Deriving Expected Loss For A Per-Occurrence Layer

Why Expected Layer Loss Requires Two Calculations

A per-occurrence excess layer pays only when an individual claim exceeds an attachment, and its payment is capped by the layer limit. Pricing its expected loss therefore requires two distinct steps: first average the layer payment over the claim severity distribution, then account for the number of claims expected during the period. The distinction matters because an expected payment per claim is not an expected annual payment.

The calculation depends on the severity distribution over the part of the loss range that can affect the layer. Losses below attachment produce no payment; losses within the layer produce a partial payment; and losses above the layer exhaust its limit. A correct integral must include all three regions, even though the first region contributes zero.

Definition: Let $X$ be a nonnegative, continuous claim severity with density $f_X(x)$. A per-occurrence layer has attachment $a\geq 0$ and limit $u>0$. Its payment on one claim is $$ Z=\min\bigl((X-a)_+,u\bigr), \qquad (y)_+=\max(y,0). $$ Equivalently, $Z=0$ for $X\leq a$, $Z=X-a$ for $a<X<a+u$, and $Z=u$ for $X\geq a+u$.

Here $a$ is the ground-up loss amount that must be exceeded before the layer pays, while $u$ is the maximum payment for one occurrence. The top of the layer is $a+u$. The claim severity $X$, attachment, and limit must use the same monetary units. For example, if severities are measured in thousands of dollars, both $a$ and $u$ must also be stated in thousands.

This tutorial assumes a continuous severity distribution, so probabilities at individual severity values are zero and expectations can be calculated with density integrals. If the severity distribution has atoms, the relevant point masses must also be included. The formulas below describe expected loss only; they do not by themselves determine a premium that includes expenses, commissions, or a risk load.

Deriving Expected Payment for One Claim

The definition of $Z$ divides the positive payments into two ranges. For severities between $a$ and $a+u$, the payment is $x-a$. Above $a+u$, it is $u$. Below $a$, it is zero. Multiplying each payment by its density and integrating over the corresponding severity range gives

\begin{aligned} E[Z] &=\int_0^\infty \min\bigl((x-a)_+,u\bigr)f_X(x)\,dx\\ &=\int_a^{a+u}(x-a)f_X(x)\,dx +u\int_{a+u}^{\infty}f_X(x)\,dx. \end{aligned}

The second integral is the probability of a severity above the top of the layer. It is multiplied by $u$, because each such claim exhausts the layer. This term is essential: integrating only through $a+u$ omits payments on claims larger than the layer top.

Result: For a nonnegative continuous severity $X$ with density $f_X$, the expected payment per claim is the within-layer integral plus the capped payment above the layer: $$ E[Z]=\int_a^{a+u}(x-a)f_X(x)\,dx+uP(X\geq a+u). $$ For a continuous distribution, $P(X\geq a+u)=P(X>a+u)$. The result does not assume a particular severity family, but evaluating it requires the relevant density and tail probability.

The structure of the formula follows directly from the payment function, rather than from a special property of a distribution. In the first integral, the payment grows with severity, beginning at zero at $a$ and reaching $u$ at the layer top. Above that point, the payment no longer grows, so the integral of the density is multiplied by a fixed amount.

Locate the layer.
Identify the attachment $a$, limit $u$, and top $a+u$, keeping all amounts in consistent units.

Write the payment by severity region.
Use zero below attachment, $x-a$ within the layer, and $u$ above the layer top.

Integrate the payment.
Integrate $(x-a)f_X(x)$ from $a$ to $a+u$, then add $u$ times the probability of exceeding $a+u$.

Allow for claim frequency.
Multiply the expected payment per claim by expected claim count when claim frequency and severities satisfy the independence and identical-distribution assumptions described below.

Worked Examples

Worked Example: Exponential Severity and Annual Claim Frequency

Hypothetical scenario. Individual claim severity $X$, in dollars, is exponential with mean $\theta=\$20{,}000$, so $f_X(x)=\frac{1}{20{,}000}e^{-x/20{,}000}$ for $x\geq0$. A per-occurrence layer attaches at $a=\$40{,}000$ and has limit $u=\$20{,}000$. Annual claim count $N$ is Poisson with mean $1.5$, independent of the severities, which are identically distributed. Find expected annual layer loss.

