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Exam FAM — Severity & Frequency Models Practice Flashcards

Thirty exam-realistic multiple-choice problems on SOA Exam FAM severity and frequency models — means, variances, raw moments and coefficient of variation of the exponential, gamma, Pareto, lognormal and Weibull; tail-weight comparison via hazard rate, mean-excess and existence of moments; identifying the $(a,b,0)$ distribution from probability ratios or a variance-to-mean test; Poisson/binomial/negative-binomial moments; zero-truncated and zero-modified probabilities; and the gamma mixture of a Poisson — each with a fully worked solution that names the common-error distractors.

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Browse all 30 problems as a list
  1. Severity distributions
    Losses follow an **exponential** distribution with mean $\theta=800$. Calculate $E[X^{3}]$. (A) $512{,}000{,}000$ (B) $1{,}024{,}000{,}000$ (C) $1{,}536{,}000{,}000$ (D) $3{,}072{,}000{,}000$ (E) $4{,}915{,}200{,}000$
    **Answer: (D).** For an exponential with mean $\theta$ the raw moments are $E[X^{k}]=k!\,\theta^{k}$. $E[X^{3}]=3!\,\theta^{3}=6(800)^{3}=6(512{,}000{,}000)=\$3{,}072{,}000{,}000$ (in cubed-dollar units). Forgetting the $k!$ factor and reporting $\theta^{3}=512{,}000{,}000$ gives distractor (A); using $2!\,\theta^{3}$ (the second-moment factor) gives $1{,}024{,}000{,}000$ (B).
  2. Severity distributions
    Losses follow a **gamma** distribution with shape $\alpha=4$ and scale $\theta=250$. Calculate the variance. (A) $62{,}500$ (B) $250{,}000$ (C) $500{,}000$ (D) $1{,}000{,}000$ (E) $4{,}000{,}000$
    **Answer: (B).** For a gamma, $E[X]=\alpha\theta$ and $\operatorname{Var}(X)=\alpha\theta^{2}$. $\operatorname{Var}(X)=4(250)^{2}=4(62{,}500)=\$250{,}000$. Using $\alpha^{2}\theta^{2}$ (squaring $\alpha$ as well) gives $16(62{,}500)=1{,}000{,}000$ (D); reporting $\theta^{2}=62{,}500$ alone gives (A); squaring the mean $(\alpha\theta)^{2}=1{,}000{,}000$ also lands on (D).
  3. Severity distributions
    Losses follow a **two-parameter Pareto** with shape $\alpha=4$ and scale $\theta=3000$. Calculate $E[X^{2}]$. (A) $1{,}000{,}000$ (B) $2{,}000{,}000$ (C) $3{,}000{,}000$ (D) $4{,}500{,}000$ (E) $9{,}000{,}000$
    **Answer: (C).** For a two-parameter Pareto, $E[X^{2}]=\dfrac{2\theta^{2}}{(\alpha-1)(\alpha-2)}$ (requires $\alpha>2$). $E[X^{2}]=\dfrac{2(3000)^{2}}{(4-1)(4-2)}=\dfrac{2(9{,}000{,}000)}{(3)(2)}=\dfrac{18{,}000{,}000}{6}=\$3{,}000{,}000$. This is the second **raw** moment, not the variance. The variance is $E[X^{2}]-E[X]^{2}=3{,}000{,}000-1000^{2}=2{,}000{,}000$ (distractor B, the common slip of reporting the variance when asked for $E[X^{2}]$). Using $(\alpha)(\alpha-1)$ in the denominator gives $\dfrac{18{,}000{,}000}{12}=1{,}500{,}000$.
  4. Severity distributions
    Losses follow a **two-parameter Pareto** with $\alpha=4$, $\theta=3000$. Calculate $P(X>6000)$. (A) $0.0123$ (B) $0.0370$ (C) $0.0625$ (D) $0.1111$ (E) $0.2500$
    **Answer: (A).** The Pareto survival function is $S(x)=\left(\dfrac{\theta}{x+\theta}\right)^{\alpha}$. $S(6000)=\left(\dfrac{3000}{6000+3000}\right)^{4}=\left(\dfrac{3000}{9000}\right)^{4}=\left(\dfrac{1}{3}\right)^{4}=\dfrac{1}{81}\approx 0.0123$. Forgetting to add $\theta$ in the denominator — using $\left(\dfrac{\theta}{x}\right)^{\alpha}=(1/2)^{4}=0.0625$ — gives (C); using exponent $\alpha-1=3$, $(1/3)^{3}\approx 0.0370$, gives (B).
