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Exam FAM — Coverage Modifications Practice Flashcards

Thirty exam-realistic multiple-choice problems on SOA Exam FAM coverage modifications — limited expected values, ordinary and franchise deductibles, policy limits with coinsurance, the per-loss versus per-payment distinction, uniform inflation, expected claim frequency, and the loss elimination ratio — each fully self-contained with a worked solution that names the candidate error behind each distractor.

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Browse all 30 problems as a list
  1. Limited expected value
    Losses follow an exponential distribution with mean $\theta=2000$. Calculate the limited expected value $E[X\wedge 1000]$. (A) $\$632.12$ (B) $\$786.94$ (C) $\$864.66$ (D) $\$1{,}213.06$ (E) $\$2{,}000.00$
    **Answer: (B).** For $X\sim\text{Exp}(\theta)$, $S(x)=e^{-x/\theta}$ and $E[X\wedge u]=\theta\big(1-e^{-u/\theta}\big)$. With $\theta=2000$ and $u=1000$: $\dfrac{u}{\theta}=0.5$, so $e^{-0.5}\approx 0.606531$. $E[X\wedge 1000]=2000(1-0.606531)=2000(0.393469)\approx \$786.94$. Using the wrong ratio $\tfrac{u}{\theta}\to\tfrac{\theta}{u}=2$ (i.e. $2000(1-e^{-2})\approx 1729$) is a sign/ratio slip; computing $\theta(1-e^{-u/\theta})$ with $\theta=1000$ instead of $2000$ gives $\$632.12$ (A). The full mean $E[X]=\$2{,}000$ (E) is the cap as $u\to\infty$, not at $u=1000$.
  2. Limited expected value
    A loss is uniform on $(0,3000)$. Calculate $E[X\wedge 1200]$. (A) $\$960.00$ (B) $\$1{,}020.00$ (C) $\$1{,}200.00$ (D) $\$1{,}320.00$ (E) $\$1{,}500.00$
    **Answer: (A).** For $X\sim U(0,b)$, $S(x)=1-\tfrac{x}{b}$ on $[0,b]$, so for $u\le b$, $E[X\wedge u]=\displaystyle\int_0^u\Big(1-\tfrac{x}{b}\Big)dx=u-\dfrac{u^2}{2b}$. With $b=3000$, $u=1200$: $E[X\wedge 1200]=1200-\dfrac{1200^2}{2(3000)}=1200-\dfrac{1{,}440{,}000}{6000}=1200-240=\$960.00$. Forgetting the $-\tfrac{u^2}{2b}$ correction and reporting $u$ itself gives $\$1{,}200.00$ (C); using $2b=3000$ (dropping the $\times 2$) gives $1200-480=\$720$. The unconditional mean $E[X]=\tfrac{b}{2}=\$1{,}500$ (E) is larger than any capped value.
  3. Limited expected value
    A loss has a Pareto distribution with $\alpha=3$ and $\theta=4000$. Calculate $E[X\wedge 2000]$. (A) $\$1{,}111.11$ (B) $\$1{,}777.78$ (C) $\$2{,}000.00$ (D) $\$2{,}222.22$ (E) $\$4{,}000.00$
    **Answer: (A).** For $X\sim\text{Pareto}(\alpha,\theta)$ with $\alpha>1$, $E[X\wedge u]=\dfrac{\theta}{\alpha-1}\left[1-\left(\dfrac{\theta}{\theta+u}\right)^{\alpha-1}\right]$. Here $\dfrac{\theta}{\alpha-1}=\dfrac{4000}{2}=2000$ and $\dfrac{\theta}{\theta+u}=\dfrac{4000}{6000}=\tfrac{2}{3}$, so $\big(\tfrac23\big)^{2}=\tfrac49\approx 0.444444$. $E[X\wedge 2000]=2000(1-0.444444)=2000(0.555556)\approx \$1{,}111.11$. Using the exponent $\alpha=3$ instead of $\alpha-1=2$ gives $2000(1-(\tfrac23)^3)=2000(0.703704)\approx\$1{,}407$; the ground-up mean $E[X]=\tfrac{\theta}{\alpha-1}=\$2{,}000$ (C) is the limit as $u\to\infty$.
  4. Limited expected value
    A discrete loss takes the values $100,300,600,1000$ with probabilities $0.5,0.2,0.2,0.1$ respectively. Calculate $E[X\wedge 400]$. (A) $\$230.00$ (B) $\$280.00$ (C) $\$320.00$ (D) $\$330.00$ (E) $\$380.00$
    **Answer: (A).** Cap each value at $400$: $\min(100,400)=100$, $\min(300,400)=300$, $\min(600,400)=400$, $\min(1000,400)=400$. $E[X\wedge 400]=100(0.5)+300(0.2)+400(0.2)+400(0.1)=50+60+80+40=\$230.00$. Forgetting to cap (using the raw values $600$ and $1000$) gives the unlimited mean $E[X]=100(0.5)+300(0.2)+600(0.2)+1000(0.1)=50+60+120+100=\$330.00$ (D). Capping at the wrong values or miscounting the tail mass produces the other distractors.
