MAY 2005 SOA EXAM C/CAS EXAM 4 SOLUTIONS
MAY 2005 SOA EXAM C/CAS 4 SOLUTIONS
Prepared by Sam Broverman, to appear in ACTEX Exam C/4 Study Guide
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1. The distribution function is J ÐBÑ œ '! 0 Ð>Ñ .> œ '! Ð">Ñ& .> œ "  Ð">Ñ% .
B
B
%
"
The Kolmogorov-Smirnov statistic is Q +BÖlJ ÐB3 Ñ  J8 ÐB
3 Ñl ß lJ ÐB3 Ñ  J8 ÐB3 Ñl×
where the maximum is taken over all data points. J8 ÐB
3 Ñ œ J8 ÐB3" Ñ .
J8 is the empirical distribution function.
B3
Þ#
Þ(
Þ*
"Þ"
"Þ$

J8 ÐB
3 Ñß J8 ÐB3 Ñ lJ ÐB3 Ñ  J8 ÐB3 Ñl
! ß Þ#
Þ&"((
Þ# ß Þ%
Þ')!$
Þ% ß Þ'
Þ&#$$
Þ' ß Þ)
Þ$%)'
Þ) ß "Þ!
Þ"'%$
J ÐB3 Ñ
Þ&"((
Þ))!$
Þ*#$$
Þ*%)'
Þ*'%$
lJ ÐB3 Ñ  J8 ÐB3Ñl
Þ$"((
Þ%)!$
Þ$#$$
Þ"%)'
Þ!$&(
The K-S statistic is .6803. With a sample of size 5, the critical values are
Level of Significance
0.10
0.05
0.025
0.01
Critical Value
.5456
.6082
.6619
.7290
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
#
Ð "Þ*'
Þ" Ñ †
Z +<ÒWÓ
ÐIÒWÓÑ#
† IÒR Ó , where W is aggregate losses and R is the frequency distribution
(negative binomial in this case). The factor 1.96 comes from the 95% probability requirement
(1.96 is the 97.5-percentile of the standard normal). With a severity random variable ] , the mean
of the compound distribution W is IÒWÓ œ IÒR Ó † IÒ] Ó and the variance is
Z +<ÒWÓ œ IÒR ] † Z +<Ò] Ó  Z +<ÒR Ó † ÐIÒ] ÓÑ# .
For this problem, IÒR Ó œ <" œ $< and Z +<ÒR Ó œ <"Ð"  "Ñ œ "#< ,
IÒ] Ó œ Ð"ÑÐÞ%Ñ  Ð"!ÑÐÞ%Ñ  Ð"!!ÑÐÞ#Ñ œ #%Þ% ,
IÒ] # Ó œ Ð"Ñ# ÐÞ%Ñ  Ð"!Ñ# ÐÞ%Ñ  Ð"!!Ñ# ÐÞ#Ñ œ #!%!Þ% , and
Z +<Ò] Ó œ #!%!Þ%  Ð#%Þ%Ñ# œ "%%&Þ!% .
Then, IÒWÓ œ Ð$<ÑÐ#%Þ%Ñ œ ($Þ#< and
Z +<ÒWÓ œ Ð$<ÑÐ"%%&Þ!%Ñ  Ð"#<ÑÐ#%Þ%Ñ# œ ""ß %(*Þ%%< .
""ß%(*Þ%%<
The standard for full credibility becomes Ð$)%Þ"'Ñ † Ð($Þ#<Ñ# † Ð$<Ñ œ #%'*Þ" .
The expected number of claims needed is at least 2470.
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Answer: E
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MAY 2005 SOA EXAM C/CAS EXAM 4 SOLUTIONS
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
"
"
"
of survival to the 5 -th time of death is Ð"  8 ÑÐ"  8" ÑâÐ"  85"
Ñ œ 85
8 , so that
8*
W8 Ð>* Ñ œ 8 . If we knew the value of 8, we could find W8 Ð>* Ñ.
s # Ñ œ ="  =# œ "  " œ $* .
The Nelson-Aalen estimate of LÐ># Ñ is LÐ>
<"
<#
8
8"
$)!
"
"
From inspection it can be seen that 8 œ #! (alternatively, 8  8" is approximately equal to
#
8 "#
#
$*
, and if we solve 8
" œ $)! , we get 8 œ "*Þ**, but 8 must be an integer).
#
Then, W8 Ð>* Ñ œ W#! Ð>* Ñ œ #!*
#! œ Þ&& .
Answer: A
4. The empirical distribution is a 2-point random variable with :Ð"Ñ œ #$ and :Ð%Ñ œ "$ .
The mean of the empirical distribution is Ð"ÑÐ #$ Ñ  Ð%ÑÐ "$ Ñ œ # Þ The third central moment of the
empirical distribution is ) œ Ð"  #Ñ$ Ð #$ Ñ  Ð%  #Ñ$ Ð "$ Ñ œ # .

The estimator is s) œ 1Ð\" ß \# ß \$ Ñ œ "$ DÐ\3  \Ñ$ for a sample \" ß \# ß \$ Þ
There are $$ œ #( 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 , \ œ " ß s) œ ! ß Ðs)  )Ñ# œ Ð!  #Ñ# œ % à

"
- two 1's and one %, 12 samples , \ œ # ß s) œ $ ÒÐ"  #Ñ$  Ð"  #Ñ$  Ð%  #Ñ$ Ó œ # ß
and Ðs)  )Ñ# œ Ð#  #Ñ# œ ! à

"
- one 1 and two %'s, 6 samples , \ œ $ ß s) œ $ ÒÐ"  $Ñ$  Ð%  $Ñ$  Ð%  $Ñ$ Ó œ  # ß
and Ðs)  )Ñ# œ Ð  #  #Ñ# œ "' à

