{
  "deckName": "Exam P — Risk & Insurance Applications",
  "examCode": "Exam P",
  "cards": [
    {
      "front": "An ordinary deductible $d$ is applied to a loss $X$. Write the insurer's payment random variable $Y$.",
      "back": "The insurer pays the excess of the loss over the deductible, with nothing paid when the loss is below $d$:\n$Y=(X-d)_{+}=\\max(X-d,0)$.\nSo $Y=0$ when $X\\leq d$ and $Y=X-d$ when $X>d$.",
      "tag": "deductibles"
    },
    {
      "front": "For a continuous nonnegative loss $X$ with survival function $S(x)$, give the tail-integral formula for the expected payment per loss under an ordinary deductible $d$.",
      "back": "$E[(X-d)_{+}]=\\int_{d}^{\\infty} S(x)\\,dx$, 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.",
      "tag": "expected-payment"
    },
    {
      "front": "Define the limited loss variable $X\\wedge u$ for a policy limit (maximum covered loss) $u$, and give the tail formula for $E[X\\wedge u]$.",
      "back": "$X\\wedge u=\\min(X,u)$: the loss is paid in full up to $u$ and capped at $u$ above it. For a nonnegative loss, $E[X\\wedge u]=\\int_{0}^{u} S(x)\\,dx$.",
      "tag": "limits"
    },
    {
      "front": "Distinguish **expected payment per loss** from **expected payment per payment** under an ordinary deductible $d$.",
      "back": "Per loss averages over all losses, including those below $d$ that pay $0$: $E[(X-d)_{+}]$.\nPer payment conditions on a payment actually being made ($X>d$): $E[X-d\\mid X>d]=\\dfrac{E[(X-d)_{+}]}{S(d)}$.\nThe per-payment figure is larger because it excludes the zero-payment cases.",
      "tag": "per-loss-vs-per-payment"
    },
    {
      "front": "Losses follow an exponential distribution with mean $\\theta=1000$. With an ordinary deductible $d=500$, find the expected payment **per loss**.",
      "back": "For an exponential loss, $S(x)=e^{-x/\\theta}$, so\n$E[(X-d)_{+}]=\\int_{500}^{\\infty} e^{-x/1000}\\,dx=1000\\,e^{-500/1000}=1000\\,e^{-0.5}$.\nThus $E[(X-d)_{+}]=1000(0.60653)\\approx 606.53$.",
      "tag": "exponential-model"
    },
    {
      "front": "Why is the expected payment **per payment** for an exponential loss with an ordinary deductible $d$ simply equal to the mean $\\theta$, regardless of $d$?",
      "back": "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: $E[(X-d)_{+}]/S(d)=\\theta e^{-d/\\theta}/e^{-d/\\theta}=\\theta$.)",
      "tag": "per-loss-vs-per-payment"
    },
    {
      "front": "What is a coinsurance factor $\\alpha$, and how does it combine with a deductible $d$ and a maximum covered loss $u$ in the standard layered payment formula?",
      "back": "Coinsurance has the insurer pay a fraction $\\alpha$ (with $0<\\alpha\\leq 1$) of the covered amount. The standard per-loss payment is\n$Y=\\alpha\\big[(X\\wedge u)-(X\\wedge d)\\big]=\\alpha\\big[\\min(X,u)-\\min(X,d)\\big]$,\nso $E[Y]=\\alpha\\big(E[X\\wedge u]-E[X\\wedge d]\\big)$.",
      "tag": "coinsurance"
    },
    {
      "front": "Express the expected cost of the **layer** from $d$ to $u$ (deductible $d$, maximum covered loss $u$) using limited expected values.",
      "back": "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 $\\alpha$.",
      "tag": "layers"
    },
    {
      "front": "Losses are exponential with mean $\\theta=1000$. A policy pays the layer between deductible $d=500$ and maximum covered loss $u=2000$. Find the expected payment per loss.",
