Deriving Layer Payments from Stop-Loss Functions

From Stop-Loss Payments to Layer Payments

A stop-loss payment above a threshold $t$ is $(X-t)_+$: the amount of a loss $X$ that exceeds $t$, where $z_+=\max(z,0)$. A capped layer pays only the part of a loss between its attachment point and its exhaustion point. The key observation is that this capped payment can be written as one stop-loss payment minus another.

This identity is useful because it converts a layer calculation into two evaluations of the stop-loss transform. It works for discrete, continuous, and mixed loss distributions, including distributions with probability mass at the attachment or exhaustion points.

Definition: For a nonnegative, integrable loss $X$, the stop-loss transform at threshold $t\geq 0$ is $\pi_X(t)=E[(X-t)_+]$. A layer with attachment point $d\geq0$ and limit $u>0$ pays $Y=\min((X-d)_+,u)$. Its exhaustion point is $d+u$: losses at or above this point produce the layer’s full payment $u$.

Here $u$ denotes the maximum payment by this layer, or its width. It is not the exhaustion threshold. For example, if $d=30$ dollars and $u=50$ dollars, the layer exhausts at a loss of $80$ dollars.

Why the Difference Identity Holds

For any fixed loss amount $x$, subtract the stop-loss payment above $d+u$ from the payment above $d$. The result is exactly the amount in the layer. Checking the three possible ranges of $x$ proves the identity, including at the boundary points.

If $x\leq d$.
Neither threshold is exceeded, so $(x-d)_+-(x-d-u)_+=0$. The layer also pays zero.

If $d<x\leq d+u$.
The first stop-loss payment is $x-d$, while the second is zero. Their difference, $x-d$, is the layer payment before exhaustion.

If $x>d+u$.
The two stop-loss payments are $x-d$ and $x-d-u$. Their difference is $u$, the layer’s full payment after exhaustion.

Thus the equality holds for every possible loss amount $x$. Taking expectations on both sides gives the expected layer payment. The assumptions needed are that $X$ is nonnegative and has a finite expectation, so the stop-loss expectations are finite.

\min((X-d)_+,u)=(X-d)_+-(X-d-u)_+

E[Y]=\pi_X(d)-\pi_X(d+u)

The second formula follows from linearity of expectation applied to the pointwise equality. It does not require $X$ to have a density or to be continuous. In particular, an atom at $d$ contributes zero to the layer, and an atom at $d+u$ contributes the full limit $u$; the identity handles both cases without a special adjustment.

Computing the Stop-Loss Transform

For a discrete loss taking values $x_j$ with probabilities $p_j$, calculate $\pi_X(t)=\sum_j p_j(x_j-t)_+$. For a continuous loss with density $f_X$, the corresponding expectation is an integral. When $X$ is nonnegative and integrable, an equivalent tail-probability form is also available.

\pi_X(t)=\int_t^\infty (x-t)f_X(x)\,dx =\int_t^\infty S_X(y)\,dy

The tail form follows by writing $(X-t)_+=\int_t^\infty \mathbf{1}_{\{X>y\}}\,dy$ and taking expectations. The integrand’s expectation is $P(X>y)=S_X(y)$; integration therefore gives the stated expression. The identity applies whether or not the loss has a density, while the density-integral form specifically assumes a continuous distribution with density $f_X$.

Worked Example: A Discrete Loss and a Capped Layer

A hypothetical loss $X$ is $0$, $40$, or $100$ dollars with probabilities $0.50$, $0.30$, and $0.20$, respectively. A layer attaches at $d=\$30$ and has limit $u=\$50$. Find its expected payment using stop-loss values, then verify the result directly. The method is to evaluate the transform at attachment and exhaustion; a common mistake is to use the limit itself as the second threshold.

Step 1: Find the exhaustion point. The layer exhausts at $d+u=\$30+\$50=\$80$, not at $\$50$.

Step 2: Evaluate the transform at attachment. At threshold $\$30$, the excesses for losses $0,40,100$ are $0,10,70$. Therefore, $\pi_X(30)=0.50(0)+0.30(10)+0.20(70)=3+14=\$17$.

Step 3: Evaluate the transform at exhaustion. At threshold $\$80$, the excesses are $0,0,20$. Thus, $\pi_X(80)=0.50(0)+0.30(0)+0.20(20)=\$4$.

Step 4: Subtract. The expected layer payment is $\pi_X(30)-\pi_X(80)=\$17-\$4=\$13.00$.

Check by direct payments. The layer pays $0$ when $X=0$, $10$ when $X=40$, and $50$ when $X=100$. Its expected payment is $0.50(0)+0.30(10)+0.20(50)=3+10=\$13.00$, agreeing with the transform calculation. Using $\pi_X(30)-\pi_X(50)$ would give $17-10=\$7$, which is wrong because $\$50$ is the layer limit, not its exhaustion point.

Continuous-Loss Example

For an exponential loss with mean $\theta$, the survival function is $S_X(x)=e^{-x/\theta}$ for $x\geq0$. The stop-loss transform can be found from the tail integral: $\pi_X(t)=\int_t^\infty e^{-y/\theta}\,dy=\theta e^{-t/\theta}$. Using this result for both thresholds gives the expected payment of a layer.