- all %'s, 1 sample , \ œ % ß s) œ ! ß Ðs)  )Ñ# œ Ð!  #Ñ# œ % .
The bootstrap estimate to the MSE of the estimator is
)
"#
'
"
"$#
Ð #( ÑÐ%Ñ  Ð #( ÑÐ!Ñ  Ð #( ÑÐ"'Ñ  Ð #( ÑÐ%Ñ œ #( œ %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: Ð"ß 4ß 4Ñ ß Ð"ß 4ß 4Ñ ß Ð4ß "ß 4Ñ ß Ð4ß "ß 4Ñ ß Ð4ß 4ß "Ñ ß Ð4ß 4ß "Ñ .
The reason that there are two samples of the form Ð"ß 4ß 4Ñ is that there are two sample points
equal to 1. If we label them 1a and 1b, then Ð"+ß 4ß 4Ñ and Ð",ß 4ß 4) should be regarded as
separate samples. The same applies to the other samples.
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Answer: E
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MAY 2005 SOA EXAM C/CAS EXAM 4 SOLUTIONS
5. The sample is of size 9. For the :-: plot, we assign smoothed percentiles to the sample points
"
"
Smoothed
"!
Percentile J8 ÐBÑ
#
$
"&
$!
&!
&"
**
"!!
#
"!
$
"!
%
"!
&
"!
'
"!
(
"!
)
"!
*
"!
We then find J ÐB3 Ñ for the fitted cdf J and the :-: plot has the smoothed percentile as the
horizontal coordinate and the fitted cdf value J ÐB3 Ñ as the vertical coordinate.
We can determine which is the correct fitted model by elimination.
For Model B, J Ð"Ñ œ "Î# , but in the plot, we have J Ð"Ñ œ ! (this is the vertical coordinate of
the first plotted point). This eliminates B.
For Model C, the uniform on Ò"ß "!!Ó, J ÐBÑ œ B"
** ß so J Ð"!!Ñ œ ", but in the plot we have
J Ð"!!Ñ œ Þ( (approx.)(this is the vertical coordinate of the last plotted point). This eliminates C.
For Model D, the cdf is J ÐBÑ œ "  /BÎ"! , so that J Ð"Ñ œ Þ!*&, but in the plot, we have
J Ð"Ñ œ ! . This eliminates D.
For Model E, J ÐBÑ œ FÐ B%!
%! Ñ , so that J Ð$!Ñ œ FÐ  Þ#&Ñ  Þ&, but in the plot we have
J Ð$!Ñ  Þ& . This eliminates Model E.
Model A can also be seen to fit by considering J ÐB3 Ñ œ "  BÞ#&
for each B3 :
3
"
"
Smoothed
"!
Percentile J8 ÐBÑ
#
$
"&
$!
&!
&"
**
"!!
#
"!
$
"!
%
"!
&
"!
'
"!
(
"!
)
"!
*
"!
"  BÞ#&
3
Þ"'
Þ#%
Þ%*
Þ&(
Þ'#
Þ'$
Þ')
Þ')
!
These are the vertical coordinates in the plot.
Answer: A
6. With 8 œ & observations, the Buhlmann credibility factor is ^ œ && @ .
+
We are told that \l) is Poisson with mean ). Therefore, the hypothetical mean is
.Ð@Ñ œ IÒ\l@Ó œ @ and the process variance is @Ð@Ñ œ Z +<Ò\l@Ó œ @.
From the form of the cdf of @, we see that @ has a Pareto distribution with α œ #Þ' and ) œ "
(not the ) of the prior - same letter, different variable).
#
"
Therefore + œ Z +<Ò.Ð@ÑÓ œ Z +<Ò@Ó œ Ð#Þ'"ÑÐ#Þ'#Ñ
 Ð #Þ'"
Ñ# œ "Þ'*#( ,
"
and @ œ IÒZ +<Ò\l@ÓÓ œ IÒ@Ó œ #Þ'"
œ Þ'#& Þ
&
Finally, ^ œ & Þ'#& œ Þ*$ .
Answer: E
"Þ'*#(
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7. Using the notation associated with the Kaplan-Meier approximation for large data sets, we
have -! œ $!! ß -" œ &!! ß -# œ "!!! ß -$ œ &!!! ß -% œ "!ß !!! .
.! œ %!! (there are 400 observations with deductible of 300) , ." œ &#& (observations with
deducible of 525) , .# œ .$ œ .% œ ! (there are no observations with deducible higher than
5000). B! œ &! , B" œ "#& ß B# œ $!! ß B$ œ $!! ß ?# œ "#! ß ?$ œ %! Þ
4
With α œ " and " œ !, the number at risk at -4 is <4 œ
s Ð-4 Ñ œ "  Š"  B! ‹âŠ"  B4" ‹ .
estimator J
<
<
!
4"
.3 
3œ!
ÐB3  ?3 Ñ , and we use the
3œ!
4"
Then with -$ œ &!!!, we have <! œ .! œ %!! ß
<" œ Ð.!  ." Ñ  ÐB!  ?! Ñ œ Ð%!!  &#&Ñ  Ð&!  !Ñ œ )(& ß
<# œ Ð.!  ."  .# Ñ  ÐB!  B"  ?!  ?" Ñ œ Ð%!!  &#&  !Ñ  Ð&!  "#&  !  !Ñ œ (&! .
s Ð&!!!Ñ œ "  Š"  B! ‹Š"  B" ‹Š"  B# ‹
Then J
œ "  Š" 
<!
<"
<#
&!
"#&
$!!
%!! ‹Š"  )(& ‹Š"  (&! ‹
œ Þ&& . Answer: E
8. The two knots are ÐB! ß C! Ñ œ Ð!ß !Ñ and ÐB" ß C" Ñ œ Ð#ß !Ñ .
The cubic spline on the two knots can be written in the form
0 ÐBÑ œ +!  ,! B  -! B#  .! B$ , ! Ÿ B Ÿ # Þ
The spline must satisfy the relationships 0 ÐB! Ñ œ C! , from which we get +! œ ! ,
and 0 ÐB" Ñ œ C" , from which we get #,!  %-!  ).! œ ! (A).
Also , 0 w Ð!Ñ œ ,! œ  #, and 0 w Ð#Ñ œ ,!  #-! Ð#Ñ  $.! Ð#Ñ# œ  #  %-!  "#.! œ # (B).
Equations A and B become %-!  ).! œ % and %-!  "#.! œ % .
Solving this system results in -! œ " and .! œ ! , along with ,! œ  # and +! œ ! .
The spline function is 0 ÐBÑ œ  #B  B# .
#
#
The squared norm measure is '! Ò0 ww ÐBÑÓ# .B œ '! ## .B œ ) .
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 α œ # and the same ) as in
the exponential distribution. If ] sample mean of \" and \# , then
>Î)
∞
T Ò]  "!Ó œ T Ò\"  \#  #!Ó œ '#! >/ # .> . 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
>ϰ
#!Î'
∞ >/>Î'
>/>Î'
>Î'
'#!
.>
œ

