Layer Loss Functions and Expected Recovery

Why Layer Loss Functions Matter

A reinsurance layer pays only for the part of a ground-up loss above an attachment, and only up to a stated limit. Pricing therefore requires translating the contract terms into a payment for each possible loss, then averaging that payment under a specified loss distribution. A small distinction in how the limit is interpreted can materially change the result.

This tutorial is aligned with the CAS Exam 9 (Risk Management for Actuaries) syllabus. It develops the limited-loss and layer-loss functions, shows how the layer payment can be represented using capped losses, and derives an expected recovery formula that works for discrete or continuous ground-up losses.

Definition: Let $X\geq 0$ be the ground-up loss for one covered claim. The limited-loss function with cap $u\geq 0$ is $L_u(x)=\min(x,u)$. For a layer with attachment $d\geq 0$ and limit amount $u>0$, the payment per loss is $Y^L=\min((X-d)_+,u)$, where $(z)_+=\max(z,0)$. The layer begins paying above $d$ and is exhausted when $X\geq d+u$. Thus $d+u$, not $u$, is the upper ground-up loss threshold.

The limit amount $u$ caps the reinsurer’s payment; it does not, under this convention, name the ground-up loss at which the layer is exhausted. The layer pays zero when $X\leq d$, pays $X-d$ when $d<X<d+u$, and pays $u$ when $X\geq d+u$.

Y^L = \begin{cases} 0, & X\leq d,\\ X-d, & d<X<d+u,\\ u, & X\geq d+u. \end{cases}

Connecting Layer Payments to Capped Losses

The limited-loss function makes the layer payment easy to express using two caps. Capping the ground-up loss at $d+u$ retains losses up to the layer’s exhaustion point; capping it at $d$ retains the part below the attachment. Their difference is exactly the layer payment.

Y^L=L_{d+u}(X)-L_d(X) =\min(X,d+u)-\min(X,d) =\min((X-d)_+,u).

This identity can be checked across the three payment ranges. If $X\leq d$, both capped losses equal $X$, so their difference is zero. If $d<X<d+u$, the first cap equals $X$ and the second equals $d$, giving $X-d$. If $X\geq d+u$, the caps equal $d+u$ and $d$, giving $u$.

Worked Example: Evaluate a Layer for a Discrete Loss Model

A hypothetical insurer has a covered-loss distribution with $P(X=0)=0.20$, $P(X=20)=0.30$, $P(X=60)=0.30$, and $P(X=100)=0.20$, with losses measured in thousands of dollars. A per-loss layer attaches at $d=30$ thousand dollars and has a limit amount $u=40$ thousand dollars. Find the payment for each outcome and the expected recovery.

Method. Apply $Y^L=\min((X-d)_+,u)$ to each possible ground-up loss, then take the probability-weighted average. This method avoids treating the limit amount as the exhaustion threshold.

The layer is exhausted at $d+u=30+40=70$ thousand dollars. The outcomes are:

If $X=0$, $Y^L=\min(0,40)=0$.
If $X=20$, $Y^L=\min(0,40)=0$.
If $X=60$, $Y^L=\min(60-30,40)=30$.
If $X=100$, $Y^L=\min(100-30,40)=40$.

Therefore, in thousands of dollars, $E[Y^L]=0(0.20)+0(0.30)+30(0.30)+40(0.20)=0+0+9+8=17$. The expected recovery is $17$ thousand dollars, or $17,000.00.

As a check using capped losses, $E[L_{70}(X)]=0(0.20)+20(0.30)+60(0.30)+70(0.20)=38$ thousand dollars. Also, $E[L_{30}(X)]=0(0.20)+20(0.30)+30(0.30)+30(0.20)=21$ thousand dollars. Their difference is $38-21=17$ thousand dollars, matching the direct calculation. The result is below the maximum possible recovery of $40$ thousand dollars.

Expected Recovery from the Survival Function

Write $S_X(x)=P(X>x)$ for the survival function of the ground-up loss. For a nonnegative random variable, the expected value of a payment can be found by integrating the probability that the payment exceeds each possible payment amount. For this layer, the event $Y^L>t$, for $0\leq t<u$, is equivalent to $X>d+t$.

Theorem: For a nonnegative ground-up loss $X$, a layer with attachment $d\geq0$ and limit amount $u>0$ has expected recovery $E[Y^L]=\int_0^u S_X(d+t)\,dt=\int_d^{d+u}S_X(x)\,dx$. This formula applies to discrete, continuous, and mixed loss distributions, provided the expectation exists. The integral form handles atoms in the distribution without requiring a continuous density.

The formula follows from the tail-integral representation. For any nonnegative random variable $Y$, its value can be written as the area under the indicator that the value exceeds a threshold: $Y=\int_0^\infty \mathbf{1}_{\{Y>t\}}\,dt$. Taking expectations gives $E[Y]=\int_0^\infty P(Y>t)\,dt$. The layer payment cannot exceed $u$, so probabilities above $u$ are zero. For $0\leq t<u$, $Y^L>t$ precisely when the ground-up loss exceeds $d+t$. Substituting and changing variables gives the layer formula.

\begin{aligned} E[Y^L] &=\int_0^u P(Y^L>t)\,dt\\ &=\int_0^u P(X>d+t)\,dt\\ &=\int_d^{d+u} S_X(x)\,dx. \end{aligned}

The same reasoning gives the expected limited loss $E[L_u(X)]=\int_0^u S_X(x)\,dx$. Applying that result to the difference of capped losses also yields $E[Y^L]=E[L_{d+u}(X)]-E[L_d(X)]$. The tail-integral approach is often especially convenient when the survival function has a simple form.

