Survival-Function Formulas For Layer Loss

Why Survival Functions Are Useful for Layer Loss

An excess layer pays only when a loss exceeds its attachment, and it stops paying once the loss reaches the top of the layer. A severity survival function describes the probability that a loss exceeds each possible amount. Integrating those probabilities over the layer gives the expected payment per loss, without requiring a full calculation of the mean loss distribution above every threshold.

This is useful for continuous severities, where a layer payment may have many possible values, and for discrete severities, where losses may have probability masses at layer boundaries. The same formulas handle both cases.

Definition: Let $X$ be a nonnegative loss, with survival function $S_X(x)=P(X>x)$. For a layer with attachment $d\geq 0$ and payment limit $u>0$, the payment per loss is $Y^L_{d,u}=\min((X-d)_+,u)$, where $(z)_+=\max(z,0)$. Here $u$ is the width of the layer: the layer pays between loss amounts $d$ and $d+u$. Its expected loss per ground-up loss is $E[Y^L_{d,u}]$.

The attachment is not itself the upper limit. For example, a layer described as $50,000 excess of $100,000 attaches at $100,000 and has a payment limit of $50,000, so its upper loss amount is $150,000. Unless stated otherwise, the calculations below are undiscounted expected losses; no interest or mortality assumption applies.

The Survival-Function Formula

For a nonnegative random variable $W$, its expectation can be written as the area under its survival function: $E[W]=\int_0^\infty P(W>t)\,dt$. To use this for a layer, consider how much the layer pays at each possible payment threshold $t$. For $0\leq t<u$, the layer pays more than $t$ exactly when the ground-up loss exceeds $d+t$.

Theorem: For a nonnegative severity $X$, attachment $d\geq0$, and layer limit $u>0$, $$ E[Y^L_{d,u}]=\int_0^u S_X(d+t)\,dt =\int_d^{d+u} S_X(x)\,dx. $$ The formula applies to continuous, discrete, and mixed severity distributions. The values of the survival function at individual endpoints do not affect the integral.

The key is to integrate over the layer's payment scale, not over all possible loss amounts. Replacing $x=d+t$ gives the second integral, whose bounds are the attachment and the top of the layer.

Set the layer bounds.
Identify the attachment $d$, payment limit $u$, and upper loss amount $d+u$.

Write the survival function.
Use $S_X(x)=P(X>x)$ for the stated severity model, or the relevant piecewise probabilities if the severity is discrete.

Integrate across the layer.
Calculate $\int_d^{d+u}S_X(x)\,dx$. Keep the loss units consistent throughout.

Check the result.
The expected layer payment must be between zero and $u$. It must also not exceed the expected payment of a lower layer that includes at least as much loss payment in every outcome.

Limited Expected Values and Stop-Loss Expectations

The same layer expectation can be obtained from limited expected values. For $z\geq0$, define $\operatorname{LEV}_X(z)=E[\min(X,z)]$. This is the expected loss capped at $z$; it is not the expected loss conditional on being below $z$.

Result: For the layer defined above, $$ E[Y^L_{d,u}] =\operatorname{LEV}_X(d+u)-\operatorname{LEV}_X(d). $$ Equivalently, if $\operatorname{SL}_X(a)=E[(X-a)_+]$ is the expected payment above attachment $a$, then $$ E[Y^L_{d,u}]=\operatorname{SL}_X(d)-\operatorname{SL}_X(d+u). $$

These identities follow by comparing the payments outcome by outcome. For any fixed loss $X$, $$ \min(X,d+u)-\min(X,d)=\min((X-d)_+,u). $$ If $X\leq d$, both sides are zero. If $d<X<d+u$, both sides equal $X-d$. If $X\geq d+u$, both sides equal $u$. Taking expectations proves the limited-expected-value identity. Also, for every loss, $$ (X-d)_+-(X-d-u)_+=\min((X-d)_+,u), $$ which proves the stop-loss identity on taking expectations. These are algebraic identities and do not require a continuous severity distribution.

Worked Calculations

Worked Example: Exponential Severity and a Finite Layer

A hypothetical insurer models the ground-up loss per claim as exponential with mean $100,000, so $S_X(x)=e^{-x/100000}$ for $x\geq0$. Calculate the expected payment per loss for a $100,000 limit excess of $50,000. There is no discounting or mortality component.

Method. The attachment is $d=\$50{,}000$, the layer width is $u=\$100{,}000$, and the upper loss amount is $d+u=\$150{,}000$. Apply the survival integral: $$ E[Y^L_{d,u}]=\int_{50{,}000}^{150{,}000} e^{-x/100{,}000}\,dx. $$

An antiderivative is $-100{,}000e^{-x/100{,}000}$, so $$ E[Y^L_{d,u}] =100{,}000\left(e^{-0.5}-e^{-1.5}\right) =100{,}000(0.60653066-0.22313016) =\$38{,}340.05. $$

As a check, $\operatorname{LEV}_X(z)=100{,}000(1-e^{-z/100{,}000})$. Thus $\operatorname{LEV}_X(150{,}000)=\$77{,}686.98$ and $\operatorname{LEV}_X(50{,}000)=\$39{,}346.93$; their difference is $38,340.05. Rounded to the nearest cent, the answer is within the layer's $100,000 limit. The common mistake avoided is integrating from $0 to $100,000 instead of from the $50,000 attachment to the $150,000 top of the layer.