/
œ !  Ò #!/'  /#!Î' Ó œ Þ"& .
¹
$'
'
+>
+>
We have used the integration by parts relationship ' >/+> .> œ >/+  /+# ; in this case + œ  "' .
Answer: D
>œ#!
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#! >Î)
10. From #9 above we have J] Ð"!Ñ œ T Ò] Ÿ "!Ó œ T Ò\"  \# Ÿ #!Ó œ '! >/)# .>
#!
Using the mle s), the estimated probability is '! >/
>Îs)
s)#
.> œ 1Ðs)Ñ .
Using the delta method, the estimated variance of 1Ðs)Ñ is Z +<Ð1Ðs)ÑÑ œ Ò1w ÐIÐs)ÑÑÓ# † Z +<Ò s) Ó .
Since s) is the mle of the mean of the exponential distribution, we know that s) œ 
B , which is the
#
Z +<Ò\Ó
sample mean of the first two observed values. Therefore, Z +<Ò s) Ó œ Z +<Ò 
BӜ
œ )#
#
(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)
s
s
.> œ ’  >/ s  />Î) “¹
œ ’  #!/s
 /#!Î) “  ’!  "“
s)
)
)
>œ!
s
#!/#!Î)
#!Îs)
s
œ"
/
œ 1Ð)Ñ .
s
'!#! >/
>œ#!
>Îs)
#
)
Then, 1 Ðs)Ñ œ 
w
s
s
#!s)/#!Î) Ð #!# Ñ#!/#!Î)
s)
s)#
 /#!Îs) Ð s#!# Ñ .
)
Using the estimate value s) œ ' , we get
1w Ð'Ñ œ 
#!Ð'Ñ/#!Î' Ð '#!# Ñ#!/#!Î'
'#
 /#!Î' Ð #!
'# Ñ œ Þ!''" .
#
Then Z +<Ð1Ðs)ÑÑ is approximately equal to Ò1w Ðs)ÑÓ# † Z +<Ò s) Ó œ Ò1w Ðs)ÑÓ# † )# , which is
#
estimated to be Ò1w Ð'ÑÓ# † '# œ ÐÞ!''"Ñ# Ð")Ñ œ 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
IÒWÓ œ IÒR Ó † IÒ] Ó and the variance is Z +<ÒWÓ œ IÒR Ó † Z +<Ò] Ó  Z +<ÒR Ó † ÐIÒ] ÓÑ # .
The hypothetical mean in this example is
IÒWl)ß -Ó œ IÒR l-Ó † IÒ] l)Ó œ - † ) and the process variance is
Z +<ÒWl-ß )Ó œ IÒR l-Ó † Z +<Ò] l)Ó  Z +<ÒR l-Ó † ÐIÒ] l)ÓÑ# œ -)#  -)# œ #-)# .
The variance of the hypothetical mean is
+ œ Z +<Ò IÒWl)ß -Ó Ó œ Z +<Ò-)Ó œ IÒÐ-)Ñ# Ó  ÐIÒ-)ÓÑ# .
Since - and ) are independent, we have IÒ-)Ó œ IÒ-Ó † IÒ)Ó œ " ‚ " ,
and IÒÐ-)Ñ# Ó œ IÒ-# )# Ó œ IÒ-# Ó † IÒ)# Ó œ Ð#ÑÐ#Ñ œ %
(since - has an exponential distribution with mean 1, the second moment of - is #ÐmeanÑ# œ #,
and since ) has a Poisson distribution with mean 1,
" œ Z +<Ò)Ó œ IÒ)# Ó  ÐIÒ)ÓÑ# œ IÒ)# Ó  Ð"Ñ# , so that IÒ)# Ó œ #).
Therefore, + œ %  " œ $ . The expected process variance is
@ œ IÒ Z +<ÒWl-ß )Ó Ó œ IÒ#-)# Ó œ #IÒ-Ó † IÒ)# Ó œ #Ð"ÑÐ#Ñ œ % . Then 5 œ +@ œ %$ .
Answer: B
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MAY 2005 SOA EXAM C/CAS EXAM 4 SOLUTIONS
12. The number of deaths R , from 100 people has a binomial distribution with 7 œ "!! and
; œ !Þ!$ . The inversion method requires knowing the distribution function. The probability
function for the binomial is T ÒR œ 5Ó œ Š 5 ‹ÐÞ!$Ñ5 ÐÞ*(Ñ"!!5 .
"!!
5
!
"
#
T ÒR œ 5Ó
Þ!%(&&
Þ"%(!(
Þ##&"&
J Ð5Ñ
Þ!%(&&
Þ"*%'#
Þ%"*((
ÞÞÞ
From ?" œ Þ#!, the simulated value of R is 8" œ #, since Þ"*%'# Ÿ Þ#  Þ%"*(( Þ
From ?# œ Þ!$, the simulated value of R is 8# œ !, since Þ!$  Þ!%(&& Þ
From ?$ œ Þ!*, the simulated value of R is 8$ œ ", since Þ!%(&& Ÿ Þ!*  Þ"*%'# Þ
The average of the simulated values is #!"
œ".
$
Answer: B
13. If ] is a mixture of \" and \# with mixing weights + and "  +, we can define the
parameter @ œ Ö"ß #×, with T Ò@ œ "Ó œ + ß T Ò@ œ #Ó œ "  + .
Then IÒ] Ó œ IÒ IÒ] l@Ó Ó œ IÒ] l@ œ "Ó † T Ò@ œ "Ó  IÒ] l@ œ #Ó † T Ò@ œ #Ó
œ IÒ\" Ó † +  IÒ\# Ó † Ð"  +Ñ
(this is the usual way the mean of a finite mixture is formulated).
Z +<Ò] Ó œ IÒ Z +<Ò] l@Ó Ó  Z +<Ò IÒ] l@Ó Ó , where
IÒ Z +<Ò] l@Ó Ó œ Z +<Ò\" Ó † +  Z +<Ò\# Ó † Ð"  +Ñ and
Z +<Ò IÒ] l@Ó Ó œ ÐIÒ\" Ó  IÒ\# ÓÑ# † +Ð"  +Ñ .
The last equality follows from the fact that if ^ is a two-point random variable
? prob. :
, then Z +<Ò^Ó œ Ð?  @Ñ# † :Ð"  :Ñ; IÒ] l@Ó is
^ œš
@ prob. "  :
IÒ\" Ó prob. +
a two-point random variable IÒ] l@Ó œ š
.
IÒ\# Ó prob. "  +
Therefore Z +<Ò] Ó œ Z +<Ò\" Ó † +  Z +<Ò\# Ó † Ð"  +Ñ  ÐIÒ\"Ó  IÒ\#ÓÑ# † +Ð"  +Ñ .
B) If \" ß \# are Poisson random variables then IÒ\" Ó œ Z +<Ò\" Ó and IÒ\# Ó œ Z +<Ò\# Ó ,
so that IÒ] Ó œ IÒ\" Ó † +  IÒ\# Ó † Ð"  +Ñ œ Z +<Ò\"Ó † +  Z +<Ò\#Ó † Ð"  +Ñ .
It follows that
Z +<Ò] Ó œ 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 .
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MAY 2005 SOA EXAM C/CAS EXAM 4 SOLUTIONS
Therefore,
IÒ] Ó œ IÒ\" Ó † +  IÒ\# Ó † Ð"  +Ñ  Z +<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 œ 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] Ó œ IÒ\" Ó † +  IÒ\# Ó † Ð"  +Ñ  Z +<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 œ Z +<Ò] Ó for a mixture of binomials.
Answer: A
14. We wish to find IÒ\# l\" œ "!Ó . This can be formulated as
'!∞ IÒ\# l-Ó † 1Ð-l\" œ "!Ñ . - , where 1Ð-l\" œ "!Ñ is the posterior density of - given
