Tail-Probability Formula for Stop-Loss Premiums

Why a Tail-Probability Formula Is Useful

An aggregate stop-loss contract pays the portion of a nonnegative aggregate loss above a specified attachment point. Its expected payment can be calculated directly from the distribution of the aggregate loss, but it can also be found by adding up the probabilities that the loss exceeds successive thresholds. That second approach is especially useful when the available information is a survival curve or a table of tail probabilities.

Here, the focus is the expected payment itself, before expenses, risk margins, or other premium loadings. All worked examples state their hypothetical model and use money in thousands of dollars where indicated.

Definition: Let $S$ be a nonnegative aggregate loss and let $d\geq 0$ be an attachment point. The stop-loss payment is $(S-d)_+=\max(S-d,0)$. Its expected value, called the stop-loss premium in this tutorial, is $\Pi(d)=E[(S-d)_+]$. The aggregate-loss survival probability at $x$ is $P(S>x)$.

The attachment point is the amount retained before the stop-loss cover begins to pay. If $S\leq d$, the payment is zero. If $S>d$, the payment is the excess $S-d$. The premium is therefore an expectation over the whole distribution, not the payment conditional on the loss exceeding $d$.

The Tail-Probability Identity

For a nonnegative loss, the stop-loss premium is the area under the survival function to the right of the attachment point. The integral is with respect to the loss amount $x$, so the units of the result are monetary units.

Theorem: If $S\geq 0$ and $d\geq 0$, then $$ \Pi(d)=E[(S-d)_+]=\int_d^\infty P(S>x)\,dx. $$ The equality holds for nonnegative random variables, including distributions with atoms. If $E[S]$ is finite, then the integral is finite. Equivalently, since $P(S>x)=1-F_S(x)$, where $F_S(x)=P(S\leq x)$, $$ \Pi(d)=\int_d^\infty \bigl(1-F_S(x)\bigr)\,dx. $$

The strict inequality in $P(S>x)$ is consistent with defining the distribution function by $F_S(x)=P(S\leq x)$. A point mass exactly at the attachment point produces no stop-loss payment. More generally, the identity remains valid at distributions with point masses: changing the integrand at isolated threshold values does not change its integral.

Why the Formula Holds

For any fixed realized loss $S=s$, the indicator $1\{s>x\}$ equals one for thresholds below $s$, and zero for thresholds at or above $s$. Consequently, the length of the interval from $d$ to $s$, when $s>d$, is exactly the excess loss. When $s\leq d$, the interval contributes nothing.

(s-d)_+=\int_d^\infty 1\{s>x\}\,dx

Apply this pointwise identity to the random variable $S$, and take expectations. Because the integrand is nonnegative, Tonelli's theorem permits the order of expectation and integration to be interchanged, even before finiteness is known:

\begin{aligned} E[(S-d)_+] &=E\left[\int_d^\infty 1\{S>x\}\,dx\right]\\ &=\int_d^\infty E[1\{S>x\}]\,dx\\ &=\int_d^\infty P(S>x)\,dx. \end{aligned}

The last equality follows because the expectation of an indicator is the probability of the event it indicates. This derivation also explains the geometric interpretation: each threshold $x$ contributes its exceedance probability to the total expected excess.

Using the Formula in Practice

The calculation depends on the form in which the aggregate-loss distribution is given. If a survival function has a closed form, integrate that function. If it is piecewise defined, split the integral at every breakpoint and evaluate each segment. If only rounded or sampled tail probabilities are available, a numerical area estimate may be appropriate, but it should be identified as an approximation rather than an exact premium.

A reliable setup is to write the attachment point as the lower limit before integrating. Then check that the integrand is a probability between zero and one, that it does not increase as $x$ increases, and that the resulting premium is nonnegative. When the mean aggregate loss is finite, another useful check is $0\leq \Pi(d)\leq E[S]$.

Worked Example: Exponential Aggregate Loss

A hypothetical portfolio has aggregate loss $S$ with an exponential distribution of mean $100$ thousand dollars. The insurer considers an unlimited stop-loss cover attaching at $100$ thousand dollars. Calculate its expected payment in thousands of dollars, to four decimal places.

Method. For an exponential loss with mean $\theta$, the survival function is $P(S>x)=e^{-x/\theta}$ for $x\geq 0$. Here $\theta=100$ thousand dollars and $d=100$ thousand dollars. The model is nonnegative and has finite mean, so the tail-integral identity applies.

Set up the integral. Measure $x$ in thousands of dollars: $$ \Pi(100)=\int_{100}^{\infty}e^{-x/100}\,dx. $$

Integrate. An antiderivative is $-100e^{-x/100}$, so $$ \Pi(100)=\left[-100e^{-x/100}\right]_{100}^{\infty} =0-(-100e^{-1}) =\frac{100}{e}. $$

Evaluate and check. Since $e\approx2.7182818$, $100/e\approx36.7879$ thousand dollars. The expected payment is therefore $36,787.94, to the nearest cent. As a check, $P(S>100)=e^{-1}$, and the exponential distribution's mean excess above 100 thousand dollars is 100 thousand dollars. Their product is $e^{-1}(100)=36.7879$ thousand dollars, agreeing with the integral. This method avoids the common mistake of integrating from zero and returning the full mean of $100,000.