      "back": "Use $E[X\\wedge a]=\\theta(1-e^{-a/\\theta})$.\n$E[X\\wedge 2000]=1000(1-e^{-2})=1000(0.86466)=864.66$.\n$E[X\\wedge 500]=1000(1-e^{-0.5})=1000(0.39347)=393.47$.\nLayer cost $=864.66-393.47\\approx 471.20$.",
      "tag": "layers"
    },
    {
      "front": "Losses are exponential with mean $\\theta=1000$. The policy has deductible $d=500$, maximum covered loss $u=2000$, and coinsurance $\\alpha=0.8$. Find the expected payment per loss.",
      "back": "The layer cost (from the previous setup) is $E[X\\wedge 2000]-E[X\\wedge 500]=864.66-393.47=471.20$.\nApplying coinsurance: $E[Y]=0.8(471.20)\\approx 376.96$.",
      "tag": "coinsurance"
    },
    {
      "front": "When a policy has a maximum covered loss (policy limit) $u$, the payment variable is **right-censored**. Describe the resulting distribution of $X\\wedge u$.",
      "back": "$X\\wedge u$ 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.",
      "tag": "censoring"
    },
    {
      "front": "Losses are exponential with mean $\\theta=1000$ and the policy limit (maximum covered loss) is $u=2000$ with no deductible. Find $E[X\\wedge u]$ and the probability mass at the cap.",
      "back": "$E[X\\wedge u]=\\theta(1-e^{-u/\\theta})=1000(1-e^{-2})=1000(0.86466)\\approx 864.66$.\nThe point mass at $u=2000$ is $\\Pr(X\\geq 2000)=S(2000)=e^{-2}\\approx 0.1353$.",
      "tag": "censoring"
    },
    {
      "front": "Losses are uniform on $[0,2000]$. With an ordinary deductible $d=500$, find the expected payment per loss.",
      "back": "For $X\\sim\\text{Uniform}(0,b)$ with $b=2000$, $S(x)=1-x/b$ on $[0,b]$.\n$E[(X-d)_{+}]=\\int_{500}^{2000}\\Big(1-\\frac{x}{2000}\\Big)dx=\\frac{(b-d)^{2}}{2b}=\\frac{(1500)^{2}}{4000}$.\nThus $E[(X-d)_{+}]=\\dfrac{2{,}250{,}000}{4000}=562.5$.",
      "tag": "expected-payment"
    },
    {
      "front": "Losses are uniform on $[0,2000]$. Find the limited expected value $E[X\\wedge u]$ for a maximum covered loss $u=1500$.",
      "back": "For $X\\sim\\text{Uniform}(0,b)$ and $u\\leq b$, $E[X\\wedge u]=\\int_{0}^{u}\\Big(1-\\frac{x}{b}\\Big)dx=u-\\frac{u^{2}}{2b}$.\nWith $u=1500,\\ b=2000$: $E[X\\wedge u]=1500-\\dfrac{1500^{2}}{4000}=1500-562.5=937.5$.",
      "tag": "limits"
    },
    {
      "front": "State the standard Exam P (single-parameter scale) **Pareto** survival function and mean, using shape $\\alpha$ and scale $\\theta$.",
      "back": "$S(x)=\\Big(\\dfrac{\\theta}{x+\\theta}\\Big)^{\\alpha}$ 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.",
      "tag": "pareto-model"
    },
    {
      "front": "Give the Pareto limited expected value $E[X\\wedge u]$ for shape $\\alpha\\neq 1$ and scale $\\theta$.",
      "back": "$E[X\\wedge u]=\\dfrac{\\theta}{\\alpha-1}\\left[1-\\Big(\\dfrac{\\theta}{u+\\theta}\\Big)^{\\alpha-1}\\right]$. Subtracting two such values, $E[X\\wedge u]-E[X\\wedge d]$, gives the expected cost of the layer from $d$ to $u$.",
      "tag": "pareto-model"
    },
    {
      "front": "Pareto losses have $\\alpha=3$ and $\\theta=2000$. With an ordinary deductible $d=1000$, find the expected payment per loss.",
      "back": "$E[(X-d)_{+}]=\\int_{d}^{\\infty} S(x)\\,dx=\\dfrac{\\theta}{\\alpha-1}\\Big(\\dfrac{\\theta}{d+\\theta}\\Big)^{\\alpha-1}$.\n$=\\dfrac{2000}{2}\\Big(\\dfrac{2000}{3000}\\Big)^{2}=1000\\cdot\\dfrac{4}{9}\\approx 444.44$.",