0 Ð"!ß-Ñ
\" œ "! . The posterior density is 1Ð-l\" œ "!Ñ œ 0 Ð"!Ñ , where the numerator is the joint
\
density of \ and -, and the denominator is the marginal probability that \ œ "!.
- "!
†" -Î"#
The joint density is 0 Ð"!ß -Ñ œ 0 Ð"!l-Ñ † 1Ð-Ñ œ / "!x
† ÒÐ!Þ%Ñ "' /-Î'  Ð!Þ'Ñ "#
/
Ó.
Þ%
Þ'
This can be written as 0 Ð"!ß -Ñ œ '‚"!x
-"! /(-Î'  "#‚"!x
-"! /"$-Î"# .
The marginal probability 0\ Ð"!Ñ is
∞
∞
Þ%
Þ'
-"! /-Î'  "#‚"!x
-"! /-Î"# Ó . 0\ Ð"!Ñ œ '! 0 Ð"!ß -Ñ . - œ '! Ò '‚"!x
Þ%
"!x
Þ'
"!x
Þ%
' ""
Þ'
""
œ '‚"!x
† Ð(Î'Ñ
 "#
† Ð "#
""  "#‚"!x † Ð"$Î"#Ñ"" œ ' † Ð ( Ñ
"$ Ñ .
∞
We have used the relationship '! >5 /+> .> œ +5x
5" for integer 5 ! .
The posterior density is
Þ%
Þ'
' ""
Þ'
"# ""
1Ð-l\" œ "!Ñ œ Ò '‚"!x
-"! /(-Î'  "#‚"!x
-"! /"$-Î"# Ó‚Ò Þ%
' † Ð ( Ñ  "# † Ð "$ Ñ Ó .
∞
The Bayesian premium IÒ\# l\" œ "!Ó is '! - † 1Ð-l\" œ "!Ñ . - .
Using the integral relationship mentioned above, we have
'!∞ - † Ò Þ% -"! /(-Î'  Þ' -"! /"$-Î"# Ó . œ
'‚"!x
Þ%
""x
'‚"!x † Ð(Î'Ñ"#
Finally,
"#‚"!x
ÐÞ%ÑÐ""Ñ
ÐÞ'ÑÐ""Ñ
Þ'
""x
 "#‚"!x † Ð"$Î"#Ñ
† Ð '( Ñ"#  "#
"# œ
'
ÐÞ%ÑÐ""Ñ ' "# ÐÞ'ÑÐ""Ñ "# "#
†Ð Ñ 
†Ð Ñ
IÒ\# l\" œ "!Ó œ ' Þ% '( "" Þ' "#"# """$ œ *Þ)) .
' †Ð ( Ñ  "# †Ð "$ Ñ
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MAY 2005 SOA EXAM C/CAS EXAM 4 SOLUTIONS
14. continued
Another approach to solving this problem follows from a careful look at the algebraic form of the
Þ%
Þ'
joint density 0 Ð"!ß -Ñ œ '‚"!x
-"! /(-Î'  "#‚"!x
-"! /"$-Î"# . We know that the posterior
density is proportional to this (as far as the parameter - is concerned), so the posterior must be a
mixture of two gamma distributions. The first gamma in the posterior has α" œ "" ß )" œ '( , and
the second gamma has α# œ "" ß )# œ "#
"$ . Therefore, the posterior density will be of the form
"! (-Î'
"! "$-Î"#
+ † -Ð ' Ñ/"" †"!x  Ð"  +Ñ † -Ð "#/Ñ"" †"!x . Since the posterior density is proportional to the joint
(
"$
Þ'
Þ%
density, the ratio of the factors Ò "#‚"!x
ÓÎÒ '‚"!x
Ó œ %$ in the joint density, must be the same as
"$"" †'"" †Ð"+Ñ
+
the ratio of the factors Ò Ð "#"+
Ó
Î
Ò
Ó
œ
.
' ""
""
"#"" †("" †+
Ñ †"!x
Ð Ñ †"!x
"$
(
"" ""
†'
"
$
Therefore, "$
"#"" †("" † Ð +  "Ñ œ % , from which we get + œ Þ$("" ß "  + œ Þ'#)* .
∞
Then IÒ\# l\" œ "!Ó is '! - † 1Ð-l\" œ "!Ñ . - is the mean of the posterior distribution,
which is the mixture of the two means +α" )"  Ð"  +Ñα# )# œ *Þ)) .
Answer: D
s
s
15. The interval is LÐ%Þ&Ñ
„ "Þ*'ÉZ s
+<ÒLÐ%Þ&ÑÓ
.
There is no indication of any censoring or truncation (monitoring is from starting date of policy to
time of first claim), so the number at risk is reduced only by the numbers of claims at the various
s
claim time points. LÐ%Þ&Ñ
œ #  "  #  # œ Þ((%' (the most recent claim time before 4.5
"#
"!
*
(
s
is time 4). The Aalen estimate of the variance of LÐ%Þ&Ñ
is
#
"
#
#
s
Zs
+<ÒLÐ%Þ&ÑÓ
œ #  #  #  # œ Þ!)*% .
The interval is Þ((%' „ "Þ*'ÈÞ!)*% œ ÐÞ")* ß "Þ$'"Ñ .
"#
"!
*
(
Answer: A
16. For any estimator s) we have the relationship Q WIs) Ð)Ñ œ Z +<Ò s) Ó  Ò ,3+=s) Ð)ÑÓ# .
From (iv) it follows that Z +<Ò s) Ó œ Ò ,3+=s Ð)ÑÓ# .
)
Since s) œ
5
5" \
, it follows that
5
5
5
"
,3+=s) Ð)Ñ œ IÒs)Ó  ) œ IÒ 5"
\Ó  ) œ 5"
IÒ\Ó  ) œ 5"
)  ) œ  5"
).
#
5
5 #
5 # )
Also, Z +<Ò s) Ó œ Z +<Ò 5"
\Ó œ Ð 5"
Ñ Z +<Ò\Ó œ Ð 5"
Ñ † #& Þ
Therefore, from Z +<Ò s) Ó œ Ò ,3+=s) Ð)ÑÓ# , we get
5 # )#
"
Ð 5"
Ñ † #& œ Ð  5"
)Ñ# , from which 5 # œ #& , and 5 œ & . Answer: D
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17. There is 8 œ " sample value B, so 
B œ B , and the Buhlmann credibility estimate for the
number of claims in Year 2 is ^B  Ð"  ^Ñ. , where ^ œ "" @ .
+
The model distribution \l" has a geometric distribution and the parameter " has a Pareto
distribution with parameters α and ) œ ".
The hypothetical mean is IÒ\l" Ó œ " and the process variance is Z +<Ò\l" Ó œ " Ð"  " Ñ .
)
"
Then . œ IÒ IÒ\l" Ó Ó œ IÒ" Ó œ α"
œ α"
,
#
+ œ Z +<Ò IÒ\l" Ó œ Z +<Ò " Ó œ IÒ" Ó  ÐIÒ"ÓÑ#
#
#
#)
) #
α)
α
œ Ðα"ÑÐ
α#Ñ  Ò α" Ó œ Ðα"Ñ# Ðα#Ñ œ Ðα"Ñ# Ðα#Ñ .
@ œ IÒ Z +<Ò\l" Ó Ó œ IÒ"Ð"  "ÑÓ œ IÒ"Ó  IÒ" #Ó
#
)
#)
"
#
α
œ α"
 Ðα"ÑÐ
α#Ñ œ α"  Ðα"ÑÐα#Ñ œ Ðα"ÑÐα#Ñ Þ
α
α
"
"
Then +@ œ Ò Ðα"ÑÐ
α#Ñ Ó‚Ò Ðα"Ñ# Ðα#Ñ Ó œ α  " ß so that ^ œ " +@ œ α Þ
The Buhlmann credibility premium is
"
^B  Ð"  ^Ñ. œ α" † B  Ð"  α" ÑÐ α"
Ñ œ B"
α .
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
