{ "deckName": "Exam FAM — Coverage Modifications", "examCode": "Exam FAM", "cards": [ { "front": "Define the **limited expected value** $E[X\\wedge u]$ and give its survival-function integral form.", "back": "$X\\wedge u=\\min(X,u)$ is the loss **capped** at the limit $u$, so $E[X\\wedge u]$ is the average payment when nothing above $u$ is covered.\nFor a nonnegative loss, $E[X\\wedge u]=\\displaystyle\\int_0^u S(x)\\,dx$, where $S(x)=P(X>x)$.\nProperties: $E[X\\wedge 0]=0$, $E[X\\wedge u]$ increases in $u$, and $E[X\\wedge u]\\uparrow E[X]$ as $u\\to\\infty$.", "tag": "Limited expected value" }, { "front": "Why is the limited expected value the single workhorse quantity for coverage-modification problems?", "back": "Almost every modified payment can be written as a difference of LEVs. A deductible $d$ removes $E[X\\wedge d]$; a maximum covered loss $u$ caps at $E[X\\wedge u]$; together the covered amount is $E[X\\wedge u]-E[X\\wedge d]$.\nBecause $E[X\\wedge u]=\\int_0^u S(x)\\,dx$, you only need the survival function (or a tabulated LEV column) to value deductibles, limits, coinsurance, and inflation in one consistent framework.", "tag": "Limited expected value" }, { "front": "For an exponential loss with mean $\\theta$, state $E[X\\wedge u]$ and evaluate it at $u=500,1000,2000$ for $\\theta=1000$.", "back": "For $X\\sim\\text{Exp}(\\theta)$, $S(x)=e^{-x/\\theta}$, so $E[X\\wedge u]=\\theta\\big(1-e^{-u/\\theta}\\big)$.\nWith $\\theta=1000$:\n$E[X\\wedge 500]=1000(1-e^{-0.5})=1000(1-0.606531)=\\$393.47$.\n$E[X\\wedge 1000]=1000(1-e^{-1})=1000(1-0.367879)=\\$632.12$.\n$E[X\\wedge 2000]=1000(1-e^{-2})=1000(1-0.135335)=\\$864.66$.\nAll lie below $E[X]=\\theta=\\$1{,}000$, approaching it as $u$ grows.", "tag": "Limited expected value" }, { "front": "A loss is **uniform on $(0,1000)$**. Find $E[X]$, $E[X\\wedge 200]$, and $E[X\\wedge 500]$.", "back": "For $X\\sim U(0,b)$, $S(x)=1-\\tfrac{x}{b}$ on $[0,b]$, so for $u\\le b$, $E[X\\wedge u]=\\int_0^u\\!\\big(1-\\tfrac{x}{b}\\big)dx = u-\\dfrac{u^2}{2b}$.\nWith $b=1000$: $E[X]=\\tfrac{b}{2}=\\$500.00$.\n$E[X\\wedge 200]=200-\\dfrac{200^2}{2000}=200-20=\\$180.00$.\n$E[X\\wedge 500]=500-\\dfrac{500^2}{2000}=500-125=\\$375.00$.", "tag": "Limited expected value" }, { "front": "A discrete loss takes values $50,150,250,500$ with probabilities $0.4,0.3,0.2,0.1$. Find $E[X]$ and $E[X\\wedge 200]$ two ways.", "back": "**Definition:** $E[X\\wedge 200]=\\sum\\min(x,200)\\,p(x)=50(0.4)+150(0.3)+200(0.2)+200(0.1)$\n$=20+45+40+20=\\$125.00$. And $E[X]=50(0.4)+150(0.3)+250(0.2)+500(0.1)=\\$165.00$.\n**Survival integral:** $S(x)=1$ on $[0,50)$, $0.6$ on $[50,150)$, $0.3$ on $[150,250)$, $0.1$ on $[250,500)$. So $E[X\\wedge 200]=50(1)+100(0.6)+50(0.3)=50+60+15=\\$125.00$. The two agree.", "tag": "Limited expected value" }, { "front": "For a **Pareto** loss with parameters $\\alpha$ and $\\theta$, state $E[X\\wedge u]$ and evaluate it at $u=1000,3000$ for $\\alpha=3,\\theta=2000$.", "back": "For $X\\sim\\text{Pareto}(\\alpha,\\theta)$ with $\\alpha>1$, $S(x)=\\big(\\tfrac{\\theta}{\\theta+x}\\big)^{\\alpha}$ and $E[X\\wedge u]=\\dfrac{\\theta}{\\alpha-1}\\left[1-\\left(\\dfrac{\\theta}{\\theta+u}\\right)^{\\alpha-1}\\right]$.