Worked Example: An Exponential Loss Layer

A hypothetical loss $X$ has an exponential distribution with mean $\theta=\$200$. A layer attaches at $d=\$100$ and has limit $u=\$100$. Find its expected payment to the nearest cent. Use the exponential stop-loss transform and remember that the second threshold is the exhaustion point.

Step 1: Identify both thresholds. The attachment point is $\$100$, and the exhaustion point is $d+u=\$200$.

Step 2: Apply the transform. Since $\pi_X(t)=200e^{-t/200}$, $\pi_X(100)=200e^{-0.5}=\$121.3061319$ and $\pi_X(200)=200e^{-1}=\$73.5758882$.

Step 3: Take the difference. $E[Y]=121.3061319-73.5758882=\$47.7302437$, which rounds to \$47.73 to the nearest cent.

Check and common mistake. The result is positive and below the layer limit of $\$100$, as it must be. Using $200e^{-0.5}-200e^{-100/200}$ would subtract the same transform value twice because it incorrectly uses the layer limit, rather than the exhaustion point, as the second threshold.

Using the Identity with a Discrete Distribution

A discrete calculation is especially useful as a check because the layer payment can be listed at every possible loss value. In the transform calculation, losses above the exhaustion point contribute to both stop-loss values; their difference is the capped amount, not the full excess above attachment.

Worked Example: A Layer on Losses Measured in Thousands

A hypothetical annual loss, measured in thousands of dollars, is $0$, $80$, $150$, or $300$, with probabilities $0.20$, $0.30$, $0.30$, and $0.20$, respectively. A layer attaches at $d=50$ and has limit $u=150$, both in thousands of dollars. Find its expected payment by the transform identity and verify directly. The key check is to use $200$, not $150$, as the exhaustion point.

Step 1: Determine the thresholds. The layer attaches at $50$ and exhausts at $d+u=50+150=200$, in thousands of dollars.

Step 2: Calculate $\pi_X(50)$. The excesses above $50$ are $0,30,100,250$. Therefore, $\pi_X(50)=0.20(0)+0.30(30)+0.30(100)+0.20(250)=9+30+50=89$ thousand dollars.

Step 3: Calculate $\pi_X(200)$. The excesses above $200$ are $0,0,0,100$. Hence, $\pi_X(200)=0.20(0)+0.30(0)+0.30(0)+0.20(100)=20$ thousand dollars.

Step 4: Subtract and verify. The expected layer payment is $89-20=69$ thousand dollars. Directly, the layer payments at losses $0,80,150,300$ are $0,30,100,150$. Their expectation is $0.20(0)+0.30(30)+0.30(100)+0.20(150)=9+30+30=69$ thousand dollars. The two methods agree.

Interpretation and Exam Checks

The stop-loss transform above attachment includes all loss above the attachment, even the portion beyond the layer’s limit. Subtracting the transform above exhaustion removes that excess portion. For a loss beyond exhaustion, the subtraction leaves exactly the layer limit; for a loss within the layer, it leaves exactly the loss amount above attachment.

A useful exam procedure is to write down the attachment and exhaustion points before calculating either transform. Then calculate each stop-loss value from the given distribution, subtract in the order $\pi_X(d)-\pi_X(d+u)$, and check that the answer lies between zero and $u$. If the answer is negative or greater than $u$, the thresholds, subtraction order, or arithmetic should be checked.

Do not confuse limit with exhaustion. Exhaustion is attachment plus limit.
Do not subtract in reverse order. Since the transform decreases as its threshold increases, $\pi_X(d)\geq\pi_X(d+u)$.
Do not omit losses beyond exhaustion. They contribute to both transforms, and their difference is the full layer limit.
Keep units consistent. If losses are stated in thousands of dollars, both transform values and the final expected payment are in thousands.

Key takeaway: For a nonnegative integrable loss $X$, the payment from a layer attaching at $d$ with limit $u$ is $(X-d)_+-(X-d-u)_+$. Its expected payment is the stop-loss transform at attachment minus the stop-loss transform at exhaustion: $\pi_X(d)-\pi_X(d+u)$.

Check Your Understanding

For each problem, identify the exhaustion point before evaluating a layer payment.

A hypothetical loss takes values $25$, $75$, and $180$ dollars with probabilities $0.40$, $0.35$, and $0.25$. A layer attaches at $\$50$ and has limit $\$80$. Find the expected payment. (Answer: $\$28.75$)
A hypothetical loss is exponential with mean $\$300$. A layer attaches at $\$100$ and has limit $\$200$. Find its expected payment to the nearest cent. (Answer: $\$104.60$)
A hypothetical loss, in thousands of dollars, is $0$, $60$, $120$, or $240$, with probabilities $0.20$, $0.30$, $0.30$, and $0.20$. A layer attaches at 40 and has limit 100. Find the expected payment in thousands of dollars. (Answer: 50 thousand dollars)
A hypothetical loss is uniformly distributed from $\$0$ to $\$200$. A layer attaches at $\$50$ and has limit $\$100$. Use the stop-loss transform to find the expected payment. (Answer: $\$50.00$)

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