mean of ). The pdf for baseline time of accident is 0! Ð>Ñ œ ") />Î) , and the baseline survival
probability is W! Ð>Ñ œ />Î) . The hazard rate for individual 4 with factors ^" and ^# is
24 Ð>Ñ œ /"" ^" "# ^# † 2! Ð>Ñ . The survival probability for that individual 4 is
" ^" "# ^#
W4 Ð>Ñ œ ÒW! Ð>ÑÓ/ "
04 Ð>Ñ œ 
W4w Ð>Ñ
œ Ð/
, and the pdf for time of accident of that individual is
"" ^" "# ^#
" ^" "# ^# "
ÑÒW! Ð>ÑÓ/ "
† 0! Ð>Ñ .
There are four individuals considered. Individual 1 has ^" œ ! ß ^# œ #! 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" Ð$Ñ œ Ð/"" ^" "# ^# ÑÒW! Ð>ÑÓ/ "
œ Ð/
ÐÞ"ÑÐ!ÑÐÞ!"ÑÐ#!Ñ
ÑÒ/
† 0! Ð>Ñ
$Î") /ÐÞ"ÑÐ!ÑÐÞ!"ÑÐ#!Ñ "
Ó
" $Î")
† ")
/
;
the log of that density is
$
$
ÐÞ"ÑÐ!Ñ  ÐÞ!"ÑÐ#!Ñ  Ð/ÐÞ"ÑÐ!ÑÐÞ!"ÑÐ#!Ñ  "ÑÐ  ")
Ñ  68Ð")Ñ  ")
œ  #Þ)*% .
Individual 2 has ^" œ ! ß ^# œ $! and the accident occurs after time 6, so this is a right
censored observation. In the likelihood function we will include the survival probability W# Ð'Ñ,
and the log of that is
'
68ÒW# Ð'ÑÓ œ Ð/"" ^" "# ^# Ñ † 68ÒW! Ð'ÑÓ œ Ð/ÐÞ"ÑÐ!ÑÐÞ!"ÑÐ$!Ñ ÑÐ  ")
Ñ œ  Þ%&! .
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18. continued
Individual 3 has ^" œ " ß ^# œ $! and the accident is observed to occur at time 7. Since this is
an observed accident time, in the likelihood function, we would include the density
0$ Ð(Ñ œ Ð/ÐÞ"ÑÐ"ÑÐÞ!"ÑÐ$!Ñ ÑÒ/(Î") Ó/
ÐÞ"ÑÐ"ÑÐÞ!"ÑÐ$!Ñ
"
" (Î")
† ")
/
;
the log of that density is
(
(
ÐÞ"ÑÐ"Ñ  ÐÞ!"ÑÐ$!Ñ  Ð/ÐÞ"ÑÐ"ÑÐÞ!"ÑÐ$!Ñ  "ÑÐ  ")
Ñ  68Ð")Ñ  ")
œ  $Þ!(!& .
Individual 4 has ^" œ " ß ^# œ %! and the accident occurs after time 8, so this is a right
censored observation. In the likelihood function we will include the survival probability W% Ð)Ñ,
and the log of that is
)
68ÒW% Ð)ÑÓ œ Ð/"" ^" "# ^# Ñ † 68ÒW! Ð)ÑÓ œ Ð/ÐÞ"ÑÐ"ÑÐÞ!"ÑÐ%!Ñ ÑÐ  ")
Ñ œ  Þ($#) Þ
The total loglikelihood function value is  #Þ)*%  Þ%&!  $Þ!("  Þ($$ œ  (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 \ œ # (2 claims in year 1), so 8 œ " and \ œ #.
The hypothetical mean is IÒ\l)Ó œ Ð!ÑÐ#)Ñ  Ð"ÑÐ)Ñ  Ð#ÑÐ"  $)Ñ œ #  &) .
The process variance is
Z +<Ò\l)Ó œ IÒ\ # l)Ó  ÐIÒ\l)ÓÑ#
œ ÒÐ!# ÑÐ#)Ñ  Ð"# ÑÐ)Ñ  Ð## ÑÐ"  $)ÑÓ  Ð#  &)Ñ# œ *)  #&)# .
Then . œ IÒ IÒ\l)Ó Ó œ IÒ#  &)Ó œ #  &IÒ)Ó œ #  &ÒÐÞ!&ÑÐÞ)Ñ  ÐÞ$ÑÐÞ#ÑÓ œ "Þ& ,
+ œ Z +<Ò IÒ\l)Ó Ó œ Z +<Ò#  &)Ó œ #&Z +<Ò)Ó œ #&ÐÞ!&  Þ$Ñ#ÐÞ)ÑÐÞ#Ñ œ Þ#& , and
@ œ IÒ Z +<Ò\l)Ó Ó œ IÒ*)  #&)# Ó œ *IÒ)Ó  #&IÒ)# Ó
œ *ÐÞ"Ñ  #&ÒÐÞ!&Ñ# ÐÞ)Ñ  ÐÞ$Ñ# ÐÞ#ÑÓ œ Þ% .
Then ^ œ
"
Þ%
" Þ#&
œ Þ$)%' , and the Buhlmann credibility estimate is
ÐÞ$)%'ÑÐ#Ñ  Ð"  Þ$)%'ÑÐ"Þ&Ñ œ "Þ'*# .
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"
21. The prior distribution is gamma with α œ & and ) œ # .
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
αw œ DB3 and )w œ 8))" . We are given 8 œ # observed values, B" œ & and B# œ $ .
"Î#
"
Therefore, the posterior distribution has parameters αw œ &  ) ß )w œ Ð#Ð"Î#ÑÑ" œ % .
The mean of the posterior distribution is αw )w œ Ð"$ÑÐ "% Ñ œ $Þ#& .
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" œ " ; 5" ÐBÑ œ # È"  ÐB  "Ñ# ! Ÿ B Ÿ # .
1
The middle curve is the kernel density function for the sample point C# œ $ ;
5$ ÐBÑ œ # È"  ÐB  $Ñ# # Ÿ B Ÿ % .
1
The curve on the right is the kernel density function for the sample point C$ œ & ;
5& ÐBÑ œ # È"  ÐB  &Ñ# % Ÿ B Ÿ ' .
1
The empirical probabilities are :Ð"Ñ œ Þ#& ß :Ð$Ñ œ Þ& (2 sample values at 3), and :Ð&Ñ œ Þ#& .
$
The kernel density estimator is
4œ"
:ÐC4 Ñ5C4 ÐBÑ œ ÐÞ#&Ñ5" ÐBÑ  ÐÞ&Ñ5$ ÐBÑ  ÐÞ#&Ñ5&ÐBÑÞ
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
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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 15.5 on page 501 of the Loss Models
book.
Answer: D
#
)
2)
24. The first and second moments of the Pareto distribution are α"
and Ðα"ÑÐ
α#Ñ .
)
The sample estimate of the first and second moments are "$!#!$&!#")")##
œ &!) œ α"
&
#
#
#
#
#
2)#
and "$! #! $&!& #") ")## œ 701,401.6 œ Ðα"ÑÐ
α#Ñ .
Dividing then second expression by the square of the first expression results in
#
2)#
)
Ðα"ÑÐα#Ñ ‚Š α" ‹
#Ðα"Ñ
("!ß%!"Þ'
œ α# œ Ð&!)Ñ#
œ #Þ(") .
Then solving for α results in α œ %Þ()'. Solving for ) results in ) œ &!)Ðα  "Ñ œ "*#$Þ# .
The limited expected value with limit 500 is
)
)
IÒ\ • &!!Ó œ Š α"
‹Ò"  Š &!!
)‹
α"
"*#$Þ#
Ó œ Š "*#$Þ#
$Þ()' ‹Ò"  Š &!!"*#$Þ# ‹
$Þ()'
Ó œ #*' .
Answer: C
s" œ
25. The credibility factor is ^
7"
7"  ss+@
, where 7" œ ""! and s
+ is given as 651.03 Þ
# 83