Method. The layer top is $a+u=\$60{,}000$. The expected payment per claim is the within-layer integral from $\$40{,}000$ to $\$60{,}000$, plus the capped payment for severities above $\$60{,}000$:

\begin{aligned} E[Z] &=\int_{40{,}000}^{60{,}000}(x-40{,}000) \frac{1}{20{,}000}e^{-x/20{,}000}\,dx +20{,}000P(X>60{,}000)\\ &=20{,}000e^{-2}\left[1-2e^{-1}\right] +20{,}000e^{-3}\\ &=20{,}000e^{-2}(1-e^{-1})\\ &\approx \$1{,}710.96. \end{aligned}

For the second line, the integral over the layer is $20{,}000e^{-2}[1-2e^{-1}]$, and the tail probability is $P(X>60{,}000)=e^{-3}$. The combined expression simplifies because $-2e^{-1}+e^{-1}=-e^{-1}$.

The expected annual claim count is $E[N]=1.5$, so expected annual layer loss is

E[L]=1.5E[Z] =1.5\left(20{,}000e^{-2}(1-e^{-1})\right) \approx \$2{,}566.45.

The result is rounded to the nearest cent only after retaining the unrounded value, approximately $\$2{,}566.4464$. Multiplying the per-claim amount after first rounding it to $\$1{,}710.96$ gives $\$2{,}566.44$; that one-cent difference is an intermediate-rounding effect. Common mistake avoided: treating the per-claim expected payment as the annual answer and failing to multiply by the expected count.

Worked Example: A Uniform Severity With a Layer Tail

Hypothetical scenario. Claim severity $X$, in thousands of dollars, is uniform on $[0,100]$, with density $f_X(x)=0.01$ on that interval. A per-occurrence layer attaches at $20$ and has limit $30$, so its top is $50$. Annual claim count has mean $4$ and is independent of identically distributed severities. Find expected annual layer loss.

Within-layer payment. For $20<x<50$, payment is $x-20$. Thus the contribution per claim from this range is

\int_{20}^{50}(x-20)(0.01)\,dx =0.01\left[\frac{(x-20)^2}{2}\right]_{20}^{50} =0.01\left(\frac{30^2}{2}-0\right) =4.5

thousand dollars. The probability that severity exceeds the layer top is $P(X>50)=(100-50)/100=0.5$. Those claims each produce the full limit, contributing $30(0.5)=15$ thousand dollars per claim. Therefore,

E[Z]=4.5+15=19.5\text{ thousand dollars per claim}, \qquad E[L]=4(19.5)=78\text{ thousand dollars annually}.

The expected annual layer loss is therefore $\$78{,}000$. As a check, the capped-tail contribution alone is $4(15)=60$ thousand dollars annually, less than the total $78$ thousand. Common mistake avoided: stopping the calculation at the top of the layer. Doing so would omit the $15$ thousand dollars per claim from severities above $50$, not merely disregard losses that the layer does not cover.

Worked Example: Lomax Severity and a Poisson Count

Hypothetical scenario. Severity $X$, in thousands of dollars, has a Lomax distribution with shape $\alpha=3$ and scale $\lambda=10$. Its density is $f_X(x)=3000/(x+10)^4$, for $x\geq0$, and its survival probability is $P(X>x)=(10/(x+10))^3$. A layer attaches at $10$ and has limit $20$. Annual claim count is Poisson with mean $0.8$, independent of the identically distributed severities. Find expected annual layer loss.

The layer top is $10+20=30$. First integrate the payment from $10$ to $30$:

\begin{aligned} \int_{10}^{30}(x-10)\frac{3000}{(x+10)^4}\,dx &=3000\int_{20}^{40}\left(\frac{1}{t^3}-\frac{20}{t^4}\right)dt\\ &=3000\left[-\frac{1}{2t^2}+\frac{20}{3t^3}\right]_{20}^{40}\\ &=3000\left[\left(-\frac{1}{3200}+\frac{20}{192000}\right) -\left(-\frac{1}{800}+\frac{20}{24000}\right)\right]\\ &=3000\left(-\frac{1}{4800}+\frac{1}{2400}\right) =0.625 \end{aligned}

The variable change is $t=x+10$, so $x-10=t-20$. The bracket is evaluated at the upper endpoint first, then the lower endpoint is subtracted. The tail probability above $30$ is $(10/40)^3=1/64$; the capped contribution is $20/64=0.3125$ thousand dollars per claim. Hence

E[Z]=0.625+0.3125=0.9375\text{ thousand dollars per claim}, \qquad E[L]=0.8(0.9375)=0.75\text{ thousand dollars}.

