Exam MAS-II · practice
Exam MAS-II — Credibility Practice Flashcards
Thirty exam-realistic multiple-choice problems on CAS Exam MAS-II greatest-accuracy credibility — Bühlmann EPV/VHM/$k$/$Z$ and the credibility premium, the Bühlmann-Straub exposure-weighted extension, nonparametric empirical-Bayes estimation from a two-way table, semiparametric Poisson estimation, and the credibility-Bayes link — each with a fully worked solution.
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Bayesian Analysis Bayesian Analysis Practice Credibility Generalized Linear Models Generalized Linear Models Practice Linear Mixed Models
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- EPV & VHMA risk structure has expected process variance $\text{EPV}=2000$ and variance of the hypothetical means $\text{VHM}=400$. Calculate the Bühlmann constant $k$. (A) $0.20$ (B) $1.60$ (C) $5.00$ (D) $400$ (E) $2000$**Answer: (C).** The Bühlmann constant is $k=\dfrac{\text{EPV}}{\text{VHM}}=\dfrac{2000}{400}=5.00$. Inverting the ratio gives the distractor $\dfrac{\text{VHM}}{\text{EPV}}=0.20$ (A). Reporting EPV or VHM alone gives (E) and (D).
- Bühlmann modelA risk structure has $\text{EPV}=2000$ and $\text{VHM}=400$, so $k=5$. A risk is observed for $n=3$ years. Calculate the Bühlmann credibility factor $Z$. (A) $0.231$ (B) $0.375$ (C) $0.600$ (D) $0.625$ (E) $0.769$**Answer: (B).** $Z=\dfrac{n}{n+k}=\dfrac{3}{3+5}=\dfrac{3}{8}=0.375$. Using $Z=\dfrac{k}{n+k}=\dfrac{5}{8}=0.625$ (the complement) gives distractor (D). Using $Z=\dfrac{n}{n+k}$ with $k$ and $n$ swapped, $\dfrac{5}{5+3}$, also yields $0.625$.
- Bühlmann modelA risk structure has $\text{EPV}=2000$, $\text{VHM}=400$ ($k=5$) and overall mean $\mu=900$. A risk observed for $n=3$ years has sample mean $\bar X=1300$. Calculate the Bühlmann credibility premium. (A) $\$1{,}050.00$ (B) $\$1{,}080.00$ (C) $\$1{,}100.00$ (D) $\$1{,}150.00$ (E) $\$1{,}250.00$**Answer: (A).** From the prior parts $Z=\dfrac{3}{3+5}=0.375$. $P_c=Z\bar X+(1-Z)\mu=0.375(1300)+0.625(900)=487.5+562.5=\$1{,}050.00$. Using the complementary weight $Z=0.625$ on $\bar X$ gives $0.625(1300)+0.375(900)=812.5+337.5=\$1{,}150.00$ (distractor D).
- Bühlmann modelFor a particular risk, $E[v(\theta)]=12{,}000$, $\text{Var}[\mu(\theta)]=750$, and the overall mean is $\mu=400$. The risk has $n=6$ years of data with $\bar X=510$. Calculate the Bühlmann credibility premium. (A) $\$420.00$ (B) $\$430.00$ (C) $\$448.00$ (D) $\$472.00$ (E) $\$510.00$**Answer: (B).** $k=\dfrac{\text{EPV}}{\text{VHM}}=\dfrac{12000}{750}=16$. $Z=\dfrac{n}{n+k}=\dfrac{6}{6+16}=\dfrac{6}{22}=0.272727$. $P_c=Z\bar X+(1-Z)\mu=\dfrac{6(510)+16(400)}{22}=\dfrac{3060+6400}{22}=\dfrac{9460}{22}=\$430.00$. Forgetting to divide and using $k=\text{VHM}/\text{EPV}$ overstates $Z$ and pushes the answer toward $\bar X=510$.
- Bühlmann modelA risk structure has $k=12$. How many years $n$ of experience are required for the Bühlmann credibility factor to reach $Z=0.40$? (A) $4.8$ (B) $7.2$ (C) $8.0$ (D) $12.0$ (E) $18.0$**Answer: (C).** Set $\dfrac{n}{n+12}=0.40$, so $n=0.40(n+12)$, giving $n-0.40n=0.40(12)$, i.e. $0.60n=4.8$ and $n=8.0$. Using $Z=0.40$ directly as $0.40\times 12=4.8$ (distractor A) skips solving the equation; $0.40\times18=7.2$ (B) inverts the algebra.
- EPV & VHMClaim counts follow a Poisson distribution with mean $\lambda$, and $\lambda$ varies across the portfolio with $\text{Gamma}$ prior having mean $E[\lambda]=0.25$ and variance $\text{Var}[\lambda]=0.05$. Calculate the Bühlmann constant $k$. (A) $0.05$ (B) $0.20$ (C) $0.25$ (D) $4.00$ (E) $5.00$**Answer: (E).** For a Poisson likelihood, $\mu(\lambda)=\lambda$ and $v(\lambda)=\lambda$ (mean equals variance). $\text{EPV}=E[v(\lambda)]=E[\lambda]=0.25$; $\text{VHM}=\text{Var}[\mu(\lambda)]=\text{Var}[\lambda]=0.05$. $k=\dfrac{\text{EPV}}{\text{VHM}}=\dfrac{0.25}{0.05}=5.00$. Inverting gives $0.20$ (B); confusing EPV with VHM and reporting $0.05/0.25=0.20$ is the same trap.
- EPV & VHMA portfolio is $70\%$ low-risk drivers ($\mu=0.10$) and $30\%$ high-risk drivers ($\mu=0.30$). Calculate the variance of the hypothetical means (VHM). (A) $0.0084$ (B) $0.0100$ (C) $0.0400$ (D) $0.1600$ (E) $0.2000$**Answer: (A).** Overall mean $\mu=0.70(0.10)+0.30(0.30)=0.16$. $\text{VHM}=E[\mu(\theta)^2]-\mu^2=[0.70(0.10^2)+0.30(0.30^2)]-0.16^2$ $=[0.70(0.01)+0.30(0.09)]-0.0256=[0.007+0.027]-0.0256=0.034-0.0256=0.0084$. Forgetting to subtract $\mu^2$ leaves $E[\mu^2]=0.034$ (not listed); using the raw difference of the two means squared, $(0.30-0.10)^2=0.04$, gives distractor (C).
- EPV & VHMA two-point risk model: with probability $0.5$ a risk has $\mu=4,\ v=8$; with probability $0.5$ it has $\mu=10,\ v=20$. A risk is observed for $n=2$ years with $\bar X=9$. Calculate the Bühlmann credibility premium. (A) $6.04$ (B) $7.00$ (C) $7.80$ (D) $8.13$ (E) $9.00$**Answer: (D).** Overall mean $\mu=0.5(4)+0.5(10)=7$. $\text{EPV}=0.5(8)+0.5(20)=14$. $\text{VHM}=0.5(4^2)+0.5(10^2)-7^2=[0.5(16)+0.5(100)]-49=[8+50]-49=58-49=9$. $k=\dfrac{14}{9}=1.5556$; $Z=\dfrac{n}{n+k}=\dfrac{2}{2+1.5556}=\dfrac{2}{3.5556}=0.5625$. $P_c=0.5625(9)+0.4375(7)=5.0625+3.0625=8.125\approx 8.13$. Inverting the constant to $k=\dfrac{\text{VHM}}{\text{EPV}}=\dfrac{9}{14}$ would overstate $Z$ and pull the estimate toward $\bar X=9$; charging the collective mean $\mu=7$ outright (distractor B) ignores the risk's experience entirely.