Exam P — Risk & Insurance Applications Practice Flashcards

**Answer: (C).** For an exponential loss the expected payment per loss under an ordinary deductible $d$ is $E[(X-d)_{+}]=\theta e^{-d/\theta}$. With $\theta=1200$ and $d=300$: $E[(X-d)_{+}]=1200\,e^{-300/1200}=1200\,e^{-0.25}=1200(0.778801)\approx 934.6$. So the expected payment per loss is about $\$935$. *(Distractor (B) just subtracts $d$ from the mean ($1200-300$); (D) is the per-payment value $\theta$, which ignores the loss-thinning; (E) divides instead of multiplying.)*

**Answer: (D).** This is a Pareto with $\alpha=3$, $\theta=2000$. The mean residual life of a Pareto is $E[X-d\mid X>d]=\dfrac{d+\theta}{\alpha-1}$. With $d=1500$, $\theta=2000$, $\alpha=3$: $\dfrac{1500+2000}{3-1}=\dfrac{3500}{2}=1750$. So the expected payment per payment is $\$1{,}750$. *(Distractor (A) is the per-loss value $1750\,S(1500)\approx 327$, which conditions wrongly; (B) is the ground-up mean $\theta/(\alpha-1)$; (C) and (E) misplace the deductible in the formula.)*

**Answer: (B).** For $X\sim\text{Uniform}(0,b)$ with $u\leq b$, $S(x)=1-x/b$ on $[0,b]$, so $E[X\wedge u]=\int_{0}^{u}\Big(1-\frac{x}{b}\Big)dx=u-\frac{u^{2}}{2b}$. With $u=3000$, $b=5000$: $E[X\wedge 3000]=3000-\dfrac{3000^{2}}{2(5000)}=3000-\dfrac{9{,}000{,}000}{10{,}000}=3000-900=2100$. So $E[X\wedge 3000]=\$2{,}100$. *(Distractor (D) is the unlimited mean $b/2=2500$; (C) halves $u$ inside the subtraction; (E) ignores the averaging and returns the cap.)*

**Answer: (B).** For $S\sim\text{Uniform}(0,b)$ with $d\leq b$, the tail integral gives $E[(S-d)_{+}]=\int_{d}^{b}\Big(1-\frac{x}{b}\Big)dx=\dfrac{(b-d)^{2}}{2b}$. With $b=8000$, $d=5000$: $E[(S-5000)_{+}]=\dfrac{(8000-5000)^{2}}{2(8000)}=\dfrac{3000^{2}}{16{,}000}=\dfrac{9{,}000{,}000}{16{,}000}=562.5$. So the net stop-loss premium is about $\$563$. *(Distractor (C) drops the factor of $2$ in the denominator; (D) uses $(b-d)/2$; (E) is the raw retention gap $b-d$ halved differently.)*

**Answer: (B).** For a mixture, the mean is the weighted average of the component means: $E[X]=\sum_i p_i\,E_i[X]=0.6(200)+0.4(700)=120+280=400$. So $E[X]=\$400$. *(Distractor (A) swaps the weights onto the wrong means; (C) uses the simple average $(200+700)/2=450$; (E) sums the means without weighting.)*

**Answer: (C).** The layer cost from $d$ to $u$ is $E[X\wedge u]-E[X\wedge d]$, where $E[X\wedge a]=\theta(1-e^{-a/\theta})$. $E[X\wedge 2500]=1000(1-e^{-2.5})=1000(1-0.082085)=917.92$. $E[X\wedge 500]=1000(1-e^{-0.5})=1000(1-0.606531)=393.47$. Layer cost $=917.92-393.47=524.45\approx\$524$. *(Distractor (B) is the $\$500$-to-$\$2{,}000$ layer; (D) is the no-limit deductible value $\theta e^{-0.5}=607$; (A) miscomputes the upper limited value.)*

**Answer: (A).** The standard per-loss payment with coinsurance $\alpha$, deductible $d$, maximum covered loss $u$ is $E[Y]=\alpha\big(E[X\wedge u]-E[X\wedge d]\big)$. Layer cost $=E[X\wedge 5000]-E[X\wedge 1000]=3200-900=2300$. Apply coinsurance: $E[Y]=0.80(2300)=1840$. So the expected payment per loss is $\$1{,}840$. *(Distractor (B) forgets the coinsurance; (C) applies coinsurance only to the upper limited value $0.8(3200)$; (E) sums the limited values then scales.)*

**Answer: (C).** After $10\%$ inflation the losses are exponential with mean $\theta'=1.10(2000)=2200$. The deductible stays at $d=1000$, so $E[(X'-d)_{+}]=\theta' e^{-d/\theta'}=2200\,e^{-1000/2200}=2200\,e^{-0.454545}$. $e^{-0.454545}=0.634736$, so $E[(X'-d)_{+}]=2200(0.634736)\approx 1396.4$. So next year's expected payment per loss is about $\$1{,}396$. *(Distractor (A) is this year's value $2000e^{-0.5}=1213$, forgetting inflation; (B) inflates only the deductible; (D) and (E) over-inflate the mean.)*