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Exam MAS-I — Credibility Practice Flashcards

Thirty exam-realistic multiple-choice problems on CAS Exam MAS-I credibility — classical limited-fluctuation full and partial standards (including the $(1+\mathrm{CV}^2)$ aggregate extension), Bühlmann EPV/VHM/$k$/$Z$ and the credibility premium, Bühlmann-Straub with exposures, and Bayesian posterior/predictive means for the Poisson-gamma, Bernoulli-beta, and normal-normal conjugate pairs — each with a fully worked solution.

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Browse all 30 problems as a list
  1. Limited fluctuation
    Claim frequency is Poisson. Full credibility is to be assigned when the observed claim count is within $\pm 5\%$ of its mean with probability $90\%$ ($z_{0.95}=1.645$). Calculate the expected number of claims required for full credibility. (A) $271$ (B) $385$ (C) $542$ (D) $1{,}082$ (E) $1{,}537$
    **Answer: (D).** For Poisson frequency the full-credibility standard (in expected claims) is $\lambda_F=\left(\dfrac{z_{(1+P)/2}}{k}\right)^{2}$ with $z=1.645$ and $k=0.05$. $\lambda_F=\left(\dfrac{1.645}{0.05}\right)^{2}=(32.9)^{2}=1082.41\approx 1082$ claims. Using $k=0.10$ instead of $0.05$ gives $\approx 271$ (A); using $z=1.960$ ($95\%$) with $k=0.10$ gives $\approx 385$ (B). The keyed value $1082$ is the standard $90\%/\pm5\%$ benchmark.
  2. Limited fluctuation
    Individual claim sizes have mean $\mu_X=5{,}000$ and standard deviation $\sigma_X=10{,}000$. Frequency is Poisson and the frequency full-credibility standard is $\lambda_F=1{,}082$ expected claims. Calculate the expected number of claims required for full credibility of **average severity**. (A) $1{,}082$ (B) $2{,}164$ (C) $4{,}328$ (D) $5{,}410$ (E) $1{,}352$
    **Answer: (C).** For severity, $n_F=\lambda_F\,\mathrm{CV}_X^{2}$ where $\mathrm{CV}_X=\dfrac{\sigma_X}{\mu_X}=\dfrac{10{,}000}{5{,}000}=2$, so $\mathrm{CV}_X^{2}=4$. $n_F=1082(4)=4328$ claims. Using $(1+\mathrm{CV}^2)=5$ (the aggregate factor) gives $5410$ (D) — that is the standard for aggregate losses, not severity. Forgetting to square the CV ($\mathrm{CV}=2$) gives $2164$ (B).
  3. Limited fluctuation
    Frequency is Poisson; individual claim sizes have mean $4{,}000$ and standard deviation $6{,}000$. The frequency full-credibility standard is $\lambda_F=1{,}082$ expected claims. Calculate the expected number of claims required for full credibility of **aggregate losses**. (A) $1{,}082$ (B) $2{,}435$ (C) $3{,}517$ (D) $4{,}328$ (E) $5{,}410$
    **Answer: (C).** For aggregate losses, $n_F=\lambda_F\,(1+\mathrm{CV}_X^{2})$. Here $\mathrm{CV}_X=\dfrac{6{,}000}{4{,}000}=1.5$, so $\mathrm{CV}_X^{2}=2.25$. $n_F=1082(1+2.25)=1082(3.25)=3516.5\approx 3517$ claims. Using only $\mathrm{CV}^2$ (dropping the $1$ for frequency) gives $1082(2.25)\approx 2435$ (B), the severity standard. The $1$ captures Poisson frequency risk; the $\mathrm{CV}^2$ captures severity risk.