  5. Ordinary deductible
    Losses are exponential with mean $\theta=1500$. An ordinary deductible of $500$ applies. Calculate the expected payment per loss, $E[(X-500)_+]$. (A) $\$716.53$ (B) $\$1{,}074.80$ (C) $\$1{,}075.00$ (D) $\$1{,}500.00$ (E) $\$425.20$
    **Answer: (B).** The per-loss expected payment is $E[(X-d)_+]=E[X]-E[X\wedge d]$. For the exponential, $E[X\wedge 500]=1500\big(1-e^{-500/1500}\big)=1500\big(1-e^{-1/3}\big)$. Since $e^{-0.333333}\approx 0.716531$, $E[X\wedge 500]=1500(0.283469)\approx 425.20$. $E[(X-500)_+]=1500-425.20=\$1{,}074.80$. By memorylessness this also equals $E[X]\cdot S(500)=1500\,e^{-1/3}=1500(0.716531)\approx \$1{,}074.80$ — a useful check. Reporting $E[X\wedge 500]=\$425.20$ (E) is the eliminated piece, not the payment; $\$1{,}500$ (D) ignores the deductible entirely.
  6. Ordinary deductible
    A loss is uniform on $(0,5000)$. An ordinary deductible of $1000$ applies. Calculate the expected payment per loss. (A) $\$1{,}600.00$ (B) $\$1{,}900.00$ (C) $\$2{,}000.00$ (D) $\$2{,}500.00$ (E) $\$900.00$
    **Answer: (A).** Per-loss payment $=E[X]-E[X\wedge d]$. For $U(0,5000)$, $E[X]=\tfrac{5000}{2}=2500$. $E[X\wedge 1000]=1000-\dfrac{1000^2}{2(5000)}=1000-\dfrac{1{,}000{,}000}{10{,}000}=1000-100=900$. $E[(X-1000)_+]=2500-900=\$1{,}600.00$. Check with $\int_{1000}^{5000}(x-1000)\tfrac{1}{5000}dx=\dfrac{(5000-1000)^2}{2(5000)}=\dfrac{4000^2}{10{,}000}=1600$. Reporting $E[X\wedge 1000]=\$900$ (E) gives the eliminated amount, not the payment; ignoring the deductible gives $E[X]=\$2{,}500$ (D).
  7. Ordinary deductible
    A discrete loss takes the values $0,200,500,1000$ with probabilities $0.4,0.3,0.2,0.1$. An ordinary deductible of $300$ applies. Calculate the expected payment per loss. (A) $\$70.00$ (B) $\$110.00$ (C) $\$160.00$ (D) $\$200.00$ (E) $\$310.00$
    **Answer: (B).** The per-loss payment is $(X-300)_+$: pay $0$ when $X\le 300$, and $X-300$ when $X>300$. Only $X=500$ and $X=1000$ exceed $300$: $(500-300)(0.2)+(1000-300)(0.1)=200(0.2)+700(0.1)=40+70=\$110.00$. The value $X=200$ contributes nothing (below the deductible). Forgetting the deductible reduction and summing $500(0.2)+1000(0.1)=\$200$ (D) ignores the $-300$ on each surviving loss. The per-payment mean (dividing by $S(300)=0.3$) would be $\dfrac{110}{0.3}\approx\$367$ — a different quantity.
  8. Per-loss vs per-payment
    Losses are exponential with mean $\theta=1000$. With an ordinary deductible of $400$, calculate the expected payment per payment (i.e. given a payment is made). (A) $\$600.00$ (B) $\$670.32$ (C) $\$1{,}000.00$ (D) $\$1{,}491.82$ (E) $\$1{,}670.32$
    **Answer: (C).** The per-payment mean is $E[Y^P]=\dfrac{E[X]-E[X\wedge d]}{S(d)}$. $E[X\wedge 400]=1000(1-e^{-0.4})=1000(1-0.670320)=329.68$, so per-loss $=1000-329.68=670.32$. And $S(400)=e^{-0.4}=0.670320$. $E[Y^P]=\dfrac{670.32}{0.670320}=\$1{,}000.00$. This equals $\theta$ by the exponential's memorylessness: given $X>d$, the excess is again $\text{Exp}(1000)$. The per-loss value $\$670.32$ (B) forgets to divide by $S(d)$ — the single most common per-loss/per-payment error.