Exam P — Risk & Insurance Applications Flashcards

The insurer pays the excess of the loss over the deductible, with nothing paid when the loss is below $d$: $Y=(X-d)_{+}=\max(X-d,0)$. So $Y=0$ when $X\leq d$ and $Y=X-d$ when $X>d$.

$E[(X-d)_{+}]=\int_{d}^{\infty} S(x)\,dx$, where $S(x)=\Pr(X>x)$. This is the area under the survival curve to the right of $d$, and it is usually faster than integrating $(x-d)f(x)$ directly.

$X\wedge u=\min(X,u)$: the loss is paid in full up to $u$ and capped at $u$ above it. For a nonnegative loss, $E[X\wedge u]=\int_{0}^{u} S(x)\,dx$.

Per loss averages over all losses, including those below $d$ that pay $0$: $E[(X-d)_{+}]$. Per payment conditions on a payment actually being made ($X>d$): $E[X-d\mid X>d]=\dfrac{E[(X-d)_{+}]}{S(d)}$. The per-payment figure is larger because it excludes the zero-payment cases.

For an exponential loss, $S(x)=e^{-x/\theta}$, so $E[(X-d)_{+}]=\int_{500}^{\infty} e^{-x/1000}\,dx=1000\,e^{-500/1000}=1000\,e^{-0.5}$. Thus $E[(X-d)_{+}]=1000(0.60653)\approx 606.53$.

The exponential is memoryless: given $X>d$, the excess $X-d$ is again exponential with the same mean $\theta$. Hence $E[X-d\mid X>d]=\theta$. (Check: $E[(X-d)_{+}]/S(d)=\theta e^{-d/\theta}/e^{-d/\theta}=\theta$.)

Coinsurance has the insurer pay a fraction $\alpha$ (with $0<\alpha\leq 1$) of the covered amount. The standard per-loss payment is $Y=\alpha\big[(X\wedge u)-(X\wedge d)\big]=\alpha\big[\min(X,u)-\min(X,d)\big]$, so $E[Y]=\alpha\big(E[X\wedge u]-E[X\wedge d]\big)$.

The expected payment per loss for the layer is $E[X\wedge u]-E[X\wedge d]$, equivalently $\int_{d}^{u} S(x)\,dx$. Coinsurance scales this by $\alpha$.

Use $E[X\wedge a]=\theta(1-e^{-a/\theta})$. $E[X\wedge 2000]=1000(1-e^{-2})=1000(0.86466)=864.66$. $E[X\wedge 500]=1000(1-e^{-0.5})=1000(0.39347)=393.47$. Layer cost $=864.66-393.47\approx 471.20$.

The layer cost (from the previous setup) is $E[X\wedge 2000]-E[X\wedge 500]=864.66-393.47=471.20$. Applying coinsurance: $E[Y]=0.8(471.20)\approx 376.96$.

$X\wedge u$ is a **mixed** random variable: it has the continuous density of $X$ on $[0,u)$, plus a probability mass at the point $u$ equal to $\Pr(X\geq u)=S(u)$. All loss outcomes above $u$ pile up at the cap.

$E[X\wedge u]=\theta(1-e^{-u/\theta})=1000(1-e^{-2})=1000(0.86466)\approx 864.66$. The point mass at $u=2000$ is $\Pr(X\geq 2000)=S(2000)=e^{-2}\approx 0.1353$.

For $X\sim\text{Uniform}(0,b)$ with $b=2000$, $S(x)=1-x/b$ on $[0,b]$. $E[(X-d)_{+}]=\int_{500}^{2000}\Big(1-\frac{x}{2000}\Big)dx=\frac{(b-d)^{2}}{2b}=\frac{(1500)^{2}}{4000}$. Thus $E[(X-d)_{+}]=\dfrac{2{,}250{,}000}{4000}=562.5$.