      "tag": "pareto-model"
    },
    {
      "front": "Pareto losses have $\\alpha=3$ and $\\theta=2000$, deductible $d=1000$. Find the expected payment **per payment**.",
      "back": "Per payment $=\\dfrac{E[(X-d)_{+}]}{S(d)}$. Here $S(1000)=\\Big(\\dfrac{2000}{3000}\\Big)^{3}=\\dfrac{8}{27}\\approx 0.2963$ and $E[(X-d)_{+}]=444.44$.\nSo per payment $=\\dfrac{444.44}{0.2963}\\approx 1500$.\n(Consistent with the Pareto mean-residual-life $\\dfrac{d+\\theta}{\\alpha-1}=\\dfrac{3000}{2}=1500$.)",
      "tag": "per-loss-vs-per-payment"
    },
    {
      "front": "Contrast an **ordinary** deductible with a **franchise** deductible of the same size $d$.",
      "back": "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 $=X\\cdot\\mathbf{1}\\{X>d\\}$.",
      "tag": "deductibles"
    },
    {
      "front": "Losses are exponential with mean $\\theta=1000$ and the policy has a franchise deductible $d=500$. Find the expected payment per loss.",
      "back": "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).\n$=1000e^{-0.5}+500e^{-0.5}=1500\\,e^{-0.5}=1500(0.60653)\\approx 909.80$.",
      "tag": "deductibles"
    },
    {
      "front": "Losses are exponential with mean $\\theta=2000$. With an ordinary deductible $d=500$ and coinsurance $\\alpha=0.8$ (no upper limit), find the expected payment per loss.",
      "back": "Without a limit, $E[(X-d)_{+}]=\\theta e^{-d/\\theta}=2000\\,e^{-500/2000}=2000\\,e^{-0.25}=2000(0.77880)=1557.60$.\nApplying coinsurance: $E[Y]=0.8(1557.60)\\approx 1246.08$.",
      "tag": "coinsurance"
    },
    {
      "front": "How does uniform inflation by a factor $(1+r)$ on the loss interact with a fixed deductible $d$? State the rule for adjusting the expected payment.",
      "back": "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 $E[((1+r)X-d)_{+}]=(1+r)\\,E\\big[(X-\\tfrac{d}{1+r})_{+}\\big]$.",
      "tag": "inflation"
    },
    {
      "front": "This year losses are exponential with mean $\\theta=1000$ and the deductible is $500$. Next year losses inflate $5\\%$ with the deductible unchanged. Find next year's expected payment per loss.",
      "back": "Inflated losses are exponential with mean $1.05(1000)=1050$. With $d=500$:\n$E[(X-d)_{+}]=1050\\,e^{-500/1050}=1050\\,e^{-0.47619}=1050(0.62115)\\approx 652.20$.\n(Up from $606.53$ this year — inflation raises layer cost faster than the loss itself because the fixed deductible shrinks in real terms.)",
      "tag": "inflation"
    },
    {
      "front": "Define a **mixture** (two-component) loss distribution and give its survival function in terms of the components.",
      "back": "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 $E[g(X)]=\\sum_{i} p_{i}\\,E_{i}[g(X)]$.",
      "tag": "mixtures"
    },
    {
      "front": "A loss is a mixture: with probability $0.7$ it is exponential with mean $100$, and with probability $0.3$ it is exponential with mean $500$. Find $E[X]$ and $\\mathrm{Var}(X)$.",
      "back": "$E[X]=0.7(100)+0.3(500)=70+150=220$.\nFor each exponential, $E[X^{2}]=2\\theta^{2}$, so $E[X^{2}]=0.7(2\\cdot100^{2})+0.3(2\\cdot500^{2})=0.7(20000)+0.3(500000)=14000+150000=164000$.\n$\\mathrm{Var}(X)=164000-220^{2}=164000-48400=115600$.",
      "tag": "mixtures"