Worked Example: Expected Recovery for a Uniform Loss

Suppose a hypothetical ground-up loss $X$, measured in thousands of dollars, is uniformly distributed on $[0,100]$. A layer attaches at $d=30$ and has limit amount $u=40$. Find its expected recovery and verify it by integrating the layer payment directly.

Method. Use the survival-function integral; then check by splitting the loss range into partial and full recovery. The check avoids overlooking the portion of losses above the exhaustion point.

For $0\leq x\leq100$, $S_X(x)=1-x/100$. The layer exhausts at $30+40=70$. Therefore,

\begin{aligned} E[Y^L] &=\int_{30}^{70}\left(1-\frac{x}{100}\right)\,dx\\ &=\left[x-\frac{x^2}{200}\right]_{30}^{70}\\ &=\left(70-\frac{70^2}{200}\right) -\left(30-\frac{30^2}{200}\right)\\ &=(70-24.5)-(30-4.5)\\ &=45.5-25.5=20. \end{aligned}

The expected recovery is $20$ thousand dollars, or $20,000.00. To verify directly, the layer pays zero for losses below 30, $x-30$ for losses from 30 to 70, and 40 for losses above 70. Since the density is $1/100$, the partial-recovery contribution is $\int_{30}^{70}(x-30)/100\,dx=8$, and the full-recovery contribution is $\int_{70}^{100}40/100\,dx=12$. Their sum is $8+12=20$ thousand dollars, as before. The result is within the layer’s maximum payment of 40 thousand dollars.

Worked Example: Expected Recovery for an Exponential Loss

A hypothetical ground-up loss $X$ has an exponential distribution with mean $\theta=50$ thousand dollars, so $S_X(x)=e^{-x/50}$ for $x\geq0$. A layer attaches at $d=40$ thousand dollars and has a limit amount $u=60$ thousand dollars. Calculate the expected recovery in dollars, rounded to the nearest cent.

Method. Integrate the exponential survival function from the attachment to the exhaustion point. This method uses the mean parameter directly and avoids confusing it with the variance.

The layer exhausts at $d+u=40+60=100$ thousand dollars. Thus,

\begin{aligned} E[Y^L] &=\int_{40}^{100}e^{-x/50}\,dx\\ &=\left[-50e^{-x/50}\right]_{40}^{100}\\ &=50(e^{-40/50}-e^{-100/50})\\ &=50(e^{-0.8}-e^{-2})\\ &\approx 50(0.449329-0.135335)\\ &\approx 15.6997\text{ thousand dollars}. \end{aligned}

Rounding only at the final step, the expected recovery is $15,699.68. The maximum possible recovery is $60,000, so the result is within the contractual bound. The probability of any recovery is $P(X>40)=e^{-0.8}\approx0.4493$, which is consistent with a positive expected recovery.

Common Limit Interpretation Trap

A frequent error is to integrate the survival function from the attachment $d$ to the limit amount $u$, as if $u$ were the upper ground-up loss threshold. Under the convention used here, $u$ is the maximum payment, so the correct upper integration bound is $d+u$.

For the uniform example, using $d=30$ and limit amount $u=40$, the incorrect calculation would be $\int_{30}^{40}(1-x/100)\,dx=32-25.5=6.5$ thousand dollars. This incorrectly treats the layer as exhausted at a ground-up loss of 40. The correct exhaustion point is $30+40=70$, and the correct expected recovery is $\int_{30}^{70}(1-x/100)\,dx=20$ thousand dollars.

Before calculating, write down both contractual quantities: the attachment $d$ and the limit amount $u$. Then state the exhaustion point $d+u$. A quick reasonableness check is that $0\leq E[Y^L]\leq u$; a result above $u$ is impossible.

Check Your Understanding

For each problem, identify the layer’s exhaustion point before calculating. Dollar answers are stated in the units given.

A hypothetical loss takes values $0,25,50,90$ thousand dollars with probabilities $0.10,0.40,0.30,0.20$, respectively. A layer attaches at $20$ thousand dollars and has a limit amount of $30$ thousand dollars. Find the payment at each loss value and the expected recovery. (Answer: payments $0,5,30,30$ thousand dollars; expected recovery $17$ thousand dollars.)
$X$ is uniform on $[0,120]$ thousand dollars. A layer attaches at $24$ thousand dollars and has a limit amount of $36$ thousand dollars. Find the expected recovery using the survival function. (Answer: $23.40$ thousand dollars.)
$X$ is exponential with mean $25$ thousand dollars. A layer attaches at $25$ thousand dollars and has a limit amount of $50$ thousand dollars. Find the expected recovery to the nearest cent. (Answer: $25(e^{-1}-e^{-3})\approx7.95$ thousand dollars, or $\$7,952.30$.)
A hypothetical ground-up loss has values $10,40,80$ thousand dollars with probabilities $0.20,0.50,0.30$, respectively. For a layer attaching at $30$ thousand dollars with a $50$-thousand-dollar limit, calculate the expected recovery both from the possible payments and from $E[L_{80}(X)]-E[L_{30}(X)]$. (Answer: $20$ thousand dollars by either method.)
A loss has $P(X=0)=0.25$, $P(X=20)=0.25$, $P(X=50)=0.30$, and $P(X=100)=0.20$, with amounts in thousands of dollars. A layer attaches at $25$ thousand dollars and has a limit amount of $50$ thousand dollars. Find the probability of a full recovery and the expected recovery. (Answer: probability of full recovery $0.20$; expected recovery $17.50$ thousand dollars.)

What You'll Learn