  4. Limited fluctuation
    Claim sizes have mean $\mu_X=2{,}500$ and variance $\sigma_X^{2}=25{,}000{,}000$. Frequency is Poisson and the frequency standard is $\lambda_F=1{,}082$. Calculate the expected number of claims required for full credibility of **aggregate losses**. (A) $1{,}082$ (B) $3{,}246$ (C) $4{,}328$ (D) $5{,}410$ (E) $11{,}902$
    **Answer: (D).** $\mathrm{CV}_X^{2}=\dfrac{\sigma_X^{2}}{\mu_X^{2}}=\dfrac{25{,}000{,}000}{2{,}500^{2}}=\dfrac{25{,}000{,}000}{6{,}250{,}000}=4$. Aggregate standard $n_F=\lambda_F(1+\mathrm{CV}_X^{2})=1082(1+4)=1082(5)=5410$ claims. Forgetting the $+1$ gives $1082(4)=4328$ (C, the severity standard). Note $\sigma_X^2$ is already the variance, so do not square it again.
  5. Limited fluctuation
    A full-credibility standard for aggregate losses is set at $99\%$ within $\pm 5\%$ ($z_{0.995}=2.576$). Frequency is Poisson and individual claim sizes have $\mathrm{CV}_X=1$. Calculate the expected number of claims required for full credibility of aggregate losses. (A) $1{,}082$ (B) $2{,}164$ (C) $2{,}654$ (D) $5{,}309$ (E) $7{,}963$
    **Answer: (D).** Frequency standard: $\lambda_F=\left(\dfrac{z}{k}\right)^{2}=\left(\dfrac{2.576}{0.05}\right)^{2}=(51.52)^{2}=2654.3$. Aggregate factor: $1+\mathrm{CV}_X^{2}=1+1^{2}=2$. $n_F=\lambda_F(1+\mathrm{CV}_X^{2})=2654.3(2)=5308.6\approx 5309$ claims. Forgetting the aggregate factor gives just $\lambda_F\approx 2654$ (C); using $z=1.645$ gives $1082\times 2=2164$ (B).
  6. Full & partial credibility
    Frequency is Poisson with $\lambda_F=1{,}082$; individual losses have $\mathrm{CV}_X=1.5$, so the aggregate standard is $n_F=3{,}517$ expected claims. A risk has generated $2{,}000$ claims. Calculate the partial credibility $Z$. (A) $0.569$ (B) $0.640$ (C) $0.754$ (D) $0.832$ (E) $1.000$
    **Answer: (C).** The square-root rule gives $Z=\sqrt{\dfrac{n}{n_F}}=\sqrt{\dfrac{2000}{3517}}=\sqrt{0.56867}\approx 0.754$. Using the (incorrect) linear rule $Z=\dfrac{n}{n_F}=0.569$ gives distractor (A). Since $n<n_F$ the risk is only partially credible, so $Z<1$; capping at $1$ (E) is wrong here.
  7. Full & partial credibility
    A risk has produced $600$ claims against a full-credibility standard of $1{,}082$ claims. Its observed pure premium is $\bar X=4{,}500$ and the manual (collective) pure premium is $M=4{,}000$. Using limited-fluctuation partial credibility, calculate the credibility premium. (A) $4{,}000$ (B) $4{,}277$ (C) $4{,}372$ (D) $4{,}500$ (E) $4{,}723$
    **Answer: (C).** $Z=\sqrt{\dfrac{n}{n_F}}=\sqrt{\dfrac{600}{1082}}=\sqrt{0.55453}\approx 0.7447$. Credibility premium $=Z\bar X+(1-Z)M = 0.7447(4500)+0.2553(4000)=3351.2+1021.2=\$4{,}372$. Using the linear weight $Z=\dfrac{600}{1082}=0.5546$ gives $0.5546(4500)+0.4454(4000)\approx 4277$ (B). The estimate lies between $M=4000$ and $\bar X=4500$, closer to $\bar X$ because $Z>0.5$.
  8. Full & partial credibility
    The full-credibility standard is $n_F=1{,}082$ claims. How many claims must a risk generate to be assigned a credibility of $Z=0.75$? (A) $406$ (B) $609$ (C) $722$ (D) $812$ (E) $1{,}082$
    **Answer: (B).** From $Z=\sqrt{n/n_F}$ we solve $n=Z^{2}\,n_F=(0.75)^{2}(1082)=0.5625(1082)=608.6\approx 609$ claims. Using the linear relation $n=Z\,n_F=0.75(1082)\approx 812$ (D) is the common error. Because $Z\propto\sqrt{n}$, reaching $75\%$ credibility needs only $(0.75)^2=56.25\%$ of the full-credibility count.