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Exam FAM — Coverage Modifications Flashcards

How deductibles, policy limits, coinsurance, and inflation transform a ground-up loss $X$ into the insurer's payment for SOA Exam FAM: the limited expected value $E[X\wedge u]=\int_0^u S(x)\,dx$ and its central role, ordinary vs. franchise deductibles, the per-loss versus per-payment distinction, the master payment formula $\alpha\big(E[X\wedge u]-E[X\wedge d]\big)$ with uniform inflation, the loss elimination ratio, and the effect of a deductible on claim frequency and on payment severity — with fully worked, self-contained calculations.

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  1. Limited expected value
    Define the **limited expected value** $E[X\wedge u]$ and give its survival-function integral form.
    $X\wedge u=\min(X,u)$ is the loss **capped** at the limit $u$, so $E[X\wedge u]$ is the average payment when nothing above $u$ is covered. For a nonnegative loss, $E[X\wedge u]=\displaystyle\int_0^u S(x)\,dx$, where $S(x)=P(X>x)$. Properties: $E[X\wedge 0]=0$, $E[X\wedge u]$ increases in $u$, and $E[X\wedge u]\uparrow E[X]$ as $u\to\infty$.
  2. Limited expected value
    Why is the limited expected value the single workhorse quantity for coverage-modification problems?
    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 $E[X\wedge u]-E[X\wedge d]$. Because $E[X\wedge u]=\int_0^u S(x)\,dx$, you only need the survival function (or a tabulated LEV column) to value deductibles, limits, coinsurance, and inflation in one consistent framework.
  3. Limited expected value
    For an exponential loss with mean $\theta$, state $E[X\wedge u]$ and evaluate it at $u=500,1000,2000$ for $\theta=1000$.
    For $X\sim\text{Exp}(\theta)$, $S(x)=e^{-x/\theta}$, so $E[X\wedge u]=\theta\big(1-e^{-u/\theta}\big)$. With $\theta=1000$: $E[X\wedge 500]=1000(1-e^{-0.5})=1000(1-0.606531)=\$393.47$. $E[X\wedge 1000]=1000(1-e^{-1})=1000(1-0.367879)=\$632.12$. $E[X\wedge 2000]=1000(1-e^{-2})=1000(1-0.135335)=\$864.66$. All lie below $E[X]=\theta=\$1{,}000$, approaching it as $u$ grows.
  4. Limited expected value
    A loss is **uniform on $(0,1000)$**. Find $E[X]$, $E[X\wedge 200]$, and $E[X\wedge 500]$.
    For $X\sim U(0,b)$, $S(x)=1-\tfrac{x}{b}$ on $[0,b]$, so for $u\le b$, $E[X\wedge u]=\int_0^u\!\big(1-\tfrac{x}{b}\big)dx = u-\dfrac{u^2}{2b}$. With $b=1000$: $E[X]=\tfrac{b}{2}=\$500.00$. $E[X\wedge 200]=200-\dfrac{200^2}{2000}=200-20=\$180.00$. $E[X\wedge 500]=500-\dfrac{500^2}{2000}=500-125=\$375.00$.
  5. Limited expected value
    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.
    **Definition:** $E[X\wedge 200]=\sum\min(x,200)\,p(x)=50(0.4)+150(0.3)+200(0.2)+200(0.1)$ $=20+45+40+20=\$125.00$. And $E[X]=50(0.4)+150(0.3)+250(0.2)+500(0.1)=\$165.00$. **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.
  6. Limited expected value
    For a **Pareto** loss with parameters $\alpha$ and $\theta$, state $E[X\wedge u]$ and evaluate it at $u=1000,3000$ for $\alpha=3,\theta=2000$.
    For $X\sim\text{Pareto}(\alpha,\theta)$ with $\alpha>1$, $S(x)=\big(\tfrac{\theta}{\theta+x}\big)^{\alpha}$ and $E[X\wedge u]=\dfrac{\theta}{\alpha-1}\left[1-\left(\dfrac{\theta}{\theta+u}\right)^{\alpha-1}\right]$. With $\alpha=3,\theta=2000$, $E[X]=\tfrac{\theta}{\alpha-1}=\$1{,}000$. $E[X\wedge 1000]=1000\big[1-(\tfrac{2000}{3000})^{2}\big]=1000(1-0.444444)=\$555.56$. $E[X\wedge 3000]=1000\big[1-(\tfrac{2000}{5000})^{2}\big]=1000(1-0.16)=\$840.00$.
