BjornvanBraak/ActuarialMathBench huggingface dataset:
Question: For a particular policy, the conditional probability of the annual number of claims given \(\Theta=\theta\), and the probability distribution of \(\Theta\) are as follows:

\begin{tabular}{|l|c|c|c|}
\hline Number of claims & 0 & 1 & 2 \\
\hline Probability & \(2 \theta\) & \(\theta\) & \(1-3 \theta\) \\
\hline
\end{tabular}

\begin{tabular}{|l|l|l|}
\hline\(\theta\) & 0.05 & 0.30 \\
\hline Probability & 0.80 & 0.20 \\
\hline
\end{tabular}

Two claims are observed in Year 1.
Calculate the Bühlmann credibility estimate of the number of claims in Year 2.

Options:
A
Less than 1.68
B
At least 1.68 , but less than 1.70
C
At least 1.70 , but less than 1.72
D
At least 1.72 , but less than 1.74
E
At least 1.74

Correct answer: B

Solution: \(E(\theta)=0.05(0.8)+0.3(0.2)=0.1\),
\(E\left(\theta^{2}\right)=0.05^{2}(0.8)+0.3^{2}(0.2)=0.02\),
\(\mu(\theta)=0(2 \theta)+1(\theta)+2(1-3 \theta)=2-5 \theta\),
\(v(\theta)=0^{2}(2 \theta)+1^{2}(\theta)+2^{2}(1-3 \theta)-(2-5 \theta)^{2}=9 \theta-25 \theta^{2}\),
\(\mu=E(2-5 \theta)=2-5(0.1)=1.5\),
\(v=E P V=E\left(9 \theta-25 \theta^{2}\right)=9(0.1)-25(0.02)=0.4\),
\(a=\operatorname{VHM}=\operatorname{Var}(2-5 \theta)=25 \operatorname{Var}(\theta)=25\left(0.02-0.1^{2}\right)=0.25\),
\(k=v / a=0.4 / 0.25=1.6, Z=\frac{1}{1+1.6}=\frac{5}{13}\),
\(P=\frac{5}{13} 2+\frac{8}{13} 1.5=1.6923\).
Question: You are given:
(i) The annual number of claims for a policyholder follows a Poisson distribution with mean \(\Lambda\).
(ii) The prior distribution of \(\Lambda\) is gamma with probability density function:
\[
f(\lambda)=\frac{(2 \lambda)^{5} e^{-2 \lambda}}{24 \lambda}, \quad \lambda>0
\]

An insured is selected at random and observed to have \(x_{1}=5\) claims during Year 1 and \(x_{2}=3\) claims during Year 2.

Calculate \(E\left[\Lambda \mid x_{1}=5, x_{2}=3\right]\).

Options:
A
\(\quad 3.00\)
B
3.25
C
3.50
D
3.75
E
4.00

Correct answer: B

Solution: \(f(\lambda \mid 5,3) \propto \frac{e^{-\lambda} \lambda}{5!} \frac{e^{-\lambda} \lambda^{3}}{3!} \frac{2^{5} \lambda^{5} e^{-2 \lambda}}{24 \lambda} \propto \lambda^{12} e^{-4 \lambda}\). This is a gamma distribution with parameters 13 and 0.25 . The expected value is \(13(0.25)=3.25\).

Alternatively, if the Poisson-gamma relationships are known, begin with the prior parameters \(\alpha=5\) and \(\beta=2\) where \(\beta=1 / \theta\) if the parameterization from Loss Models is considered. Then the posterior parameters are \(\alpha^{\prime}=5+5+3=13\) where the second 5 and the 3 are the observations and \(\beta^{\prime}=2+2=4\) where the second 2 is the number of observations. The posterior mean is then \(13 / 4=3.25\).
Question: You are given:
(i) The annual number of claims on a given policy has the geometric distribution with parameter \(\beta\).
(ii) One-third of the policies have \(\beta=2\), and the remaining two-thirds have \(\beta=5\).