"
734 Ð\34  \ 3 Ñ# •
s@ œ Ð8" "ÑÐ8# "Ñ † ”
3œ" 4œ"
œ
"
Ð#"ÑÐ#"Ñ
† Ò&!Ð#!!  ##(Þ#(Ñ#  '!Ð#&!  ##(Þ#(Ñ#
 "!!Ð"'!  "()Þ*&Ñ#  *!Ð#!!  "()Þ*&Ñ# Ó œ ("ß *)&Þ' .
s" œ
Then ^
""!
""! ("ß*)&Þ'
'&"Þ!$
œ Þ%** Þ
Answer: B
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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"!! Ð"ß !!!Ñ œ Þ"' . Similarly, J"!! Ð$ß !!!Ñ œ Þ$) ß J"!! Ð&ß !!!Ñ œ Þ'$ ß J"!! Ð"!ß !!!Ñ œ Þ)"ß
etc. We wish to find T Ò#ß !!!  \ 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"!! Ð#ß !!!Ñ œ Þ'''  Þ#(! œ Þ$*' . Answer: B
α†"!ß!!!α
27. The pdf for the Pareto with ) œ "!ß !!! is 0 ÐBÑ œ ÐB"!ß!!!Ñα"
and the cdf is J ÐBÑ œ "  Š B"!ß!!! ‹ , so that "  J ÐBÑ œ Š B"!ß!!! ‹ Þ
"!ß!!!
α
"!ß!!!
α
The log of the density is 68 0 ÐBÑ œ 68 α  α 68 "!ß !!!  Ðα  "Ñ 68ÐB  "!ß !!!Ñ ,
and the log of "  J ÐBÑ is 68Ð"  J ÐBÑÑ œ α 68Š B"!ß!!! ‹ .
"!ß!!!
The derivative with respect to α of 68 0 ÐBÑ is
.
.α
68 0 ÐBÑ œ α"  68 "!ß !!!  68ÐB  "!ß !!!Ñ ,
and the derivative with respect to α of 68Ð"  J ÐBÑÑ is 68Š B"!ß!!! ‹
"!ß!!!
For each of the 3 losses of amount 750 with deductible 200, the likelihood function has a factor of
0 Ð(&!Ñ
the conditional density 0 Ð(&!l\  #!!Ñ œ "J Ð#!!Ñ , and the derivative with respect to α of the
log of the conditional density is
.
"
. α Ò68 0 Ð(&!Ñ  68Ð"  J Ð#!!ÑÑÓ œ α
œ α"  68 "!ß #!!  68Ð"!ß (&!Ñ .
 68 "!ß !!!  68Ð"!ß (&!Ñ  68Š "!ß#!! ‹
"!ß!!!
For each of the 3 losses of amount 200 with no deductible (and policy limit 10,000), the
likelihood function has a factor of 0 Ð#!!Ñ , and the derivative of the log is
"
α
 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 0 Ð$!!Ñ , and the derivative of the log is
"
α
 68 "!ß !!!  68Ð"!ß $!!Ñ .
For each of the 6 losses above 10,000 with non deductible and policy limit 10,000, the likelihood
factor is "  J Ð"!ß !!!Ñ, and the derivative of the log of the likelihood factor is 68Š #!ß!!! ‹ .
"!ß!!!
For each of the 4 losses of amount 400 with deductible 300, the likelihood function has a factor of
0 Ð%!!Ñ
the conditional density 0 Ð%!!l\  $!!Ñ œ "J Ð$!!Ñ , and the derivative with respect to α of the
log of the conditional density is
.
"
. α Ò68 0 Ð%!!Ñ  68Ð"  J Ð$!!ÑÑÓ œ α
œ α"  68 "!ß $!!  68Ð"!ß %!!Ñ .
 68 "!ß !!!  68Ð"!ß %!!Ñ  68Š "!ß$!! ‹
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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 α:
$’ α"  68 "!ß #!!  68 "!ß (&!“  $’ α"  68 "!ß !!!  68Ð"!ß #!!Ñ“
 %’ α"  68 "!ß !!!  68Ð"!ß $!!Ñ“  ' ’68 "!ß !!!  68Ð#!ß !!!Ñ“
 %’ α"  68 "!ß $!!  68Ð"!ß %!!Ñ“
œ "%
α  "$ 68 "!ß !!!  $ 68 "!ß (&!  ' 68 #!ß !!!  % 68 "!ß %!! œ ! .
Solving for α results in α œ $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$!Ð"Ñ"&Ð#Ñ%Ð$Ñ"Ð%Ñ