  7. Ordinary deductible
    Define an **ordinary deductible** $d$ and the **per-loss** random variable it induces.
    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 $X>d$. "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 $F(d)=P(X\le d)$.
  8. Ordinary deductible
    Give the LEV formula for the expected **per-loss** payment under an ordinary deductible $d$.
    $E[(X-d)_{+}]=E[X]-E[X\wedge d]$. Intuition: $E[X]$ 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. Equivalently $E[(X-d)_{+}]=\int_d^\infty S(x)\,dx$.
  9. Per-loss vs per-payment
    Define the **per-payment** variable $Y^{P}$ for an ordinary deductible $d$ and give its mean.
    $Y^{P}=X-d\mid X>d$ 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 $0$. Its 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.
  10. Per-loss vs per-payment
    Distinguish **per-loss** ($Y^{L}$) and **per-payment** ($Y^{P}$) variables, and state the relationship between their means.
    **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. **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**. Relationship: $E[Y^{L}]=S(d)\,E[Y^{P}]$. Since $S(d)\le 1$, the per-payment mean is always at least the per-loss mean.
  11. Per-loss vs per-payment
    An **exponential** loss has mean $\theta=1000$. With an ordinary deductible $d=500$, find the per-loss and per-payment expected payments.
    $E[X]=1000$ and $E[X\wedge 500]=1000(1-e^{-0.5})=393.47$. **Per-loss:** $E[(X-500)_{+}]=E[X]-E[X\wedge 500]=1000-393.47=\$606.53$. Also $S(500)=e^{-0.5}=0.606531$. **Per-payment:** $E[Y^{P}]=\dfrac{606.53}{0.606531}=\$1{,}000.00$. The per-payment mean equals $\theta$ — the exponential's **memorylessness**: given $X>d$, the excess $X-d$ is again $\text{Exp}(\theta)$.
  12. Per-loss vs per-payment
    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.
    $E[X]=0(0.5)+100(0.2)+300(0.2)+600(0.1)=\$140.00$. Per-loss: $E[(X-200)_{+}]=0+0+(300-200)(0.2)+(600-200)(0.1)=20+40=\$60.00$. $S(200)=P(X>200)=0.2+0.1=0.30$. Per-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. $\tfrac{0.1}{0.3}$: $100(\tfrac23)+400(\tfrac13)=200$.)
  13. Per-loss vs per-payment
    A **Pareto** loss has $\alpha=3,\theta=2000$. With an ordinary deductible $d=1000$, find the per-loss and per-payment expected payments.
    $E[X]=\tfrac{2000}{2}=1000$ and $E[X\wedge 1000]=1000\big[1-(\tfrac{2000}{3000})^{2}\big]=555.56$. **Per-loss:** $E[(X-1000)_{+}]=1000-555.56=\$444.44$. $S(1000)=(\tfrac{2000}{3000})^{3}=0.296296$. **Per-payment:** $E[Y^{P}]=\dfrac{444.44}{0.296296}=\$1{,}500.00$. Unlike the exponential, the Pareto's per-payment mean **exceeds** the ground-up mean — heavy tails make surviving losses larger.
  14. Franchise deductible
    Define a **franchise deductible** $d$ and contrast its payment rule with the ordinary deductible.
    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 $X>d$. Vs. 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 $d$.
  15. Franchise deductible
    Give the expected **per-loss** and **per-payment** cost of a **franchise** deductible $d$ in terms of the ordinary deductible.
    Because a franchise claim pays $d$ more than the ordinary claim exactly when $X>d$: **Per-loss:** $E[Y^{L}_{\text{fr}}]=E[(X-d)_{+}]+d\,S(d)=\big(E[X]-E[X\wedge d]\big)+d\,S(d)$. **Per-payment:** $E[Y^{P}_{\text{fr}}]=\dfrac{E[X]-E[X\wedge d]}{S(d)}+d$. In both cases you take the ordinary-deductible result and add back the deductible (weighted by $S(d)$ for per-loss).