A randomly selected policy had two claims in Year 1.
Calculate the Bayesian expected number of claims for the selected policy in Year 2.

Options:
A
3.4
B
3.6
C
3.8
D
4.0
E
4.2

Correct answer: C

Solution: For the geometric distribution, \(\operatorname{Pr}\left(X_{1}=2 \mid \beta\right)=\frac{\beta^{2}}{(1+\beta)^{3}}\) and the expected value is \(\beta\).
\(\operatorname{Pr}\left(\beta=2 \mid X_{1}=2\right)=\frac{\operatorname{Pr}\left(X_{1}=2 \mid \beta=2\right) \operatorname{Pr}(\beta=2)}{\operatorname{Pr}\left(X_{1}=2 \mid \beta=2\right) \operatorname{Pr}(\beta=2)+\operatorname{Pr}\left(X_{1}=2 \mid \beta=5\right) \operatorname{Pr}(\beta=5)}\)
\(=\frac{\frac{4}{27} \frac{1}{3}}{\frac{4}{27} \frac{1}{3}+\frac{25}{216} \frac{2}{3}}=0.39024\).
The expected value is then \(0.39024(2)+0.60976(5)=3.83\).
Question: The length of time, in years, that a person will remember an actuarial statistic is modeled by an exponential distribution with mean \(1 / Y\). In a certain population, \(Y\) has a gamma distribution with \(\alpha=\theta=2\).

Calculate the probability that a person drawn at random from this population will remember an actuarial statistic less than \(1 / 2\) year.

Options:
A
0.125
B
0.250
C
0.500
D
0.750
E
0.875

Correct answer: D

Solution: Let \(X\) be the random variable for when the statistic is forgotten. Then \(F_{X}(x \mid y)=1-e^{-x y}\) For the unconditional distribution of \(X\), integrate with respect to y
\(F_{X}(x)=\int_{0}^{\infty}\left(1-e^{-x y}\right) \frac{1}{\Gamma(2) y}\left(\frac{y}{2}\right)^{2} e^{-y / 2} d y=1-\frac{1}{4} \int_{0}^{\infty} y e^{-y(x+1 / 2)} d y=1-\frac{1}{4(x+1 / 2)^{2}}\)
\(F(1 / 2)=1-\frac{1}{4(1 / 2+1 / 2)^{2}}=0.75\)
There are various ways to evaluate the integral (1) Integration by parts. (2) Recognize that \(\int_{0}^{\infty} y(x+1 / 2) e^{-y(x+1 / 2)} d y\) is the expected value of an exponential random variable with mean \((x+1 / 2)^{-1}\). (3) Recognize that \(\Gamma(2)(x+1 / 2)^{2} y e^{-y(x+1 / 2)}\) is the density function for a gamma random variable with \(\alpha=2\) and \(\theta=(x+1 / 2)^{-1}\), so it would integrate to 1 .
Question: In a given week, the number of projects that require you to work overtime has a geometric distribution with \(\beta=2\). For each project, the distribution of the number of overtime hours in the week is the following:

\begin{tabular}{ccc}
\(x\) & & \(f(x)\) \\
\cline { 1 - 1 } \cline { 3 - 3 } 5 & & 0.2 \\
10 & & 0.3 \\
20 & & 0.5
\end{tabular}

The number of projects and number of overtime hours are independent. You will get paid for overtime hours in excess of 15 hours in the week.

Calculate the expected number of overtime hours for which you will get paid in the week.

Options:
A
18.5
B
18.8
C
22.1
D
26.2
E
28.0

Correct answer: B

Solution: Frequency is geometric with \(\beta=2\), so \(p_{0}=1 / 3, p_{1}=2 / 9, p_{2}=4 / 27\).