\nWith $\\alpha=3,\\theta=2000$, $E[X]=\\tfrac{\\theta}{\\alpha-1}=\\$1{,}000$.\n$E[X\\wedge 1000]=1000\\big[1-(\\tfrac{2000}{3000})^{2}\\big]=1000(1-0.444444)=\\$555.56$.\n$E[X\\wedge 3000]=1000\\big[1-(\\tfrac{2000}{5000})^{2}\\big]=1000(1-0.16)=\\$840.00$.", "tag": "Limited expected value" }, { "front": "Define an **ordinary deductible** $d$ and the **per-loss** random variable it induces.", "back": "Under an ordinary deductible $d$, the insurer pays the excess of the loss over $d$, and nothing if the loss is below $d$. The per-loss payment is $Y^{L}=(X-d)_{+}$, equal to $0$ when $X\\le d$ and $X-d$ when $X>d$.\n\"Per-loss\" means the variable is defined over **every** loss, including those that produce a zero payment — so $Y^{L}$ has a point mass at $0$ of size $F(d)=P(X\\le d)$.", "tag": "Ordinary deductible" }, { "front": "Give the LEV formula for the expected **per-loss** payment under an ordinary deductible $d$.", "back": "$E[(X-d)_{+}]=E[X]-E[X\\wedge d]$.\nIntuition: $E[X]$ is the full ground-up cost; $E[X\\wedge d]$ is the part the deductible eliminates (everything up to $d$). The difference is the insurer's average payment **per loss**, averaging in the zero payments from small losses.\nEquivalently $E[(X-d)_{+}]=\\int_d^\\infty S(x)\\,dx$.", "tag": "Ordinary deductible" }, { "front": "Define the **per-payment** variable $Y^{P}$ for an ordinary deductible $d$ and give its mean.", "back": "$Y^{P}=X-d\\mid X>d$ is the payment **conditioned on a payment occurring** (i.e. on $X>d$). It excludes the losses below $d$ that produce no claim, so it has **no** mass at $0$.\nIts mean is $E[Y^{P}]=\\dfrac{E[(X-d)_{+}]}{S(d)}=\\dfrac{E[X]-E[X\\wedge d]}{S(d)}$, which is the per-loss expectation \"grossed up\" by dividing out the probability $S(d)=P(X>d)$ that a payment happens.", "tag": "Per-loss vs per-payment" }, { "front": "Distinguish **per-loss** ($Y^{L}$) and **per-payment** ($Y^{P}$) variables, and state the relationship between their means.", "back": "**Per-loss** $Y^{L}=(X-d)_{+}$ is defined for every loss and includes a mass at $0$; its mean drives the insurer's expected **aggregate** cost when frequency counts all losses.\n**Per-payment** $Y^{P}=(X-d\\mid X>d)$ conditions on a payment being made; its mean and distribution describe the **size of an actual claim**.\nRelationship: $E[Y^{L}]=S(d)\\,E[Y^{P}]$. Since $S(d)\\le 1$, the per-payment mean is always at least the per-loss mean.", "tag": "Per-loss vs per-payment" }, { "front": "An **exponential** loss has mean $\\theta=1000$. With an ordinary deductible $d=500$, find the per-loss and per-payment expected payments.", "back": "$E[X]=1000$ and $E[X\\wedge 500]=1000(1-e^{-0.5})=393.47$.\n**Per-loss:** $E[(X-500)_{+}]=E[X]-E[X\\wedge 500]=1000-393.47=\\$606.53$.\nAlso $S(500)=e^{-0.5}=0.606531$.\n**Per-payment:** $E[Y^{P}]=\\dfrac{606.53}{0.606531}=\\$1{,}000.00$.