  16. Franchise deductible
    An **exponential** loss has $\theta=1000$. For a **franchise** deductible $d=500$, find the expected per-loss and per-payment payments.
    Ordinary results: $E[(X-500)_{+}]=606.53$, $S(500)=e^{-0.5}=0.606531$. **Franchise per-loss:** $E[Y^{L}_{\text{fr}}]=606.53+500(0.606531)=606.53+303.27=\$909.80$. **Franchise per-payment:** $E[Y^{P}_{\text{fr}}]=\dfrac{606.53}{0.606531}+500=1000+500=\$1{,}500.00$. Each exceeds the ordinary-deductible counterpart by $d\,S(d)$ and by $d$, respectively.
  17. Franchise deductible
    A loss is **uniform on $(0,1000)$**. For a **franchise** deductible $d=200$, find the expected per-loss payment.
    For $U(0,1000)$: $E[X]=500$, $E[X\wedge 200]=200-\tfrac{200^2}{2000}=180$, so $E[(X-200)_{+}]=500-180=320$. $S(200)=1-\tfrac{200}{1000}=0.80$. **Franchise per-loss:** $E[Y^{L}_{\text{fr}}]=E[(X-200)_{+}]+d\,S(200)=320+200(0.80)=320+160=\$480.00$. (Check directly: $\int_{200}^{1000} x\cdot\tfrac{1}{1000}\,dx=\tfrac{1000^2-200^2}{2000}=480$. Agrees.)
  18. Limits & coinsurance
    Define the **policy limit** and the **maximum covered loss** $u$, and how they relate when there is a deductible $d$.
    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. With 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 $L=u-d$. Don't confuse the two: the limit caps the **payment**; $u$ caps the **loss**.
  19. Limits & coinsurance
    Define the **coinsurance factor** $\alpha$ and where it enters the payment.
    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 $(1-\alpha)$. Coinsurance 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 $\alpha(u-d)$.
  20. Limits & coinsurance
    State the **master per-loss formula** for expected cost with coinsurance $\alpha$, ordinary deductible $d$, and maximum covered loss $u$.
    $E[\text{payment}]=\alpha\big(E[X\wedge u]-E[X\wedge d]\big)$. Reading 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. Special cases: deductible only ($u\to\infty$): $\alpha(E[X]-E[X\wedge d])$; limit only ($d=0$): $\alpha\,E[X\wedge u]$.
  21. Limits & coinsurance
    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.
    Maximum covered loss $u=d+\text{limit}=250+2000=2250$. $E[X\wedge 2250]=1000(1-e^{-2.25})=1000(1-0.105399)=894.60$. $E[X\wedge 250]=1000(1-e^{-0.25})=1000(1-0.778801)=221.20$. Payment $=\alpha\big(E[X\wedge u]-E[X\wedge d]\big)=0.80(894.60-221.20)=0.80(673.40)=\$538.72$.
  22. Limits & coinsurance
    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.
    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 $\tfrac{\theta}{\alpha_P-1}=1000$: $E[X\wedge 10500]=1000\big[1-(\tfrac{2000}{12500})^{2}\big]=1000(1-0.0256)=974.40$. $E[X\wedge 500]=1000\big[1-(\tfrac{2000}{2500})^{2}\big]=1000(1-0.64)=360.00$. Payment $=0.90(974.40-360.00)=0.90(614.40)=\$552.96$.
  23. Limits & coinsurance
    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.
    Maximum covered loss $u=100+400=500$. For $U(0,1000)$, $E[X\wedge u]=u-\tfrac{u^2}{2000}$: $E[X\wedge 500]=500-\tfrac{250000}{2000}=500-125=375.00$. $E[X\wedge 100]=100-\tfrac{10000}{2000}=100-5=95.00$. Payment $=0.75(375.00-95.00)=0.75(280.00)=\$210.00$.
  24. Inflation & LER
    Define the **loss elimination ratio** $\text{LER}(d)$ for an ordinary deductible $d$.
    $\text{LER}(d)=\dfrac{E[X\wedge d]}{E[X]}=\dfrac{E[X]-E[(X-d)_{+}]}{E[X]}$. It 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.