Convolutions of \(f_{X}(x)\) needed are

\begin{tabular}{ccc}
\(x\) & \(f(x)\) & \(f^{* 2}(x)\) \\
\hline 5 & 0.2 & 0 \\
10 & 0.3 & 0.04
\end{tabular}
\[
\begin{aligned}
& f_{S}(0)=1 / 3=0.333, f_{S}(5)=(2 / 9)(0.2)=0.044, f_{S}(10)=(2 / 9)(0.3)+(4 / 27)(0.04)=0.073 \\
& E(X)=0.2(5)+0.3(10)+0.5(20)=14 \\
& E(S)=2 E(X)=28 \\
& E\left[(S-15)_{+}\right]=E(S)-5[1-F(0)]-5[1-F(5)]-5[1-F(10)] \\
& \quad=28-5(1-0.333)-5(1-0.333-0.044)-5(1-0.333-0.044-0.073)=18.81
\end{aligned}
\]

Alternatively,
\[
\begin{aligned}
& E\left[(S-15)_{+}\right]=E(S)-15+15 f_{S}(0)+10 f_{S}(5)+5 f_{S}(10) \\
& \quad=28-15+15(0.333)+10(0.044)+5(0.073)=18.81
\end{aligned}
\]
Question: For an insurance portfolio:
(i) The number of claims has the probability distribution

\begin{tabular}{ccc}
\(n\) & \(p_{n}\) \\
\hline 0 & & 0.1 \\
1 & & 0.4 \\
2 & & 0.3 \\
3 & & 0.2
\end{tabular}
(ii) Each claim amount has a Poisson distribution with mean 3; and
(iii) The number of claims and claim amounts are mutually independent. Calculate the variance of aggregate claims.

Options:
A
4.8
B
\(\quad 6.4\)
C
8.0
D
10.2
E
\(\quad 12.4\)

Correct answer: E

Solution: \begin{tabular}{|l|l|l|l|}
\hline N & \(p_{n}\) & \(n p_{n}\) & \(n^{2} p_{n}\) \\
\hline 0 & 0.1 & 0 & 0 \\
\hline 1 & 0.4 & 0.4 & 0.4 \\
\hline 2 & 0.3 & 0.6 & 1.2 \\
\hline 3 & 0.2 & 0.6 & 1.8 \\
\hline & & \(E(N)=1.6\) & \(E\left(N^{2}\right)=3.4\) \\
\hline
\end{tabular}
\(\operatorname{Var}(N)=3.4-1.6^{2}=0.84\)
\(E(X)=\operatorname{Var}(X)=\lambda=3\)
\(\operatorname{Var}(S)=E(N) \operatorname{Var}(X)+E(X)^{2} \operatorname{Var}(N)=1.6(3)+3^{2}(0.84)=12.36\)
Question: You are given:
(i) The conditional distribution of the number of claims per policyholder is Poisson with mean \(\lambda\).
(ii) The variable \(\lambda\) has a gamma distribution with parameters \(\alpha\) and \(\theta\).
(iii) For policyholders with 1 claim in Year 1, the credibility estimate for the number of claims in Year 2 is 0.15 .
(iv) For policyholders with an average of 2 claims per year in Year 1 and Year 2, the credibility estimate for the number of claims in Year 3 is 0.20 .

Calculate \(\theta\).

Options:
A
Less than 0.02
B
At least 0.02 , but less than 0.03
C
At least 0.03 , but less than 0.04
D
At least 0.04 , but less than 0.05
E
At least 0.05

Correct answer: A

Solution: \(E(X \mid \lambda)=\operatorname{Var}(X \mid \lambda)=\lambda\)
\(\mu=v=E(\lambda)=\alpha \theta ; a=\operatorname{Var}(\lambda)=\alpha \theta^{2} ; k=v / a=1 / \theta\)
\(Z=\frac{n}{n+1 / \theta}=\frac{n \theta}{n \theta+1}\)
\(0.15=\frac{\theta}{\theta+1}(1)+\frac{1}{\theta+1} \mu=\frac{\theta+\mu}{\theta+1}\)
\(0.20=\frac{2 \theta}{2 \theta+1}(2)+\frac{1}{2 \theta+1} \mu=\frac{4 \theta+\mu}{2 \theta+1}\)
From the first equation,
\(0.15 \theta+0.15=\theta+\mu\) and so \(\mu=0.15-0.85 \theta\)
Then the second equation becomes
\(0.4 \theta+0.2=4 \theta+0.15-0.85 \theta\)
\(0.05=2.75 \theta ; \theta=0.01818\)
Question: You are given:
(i) A region is comprised of three territories. Claims experience for Year 1 is as follows:

\begin{tabular}{|c|c|c|}
\hline Territory & Number of Insureds & Number of Claims \\
\hline A & 10 & 4 \\
\hline B & 20 & 5 \\
\hline C & 30 & 3 \\
\hline
\end{tabular}
(ii) The number of claims for each insured each year has a Poisson distribution.
(iii) Each insured in a territory has the same expected claim frequency.
(iv) The number of insureds is constant over time for each territory.

Calculate the Bühlmann-Straub empirical Bayes estimate of the credibility factor \(Z\) for Territory A.

Options:
A
Less than 0.4
B
At least 0.4 , but less than 0.5
C
At least 0.5 , but less than 0.6
D
At least 0.6 , but less than 0.7
E
At least 0.7

Correct answer: A

Solution: \(\hat{\mu}=\bar{x}=12 / 60=0.2, E V P V=\hat{v}=\bar{x}=0.2\)
\(V H M=\hat{a}=\frac{10(0.4-0.2)^{2}+20(0.25-0.2)^{2}+30(0.1-0.2)^{2}-(3-1)(0.2)}{60-\frac{10^{2}+20^{2}+30^{2}}{60}}=0.009545\)
\(\hat{k}=20.9524 ; \quad Z=\frac{10}{10+20.9524}=0.323\)
Question: An insurance company sells three types of policies with the following characteristics:

\begin{tabular}{|c|c|c|}
\hline Type of Policy & \begin{tabular}{c} 
Proportion of Total \\
Policies
\end{tabular} & Annual Claim Frequency \\
\hline I & \(5 \%\) & Poisson with mean 0.25 \\
\hline II & \(20 \%\) & Poisson with mean 0.50 \\
\hline III & \(75 \%\) & Poisson with mean 1.00 \\
\hline
\end{tabular}

A randomly selected policyholder is observed to have a total of one claim for Year 1 through
Year 4.

For the same policyholder, calculate the Bayesian estimate of the expected number of claims in Year 5.

Options:
A
Less than 0.4
B
At least 0.4 , but less than 0.5
C
At least 0.5 , but less than 0.6
D
At least 0.6 , but less than 0.7
E
At least 0.7

Correct answer: D

Solution: Let \(E\) be the even of having 1 claim in the first four years. In four years, the total number of claims is Poisson(4 \(\lambda\) ).
\(\operatorname{Pr}(\) Type \(I \mid E)=\frac{\operatorname{Pr}(E \mid \text { Type I }) \operatorname{Pr}(\text { Type I })}{\operatorname{Pr}(E)}=\frac{e^{-1}(0.05)}{\operatorname{Pr}(E)}=\frac{0.01839}{\operatorname{Pr}(E)}=0.14427\)
\(\operatorname{Pr}(\) Type II \(\mid E)=\frac{e^{-2}(2)(0.2)}{\operatorname{Pr}(E)}=\frac{0.05413}{\operatorname{Pr}(E)}=0.42465\)
\(\operatorname{Pr}(\) Type III \(\mid E)=\frac{e^{-4}(4)(0.75)}{\operatorname{Pr}(E)}=\frac{0.05495}{\operatorname{Pr}(E)}=0.43108\)
Note: \(\operatorname{Pr}(E)=0.01839+.05413+.05495=0.12747\)
The Bayesian estimate of the number of claims in Year 5 is:
\(0.14427(0.25)+0.42465(0.5)+0.43108(1)=0.67947\).
Question: You have observed the following three loss amounts:
\[
\begin{array}{lll}
186 & 91 & 66
\end{array}
\]

Seven other amounts are known to be less than or equal to 60. Losses follow an inverse exponential with distribution function
\[
F(x)=e^{-\theta / x}, \quad x>0
\]

Calculate the maximum likelihood estimate of the population mode.