.
œ Þ(' and
s œ s@ œ \ œ
"!!
#


"
"
+ œ "!!" ÒD\3#  "!!\ Ó  \ œ ** Ò"%#  "!!ÐÞ('Ñ# Ó  Þ(' œ Þ!*!* .
s
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
^œ
"
Þ('
" Þ!*!*
œ Þ"!') . The estimate for the expected number of claims in the next 2-year period
for a policyholder who had 1 in claim in the first 2-year period is
ÐÞ"!')ÑÐ"Ñ  Ð"  Þ"!')ÑÐÞ('Ñ œ 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 œ Þ$*$ Þ
Answer: C
29. The baseline hazard rate is parametric so we can formulate a full (no partial) likelihood
#
>
function. We find L! Ð>Ñ œ '! 2! ÐBÑ .B œ > Þ Then for baseline survival, the survival
)
probability is W! Ð>Ñ œ /
L! Ð>Ñ
œ/
># Î)
># Î)
, and the density is 0! Ð>Ñ œ  W!w Ð>Ñ œ #>
. Baseline
)/
is loss amount distribution for Class A. For Class B,
"
># / "
># /" Î)
the hazard rate is 2F ÐBÑ œ /" 2! ÐBÑ œ #B/
,
) , and LF Ð>Ñ œ ) ß WF Ð>Ñ œ /
"
#
"
> / Î)
and 0F Ð>Ñ œ #>/
. The likelihood function for the given data, with Class A losses of 1
) /
and 3 and Class B losses of 2 and 4 is
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29. continued
"
"
"
"
P œ 0! Ð"Ñ † 0! Ð$Ñ † 0F Ð#Ñ † 0F Ð%Ñ œ #) /"Î) † )' /*Î) † %/) /%/ Î) † )/) /"'/ Î)
"
œ $)%
/#" Ð"!#!/ ÑÎ) Þ
)%
"
The loglikelihood is 68 P œ 68 $)%  %68 )  #"  "!#!/
.
)
Taking partial derivatives and setting equal to 0, we get
"
`
`)
"
68 P œ  %)  "!#!/
œ ! and ``" 68 P œ #  #!/
œ!.
)#
)
From the second equation we have ) œ "!/" , and then substituting this into the first equation,
"
%
"!#!/
we have  "!/
"  "!!/#" œ ! .
Solving for /" results in /" œ "# , so that " œ 68 "# œ  Þ'*$ Þ Answer: A
30. Notice that at the three knots we have 2Ð  #Ñ œ  $# ß 2Ð!Ñ œ ! ß 2Ð#Ñ œ $# ,
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 C œ 7B  5 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
Ò  # ß !Ó is of the form
0! ÐBÑ œ +!  ,! ÐB  #Ñ  -! ÐB  #Ñ#  .! ÐB  #Ñ$ , and 0!ww ÐBÑ œ #-!  '.!ÐB  #Ñ .
For a natural spline one 3 knots, we have 7! œ 7# œ ! and
#Ð2!  2" Ñ7" œ 'Š #2
"
C C"

C" C!
2! ‹
, so that #Ð#  #Ñ7" œ 'Š $#!
# 
7
Solving for 7" results in 7" œ ! . We use the relationships -4 œ #4
to get -! œ ! and .! œ ! . Therefore, 0!ww Ð  !Þ&Ñ œ ! .
!Ð$#Ñ
‹.
#
7 74
and .4 œ 4"
'24
Answer: C
Þ#
Þ) ÐBÎ)Ñ
31. The pdf of this Weibull with 7 œ Þ# is 0 ÐBÑ œ Þ#B )/Þ#
, its log is
B Þ#
68 0 ÐBÑ œ 68 Þ#  Þ) 68 B  Þ# 68 )  Ð ) Ñ , and the derivative of 68 0 ÐBÑ with respect to ) is
.
.)
Þ#
Þ#B
68 0 ÐBÑ œ  Þ#
)  )"Þ# .
Þ#
The derivative of 68Ð"  J ÐBÑÑ is ..) 68Ð"  J ÐBÑÑ œ ..) Ò  Ð B) ÑÞ# Ó œ Þ#B
)"Þ# .
For the given data points, the derivative of the loglikelihood function is the sum of the derivatives
of 68 0 ÐBÑ for the four given claim amounts plus 2 ‚ the derivative of 68Ò"  J Ð"!!!ÑÓ .
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MAY 2005 SOA EXAM C/CAS EXAM 4 SOLUTIONS
31. continued
.
.)
Þ#‚"$!Þ#
Þ#
Þ#‚#%!Þ#
68 P œ  Þ#



 Þ#
"Þ#
)
)
)
)"Þ#
)
Þ#
Þ#
Þ#‚&%!Þ#
Þ#‚"!!!Þ#
 Þ#‚$!!