  25. Inflation & LER
    An **exponential** loss has $\theta=1000$. Find $\text{LER}(500)$ and interpret it.
    $E[X\wedge 500]=1000(1-e^{-0.5})=393.47$ and $E[X]=1000$. $\text{LER}(500)=\dfrac{393.47}{1000}=0.39347$. A $\$500$ 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 $\$1{,}000$.
  26. Inflation & LER
    A loss takes values $50,150,250,500$ with probabilities $0.4,0.3,0.2,0.1$. Find $\text{LER}(200)$.
    $E[X]=50(0.4)+150(0.3)+250(0.2)+500(0.1)=20+45+50+50=\$165.00$. $E[X\wedge 200]=50(0.4)+150(0.3)+200(0.2)+200(0.1)=20+45+40+20=\$125.00$. $\text{LER}(200)=\dfrac{125}{165}=0.75758$. The $\$200$ deductible eliminates about **75.8%** of expected losses — large here because the bulk of probability sits at or below $200$.
  27. Inflation & LER
    How does **uniform inflation** at rate $r$ change the loss, and what happens to a fixed deductible $d$ and fixed limit?
    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. Key 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 $(1+r)$.
  28. Inflation & LER
    State the **master per-loss formula with uniform inflation** at rate $r$ (deductible $d$, max covered loss $u$, coinsurance $\alpha$).
    $E[\text{payment}]=\alpha(1+r)\left(E\!\left[X\wedge \tfrac{u}{1+r}\right]-E\!\left[X\wedge \tfrac{d}{1+r}\right]\right)$, 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 $\alpha\big(E[X\wedge u]-E[X\wedge d]\big)$.
  29. Inflation & LER
    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.
    Rescale thresholds: $\tfrac{u}{1.1}=\tfrac{2250}{1.1}=2045.45$, $\tfrac{d}{1.1}=\tfrac{250}{1.1}=227.27$. $E[X\wedge 2045.45]=1000(1-e^{-2.04545})=1000(1-0.129321)=870.68$. $E[X\wedge 227.27]=1000(1-e^{-0.22727})=1000(1-0.796703)=203.30$. Payment $=0.80(1.10)(870.68-203.30)=0.88(667.38)=\$587.30$. This exceeds the pre-inflation $\$538.72$ — inflation raises the insurer's cost more than $10\%$ because the fixed deductible erodes.
  30. Inflation & LER
    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.
    Rescale: $\tfrac{u}{1.05}=\tfrac{10500}{1.05}=10{,}000$, $\tfrac{d}{1.05}=\tfrac{500}{1.05}=476.19$. Using $E[X\wedge t]=1000\big[1-(\tfrac{2000}{2000+t})^{2}\big]$: $E[X\wedge 10000]=1000\big[1-(\tfrac{2000}{12000})^{2}\big]=1000(1-0.027778)=972.22$. $E[X\wedge 476.19]=1000\big[1-(\tfrac{2000}{2476.19})^{2}\big]=1000(1-0.652367)=347.63$. Payment $=0.90(1.05)(972.22-347.63)=0.945(624.59)=\$590.24$. Up from the pre-inflation $\$552.96$.
  31. Inflation & LER
    Does the **loss elimination ratio** of a fixed deductible rise or fall under inflation? Show it for an exponential loss.
    It **falls**. Inflation makes losses larger relative to the fixed $d$, so the deductible eliminates a *smaller* fraction. Example: $X\sim\text{Exp}(\theta)$, $d=500$. Before: $\theta=1000$, $\text{LER}=1-e^{-500/1000}=0.39347$. After $20\%$ inflation, $X'\sim\text{Exp}(1200)$: $\text{LER}=\dfrac{E[X'\wedge 500]}{E[X']}=1-e^{-500/1200}=1-e^{-0.41667}=0.34076$. The ratio dropped from $39.35\%$ to $34.08\%$ — the insurer now pays a larger share, so claim costs rise faster than inflation alone.
  32. Per-loss vs per-payment
    How does an ordinary deductible $d$ change the **expected number of payments** (claim frequency)?
    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 $E[N_{P}]=E[N]\cdot S(d)=E[N]\cdot P(X>d)$. The 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.)
  33. Per-loss vs per-payment
    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.