Options:
A
Less than 11
B
At least 11, but less than 16
C
At least 16, but less than 21
D
At least 21, but less than 26
E
At least 26

Correct answer: A

Solution: The density function is \(f(x)=\theta x^{-2} e^{-\theta / x}\) and the likelihood function is
\(L(\theta)=\theta\left(186^{-2}\right) e^{-\theta / 186} \theta\left(91^{-2}\right) e^{-\theta / 91} \theta\left(66^{-2}\right) e^{-\theta / 66}\left(e^{-\theta / 60}\right)^{7} \propto \theta^{3} e^{-0.148184 \theta}\)
\(l(\theta)=\ln L(\theta)=3 \ln (\theta)-0.148184 \theta\)
\(l^{\prime}(\theta)=3 \theta^{-1}-0.148184=0\)
\(\theta=3 / 0.148184=20.25\).
The mode is \(\theta / 2=20.25 / 2=10.125\).
Question: For a group of policies, you are given:
(i) The annual loss on an individual policy follows a gamma distribution with parameters \(\alpha=4\) and \(\theta\).
(ii) The prior distribution of \(\theta\) has mean 600 .
(iii) A randomly selected policy had losses of 1400 in Year 1 and 1900 in Year 2.
(iv) Loss data for Year 3 was misfiled and unavailable.
(v) Based on the data in (iii), the Bühlmann credibility estimate of the loss on the selected policy in Year 4 is 1800.
(vi) After the estimate in (v) was calculated, the data for Year 3 was located. The loss on the selected policy in Year 3 was 2763.

Calculate the Bühlmann credibility estimate of the loss on the selected policy in Year 4 based on the data for Years 1, 2 and 3.

Options:
A
Less than 1850
B
At least 1850 , but less than 1950
C
At least 1950, but less than 2050
D
At least 2050, but less than 2150
E
At least 2150

Correct answer: D

Solution: We have \(\mu(\theta)=4 \theta\) and \(\mu=4 E(\theta)=4(600)=2400\). The average loss for Years 1 and 2 is 1650 and so \(1800=Z(1650)+(1-Z)(2400)\) which gives \(Z=0.8\). Because there were two years, \(Z=0.8=2 /(2+k)\) which gives \(k=0.5\).

For three years, the revised value is \(Z=3 /(3+0.5)=6 / 7\) and the revised credibility estimate (using the new sample mean of 2021), (6/7)(2021) \(+(1 / 7)(2400)=2075.14\).
Question: You are given:
(i) A portfolio consists of 100 identically and independently distributed risks.
(ii) The number of claims for each risk follows a Poisson distribution with mean \(\lambda\).
(iii) The prior distribution of \(\lambda\) is:
\[
\pi(\lambda)=\frac{(50 \lambda)^{4} e^{-50 \lambda}}{6 \lambda}, \quad \lambda>0
\]

During Year 1, the following loss experience is observed:

\begin{tabular}{|c|c|}
\hline Number of Claims & Number of Risks \\
\hline 0 & 90 \\
\hline 1 & 7 \\
\hline 2 & 2 \\
\hline 3 & 1 \\
\hline Total & 100 \\
\hline
\end{tabular}

Calculate the Bayesian expected number of claims for the portfolio in Year 2.

Options:
A
8
B
10
C
11
D
12
E
14

Correct answer: D

Solution: The posterior distribution is
\(\pi(\lambda \mid\) data \() \propto\left(e^{-\lambda}\right)^{90}\left(\lambda e^{-\lambda}\right)^{7}\left(\lambda^{2} e^{-\lambda}\right)^{2}\left(\lambda^{3} e^{-\lambda}\right) \frac{\lambda^{4} e^{-50 \lambda}}{\lambda}=\lambda^{17} e^{-150 \lambda}\) which is a gamma distribution with parameters 18 and 1/150. For one risk, the estimated value is the mean, 18/150. For 100 risks it is \(100(18) / 150=12\).