\nThe per-payment mean equals $\\theta$ — the exponential's **memorylessness**: given $X>d$, the excess $X-d$ is again $\\text{Exp}(\\theta)$.", "tag": "Per-loss vs per-payment" }, { "front": "A **discrete** loss takes values $0,100,300,600$ with probabilities $0.5,0.2,0.2,0.1$. With an ordinary deductible $d=200$, find the per-loss and per-payment means.", "back": "$E[X]=0(0.5)+100(0.2)+300(0.2)+600(0.1)=\\$140.00$.\nPer-loss: $E[(X-200)_{+}]=0+0+(300-200)(0.2)+(600-200)(0.1)=20+40=\\$60.00$.\n$S(200)=P(X>200)=0.2+0.1=0.30$.\nPer-payment: $E[Y^{P}]=\\dfrac{60}{0.30}=\\$200.00$. (Equivalently, $Y^{P}$ pays $100$ w.p. $\\tfrac{0.2}{0.3}$ and $400$ w.p. $\\tfrac{0.1}{0.3}$: $100(\\tfrac23)+400(\\tfrac13)=200$.)", "tag": "Per-loss vs per-payment" }, { "front": "A **Pareto** loss has $\\alpha=3,\\theta=2000$. With an ordinary deductible $d=1000$, find the per-loss and per-payment expected payments.", "back": "$E[X]=\\tfrac{2000}{2}=1000$ and $E[X\\wedge 1000]=1000\\big[1-(\\tfrac{2000}{3000})^{2}\\big]=555.56$.\n**Per-loss:** $E[(X-1000)_{+}]=1000-555.56=\\$444.44$.\n$S(1000)=(\\tfrac{2000}{3000})^{3}=0.296296$.\n**Per-payment:** $E[Y^{P}]=\\dfrac{444.44}{0.296296}=\\$1{,}500.00$.\nUnlike the exponential, the Pareto's per-payment mean **exceeds** the ground-up mean — heavy tails make surviving losses larger.", "tag": "Per-loss vs per-payment" }, { "front": "Define a **franchise deductible** $d$ and contrast its payment rule with the ordinary deductible.", "back": "A franchise deductible pays **nothing** if $X\\le d$, but pays the **entire** loss $X$ (not $X-d$) once $X>d$: $Y^{L}=0$ for $X\\le d$ and $Y^{L}=X$ for $X>d$.\nVs. the ordinary deductible, which pays $X-d$ above the threshold. The franchise payment is exactly $d$ larger than the ordinary payment whenever a claim is made, so it \"refunds the deductible\" at the moment the loss crosses $d$.", "tag": "Franchise deductible" }, { "front": "Give the expected **per-loss** and **per-payment** cost of a **franchise** deductible $d$ in terms of the ordinary deductible.", "back": "Because a franchise claim pays $d$ more than the ordinary claim exactly when $X>d$:\n**Per-loss:** $E[Y^{L}_{\\text{fr}}]=E[(X-d)_{+}]+d\\,S(d)=\\big(E[X]-E[X\\wedge d]\\big)+d\\,S(d)$.\n**Per-payment:** $E[Y^{P}_{\\text{fr}}]=\\dfrac{E[X]-E[X\\wedge d]}{S(d)}+d$.\nIn both cases you take the ordinary-deductible result and add back the deductible (weighted by $S(d)$ for per-loss).", "tag": "Franchise deductible" }, { "front": "An **exponential** loss has $\\theta=1000$. For a **franchise** deductible $d=500$, find the expected per-loss and per-payment payments.", "back": "Ordinary results: $E[(X-500)_{+}]=606.53$, $S(500)=e^{-0.5}=0.606531$.\n**Franchise per-loss:** $E[Y^{L}_{\\text{fr}}]=606.53+500(0.606531)=606.53+303.27=\\$909.80$.\n**Franchise per-payment:** $E[Y^{P}_{\\text{fr}}]=\\dfrac{606.53}{0.606531}+500=1000+500=\\$1{,}500.00$.