    $S(500)=\big(\tfrac{2000}{2500}\big)^{3}=0.8^{3}=0.512$. Expected payments: $E[N_P]=10(0.512)=\mathbf{5.12}$ per year. Expected payment **per loss**: $E[(X-500)_{+}]=E[X]-E[X\wedge 500]=1000-1000\big[1-(\tfrac{2000}{2500})^{2}\big]=1000-360=640$. Expected **aggregate**: $E[N]\cdot E[(X-d)_{+}]=10(640)=\$6{,}400.00$. (Check via per-payment: $5.12\times \tfrac{640}{0.512}=5.12\times 1250=\$6{,}400.00$.)
  34. Per-loss vs per-payment
    How does an ordinary deductible reshape the **payment-severity distribution**? Distinguish per-loss from per-payment.
    **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. **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 $0$.
  35. Per-loss vs per-payment
    For an **exponential** loss with $\theta=500$ and ordinary deductible $d=300$, identify the per-payment distribution and give its mean and variance.
    By **memorylessness**, the conditional excess $Y^{P}=X-300\mid X>300$ is again **exponential with the same mean** $\theta=500$. Hence $E[Y^{P}]=\theta=\$500.00$ and $\operatorname{Var}(Y^{P})=\theta^{2}=500^{2}=250{,}000$, so the standard deviation is $\$500.00$. The deductible leaves the per-payment severity completely unchanged — a property unique to the exponential among the common severities.
  36. Per-loss vs per-payment
    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.
    For $U(0,4000)$: $E[X]=2000$. $E[X\wedge 1000]=1000-\dfrac{1000^2}{2(4000)}=1000-125=875$. **Per-loss:** $E[(X-1000)_{+}]=2000-875=\$1{,}125.00$. $S(1000)=1-\tfrac{1000}{4000}=0.75$. **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 $1500$.)
  37. Limits & coinsurance
    Why does applying a **policy limit** to the per-payment variable require care, and what is the maximum the per-payment variable can pay?
    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 $u-d$. Its 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.
  38. Limits & coinsurance
    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.
    $E[X\wedge 5000]=2000(1-e^{-2.5})=2000(1-0.082085)=1835.83$. $E[X\wedge 500]=2000(1-e^{-0.25})=2000(1-0.778801)=442.40$. **Per-loss:** $E[X\wedge u]-E[X\wedge d]=1835.83-442.40=\$1{,}393.43$. $S(500)=e^{-0.25}=0.778801$. **Per-payment:** $\dfrac{1393.43}{0.778801}=\$1{,}789.20$.
  39. Limits & coinsurance
    Summarize the **build-up of the modified payment** from ground-up loss $X$ to insurer payment, in the correct order of operations.
    Apply the modifications in this order: 1. **Deductible** $d$: pay $(X-d)_{+}$ (ordinary) — removes small losses. 2. **Maximum covered loss / limit** $u$: cap the loss, so the covered amount is $(X\wedge u-d)_{+}$, maxing out at $u-d$. 3. **Coinsurance** $\alpha$: pay a fraction $\alpha$ of what remains, capping the payment at $\alpha(u-d)$. 4. **Inflation** $r$: replace $X$ by $(1+r)X$, i.e. rescale fixed thresholds by $\tfrac{1}{1+r}$. Expected per-loss cost: $\alpha(1+r)\big(E[X\wedge\tfrac{u}{1+r}]-E[X\wedge\tfrac{d}{1+r}]\big)$.
  40. Franchise deductible
    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.
    $E[X\wedge 300]=1000(1-e^{-0.3})=1000(1-0.740818)=259.18$, so ordinary per-loss $=1000-259.18=740.82$. $S(300)=e^{-0.3}=0.740818$. **(a) Ordinary:** $\$740.82$. **(b) Franchise:** add $d\,S(d)=300(0.740818)=222.25$, giving $740.82+222.25=\$963.07$. The franchise deductible costs the insurer $\$222.25$ more per loss — it refunds the full deductible on every claim that exceeds $\$300$.
  41. Franchise deductible
    Why does a deductible always **reduce** the variance contribution from small losses but a franchise deductible introduces a **jump** at $d$?
    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 $X\downarrow d$). A **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 $d$.