Alternatively,
The prior distribution is gamma with \(\alpha=4\) and \(\beta=50\). The posterior will continue to be gamma, with \(\alpha^{\prime}=\alpha+\) no. of claims \(=4+14=18\) and \(\beta^{\prime}=\beta+\) no. of exposures \(=50+100=150\). Mean of the posterior is \(\alpha / \beta=18 / 150=0.12\).
Expected number of claims for the portfolio is \(0.12(100)=12\).
Question: (i) The annual number of claims for an individual risk follows a Poisson distribution with mean \(\lambda\).
(ii) For \(75 \%\) of the risks, \(\lambda=1\).
(iii) For \(25 \%\) of the risks, \(\lambda=3\).

A randomly selected risk had \({ }_{r}\) claims in Year 1. The Bayesian estimate of this risk's expected number of claims in Year 2 is 2.98 .

Calculate the Bühlmann credibility estimate of the expected number of claims for this risk in Year 2.

Options:
A
Less than 1.9
B
At least 1.9, but less than 2.3
C
At least 2.3 , but less than 2.7
D
At least 2.7, but less than 3.1
E
At least 3.1

Correct answer: E

Solution: \[
\begin{aligned}
\operatorname{Pr}(\lambda & \left.=1 \mid X_{1}=r\right)=\frac{\operatorname{Pr}\left(X_{1}=r \mid \lambda=1\right) \operatorname{Pr}(\lambda=1)}{\operatorname{Pr}\left(X_{1}=r \mid \lambda=1\right) \operatorname{Pr}(\lambda=1)+\operatorname{Pr}\left(X_{1}=r \mid \lambda=3\right) \operatorname{Pr}(\lambda=3)} \\
& =\frac{\frac{e^{-1}}{r!}(0.75)}{\frac{e^{-1}}{r!}(0.75)+\frac{e^{-3} 3^{r}}{r!}(0.25)}=\frac{0.2759}{0.2759+0.1245\left(3^{r}\right)} .
\end{aligned}
\]

Then,
\(2.98=\frac{0.2759}{0.2759+0.1245\left(3^{r}\right)}(1)+\frac{0.1245\left(3^{r}\right)}{0.2759+0.1245\left(3^{r}\right)}(3)=\frac{0.2759+0.3735\left(3^{r}\right)}{0.2759+0.1245\left(3^{r}\right)}\).
Rearrange to obtain
\(0.82218+0.037103\left(3^{r}\right)=0.2759+0.03735\left(3^{r}\right)\)
\(0.54628=0.00025\left(3^{r}\right)\)
\(r=7\).
Because the risks are Poisson:
\(\mu=v=E(\lambda)=0.75(1)+0.25(3)=1.5\)
\(a=\operatorname{Var}(\lambda)=0.75(1)+0.25(9)-2.25=0.75\)
\(Z=\frac{1}{1+1.5 / 0.75}=1 / 3\)
and the estimate is \((1 / 3)(7)+(2 / 3)(1.5)=3.33\).
Question: You are given:
(i) The number of claims made by an individual in any given year has a binomial distribution with parameters \(m=4\) and \(q\).
(ii) The prior distribution of \(q\) has probability density function
\[
\pi(q)=6 q(1-q), \quad 0<q<1 .
\]
(iii) Two claims are made in a given year.

Calculate the mode of the posterior distribution of \(q\).

Options:
A
0.17
B
0.33
C
0.50
D
0.67
E
0.83

Correct answer: C

Solution: \(\pi(q \mid 2)=6 q^{2}(1-q)^{2} 6 q(1-q) \propto q^{3}(1-q)^{3}\)
The mode can be determined by setting the derivative equal to zero.
\(\pi^{\prime}(q \mid 2) \propto 3 q^{2}(1-q)^{3}-3 q^{3}(1-q)^{2}=0\)
\((1-q)-q=0\)
\(q=0.5\).
Question: A compound Poisson distribution has \(\lambda=5\) and claim amount distribution as follows:

\begin{tabular}{ccc}
\(x\) & & \(p(x)\) \\
\cline { 1 - 1 } \cline { 3 - 3 } 100 & & 0.80 \\
500 & & 0.16 \\
1000 & & 0.04
\end{tabular}

Calculate the probability that aggregate claims will be exactly 600 .