    },
    {
      "front": "A claim occurs with probability $0.2$; given a claim, the loss is exponential with mean $5000$ (otherwise the loss is $0$). Find $E[X]$ and $\\mathrm{Var}(X)$ of the loss.",
      "back": "This is a mixture with a point mass at $0$. $E[X]=0.2(5000)=1000$.\n$E[X^{2}]=0.2\\,(2\\cdot5000^{2})=0.2(50{,}000{,}000)=10{,}000{,}000$.\n$\\mathrm{Var}(X)=10{,}000{,}000-1000^{2}=9{,}000{,}000$ (so $\\mathrm{SD}=3000$).",
      "tag": "mixtures"
    },
    {
      "front": "Define a **stop-loss** insurance on an aggregate loss $S$ with retention $d$, and give its expected cost.",
      "back": "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 $S$.",
      "tag": "stop-loss"
    },
    {
      "front": "Aggregate losses $S$ are uniform on $[0,10]$ (in millions). Find the net stop-loss premium $E[(S-d)_{+}]$ for retention $d=4$.",
      "back": "$E[(S-d)_{+}]=\\int_{4}^{10}\\Big(1-\\frac{x}{10}\\Big)dx=\\dfrac{(10-4)^{2}}{2(10)}=\\dfrac{36}{20}=1.8$ (million).\nThis uses the uniform tail formula $\\int_{d}^{b}(1-x/b)\\,dx=(b-d)^{2}/(2b)$.",
      "tag": "stop-loss"
    },
    {
      "front": "Two stop-loss retentions $d_{1}<d_{2}$ satisfy the identity $E[(S-d_{1})_{+}]-E[(S-d_{2})_{+}]=\\,?$ Interpret it.",
      "back": "$E[(S-d_{1})_{+}]-E[(S-d_{2})_{+}]=E[S\\wedge d_{2}]-E[S\\wedge d_{1}]=\\int_{d_{1}}^{d_{2}} S_{S}(x)\\,dx$. It is the expected cost of the **layer** of aggregate losses between $d_{1}$ and $d_{2}$.",
      "tag": "stop-loss"
    },
    {
      "front": "Derive the relationship $E[(X-d)_{+}]=E[X]-E[X\\wedge d]$ and say why it is useful.",
      "back": "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.",
      "tag": "expected-payment"
    },
    {
      "front": "Losses are exponential with mean $\\theta=1000$ and an ordinary deductible $d=500$ applies. Find the **variance** of the payment per loss $Y=(X-d)_{+}$.",
      "back": "By memorylessness, $E[Y]=\\theta e^{-d/\\theta}=1000e^{-0.5}=606.53$ and $E[Y^{2}]=2\\theta^{2}e^{-d/\\theta}=2(1000)^{2}e^{-0.5}=2{,}000{,}000(0.60653)=1{,}213{,}061$.\n$\\mathrm{Var}(Y)=1{,}213{,}061-606.53^{2}=1{,}213{,}061-367{,}879\\approx 845{,}182$.",
      "tag": "expected-payment"
    },
    {
      "front": "Why must the payment-per-loss random variable under a deductible be treated as a **mixed** distribution when computing its variance, and how do you handle the mass at zero?",
      "back": "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 $\\Pr(Y=0)$.",
      "tag": "censoring"
    },
    {
      "front": "A policy applies coinsurance $\\alpha$, then a maximum covered loss $u$, on a loss $X$ with no deductible. Is the expected payment $\\alpha\\,E[X\\wedge u]$ or $E[X\\wedge(\\alpha u)]$? Explain the ordering.",
      "back": "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.",
      "tag": "coinsurance"
    },
    {
      "front": "Pareto losses have $\\alpha=3$ and $\\theta=2000$. Find the expected cost of the layer from $d=0$ up to a maximum covered loss $u=3000$ (i.e. $E[X\\wedge 3000]$).",
      "back": "$E[X\\wedge u]=\\dfrac{\\theta}{\\alpha-1}\\left[1-\\Big(\\dfrac{\\theta}{u+\\theta}\\Big)^{\\alpha-1}\\right]=\\dfrac{2000}{2}\\left[1-\\Big(\\dfrac{2000}{5000}\\Big)^{2}\\right]$.\n$=1000\\big[1-0.16\\big]=1000(0.84)=840$.",
      "tag": "pareto-model"
    }
  ]
}