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Exam FAM — Credibility Practice Flashcards

Thirty exam-realistic multiple-choice problems on SOA Exam FAM credibility — limited-fluctuation full-credibility standards for frequency, severity, and aggregate losses, the square-root partial-credibility rule, Bühlmann EPV/VHM/$k$/$Z$ from explicit risk structures, the Bühlmann–Straub model with varying exposures, and the Bayesian-equals-Bühlmann conjugate cases — each with a fully worked solution that names the common candidate error behind each distractor.

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  1. Full credibility standard
    Under limited-fluctuation credibility, claim counts are Poisson. Calculate the full-credibility standard for frequency (expected number of claims) at a $90\%$ probability of being within $\pm5\%$ of the true mean. (A) $1{,}082$ (B) $271$ (C) $384$ (D) $1{,}537$ (E) $4{,}330$
    **Answer: (A).** The frequency standard is $\lambda_F=\left(\dfrac{z_p}{k}\right)^2$. For $p=0.90$ the two-sided percentile is $z_p=z_{0.95}=1.645$, and the relative tolerance is $k=0.05$. $\lambda_F=\left(\dfrac{1.645}{0.05}\right)^2=(32.9)^2\approx 1082.41$. So about $1{,}082$ expected claims are needed. Using the $95\%$ value $z=1.960$ by mistake gives $(1.960/0.05)^2\approx 1537$ (distractor D). Using $k=0.10$ instead of $0.05$ gives $(1.645/0.10)^2\approx 271$ (B). Quadrupling for $k=0.025$ gives $\approx 4330$ (E).
  2. Full credibility standard
    Claim counts are Poisson. Calculate the full-credibility frequency standard (expected number of claims) at a $95\%$ probability of being within $\pm5\%$ of the mean. Use $z_{0.975}=1.960$. (A) $1{,}082$ (B) $1{,}537$ (C) $1{,}600$ (D) $2{,}165$ (E) $6{,}147$
    **Answer: (B).** $\lambda_F=\left(\dfrac{z_p}{k}\right)^2$. For $p=0.95$ the two-sided standard-normal percentile is $z_p=z_{0.975}=1.960$, and $k=0.05$. $\lambda_F=\left(\dfrac{1.960}{0.05}\right)^2=(39.2)^2\approx 1536.64$. So about $1{,}537$ expected claims. Using the $90\%$ value $z=1.645$ gives the $1{,}082$ standard (distractor A). Mistaking $p=0.95$ for the one-sided $z=1.645$ is the classic percentile error.
  3. Full credibility standard
    An insurer writes policies each expected to produce $\lambda=0.04$ claims per year. Claim counts are Poisson. Using a $90\%/\pm5\%$ standard, calculate the number of policy-years of exposure needed for full credibility of frequency. (A) $43$ (B) $1{,}082$ (C) $5{,}412$ (D) $21{,}648$ (E) $27{,}060$
    **Answer: (E).** The frequency claim standard is $\lambda_F=(1.645/0.05)^2\approx 1082.41$ expected claims. Each policy-year produces $\lambda=0.04$ expected claims, so the required exposure is $\dfrac{\lambda_F}{\lambda}=\dfrac{1082.41}{0.04}\approx 27060.25$ policy-years. Dividing by $0.20$ instead of $0.04$ gives the $5{,}412$ distractor (C); multiplying $1082.41\times0.04\approx 43$ (A) inverts the relationship; $1{,}082$ (B) is the claim standard itself, not exposures.
  4. Full credibility standard
    Claim sizes have mean $\mu_X=\$800$ and standard deviation $\sigma_X=\$2{,}000$. Using a $90\%/\pm5\%$ standard, calculate the number of claims required for full credibility of severity. (A) $173$ (B) $1{,}082$ (C) $2{,}706$ (D) $5{,}412$ (E) $6{,}765$
    **Answer: (E).** The severity standard is $n_F=\left(\dfrac{z_p}{k}\right)^2\left(\dfrac{\sigma_X}{\mu_X}\right)^2=\lambda_{F,0}\,(\text{CV}_X)^2$. Basic frequency standard: $\lambda_{F,0}=(1.645/0.05)^2\approx 1082.41$. Coefficient of variation: $\text{CV}_X=\dfrac{2000}{800}=2.5$, so $(\text{CV}_X)^2=6.25$. $n_F=1082.41\times 6.25\approx 6765.06$ claims. Forgetting the CV factor leaves $1{,}082$ (B). Using CV instead of CV$^2$ ($\times2.5$) gives $\approx 2706$ (C); using $\lambda_{F,0}/6.25\approx 173$ inverts the dispersion (A).