\nEach exceeds the ordinary-deductible counterpart by $d\\,S(d)$ and by $d$, respectively.", "tag": "Franchise deductible" }, { "front": "A loss is **uniform on $(0,1000)$**. For a **franchise** deductible $d=200$, find the expected per-loss payment.", "back": "For $U(0,1000)$: $E[X]=500$, $E[X\\wedge 200]=200-\\tfrac{200^2}{2000}=180$, so $E[(X-200)_{+}]=500-180=320$.\n$S(200)=1-\\tfrac{200}{1000}=0.80$.\n**Franchise per-loss:** $E[Y^{L}_{\\text{fr}}]=E[(X-200)_{+}]+d\\,S(200)=320+200(0.80)=320+160=\\$480.00$.\n(Check directly: $\\int_{200}^{1000} x\\cdot\\tfrac{1}{1000}\\,dx=\\tfrac{1000^2-200^2}{2000}=480$. Agrees.)", "tag": "Franchise deductible" }, { "front": "Define the **policy limit** and the **maximum covered loss** $u$, and how they relate when there is a deductible $d$.", "back": "The **policy limit** is the largest amount the insurer will **pay** on a loss. The **maximum covered loss** $u$ is the largest loss to which coverage applies — the point at which the payment maxes out.\nWith an ordinary deductible $d$ and policy limit $L$, the payment hits the cap when $X-d=L$, i.e. when $X=L+d$. So $u=L+d$, and the maximum payment $L=u-d$.\nDon't confuse the two: the limit caps the **payment**; $u$ caps the **loss**.", "tag": "Limits & coinsurance" }, { "front": "Define the **coinsurance factor** $\\alpha$ and where it enters the payment.", "back": "After the deductible removes losses below $d$ and the maximum covered loss $u$ caps the loss, the insurer pays only a fraction $\\alpha$ (with $0<\\alpha\\le 1$) of the remaining covered amount; the policyholder retains $(1-\\alpha)$.\nCoinsurance multiplies the **already-modified** loss, so it scales the whole covered layer: $\\alpha\\big(E[X\\wedge u]-E[X\\wedge d]\\big)$. The maximum payment becomes $\\alpha(u-d)$.", "tag": "Limits & coinsurance" }, { "front": "State the **master per-loss formula** for expected cost with coinsurance $\\alpha$, ordinary deductible $d$, and maximum covered loss $u$.", "back": "$E[\\text{payment}]=\\alpha\\big(E[X\\wedge u]-E[X\\wedge d]\\big)$.\nReading it left to right: cap the loss at $u$ (limit), subtract the part eliminated by the deductible $E[X\\wedge d]$, then apply coinsurance $\\alpha$. Order matters in the *rule* (deductible first, then limit, then coinsurance), but because both caps are LEVs the expectation collapses to this clean difference.\nSpecial cases: deductible only ($u\\to\\infty$): $\\alpha(E[X]-E[X\\wedge d])$; limit only ($d=0$): $\\alpha\\,E[X\\wedge u]$.", "tag": "Limits & coinsurance" }, { "front": "An **exponential** loss has $\\theta=1000$. A policy has deductible $d=250$, policy limit $2000$, and coinsurance $\\alpha=0.80$. Find the expected per-loss payment.", "back": "Maximum covered loss $u=d+\\text{limit}=250+2000=2250$.\n$E[X\\wedge 2250]=1000(1-e^{-2.25})=1000(1-0.105399)=894.60$.\n$E[X\\wedge 250]=1000(1-e^{-0.25})=1000(1-0.778801)=221.20$.