  5. Severity distributions
    Losses follow a **Weibull** distribution with shape $\tau=2$ and scale $\theta=1000$. Calculate $P(X>2000)$. (A) $0.0183$ (B) $0.1353$ (C) $0.1813$ (D) $0.3679$ (E) $0.8647$
    **Answer: (A).** The Weibull survival function is $S(x)=e^{-(x/\theta)^{\tau}}$. $S(2000)=e^{-(2000/1000)^{2}}=e^{-2^{2}}=e^{-4}\approx 0.0183$. Dropping the exponent $\tau$ (treating it like an exponential, $e^{-x/\theta}=e^{-2}\approx 0.1353$) gives (B); using $(x/\theta)^{\tau}$ without squaring, $e^{-2}$, also lands on (B); $e^{-1}\approx 0.3679$ comes from $e^{-(x/\theta)}$ with $x=\theta$.
  6. Severity distributions
    Losses follow a **lognormal** distribution with parameters $\mu=8$ and $\sigma=0.5$. Using $\Phi(1.03)\approx 0.8485$, calculate $P(X\le 5000)$. (A) $0.5793$ (B) $0.6915$ (C) $0.8485$ (D) $0.9332$ (E) $0.9772$
    **Answer: (C).** For a lognormal, $F(x)=\Phi\!\left(\dfrac{\ln x-\mu}{\sigma}\right)$. $\ln 5000\approx 8.51719$, so $z=\dfrac{8.51719-8}{0.5}=\dfrac{0.51719}{0.5}\approx 1.0344$. $P(X\le 5000)=\Phi(1.03)\approx 0.8485$. Forgetting to divide by $\sigma$ (using $z=\ln x-\mu=0.517$, $\Phi(0.52)\approx 0.6985$) lands near distractor (B); standardizing $x$ itself instead of $\ln x$ is the other classic error.
  7. Moments & CV
    A **gamma** severity has shape $\alpha=3$ and scale $\theta=500$. Calculate its coefficient of variation. (A) $0.3333$ (B) $0.5774$ (C) $0.7071$ (D) $1.0000$ (E) $1.7321$
    **Answer: (B).** For a gamma, $E[X]=\alpha\theta=1500$ and $\operatorname{Var}(X)=\alpha\theta^{2}=3(500)^{2}=750{,}000$, so $\text{SD}=\sqrt{750{,}000}\approx 866.03$. $\text{CV}=\dfrac{866.03}{1500}\approx 0.5774$. The shortcut is $\text{CV}=\dfrac{1}{\sqrt{\alpha}}=\dfrac{1}{\sqrt{3}}\approx 0.5774$ — scale-free, depending only on the shape. Using $\dfrac{1}{\alpha}=0.3333$ (forgetting the square root) gives (A); the exponential value $\text{CV}=1$ is (D).
  8. Moments & CV
    A **Pareto** severity has $\alpha=3$, $\theta=2000$. Calculate its coefficient of variation. (A) $0.5774$ (B) $1.0000$ (C) $1.4142$ (D) $1.7321$ (E) $3.0000$
    **Answer: (D).** $E[X]=\dfrac{\theta}{\alpha-1}=\dfrac{2000}{2}=1000$. $E[X^{2}]=\dfrac{2\theta^{2}}{(\alpha-1)(\alpha-2)}=\dfrac{2(2000)^{2}}{(2)(1)}=4{,}000{,}000$, so $\operatorname{Var}(X)=4{,}000{,}000-1000^{2}=3{,}000{,}000$ and $\text{SD}=\sqrt{3{,}000{,}000}\approx 1732.05$. $\text{CV}=\dfrac{1732.05}{1000}\approx 1.7321>1$, confirming a heavy tail. Note $\text{CV}^{2}=\dfrac{\operatorname{Var}(X)}{E[X]^{2}}=\dfrac{3{,}000{,}000}{1{,}000{,}000}=3$, so $\text{CV}=\sqrt{3}\approx 1.7321$. Reporting $\text{CV}^{2}=3$ itself (skipping the square root) gives the distractor $3.0000$ (E); a CV of $1$ (B) is the exponential's value, and $\sqrt{2}\approx 1.4142$ (C) comes from mis-cancelling the $(\alpha-1)(\alpha-2)$ factors.