Worked Example: A Piecewise-Linear Survival Curve

A hypothetical aggregate loss $S$, measured in thousands of dollars, has the following survival function: $$ P(S>x)= \begin{cases} 1-0.005x, & 0\leq x\leq 100,\\ 0.5\left(\dfrac{300-x}{200}\right), & 100<x\leq 300,\\ 0, & x>300. \end{cases} $$ The survival curve is continuous at 100, decreases to zero at 300, and is zero beyond 300. Calculate the expected payment for an unlimited stop-loss cover attaching at $d=50$ thousand dollars.

Method. The lower limit is the attachment point, 50. Since the survival function changes formula at 100, split the integral there: $$ \Pi(50)=\int_{50}^{100}(1-0.005x)\,dx +\int_{100}^{300}0.5\left(\frac{300-x}{200}\right)\,dx. $$

First segment. The survival probabilities at 50 and 100 are $1-0.005(50)=0.75$ and $1-0.005(100)=0.50$. The segment is linear, so its area is the width times the average endpoint height: $$ (100-50)\frac{0.75+0.50}{2} =50(0.625)=31.25 $$ thousand dollars.

Second segment. The survival probabilities at 100 and 300 are 0.50 and 0. The area is $$ (300-100)\frac{0.50+0}{2} =200(0.25)=50 $$ thousand dollars.

Add and check. Thus $\Pi(50)=31.25+50=81.25$ thousand dollars, or $81,250.00. The area is positive and does not exceed the full mean loss, which is the area under the complete survival curve: $100(1+0.5)/2+200(0.5+0)/2=75+50=125$ thousand dollars. A common error is to omit the first segment from 50 to 100, which would understate the premium by $31,250.

Worked Example: A Pareto-Tail Aggregate Loss

A hypothetical aggregate loss $S$, measured in thousands of dollars, has survival function $P(S>x)=(1+x/100)^{-3}$ for $x\geq0$. This is a Pareto-type model with scale parameter 100 thousand dollars and tail exponent 3. Calculate the expected stop-loss payment when $d=100$ thousand dollars, to the nearest cent.

Method. Substitute the stated survival function into the tail integral. The exponent is greater than one, so the tail integral converges: $$ \Pi(100)=\int_{100}^{\infty}\left(1+\frac{x}{100}\right)^{-3}\,dx. $$

Change variables. Let $u=1+x/100$, so $dx=100\,du$. When $x=100$, $u=2$, and as $x$ tends to infinity, $u$ tends to infinity. Therefore $$ \Pi(100)=100\int_2^\infty u^{-3}\,du =100\left[-\frac{1}{2u^2}\right]_2^\infty =100\left(0+\frac{1}{8}\right) =12.5 $$ thousand dollars.

Check. The exceedance probability at the attachment is $P(S>100)=2^{-3}=0.125$. Dividing the expected excess, 12.5 thousand dollars, by 0.125 gives a conditional mean excess of 100 thousand dollars; multiplying it back by the exceedance probability recovers 12.5 thousand dollars. The expected payment is $12,500.00. This calculation avoids treating the tail exponent as the scale or integrating the model below the attachment point.

Common Exam Trap: Confusing Tail Probability With the Premium

The probability of exceeding the attachment point is not itself the stop-loss premium. It is dimensionless, whereas the premium is an amount of money. The premium accumulates the exceedance probabilities over all thresholds above $d$, with each increment measured in loss units.

For the exponential example, $P(S>100)=e^{-1}\approx0.3679$. Reporting 0.3679 as the premium confuses a probability with a monetary expected payment. Another incorrect calculation is $\int_0^\infty P(S>x)\,dx$, which equals $E[S]$, not $E[(S-d)_+]$ when $d>0$. For that example, the wrong lower limit gives 100 thousand dollars; using the correct lower limit gives $100/e\approx36.7879$ thousand dollars.

If $P(S>d)>0$, one may also write $$ \Pi(d)=P(S>d)E[S-d\mid S>d]. $$ This follows because $(S-d)_+$ is zero unless $S>d$. It is an alternative expectation calculation, not a replacement for the tail integral. In particular, the conditional excess alone is not the unconditional stop-loss premium.

Key takeaway: For a nonnegative aggregate loss, the expected unlimited stop-loss payment at attachment $d$ is the area $\int_d^\infty P(S>x)\,dx$. Start at the attachment, use the unconditional survival probability, and preserve the loss units throughout the integration.

Check Your Understanding

For each hypothetical aggregate-loss model, use the tail-probability identity. Money is in thousands of dollars unless otherwise stated.

$S$ is exponential with mean $50,000 and the attachment is $25,000. Find the expected stop-loss payment to the nearest cent. (Answer: $30,326.53)
The survival function is $P(S>x)=1-x/200$ for $0\leq x\leq200$, and zero for $x>200$. Find the expected payment at an attachment of 50. (Answer: $56,250.00)
A Pareto-type aggregate loss has $P(S>x)=(1+x/50)^{-2}$ for $x\geq0$. Find the expected payment at $d=50$. (Answer: $25,000.00)
A continuous aggregate-loss model has survival function $1-0.004x$ for $0\leq x\leq100$, $0.6-0.004(x-100)$ for $100<x\leq250$, and zero thereafter. Find the expected payment at $d=50$. (Answer: $80,000.00)
With probability 0.25 a hypothetical aggregate loss is zero; otherwise it is exponential with mean $40,000. Find the expected payment at $d=$40,000$, to the nearest cent. (Answer: $11,036.38)

What You'll Learn