\nPayment $=\\alpha\\big(E[X\\wedge u]-E[X\\wedge d]\\big)=0.80(894.60-221.20)=0.80(673.40)=\\$538.72$.", "tag": "Limits & coinsurance" }, { "front": "A **Pareto** loss has $\\alpha_{P}=3,\\theta=2000$. A policy has deductible $d=500$, policy limit $10{,}000$, and coinsurance $\\alpha=0.90$. Find the expected per-loss payment.", "back": "Maximum covered loss $u=500+10{,}000=10{,}500$. Using $E[X\\wedge u]=\\tfrac{\\theta}{\\alpha_P-1}\\big[1-(\\tfrac{\\theta}{\\theta+u})^{\\alpha_P-1}\\big]$ with $\\tfrac{\\theta}{\\alpha_P-1}=1000$:\n$E[X\\wedge 10500]=1000\\big[1-(\\tfrac{2000}{12500})^{2}\\big]=1000(1-0.0256)=974.40$.\n$E[X\\wedge 500]=1000\\big[1-(\\tfrac{2000}{2500})^{2}\\big]=1000(1-0.64)=360.00$.\nPayment $=0.90(974.40-360.00)=0.90(614.40)=\\$552.96$.", "tag": "Limits & coinsurance" }, { "front": "A loss is **uniform on $(0,1000)$**. A policy pays $\\alpha=0.75$ coinsurance with deductible $d=100$ and policy limit $400$. Find the expected per-loss payment.", "back": "Maximum covered loss $u=100+400=500$. For $U(0,1000)$, $E[X\\wedge u]=u-\\tfrac{u^2}{2000}$:\n$E[X\\wedge 500]=500-\\tfrac{250000}{2000}=500-125=375.00$.\n$E[X\\wedge 100]=100-\\tfrac{10000}{2000}=100-5=95.00$.\nPayment $=0.75(375.00-95.00)=0.75(280.00)=\\$210.00$.", "tag": "Limits & coinsurance" }, { "front": "Define the **loss elimination ratio** $\\text{LER}(d)$ for an ordinary deductible $d$.", "back": "$\\text{LER}(d)=\\dfrac{E[X\\wedge d]}{E[X]}=\\dfrac{E[X]-E[(X-d)_{+}]}{E[X]}$.\nIt is the **fraction of ground-up expected loss eliminated** by introducing a deductible $d$ — the proportional saving to the insurer. $\\text{LER}(0)=0$ (no saving) and $\\text{LER}(d)\\to 1$ as $d\\to\\infty$. The complement $1-\\text{LER}(d)=\\tfrac{E[(X-d)_+]}{E[X]}$ is the fraction of expected loss the insurer still pays.", "tag": "Inflation & LER" }, { "front": "An **exponential** loss has $\\theta=1000$. Find $\\text{LER}(500)$ and interpret it.", "back": "$E[X\\wedge 500]=1000(1-e^{-0.5})=393.47$ and $E[X]=1000$.\n$\\text{LER}(500)=\\dfrac{393.47}{1000}=0.39347$.\nA $\\$500$ deductible eliminates about **39.3%** of the insurer's expected loss cost. Equivalently the insurer still expects to pay $60.7\\%$ of ground-up losses, matching the per-loss $\\$606.53$ out of $\\$1{,}000$.", "tag": "Inflation & LER" }, { "front": "A loss takes values $50,150,250,500$ with probabilities $0.4,0.3,0.2,0.1$. Find $\\text{LER}(200)$.", "back": "$E[X]=50(0.4)+150(0.3)+250(0.2)+500(0.1)=20+45+50+50=\\$165.00$.\n$E[X\\wedge 200]=50(0.4)+150(0.3)+200(0.2)+200(0.1)=20+45+40+20=\\$125.00$.\n$\\text{LER}(200)=\\dfrac{125}{165}=0.75758$.\nThe $\\$200$ deductible eliminates about **75.8%** of expected losses — large here because the bulk of probability sits at or below $200$.", "tag": "Inflation & LER" }, { "front": "How does **uniform inflation** at rate $r$ change the loss, and what happens to a fixed deductible $d$ and fixed limit?", "back": "If every loss inflates uniformly, the new loss is $X'=(1+r)X$. The **policy terms ($d$, $u$) are fixed in dollars**, so relative to the larger losses they become *smaller* — the deductible erodes and the limit binds more often.\nKey scaling identity: $E[X'\\wedge u]=(1+r)\\,E\\!