  5. Full credibility standard
    Claim counts are Poisson; claim severities have mean $\mu_X=\$2{,}500$ and standard deviation $\sigma_X=\$5{,}000$. Using a $90\%/\pm5\%$ standard, calculate the number of claims needed for full credibility of aggregate losses. (A) $5{,}412$ (B) $4{,}330$ (C) $1{,}082$ (D) $6{,}494$ (E) $43{,}296$
    **Answer: (A).** The aggregate (pure-premium) standard with Poisson frequency is $n_F=\lambda_{F,0}\big(1+(\text{CV}_X)^2\big)$. $\lambda_{F,0}=(1.645/0.05)^2\approx 1082.41$. $\text{CV}_X=\dfrac{5000}{2500}=2$, so $(\text{CV}_X)^2=4$ and $1+(\text{CV}_X)^2=5$. $n_F=1082.41\times 5\approx 5412.05$ claims. Using only $(\text{CV}_X)^2=4$ (the severity formula, dropping the $+1$) gives $\approx 4330$ (B). The bare frequency standard $1{,}082$ (C) ignores severity entirely.
  6. Full credibility standard
    Claim counts are Poisson with $\lambda=0.10$ claims per exposure. Severity has mean $\$3{,}000$ and standard deviation $\$3{,}000$. Using a $90\%/\pm5\%$ standard, calculate the number of exposures (policy-years) needed for full credibility of aggregate losses. (A) $2{,}165$ (B) $10{,}824$ (C) $16{,}236$ (D) $21{,}648$ (E) $27{,}060$
    **Answer: (D).** First the aggregate claim standard: $\text{CV}_X=\dfrac{3000}{3000}=1$, so $1+(\text{CV}_X)^2=2$. $n_F=\lambda_{F,0}\big(1+(\text{CV}_X)^2\big)=1082.41\times 2\approx 2164.82$ claims. Convert claims to exposures by dividing by $\lambda=0.10$: $\dfrac{n_F}{\lambda}=\dfrac{2164.82}{0.10}\approx 21648.2$ exposures. Leaving the answer in claims gives $\approx 2165$ (A). Using only the frequency standard $1082.41/0.10\approx 10824$ (B) ignores severity. Using $\lambda_{F,0}\times1.5$ for a mistaken CV term gives $\approx 16236$ (C).
  7. Partial credibility
    A territory has accumulated $600$ claims; the full-credibility standard for frequency is $1{,}082$ claims. Using the square-root rule, calculate the limited-fluctuation partial-credibility factor $Z$. (A) $0.745$ (B) $0.555$ (C) $0.308$ (D) $0.804$ (E) $1.000$
    **Answer: (A).** The square-root rule assigns $Z=\sqrt{\dfrac{n}{n_F}}$ (capped at $1$). $Z=\sqrt{\dfrac{600}{1082}}=\sqrt{0.554529}\approx 0.74467$. So $Z\approx 0.745$. Using the linear rule $Z=n/n_F=0.555$ (distractor B) understates credibility — the standard error shrinks like $1/\sqrt n$, so $Z$ scales with $\sqrt{n/n_F}$, not $n/n_F$. Squaring instead gives $0.555^2\approx 0.308$ (C).
  8. Partial credibility
    A class has $400$ observed claims against a full-credibility standard of $1{,}082$ claims. Its observed mean pure premium is $\$340$ and the manual (collective) rate is $\$300$. Using limited-fluctuation credibility, calculate the credibility-weighted estimate. (A) $\$300.00$ (B) $\$314.79$ (C) $\$324.32$ (D) $\$334.79$ (E) $\$340.00$
    **Answer: (C).** Partial credibility: $Z=\sqrt{\dfrac{n}{n_F}}=\sqrt{\dfrac{400}{1082}}=\sqrt{0.369686}\approx 0.60802$. Estimate $=Z\bar X+(1-Z)M=0.60802(340)+0.39198(300)$ $=206.727+117.594\approx \$324.32$. Using the linear rule $Z=400/1082\approx 0.36969$ gives $0.36969(340)+0.63031(300)\approx \$314.79$ (distractor B), which under-weights the data. Taking the manual rate alone gives $\$300.00$ (A).