favored is the one with the largest value of j <#68 8 .
The Bayesian estimate for the second year is
IÒ\ l\ œ <Ó œ IÒ\ l œ "Ó † T Ð œ "l\ œ <Ñ IÒ\ l œ $Ó † T Ð œ $l\ œ <Ñ# " #- - " #- - "
œ Ð"Ñ † T Ð œ "l\ œ <Ñ Ð$Ñ † T Ð œ $l\ œ <Ñ œ #Þ*)- " - " .
If we denote T Ð œ "l\ œ <Ñ œ -- " , then T Ð œ $l\ œ <Ñ œ " -- " , and then - $Ð" -Ñ œ #Þ*) gives us - œ T Ð œ "l\ œ <Ñ œ Þ!"- .
NOVEMBER 2006 SOA EXAM C/CAS EXAM 4 SOLUTIONS
©S. Broverman, 2006 www.sambroverman.com The Buhlmann credibility estimate is ^\ Ð" ^Ñ .
24. There are 9 distinct observed values, The empirical probabilities are :ÐCÑ œ Þ" for all -C values except :Ð$&Ñ œ Þ# . We wish to estimate T ÐX %!Ñ œ " J Ð%!Ñ .
25. The method of extrapolation for a natural cubic spline uses a straight line extrapolation from the right endpoint of the data set. The line has the same slope as the spline at the right
endpoint.0 ÐBÑ œ $ÐB "Ñ Ð ÑÐB "Ñw $% # for " Ÿ B Ÿ $ , so 0 Ð$Ñ œ $w .
From the definition of variance, we know that
Z +<Ò\Ó œ IÒ\ Ó ÐIÒ\ÓÑ # # , so that IÒ\ Ó œ Z +<Ò\Ó ÐIÒ\ÓÑ# # .
It follows that IÒ\ Ó œ# )# #œ # for the exponential distribution. Answer: A
8 8
8"
) Ð Ñ)
NOVEMBER 2006 SOA EXAM C/CAS EXAM 4 SOLUTIONS
27. There are < œ $ policyholders, and 8 œ % observations for each policyholder.
We use the equal sample size version of nonparametric empirical Bayes estimation.
\ œ $ ß \ œ & ß \ œ % ß \ œ %
28. The numbers at risk for each covariate class at each data point Ÿ $# are:
## #& #( #) $"
D œ !: & % $ $ #
D œ ": & & & % % There is a single observation at each data point.
The estimate of L Ð$#Ñ! is &&/" " %&/" " $&/" " $%/" " #%/" " .
Substituting " œ " into this expression gives a value of 1.036 . Answer: C
29. The prior distribution is a beta distribution (with ) œ "), with + œ # and , œ #. The Bayesian combination of a beta prior distribution for with parameters and , and a; + , binomial model distribution for with parameters and , and with observed values\ 7 ; 8 is maximized. This occurs where log pdf is maximized.
Log pdf of posterior is 68 "%! $68 ; 68Ð" ;Ñ , and the derivative is $; ";$ .
Setting the derivative to 0 and solving for results in ; ; œ Þ& . This could have been anticipated from the form of the posterior cdf which is symmetric on the interval Ð!ß "Ñ. Answer: C
30. R (number of claims) is Poisson and we are told that the standard for full credibility for R based on total number of claims (not total number of exposures of R) is 2000. This standard is
NOVEMBER 2006 SOA EXAM C/CAS EXAM 4 SOLUTIONS
©S. Broverman, 2006 www.sambroverman.com
The standard is based on expected number of claims (not expected exposures of ), and thatW standard is ÐÞ!&D: Ñ#†ÐIÒWÓÑZ +<ÒWÓ# † IÒR Ó . Since R is Poisson, Z +<ÒWÓ œ IÒR Ó † IÒ\ Ó# , and using The simulated dental charges, and related reimbursements are:
? œ !Þ$! œ " /Þ!!"B p B œ $&'Þ'( , 257.67‚.8œ206 is reimbursed ,
33. The likelihood function is the product of probabilities for each interval.
The probability for the interval B Ÿ "! is T Ð\ Ÿ "!Ñ œ J Ð"!Ñ œ " "!) œ "!"!) .
NOVEMBER 2006 SOA EXAM C/CAS EXAM 4 SOLUTIONS
34. For the exponential distribution with mean , ) WÐ"&!!Ñ œ /"&!!Î) .
If is the mle of , then the mle of s) ) WÐ"&!!Ñ is /"&!!Îs), and according to the delta method, the variance pf this estimate is Ð . / Ñ † Z +<Ò Ós .
.) "&!!Î) # )
For the exponential distribution, the mle of is ) \ œ "!!! , and the variance is
Z +<Ò\Ó œ Z +<Ò\Ó œ 8 œ ' sœ "!!! s
8 )8# . Since and ) , the estimated variance of is) "!!!' # .
. "&!!
.) /"&!!Î) œ /"&!!Î) †Ð )# Ñ. Applying the delta method, the variance of the estimate of
WÐ"&!!Ñ is Ò/"&!!Î)†Ð"&!!ÑÓ †# , which is estimated to be
8
) )
#
#
Ò/"&!!Î"!!!†Ð"&!!ÑÓ †# "!!! œ Þ!")(
"!!!# '
# . Answer: A
35. Using linearity (uniform distribution) within each interval, we have J Ð*!Ñ œ8 $'Þ%B8 œ Þ#" , so that $' Þ%B œ Þ#"8.
Similarly, J Ð#"!Ñ œ8 $'BÞ%C8 œ Þ&" , so that $' B Þ'C œ Þ&"8 .
The total number of losses is 8 œ $' B C )% )! œ #!! B C œ 8. The first two equations become $' Þ%B œ Þ#"Ð#!! B CÑ œ %# Þ#"B Þ#"C and $' B Þ'C œ Þ&"Ð#!! B CÑ œ "!# Þ&"B Þ&"C .
Solving these last two equations for and results in B C B œ "#! and C œ )! . Answer: A