\\big[X\\wedge \\tfrac{u}{1+r}\\big]$. You divide the **dollar threshold** by $(1+r)$ to express it on the original $X$ scale, then multiply the LEV back up by $(1+r)$.", "tag": "Inflation & LER" }, { "front": "State the **master per-loss formula with uniform inflation** at rate $r$ (deductible $d$, max covered loss $u$, coinsurance $\\alpha$).", "back": "$E[\\text{payment}]=\\alpha(1+r)\\left(E\\!\\left[X\\wedge \\tfrac{u}{1+r}\\right]-E\\!\\left[X\\wedge \\tfrac{d}{1+r}\\right]\\right)$, where the LEVs are computed on the **original** (pre-inflation) loss $X$. The $(1+r)$ outside accumulates the layer; the $\\tfrac{1}{1+r}$ inside rescales the fixed dollar thresholds onto $X$'s scale. With $r=0$ it reduces to $\\alpha\\big(E[X\\wedge u]-E[X\\wedge d]\\big)$.", "tag": "Inflation & LER" }, { "front": "An **exponential** loss has $\\theta=1000$, deductible $d=250$, policy limit $2000$ ($u=2250$), coinsurance $\\alpha=0.80$. Apply **10% uniform inflation** and find the new expected per-loss payment.", "back": "Rescale thresholds: $\\tfrac{u}{1.1}=\\tfrac{2250}{1.1}=2045.45$, $\\tfrac{d}{1.1}=\\tfrac{250}{1.1}=227.27$.\n$E[X\\wedge 2045.45]=1000(1-e^{-2.04545})=1000(1-0.129321)=870.68$.\n$E[X\\wedge 227.27]=1000(1-e^{-0.22727})=1000(1-0.796703)=203.30$.\nPayment $=0.80(1.10)(870.68-203.30)=0.88(667.38)=\\$587.30$.\nThis exceeds the pre-inflation $\\$538.72$ — inflation raises the insurer's cost more than $10\\%$ because the fixed deductible erodes.", "tag": "Inflation & LER" }, { "front": "A **Pareto** loss has $\\alpha_{P}=3,\\theta=2000$, deductible $d=500$, limit $10{,}000$ ($u=10{,}500$), coinsurance $\\alpha=0.90$. Apply **5% inflation** and find the new expected per-loss payment.", "back": "Rescale: $\\tfrac{u}{1.05}=\\tfrac{10500}{1.05}=10{,}000$, $\\tfrac{d}{1.05}=\\tfrac{500}{1.05}=476.19$. Using $E[X\\wedge t]=1000\\big[1-(\\tfrac{2000}{2000+t})^{2}\\big]$:\n$E[X\\wedge 10000]=1000\\big[1-(\\tfrac{2000}{12000})^{2}\\big]=1000(1-0.027778)=972.22$.\n$E[X\\wedge 476.19]=1000\\big[1-(\\tfrac{2000}{2476.19})^{2}\\big]=1000(1-0.652367)=347.63$.\nPayment $=0.90(1.05)(972.22-347.63)=0.945(624.59)=\\$590.24$.\nUp from the pre-inflation $\\$552.96$.", "tag": "Inflation & LER" }, { "front": "Does the **loss elimination ratio** of a fixed deductible rise or fall under inflation? Show it for an exponential loss.", "back": "It **falls**. Inflation makes losses larger relative to the fixed $d$, so the deductible eliminates a *smaller* fraction.\nExample: $X\\sim\\text{Exp}(\\theta)$, $d=500$. Before: $\\theta=1000$, $\\text{LER}=1-e^{-500/1000}=0.39347$. After $20\\%$ inflation, $X'\\sim\\text{Exp}(1200)$: $\\text{LER}=\\dfrac{E[X'\\wedge 500]}{E[X']}=1-e^{-500/1200}=1-e^{-0.41667}=0.34076$.\nThe ratio dropped from $39.35\\%$ to $34.08\\%$ — the insurer now pays a larger share, so claim costs rise faster than inflation alone.", "tag": "Inflation & LER" }, { "front": "How does an ordinary deductible $d$ change the **expected number of payments** (claim frequency)?", "back": "Only losses exceeding $d$ generate a payment. If $N$ is the number of **ground-up losses** with $E[N]$ expected, then the number of **payments** has expectation $E[N_{P}]=E[N]\\cdot S(d)=E[N]\\cdot P(X>d)$.