#
†
œ!
)"Þ#
)
)"Þ#
)"Þ#
(note that the factor .2 cancels from all terms).
Solving for ) results in the mle estimate ) œ $$#' .
An alternative solution makes use of the relation between the Weibull and exponential
distributions. If \ has a Weibull distribution with 7 known and ) unknown, then ^ œ \ 7 has
an exponential distribution with mean )7 . If B" ß ß ß ÞB8 are sample values from a Weibull
distribution with 7 known, then D" œ B7" ß Þ Þ Þ ß D8 œ B78 behave like a sample from an
exponential distribution with mean )7 (this applies to known B-amounts and limit amounts also).
We are given 7 œ Þ#, so the values "$!Þ# ß #%!Þ# ß $!!Þ# ß &%!Þ# and the two limit values of "!!!Þ#
are from an exponential distribution with mean )Þ# . When a sample from an exponential
distribution has uncensored values and censored (limit) values, the mle of the exponential mean
sum of all observed values (including limit values)
. In this case, the mle of )Þ# is
number of uncensored values
"$!Þ# #%!Þ# $!!Þ# &%!Þ# #‚"!!!Þ#
œ &Þ!'$ , so that the mle of ) is Ð&Þ!'$Ñ& œ
%
is
$$#( .
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\" œ 8Ó œ '" - † 1Ð-l\" œ 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&
first year is 0, then Ð"  ^Ñ. œ Þ& . Since " Ÿ . Ÿ % , it follows that Þ&
% Ÿ . Ÿ Þ& , and then
since "  ^ œ Þ&
. , we have Þ& Ÿ ^ Ÿ Þ)(& . But then, if \ œ * 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 \ œ * in the first year.
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MAY 2005 SOA EXAM C/CAS EXAM 4 SOLUTIONS
33. The hypothesized probabilities for the intervals are J Ð#Ñ œ Þ!$& ß J Ð&Ñ œ Þ"$! ß
J Ð(Ñ œ Þ'$! ß J Ð)Ñ œ Þ)$! ß J Ð∞Ñ œ ". Therefore, T Ò\ Ÿ #Ó œ Þ!$& ß T Ò#  \ Ÿ &Ó œ Þ!*&ß
T Ò&  \ Ÿ (Ó œ Þ&!! ß T Ò(  \ Ÿ )Ó œ Þ#!! ß T Ò)  \Ó œ Þ"(! . For the 300 observations,
the expected numbers for each interval, based on the hypothesized distribution are
Interval
Ò! ß #Ó
Ð# ß &Ó
Ð& ß (Ó
Ð( ß )Ó
Ð) ß ∞Ñ
Expected Number of Observations
$!!ÐÞ!$&Ñ œ "!Þ&
$!!ÐÞ!*&Ñ œ #)Þ&
$!!ÐÞ&!!Ñ œ "&!
$!!ÐÞ#!!Ñ œ '!
$!!ÐÞ"(!Ñ œ &"
Actual Number of Observations
&
%#
"$(
''
&!
The chi-square statistic is
Uœ
Ð&"!Þ&Ñ#
"!Þ&

Ð%##)Þ&Ñ#
#)Þ&

Ð"$("&!Ñ#
"&!

Ð'''!Ñ#
'!

Ð&"&!Ñ#
&"
œ ""Þ!## .
The degrees of freedom in this chi-square test is &  " œ % (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 œ *Þ%)) (95-th percentile) and ;Þ!#&
Ð%Ñ œ ""Þ"%$ (97.5-percentile).
The null hypothesis that the given data comes from the known distribution is rejected at the 5%
level of significance because U œ ""Þ!##  *Þ%)), but the hypothesis is no rejected at the 2.5%
level of significance.
Answer: C
34. The cdf of the lognormal distribution is J ÐBÑ œ FŠ
68 B  .
‹
5
. According to the inversion
method, given uniform number ?, we solve for B from ? œ J ÐBÑ œ FŠ
68 B  .
‹
5
.
 &Þ'
? œ Þ'"(* œ FŠ 68 BÞ(&
‹ ; from the standard normal distribution table, we see that
 &Þ'
FÐÞ$Ñ œ Þ'"(* ; therefore, 68 BÞ(&
œ Þ$ , so that the simulated value of B is 338.66 .
? œ Þ%'!# œ "  Þ&$*) and Þ&$*) œ FÐÞ"Ñ , so that Þ%'!# œ FÐ  Þ"Ñ, and then
68 B  &Þ'
Þ(&
œ  Þ" and the simulated value of B is 250.89 .
? œ Þ*%&# œ FÐ"Þ'Ñ p B œ )*(Þ)& , so the policy limit of 400 is applied.
? œ Þ!)!) œ FÐ  "Þ'%Ñ p B œ *%Þ'$ .
? œ Þ())" œ FÐÞ)Ñ p B œ %*#Þ(& , and the policy limit of 400 is applied.
? œ Þ%#!( œ FÐ  Þ#Ñ p B œ #$#Þ(' .
The average payment per claim is $$)Þ''#&!Þ)*%!!*%Þ'$%!!#$#Þ('
œ #)' .
'
Answer: A
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35. We wish to find IÒ\# l\" œ #Ó .
Using the Bayesian methodology, this can be formulated as
IÒ\# l" œ #Ó † T Ò" œ #l\" œ #Ó  IÒ\# l" œ &Ó † T Ò" œ &l\" œ #Ó .
The mean of the geometric distribution with parameter " is " , so
IÒ\# l" œ #Ó œ # and IÒ\# l" œ &Ó œ & .
The conditional probabilities can be found in the following way:
T Ð" œ #Ñ œ "$
T Ð" œ &Ñ œ #$
#
#
#
%
T Ò\" œ #l" œ #Ó œ Ð"#Ñ
$ œ #(
&
#&
T Ò\" œ #l" œ &Ó œ Ð"&Ñ
$ œ #"'
Ì
Ì
T Ò\" œ # ∩ " œ #Ó
T Ò\" œ # ∩ " œ &Ó
œ T Ò\" œ #l" œ #Ó † T Ð" œ #Ñ œ
%
)"
&!
œ T Ò\" œ #l" œ &Ó † T Ð" œ &Ñ œ '%)
Ì
%
&!
T Ò\" œ #Ó œ T Ò\" œ # ∩ " œ #Ó  T Ò\" œ # ∩ " œ &Ó œ )"
 '%)
œ Þ"#'&%
Ì
T Ò" œ #l\" œ #Ó œ
T Ò\" œ#∩" œ#Ó
T Ò\" œ#Ó
œ Þ"#'&% œ Þ$*!
%Î)"
T Ò\" œ#∩" œ&Ó
T Ò\" œ#Ó
œ Þ"#'&% œ Þ'"! .
and
T Ò" œ #l\" œ &Ó œ
&!Î'%)
Then, IÒ\# l\" œ #Ó œ Ð#ÑÐÞ$*Ñ  Ð&ÑÐÞ'"Ñ œ $Þ)$ .
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Answer: C
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