Expected Layer Loss Using Limited Expected Values

From a Layer Payment to a Limited Expected Value

A layer covers only the portion of a loss between an attachment point and an exhaustion point. The insurer pays nothing below the attachment point, pays the loss above it within the layer, and pays no more after the loss reaches the exhaustion point. Although this payment can be written directly, its expectation can often be calculated more simply by taking the difference of two limited expected values.

Let $X$ be a nonnegative loss, and let $a$ and $b$ be the layer’s attachment and exhaustion points, with $0\leq a\leq b$. There is no interest discounting in the calculations below; equivalently, take $i=0$. Mortality is not part of these severity models.

Definition: For a nonnegative loss $X$, the limited expected value at cap $u\geq 0$ is $\operatorname{LEV}_X(u)=E[\min(X,u)]$. It is the expected amount of loss remaining after the loss is capped at $u$.

The layer payment per loss is $Y^L_{a,b}=\min((X-a)_+,b-a)$, where $(z)_+=\max(z,0)$. The layer has width $b-a$: this is its maximum payment, while $b$ is the exhaustion point measured on the original-loss scale.

Theorem: For any nonnegative loss $X$ with a finite expectation, and any $0\leq a\leq b$, the expected payment for the layer attaching at $a$ and exhausting at $b$ is $E[Y^L_{a,b}]=\operatorname{LEV}_X(b)-\operatorname{LEV}_X(a)$.

Why the Difference Formula Holds

The result follows by comparing two capped versions of the same loss, for each possible value of $X$. The difference between the cap at $b$ and the cap at $a$ is exactly the portion of the loss in the layer.

Loss at or below attachment.
If $X\leq a$, then $\min(X,b)-\min(X,a)=X-X=0$. The layer payment is also zero.

Loss inside the layer.
If $a<X<b$, then $\min(X,b)-\min(X,a)=X-a$. This is the loss above the attachment point, which the layer pays.

Loss at or above exhaustion.
If $X\geq b$, then $\min(X,b)-\min(X,a)=b-a$. The layer pays its full width.

Take expectations.
The pointwise identity is $Y^L_{a,b}=\min(X,b)-\min(X,a)$. Taking expectations of both sides gives $E[Y^L_{a,b}]=E[\min(X,b)]-E[\min(X,a)]$, which is the stated limited-expected-value formula.

This justification does not require a continuous distribution or a density. It also covers losses exactly at $a$ or $b$: at $a$, the payment is zero; at $b$, the layer is fully exhausted. The result applies to any nonnegative severity distribution for which the expectations exist.

Calculating Limited Expected Values

The layer formula reduces the problem to calculating $\operatorname{LEV}_X(u)$ at two caps. The calculation depends on the severity model. In each worked example, first identify the distribution and its parameters, then evaluate the limited expected value at the exhaustion point and subtract the value at the attachment point.

Worked Example: An Exponential Loss Layer

A hypothetical insurer covers an individual loss $X$ with a layer attaching at $a=\$500$ and exhausting at $b=\$1{,}500$. Assume $X$ has an exponential distribution with mean $\theta=\$1{,}000$, so $S_X(x)=e^{-x/1000}$ for $x\geq0$. There is no discounting ($i=0$); mortality is irrelevant.

Method. Use the limited expected value for an exponential loss and take the difference at the exhaustion and attachment points. For an exponential loss with mean $\theta$, the limited expected value is $\theta(1-e^{-u/\theta})$.

At $b=\$1{,}500$, $\operatorname{LEV}_X(1500)=1000(1-e^{-1.5})=1000(1-0.22313016)=\$776.86984$.

At $a=\$500$, $\operatorname{LEV}_X(500)=1000(1-e^{-0.5})=1000(1-0.60653066)=\$393.46934$.

Therefore, $E[Y^L_{500,1500}]=776.86984-393.46934=\$383.40050$, or \$383.40 to the nearest cent. As a check, this is below the layer’s maximum payment of $1500-500=\$1{,}000$.

Common mistake avoided: the exhaustion cap is $\$1{,}500$, not the layer width of $\$1{,}000$. Subtracting $\operatorname{LEV}_X(500)$ from $\operatorname{LEV}_X(1000)$ would instead give $632.12056-393.46934=\$238.65$, which is not the expected payment for this layer.

Worked Example: A Uniform Loss Layer

Suppose a hypothetical equipment policy pays a layer attaching at $a=\$400$ and exhausting at $b=\$1{,}200$. The loss $X$ is uniform on $[0,\$2{,}000]$, with density $f_X(x)=1/2000$ on that interval. There is no discounting ($i=0$); mortality is irrelevant.

Method. For $0\leq u\leq2000$, calculate the expected capped loss directly: $\operatorname{LEV}_X(u)=\int_0^u x(1/2000)\,dx+u\int_u^{2000}(1/2000)\,dx$. The first term accounts for losses below the cap; the second accounts for losses at or above it.