\nThe deductible **thins** the loss count by the survival probability at $d$. (If $N$ is Poisson, $N_P$ is again Poisson with mean $\\lambda S(d)$ — a classic thinning result.)", "tag": "Per-loss vs per-payment" }, { "front": "Losses follow **Poisson** with mean $\\lambda=10$ per year and severity **Pareto** $\\alpha=3,\\theta=2000$. With a deductible $d=500$, find the expected number of **payments** per year and the expected aggregate payment.", "back": "$S(500)=\\big(\\tfrac{2000}{2500}\\big)^{3}=0.8^{3}=0.512$.\nExpected payments: $E[N_P]=10(0.512)=\\mathbf{5.12}$ per year.\nExpected payment **per loss**: $E[(X-500)_{+}]=E[X]-E[X\\wedge 500]=1000-1000\\big[1-(\\tfrac{2000}{2500})^{2}\\big]=1000-360=640$.\nExpected **aggregate**: $E[N]\\cdot E[(X-d)_{+}]=10(640)=\\$6{,}400.00$.\n(Check via per-payment: $5.12\\times \\tfrac{640}{0.512}=5.12\\times 1250=\\$6{,}400.00$.)", "tag": "Per-loss vs per-payment" }, { "front": "How does an ordinary deductible reshape the **payment-severity distribution**? Distinguish per-loss from per-payment.", "back": "**Per-loss** $Y^{L}=(X-d)_{+}$: same shape as $X$ shifted left by $d$ on $(d,\\infty)$, with all probability below $d$ piled into a **point mass at $0$** of size $F(d)$. It is a mixed (part-discrete, part-continuous) variable.\n**Per-payment** $Y^{P}=X-d\\mid X>d$: the conditional excess distribution. Its CDF is $F_{Y^P}(y)=\\dfrac{F(d+y)-F(d)}{S(d)}$ and density $f_{Y^P}(y)=\\dfrac{f(d+y)}{S(d)}$ for $y>0$ — a renormalized tail with **no** mass at $0$.", "tag": "Per-loss vs per-payment" }, { "front": "For an **exponential** loss with $\\theta=500$ and ordinary deductible $d=300$, identify the per-payment distribution and give its mean and variance.", "back": "By **memorylessness**, the conditional excess $Y^{P}=X-300\\mid X>300$ is again **exponential with the same mean** $\\theta=500$.\nHence $E[Y^{P}]=\\theta=\\$500.00$ and $\\operatorname{Var}(Y^{P})=\\theta^{2}=500^{2}=250{,}000$, so the standard deviation is $\\$500.00$.\nThe deductible leaves the per-payment severity completely unchanged — a property unique to the exponential among the common severities.", "tag": "Per-loss vs per-payment" }, { "front": "A $\\$1{,}000$ ordinary deductible applies to losses uniform on $(0,4000)$. Find the per-loss expected payment, $S(d)$, and the per-payment mean.", "back": "For $U(0,4000)$: $E[X]=2000$. $E[X\\wedge 1000]=1000-\\dfrac{1000^2}{2(4000)}=1000-125=875$.\n**Per-loss:** $E[(X-1000)_{+}]=2000-875=\\$1{,}125.00$.\n$S(1000)=1-\\tfrac{1000}{4000}=0.75$.\n**Per-payment:** $E[Y^{P}]=\\dfrac{1125}{0.75}=\\$1{,}500.00$. (Indeed $X-1000\\mid X>1000$ is uniform on $(0,3000)$ with mean $1500$.)", "tag": "Per-loss vs per-payment" }, { "front": "Why does applying a **policy limit** to the per-payment variable require care, and what is the maximum the per-payment variable can pay?", "back": "The per-payment variable already conditions on $X>d$, so a maximum covered loss $u$ caps the payment at $u-d$ (times $\\alpha$ if there is coinsurance). The conditional payment is $\\big((X\\wedge u)-d\\mid X>d\\big)$, bounded above by $u-d$.