Options:
A
0.022
B
0.038
C
0.049
D
0.060
E
\(\quad 0.070\)

Correct answer: D

Solution: 600 can be obtained only 2 ways, from \(500+100\) or from \(6(100)\).
Since \(\lambda=5\) and \(p(100)=0.8, p(500)=0.16\).
\(\operatorname{Pr}(6\) claims for 100\()=\frac{e^{-5} 5^{6}}{6!}(0.8)^{6}=0.03833\) or \(3.83 \%\)
\(\operatorname{Pr}(500+100)=\frac{e^{-5} 5^{2}}{2!}\left[(0.8)^{1}(0.16)^{1}(2)\right]=0.02156=2.16 \%\)
The factor of 2 inside the bracket is because you could get a 500 then 100 or you could get a 100 then 500 . Total is \(3.83 \%+2.16 \%=5.99 \%\).
Question: Five models are fitted to a sample of \(n=260\) observations with the following results:

\begin{tabular}{|c|c|c|}
\hline Model & Number of Parameters & Loglikelihood \\
\hline I & 1 & -414 \\
\hline II & 2 & -412 \\
\hline III & 3 & -411 \\
\hline IV & 4 & -409 \\
\hline V & 6 & -409 \\
\hline
\end{tabular}
Determine the model favored by the Akaike Information criterion.

Options:
A
I
B
II
C
III
D
IV
E
\(\quad \mathrm{V}\)

Correct answer: D

Solution: (This question is effective with the October 2016 syllabus.) The deduction to get the AIC is \(r\), where \(r\) is the number of parameters. The AIC values are then \(-415,-414,-414,-413\), and -415 . The largest value is the fourth one, so model IV is to be selected.
Question: You are given the following information:
i)

\begin{tabular}{|l|c|c|c|c|}
\hline \multirow{3}{*}{\begin{tabular}{l} 
Accident \\
Year
\end{tabular}} & \multicolumn{4}{|c|}{\begin{tabular}{c} 
Cumulative Paid Losses through \\
Development Month
\end{tabular}} \\
\cline { 2 - 5 } & \(\mathbf{1 2}\) & \(\mathbf{2 4}\) & \(\mathbf{3 6}\) & \(\mathbf{4 8}\) \\
\hline AY5 & 27,000 & 49,000 & 65,000 & 72,000 \\
\hline AY6 & 28,000 & 57,000 & 71,000 & \\
\hline AY7 & 33,000 & 65,000 & & \\
\hline AY8 & 35,000 & & & \\
\hline
\end{tabular}
ii)

\begin{tabular}{|l|c|}
\hline Interval & \begin{tabular}{c} 
Selected Age-to-Age Paid Loss \\
Development Factors
\end{tabular} \\
\hline \(12-24\) months & 2.00 \\
\hline \(24-36\) months & 1.20 \\
\hline \(36-48\) months & 1.15 \\
\hline \(48-\) ultimate & 1.00 \\
\hline
\end{tabular}
iii) The interest rate is \(5.0 \%\) per annum effective.

Calculate the ratio of discounted reserves to undiscounted reserves as of December 31, CY8.

Options:
A
0.93
B
\(\quad 0.94\)
C
0.95
D
0.96
E
\(\quad 0.97\)

Correct answer: C

Solution: The completed triangle values are AY6: \(48-71,000(1.15)=81,650\); AY7: \(36: 65,000(1.2)=\) 78,000 and 48: 78,000(1.15) = 89,700; and AY8: \(24-35,000(2)=70,000,36-70,000(1.2)=\) 84,000 , and \(48-84,000(1.15)=96,600\). The undiscounted reserve is: \((81,650-71,000)+(89,700-65,000)+(96,600-35,000)=96,950\).