For $X\sim\text{Uniform}(0,b)$ and $u\leq b$, $E[X\wedge u]=\int_{0}^{u}\Big(1-\frac{x}{b}\Big)dx=u-\frac{u^{2}}{2b}$. With $u=1500,\ b=2000$: $E[X\wedge u]=1500-\dfrac{1500^{2}}{4000}=1500-562.5=937.5$.

$S(x)=\Big(\dfrac{\theta}{x+\theta}\Big)^{\alpha}$ for $x>0$. The mean exists when $\alpha>1$ and equals $E[X]=\dfrac{\theta}{\alpha-1}$. The Pareto is heavy-tailed, so high deductibles and limits matter.

$E[X\wedge u]=\dfrac{\theta}{\alpha-1}\left[1-\Big(\dfrac{\theta}{u+\theta}\Big)^{\alpha-1}\right]$. Subtracting two such values, $E[X\wedge u]-E[X\wedge d]$, gives the expected cost of the layer from $d$ to $u$.

$E[(X-d)_{+}]=\int_{d}^{\infty} S(x)\,dx=\dfrac{\theta}{\alpha-1}\Big(\dfrac{\theta}{d+\theta}\Big)^{\alpha-1}$. $=\dfrac{2000}{2}\Big(\dfrac{2000}{3000}\Big)^{2}=1000\cdot\dfrac{4}{9}\approx 444.44$.

Per payment $=\dfrac{E[(X-d)_{+}]}{S(d)}$. Here $S(1000)=\Big(\dfrac{2000}{3000}\Big)^{3}=\dfrac{8}{27}\approx 0.2963$ and $E[(X-d)_{+}]=444.44$. So per payment $=\dfrac{444.44}{0.2963}\approx 1500$. (Consistent with the Pareto mean-residual-life $\dfrac{d+\theta}{\alpha-1}=\dfrac{3000}{2}=1500$.)

Both pay nothing when $X\leq d$. Above $d$: an ordinary deductible pays $X-d$, whereas a franchise deductible pays the **full** loss $X$ (the deductible "disappears" once the threshold is crossed). So franchise payment $=X\cdot\mathbf{1}\{X>d\}$.

Franchise payment per loss $=E[(X-d)_{+}]+d\,S(d)$ (the ordinary-deductible amount plus the $d$ that is no longer withheld on paid claims). $=1000e^{-0.5}+500e^{-0.5}=1500\,e^{-0.5}=1500(0.60653)\approx 909.80$.

Without a limit, $E[(X-d)_{+}]=\theta e^{-d/\theta}=2000\,e^{-500/2000}=2000\,e^{-0.25}=2000(0.77880)=1557.60$. Applying coinsurance: $E[Y]=0.8(1557.60)\approx 1246.08$.

Inflation scales the loss but **not** the deductible (unless the problem says the deductible is indexed). If $X$ inflates to $(1+r)X$, then for an exponential-type model evaluate the layer using the inflated mean. In general $E[((1+r)X-d)_{+}]=(1+r)\,E\big[(X-\tfrac{d}{1+r})_{+}\big]$.

Inflated losses are exponential with mean $1.05(1000)=1050$. With $d=500$: $E[(X-d)_{+}]=1050\,e^{-500/1050}=1050\,e^{-0.47619}=1050(0.62115)\approx 652.20$. (Up from $606.53$ this year — inflation raises layer cost faster than the loss itself because the fixed deductible shrinks in real terms.)

A mixture draws from component $i$ with mixing weight $p_{i}$ (with $\sum p_{i}=1$). Its survival function is the weighted average $S(x)=\sum_{i} p_{i}\,S_{i}(x)$, and likewise $f(x)=\sum_{i} p_{i} f_{i}(x)$ and $E[g(X)]=\sum_{i} p_{i}\,E_{i}[g(X)]$.