\nIts mean is $\\dfrac{E[X\\wedge u]-E[X\\wedge d]}{S(d)}$ — the per-loss covered layer divided by $S(d)$. Forgetting the $S(d)$ denominator is the classic error: per-loss and per-payment differ exactly by that factor.", "tag": "Limits & coinsurance" }, { "front": "An **exponential** loss has $\\theta=2000$. A policy has deductible $d=500$, maximum covered loss $u=5000$, coinsurance $\\alpha=1$. Find both the **per-loss** and **per-payment** expected payments.", "back": "$E[X\\wedge 5000]=2000(1-e^{-2.5})=2000(1-0.082085)=1835.83$.\n$E[X\\wedge 500]=2000(1-e^{-0.25})=2000(1-0.778801)=442.40$.\n**Per-loss:** $E[X\\wedge u]-E[X\\wedge d]=1835.83-442.40=\\$1{,}393.43$.\n$S(500)=e^{-0.25}=0.778801$.\n**Per-payment:** $\\dfrac{1393.43}{0.778801}=\\$1{,}789.20$.", "tag": "Limits & coinsurance" }, { "front": "Summarize the **build-up of the modified payment** from ground-up loss $X$ to insurer payment, in the correct order of operations.", "back": "Apply the modifications in this order:\n1. **Deductible** $d$: pay $(X-d)_{+}$ (ordinary) — removes small losses.\n2. **Maximum covered loss / limit** $u$: cap the loss, so the covered amount is $(X\\wedge u-d)_{+}$, maxing out at $u-d$.\n3. **Coinsurance** $\\alpha$: pay a fraction $\\alpha$ of what remains, capping the payment at $\\alpha(u-d)$.\n4. **Inflation** $r$: replace $X$ by $(1+r)X$, i.e. rescale fixed thresholds by $\\tfrac{1}{1+r}$.\nExpected per-loss cost: $\\alpha(1+r)\\big(E[X\\wedge\\tfrac{u}{1+r}]-E[X\\wedge\\tfrac{d}{1+r}]\\big)$.", "tag": "Limits & coinsurance" }, { "front": "An **exponential** loss has $\\theta=1000$. Compare the insurer's expected per-loss payment under (a) an **ordinary** $\\$300$ deductible and (b) a **franchise** $\\$300$ deductible.", "back": "$E[X\\wedge 300]=1000(1-e^{-0.3})=1000(1-0.740818)=259.18$, so ordinary per-loss $=1000-259.18=740.82$. $S(300)=e^{-0.3}=0.740818$.\n**(a) Ordinary:** $\\$740.82$.\n**(b) Franchise:** add $d\\,S(d)=300(0.740818)=222.25$, giving $740.82+222.25=\\$963.07$.\nThe franchise deductible costs the insurer $\\$222.25$ more per loss — it refunds the full deductible on every claim that exceeds $\\$300$.", "tag": "Franchise deductible" }, { "front": "Why does a deductible always **reduce** the variance contribution from small losses but a franchise deductible introduces a **jump** at $d$?", "back": "An ordinary deductible truncates and shifts: payments below $d$ collapse to $0$, so the payment variable is continuous at the threshold ($(X-d)_+\\to 0$ as $X\\downarrow d$).\nA **franchise** deductible pays $0$ for $X\\le d$ but jumps to paying the **full** $X\\ge d$ at the threshold — a discontinuity of size $d$ in the payment as $X$ crosses $d$. This jump adds variability and is why the franchise's expected cost carries the extra $d\\,S(d)$ term: the payment is never between $0$ and $d$.", "tag": "Franchise deductible" } ] }