Expected annual layer loss is $\$750.00$. The result is consistent with the layer limit: the mean payment per claim is below $20$ thousand dollars. Common mistake avoided: reversing the order of the antiderivative endpoints. The definite integral is the value at $40$ minus the value at $20$, which here is positive.

From Expected Payment Per Claim to Annual Loss

Let $N$ be the number of claims in a period, and let $Z_j$ be the layer payment on claim $j$. Annual layer loss is $L=\sum_{j=1}^{N}Z_j$. Assume that $N$ is independent of the severities and that each claim has the same severity distribution, so each $Z_j$ has expected value $E[Z]$. Conditional on observing $N=n$ claims, linearity of expectation gives $E[L\mid N=n]=nE[Z]$. Taking expectation over $N$ gives

E[L]=E\bigl[E[L\mid N]\bigr] =E[N]E[Z].

This result does not require $N$ to be Poisson; only its mean is needed under the stated assumptions. The Poisson assumption in an example specifies a frequency distribution with mean $\lambda$, so $E[N]=\lambda$. If the count is independent of the severities, the annual expected layer loss is $\lambda E[Z]$.

If severity distributions vary by claim type, the simple product $E[N]E[Z]$ may not apply as written. For example, if different claim types have different expected layer payments, expected annual loss must account for the expected number of claims of each type and its corresponding expected payment. Dependence between frequency and severity can also invalidate the simple product; in that case, the conditional expected payment given the claim count or other relevant conditions must be considered.

Common Checks and Exam Traps

Omitting the capped tail. The integral over $a$ to $a+u$ is not the full expected payment. Add $uP(X>a+u)$.
Using the wrong upper bound. The top of a layer is attachment plus limit, $a+u$, not the limit $u$ measured from ground up.
Applying frequency too early or not at all. First calculate payment per claim; then multiply by expected claim count under the stated assumptions.
Mixing units. A severity expressed in thousands must be paired with attachment and limit in thousands. Convert the final expected amount to dollars if needed.
Reversing integration endpoints. A definite integral from $a$ to $a+u$ is the antiderivative at $a+u$ minus its value at $a$. A negative answer for a nonnegative payment integral signals an error.

Key takeaway: Expected layer loss per claim is obtained by integrating the actual layer payment over the severity distribution, including the full-limit payment above the layer top. Under independent frequency and identically distributed severities, multiply that per-claim expectation by the expected number of claims to obtain expected annual layer loss.

Check Your Understanding

For each hypothetical portfolio, calculate the expected payment per claim first, then use the expected claim count to find annual expected layer loss.

Severity is uniform on $[0,80]$ thousand dollars. A layer attaches at $10$ thousand and has limit $20$ thousand. Expected annual claim count is $2$, independent of severity. Find expected annual layer loss. (Answer: $\$30{,}000$)
Severity is exponential with mean $\$10{,}000$. A layer attaches at $\$10{,}000$ and has limit $\$10{,}000$. Expected annual claim count is $0.4$, independent of severity. Find expected annual layer loss, rounded to cents. (Answer: $\$930.18$)
Severity, in thousands of dollars, is Lomax with shape $2$, scale $5$, density $50/(x+5)^3$, and survival probability $(5/(x+5))^2$. A layer attaches at $5$ and has limit $5$. Expected annual claim count is $0.6$, independent of severity. Find expected annual layer loss. (Answer: $\$500.00$)
Severity is uniform on $[0,120]$ thousand dollars. A layer attaches at $30$ thousand and has limit $40$ thousand. Expected annual claim count is $1.25$, independent of severity. Find expected annual layer loss, rounded to the nearest dollar. (Answer: $\$29{,}167$)

What You'll Learn