$E[X]=0.7(100)+0.3(500)=70+150=220$. For each exponential, $E[X^{2}]=2\theta^{2}$, so $E[X^{2}]=0.7(2\cdot100^{2})+0.3(2\cdot500^{2})=0.7(20000)+0.3(500000)=14000+150000=164000$. $\mathrm{Var}(X)=164000-220^{2}=164000-48400=115600$.

This is a mixture with a point mass at $0$. $E[X]=0.2(5000)=1000$. $E[X^{2}]=0.2\,(2\cdot5000^{2})=0.2(50{,}000{,}000)=10{,}000{,}000$. $\mathrm{Var}(X)=10{,}000{,}000-1000^{2}=9{,}000{,}000$ (so $\mathrm{SD}=3000$).

Stop-loss reimburses the excess of aggregate losses over a retention $d$: payment $=(S-d)_{+}$. The expected (net) stop-loss premium is $E[(S-d)_{+}]=\int_{d}^{\infty} S_{S}(x)\,dx$, where $S_{S}$ is the survival function of the aggregate $S$. It is structurally identical to an ordinary deductible applied to $S$.

$E[(S-d)_{+}]=\int_{4}^{10}\Big(1-\frac{x}{10}\Big)dx=\dfrac{(10-4)^{2}}{2(10)}=\dfrac{36}{20}=1.8$ (million). This uses the uniform tail formula $\int_{d}^{b}(1-x/b)\,dx=(b-d)^{2}/(2b)$.

$E[(S-d_{1})_{+}]-E[(S-d_{2})_{+}]=E[S\wedge d_{2}]-E[S\wedge d_{1}]=\int_{d_{1}}^{d_{2}} S_{S}(x)\,dx$. It is the expected cost of the **layer** of aggregate losses between $d_{1}$ and $d_{2}$.

Since $X=(X\wedge d)+(X-d)_{+}$ for every outcome (split at $d$), taking expectations gives $E[X]=E[X\wedge d]+E[(X-d)_{+}]$, hence $E[(X-d)_{+}]=E[X]-E[X\wedge d]$. It lets you get the deductible payment from the (often tabulated) limited expected value without a new integral.

By memorylessness, $E[Y]=\theta e^{-d/\theta}=1000e^{-0.5}=606.53$ and $E[Y^{2}]=2\theta^{2}e^{-d/\theta}=2(1000)^{2}e^{-0.5}=2{,}000{,}000(0.60653)=1{,}213{,}061$. $\mathrm{Var}(Y)=1{,}213{,}061-606.53^{2}=1{,}213{,}061-367{,}879\approx 845{,}182$.

When $X\leq d$ the payment is exactly $0$, giving a probability mass $\Pr(X\leq d)=F(d)$ at $Y=0$; above $d$ the payment is continuous. Compute moments as $E[Y^{k}]=\int_{d}^{\infty}(x-d)^{k} f(x)\,dx$ (the mass at $0$ contributes nothing to $E[Y^{k}]$), then $\mathrm{Var}(Y)=E[Y^{2}]-(E[Y])^{2}$. The mass at $0$ still matters for probabilities like $\Pr(Y=0)$.

Order matters. If the limit $u$ caps the **ground-up loss** before coinsurance, the payment is $\alpha\min(X,u)$ and $E[Y]=\alpha\,E[X\wedge u]$. If instead the limit caps the **insurer's payment** (a benefit limit $u$ on $\alpha X$), then $Y=\min(\alpha X,u)=\alpha\min(X,u/\alpha)$ and $E[Y]=\alpha\,E[X\wedge(u/\alpha)]$. Read which quantity the cap applies to.

$E[X\wedge u]=\dfrac{\theta}{\alpha-1}\left[1-\Big(\dfrac{\theta}{u+\theta}\Big)^{\alpha-1}\right]=\dfrac{2000}{2}\left[1-\Big(\dfrac{2000}{5000}\Big)^{2}\right]$. $=1000\big[1-0.16\big]=1000(0.84)=840$.