Evaluating the integrals gives $\operatorname{LEV}_X(u)=u^2/4000+u(2000-u)/2000=u-u^2/4000$.

At $b=\$1{,}200$, $\operatorname{LEV}_X(1200)=1200-1{,}440{,}000/4000=1200-360=\$840$.

At $a=\$400$, $\operatorname{LEV}_X(400)=400-160{,}000/4000=400-40=\$360$.

Thus $E[Y^L_{400,1200}]=840-360=\$480$. This is less than the layer maximum of $\$800$, as required. The calculation uses the cap points on the original-loss scale, not the amount of loss paid by the layer.

Common mistake avoided: losses above $\$1{,}200$ still contribute $\$1{,}200$ to $\operatorname{LEV}_X(1200)$; they are not omitted from the capped expectation.

Worked Example: A Lomax Loss Layer

A hypothetical specialty insurer considers a layer attaching at $a=\$1{,}000$ and exhausting at $b=\$3{,}000$. Assume $X$ has a Lomax distribution with scale $\theta=\$1{,}000$ and shape $\alpha=2$, so $S_X(x)=(1+x/1000)^{-2}$ for $x\geq0$. The mean is finite. There is no discounting ($i=0$); mortality is irrelevant.

Method. The limited expected value for this Lomax model is $\operatorname{LEV}_X(u)=\frac{\theta}{\alpha-1}\left[1-(1+u/\theta)^{1-\alpha}\right]$, for $\alpha>1$. Here $\theta/(\alpha-1)=\$1{,}000$ and $1-\alpha=-1$.

At $b=\$3{,}000$, $\operatorname{LEV}_X(3000)=1000[1-(1+3)^{-1}]=1000(1-1/4)=\$750$.

At $a=\$1{,}000$, $\operatorname{LEV}_X(1000)=1000[1-(1+1)^{-1}]=1000(1-1/2)=\$500$.

The expected layer payment is $750-500=\$250$. It is nonnegative and below the layer width of $\$2{,}000$, which provides a basic check on the result.

Common mistake avoided: the scale parameter is $\$1{,}000$, not the mean of every Lomax model. Here the mean equals the scale only because the shape parameter is $2$; using the wrong parameterization can change both limited expected values.

Interpretation and Checks

The limited-expected-value difference is useful because it converts a layer into the marginal increase in expected capped loss when the cap rises from $a$ to $b$. It separates the expected payment calculation from the specific wording used to describe a layer, provided the attachment and exhaustion points are correctly identified.

Several checks help detect errors:

Nonnegativity: since $b\geq a$, $\operatorname{LEV}_X(b)\geq\operatorname{LEV}_X(a)$, so the expected layer payment cannot be negative.
Maximum-payment bound: $0\leq E[Y^L_{a,b}]\leq b-a$, because every individual payment lies between zero and the layer width.
Zero-width layer: if $a=b$, the payment is always zero and the difference of limited expected values is zero.
Layer from the ground up: if $a=0$, then $\operatorname{LEV}_X(0)=0$, so the expected payment is $\operatorname{LEV}_X(b)$.
Exhaustion beyond all possible losses: if $X\leq b$ with probability one, then $\operatorname{LEV}_X(b)=E[X]$; the formula still subtracts the expected loss capped at the attachment point.

The key exam distinction is between a cap on the original loss and a cap on the layer payment. The layer attaching at $a$ and exhausting at $b$ has width $b-a$, but its expected payment is formed from limited expected values at $a$ and $b$. Replacing $b$ by $b-a$ inside the first limited expected value generally calculates a different quantity.

Key takeaway: To find the expected payment for a layer attaching at $a$ and exhausting at $b$, calculate the expected loss capped at $b$, subtract the expected loss capped at $a$, and check that the result lies between zero and $b-a$.

Check Your Understanding

For each hypothetical model, identify the two limited expected values required, then calculate the expected layer payment. There is no discounting ($i=0$) in these questions; mortality is irrelevant.

A loss is exponential with mean $\$800$. Find the expected payment for a layer attaching at $\$0$ and exhausting at $\$800$. (Answer: $\$505.70$)
A loss is uniform on $[0,\$3{,}000]$. Find the expected payment for a layer attaching at $\$500$ and exhausting at $\$2{,}000$. (Answer: $\$875.00$)
A loss has a Lomax distribution with scale $\theta=\$500$ and shape $\alpha=3$. Find the expected payment for a layer attaching at $\$500$ and exhausting at $\$1{,}500$. (Answer: $\$46.88$)
A loss is exponential with mean $\$1{,}200$. Calculate the expected payment for a layer attaching at $\$600$ and exhausting at $\$2{,}400$, and check it against the layer maximum. (Answer: $\$565.43$; maximum payment $\$1{,}800$)
A loss is uniform on $[0,\$4{,}000]$. Find the expected payment for a layer attaching at $\$1{,}000$ and exhausting at $\$5{,}000$. (Answer: $\$1{,}125.00$)

What You'll Learn