The increments are AY6: 10,650 ; AY7: 13,000 and 11,700 ; and AY8: \(35,000,14,000\), and 12,600 . By development month the totals are next 12 months: \(58,650,12-24\) months: 25,700 , and \(24-36\) months: 12,600 . The discounted value is \(58,650 / 1.05^{0.5}+25,700 / 1.05^{1.5}+ 12,600 / 1.05^{2.5}=92,276\).

The ratio is \(92,276 / 96,950=0.95\).
Question: The following developed losses evaluated at various maximum loss sizes are given:
- The total losses limited at 50,000 from all policies with a policy limit of 50,000 or more is \(22,000,000\).
- The total losses limited at 50,000 from all policies with a policy limit of 250,000 or more is \(14,000,000\).
- The total losses limited at 250,000 from all policies with a policy limit of 250,000 or more is \(25,000,000\).

The base rate at the 50,000 basic limit is 300 per exposure unit, consisting of 240 pure premium, 30 fixed expense, and 30 variable expense.

Calculate the rate at the 250,000 limit.

Options:
A
370
B
400
C
450
D
480
E
510

Correct answer: E

Solution: The increased limits factor is \(25,000,000 / 14,000,000\) and so the pure premium is \(240(25 / 14)=\) 428.57. The fixed expense stays at 30 . The variable expense is \(30 / 300=0.1\) of the rate. So the rate is the solution to \(0.1 r+30+428.57=r\) and thus \(r=458.57 / 0.9=509.52\).
Question: The information associated with the maximum likelihood estimator of a parameter \(\theta\) is \(4 n\), where \(n\) is the number of observations.

Calculate the asymptotic variance of the maximum likelihood estimator of \(2 \theta\).

Options:
A
\(\quad 1 /(2 n)\)
B
\(\quad 1 / n\)
C
\(4 / n\)
D
\(8 n\)
E
\(16 n\)

Correct answer: B

Solution: From the information, the asymptotic variance of \(\hat{\theta}\) is \(1 / 4 n\). Then
\(\operatorname{Var}(2 \hat{\theta})=4 \operatorname{Var}(\hat{\theta})=4(1 / 4 n)=1 / n\)
Question: A random sample of size \(n\) is drawn from a distribution with probability density function:
\[
f(x)=\frac{\theta}{(\theta+x)^{2}}, \quad 0<x<\infty, \theta>0
\]

Calculate the asymptotic variance of the maximum likelihood estimator of \(\theta\).

Options:
A
\(\frac{3 \theta^{2}}{n}\)
B
\(\frac{1}{3 n \theta^{2}}\)
C
\(\frac{3}{n \theta^{2}}\)
D
\(\frac{n}{3 \theta^{2}}\)
E
\(\frac{1}{3 \theta^{2}}\)

Correct answer: A

Solution: \(\ln f(x)=\ln \theta-2 \ln (\theta+x)\)
\(\frac{\partial \ln f(x)}{\partial \theta}=\frac{1}{\theta}-\frac{2}{\theta+x}\)
\(\frac{\partial^{2} \ln f(x)}{\partial \theta^{2}}=-\frac{1}{\theta^{2}}+\frac{2}{(\theta+x)^{2}}\)
\(E\left[\frac{\partial^{2} \ln f(x)}{\partial \theta^{2}}\right]=-\frac{1}{\theta^{2}}+\int_{0}^{\infty} \frac{2 \theta}{(\theta+x)^{4}} d x=-\frac{1}{\theta^{2}}+\left.\left[-\frac{2 \theta}{3(\theta+x)^{3}}\right]\right|_{0} ^{\infty}=-\frac{1}{\theta^{2}}+\frac{2}{3 \theta^{2}}=-\frac{1}{3 \theta^{2}}\)
\(I(\theta)=\frac{n}{3 \theta^{2}} ; \quad \operatorname{Var}=\frac{3 \theta^{2}}{n}\)