Worked Example: A Discrete Severity With Boundary Losses

For a hypothetical equipment portfolio, suppose loss $X$ has this distribution: $0 with probability 0.20, $50,000 with probability 0.30, $150,000 with probability 0.30, and $300,000 with probability 0.20. Find the expected payment per loss for a $150,000 limit excess of $50,000. No interest or mortality assumption is involved.

Method 1: Calculate layer payments by outcome. For losses of $0 and $50,000, the payment is zero. At a loss of $150,000, the payment is $\min(\$150{,}000-\$50{,}000,\$150{,}000)=\$100{,}000$. At a loss of $300,000, it is capped at $150,000. Therefore, $$ E[Y^L_{d,u}] =0.20(0)+0.30(0)+0.30(\$100{,}000)+0.20(\$150{,}000) =\$60{,}000. $$

Method 2: Integrate the survival function. Here $d=\$50{,}000$ and $d+u=\$200{,}000$. For $50{,}000\leq x<150{,}000$, $S_X(x)=P(X>x)=0.30+0.20=0.50$. For $150{,}000\leq x<200{,}000$, $S_X(x)=0.20$. Hence $$ \int_{50{,}000}^{200{,}000}S_X(x)\,dx =(\$100{,}000)(0.50)+(\$50{,}000)(0.20) =\$50{,}000+\$10{,}000 =\$60{,}000. $$

The two methods agree. The expected payment is below the $150,000 limit. A loss exactly at the $150,000 boundary contributes to the payment, but changing the survival function at that single point does not change the integral. A common error is to count only losses strictly above $150,000; that would omit the $150,000 loss's $100,000 layer payment.

Worked Example: Pareto-Type Severity Using Limited Expected Values

Suppose a hypothetical liability severity in thousands of dollars has survival function $S_X(x)=(1+x/100)^{-2}$ for $x\geq0$. Calculate the expected payment per loss for a limit of $200,000 excess of $100,000. The distribution parameters are scale $\theta=100$ thousand dollars and shape $\alpha=2$; no discounting or mortality applies.

Method. Work in thousands of dollars, so $d=100$, $u=200$, and the upper loss amount is $300$. Integrate the survival function: $$ E[Y^L_{d,u}] =\int_{100}^{300}\left(1+\frac{x}{100}\right)^{-2}dx =\int_{100}^{300}\frac{10000}{(100+x)^2}\,dx. $$

An antiderivative is $-10000/(100+x)$. Therefore, $$ E[Y^L_{d,u}] =10000\left(\frac{1}{200}-\frac{1}{400}\right) =10000(0.005-0.0025) =25 $$ thousand dollars, or $25,000.00.

Check with limited expected values: $\operatorname{LEV}_X(z)=100z/(100+z)$ thousand dollars. Thus $\operatorname{LEV}_X(300)=\$75{,}000$ and $\operatorname{LEV}_X(100)=\$50{,}000$, giving a difference of $25,000. The answer is positive and below the $200,000 limit. The common mistake avoided is treating the stated $200,000 limit as the upper loss amount; the layer actually extends to $300,000.

Interpretation and Exam Traps

A survival-integral result is an expected payment per ground-up loss, not automatically an annual layer loss or a premium. Turning it into an annual amount requires information about claim frequency and the annual loss structure. Keep the calculation here at the per-loss level.

Confusing limit with upper loss amount. A limit $u$ is the maximum payment; the layer's upper loss amount is $d+u$.
Using the wrong integration interval. The survival integral starts at $d$ and ends at $d+u$, not at zero and $u$.
Applying the deductible after capping. The payment is $\min((X-d)_+,u)$, not $(\min(X,u)-d)_+$. The latter can describe a different payment structure.
Dropping probabilities at atoms. With discrete losses, compute $P(X>x)$ on each interval or use the outcome-by-outcome payment. Do not replace the survival function by a cumulative distribution function.
Mixing units. If $x$, $d$, and $u$ are in thousands of dollars, the integral is in thousands of dollars as well.

A useful reasonableness check is that $0\leq E[Y^L_{d,u}]\leq u$. Increasing the attachment while keeping the limit fixed cannot increase the payment for any individual loss, so it cannot increase the expected payment. Increasing the limit while keeping the attachment fixed cannot decrease it.

Check Your Understanding

For each problem, use a survival integral, a limited-expected-value difference, or direct outcome payments as appropriate. Answers are shown for checking.

A hypothetical severity is exponential with mean $80,000. Find the expected payment per loss for a limit of $80,000 excess of $40,000. (Answer: $30,672.04)
A discrete loss is $0, $40,000, $100,000, or $250,000 with respective probabilities 0.40, 0.20, 0.25, and 0.15. Find expected payment per loss for a $100,000 limit excess of $75,000. (Answer: $21,250)
A severity model has $\operatorname{LEV}_X(100{,}000)=\$42{,}000$ and $\operatorname{LEV}_X(250{,}000)=\$78{,}000$. Find the expected payment per loss for a $150,000 limit excess of $100,000. (Answer: $36,000)
In thousands of dollars, a severity survival function is 0.40 on $20\leq x<50$ and 0.10 on $50\leq x<80$. Find expected payment per loss for a $60,000 limit excess of $20,000. (Answer: $15,000)
For a hypothetical severity, the expected payment above $80,000 is $55,000, and the expected payment above $200,000 is $18,000. Find the expected payment per loss for a $120,000 limit excess of $80,000. (Answer: $37,000)

What You'll Learn