Mean Severity from the Survival Function

Why the Survival Curve Gives the Mean

A severity model can be described by the probability that a claim exceeds each possible amount. That information is especially useful when a problem supplies a survival curve or exceedance probabilities rather than a density or a complete probability table. For a nonnegative claim severity, the mean has a direct geometric interpretation: it is the area under the survival curve.

The area interpretation applies to discrete, continuous, and mixed severity distributions. In a discrete model, the survival curve is a step function; in a continuous model, it may be a smooth curve. In either case, the same integral gives the expected severity, provided the integral is interpreted as possibly infinite.

Definition: For a nonnegative severity $X$, define its survival function for $x \ge 0$ by $S_X(x)=P(X>x)$. The value $S_X(x)$ is the probability that a claim exceeds amount $x$. The mean severity is $$ E[X]=\int_0^\infty S_X(x)\,dx, $$ where the integral may be infinite. Thus $E[X]$ is finite exactly when the area under the survival curve is finite.

The integral uses the same monetary units as the severity. If $x$ is measured in dollars, then $S_X(x)$ is dimensionless and integrating over dollars produces an answer in dollars. When the survival curve is given in thousands of dollars, the resulting mean is also in thousands of dollars unless converted.

Deriving the Tail-Integral Formula

The formula follows by expressing each realized severity as the total length of the range of thresholds it exceeds. For any fixed nonnegative value $X$, the indicator $1_{\{X>x\}}$ equals 1 for thresholds below $X$, and 0 for thresholds at or above $X$. Therefore, the area of that indicator over all nonnegative thresholds is exactly $X$.

Write severity as an area.
For every realized value $X\ge 0$, $X=\int_0^\infty 1_{\{X>x\}}\,dx$. The integrand is 1 over an interval of length $X$, and 0 afterward.

Take expectations.
Because the indicator is nonnegative, Tonelli's theorem permits exchanging expectation and integration, including when the result is infinite.

Recognize the survival probability.
The expectation of the indicator is $E[1_{\{X>x\}}]=P(X>x)=S_X(x)$. Hence $E[X]=\int_0^\infty S_X(x)\,dx$.

E[X] =E\left[\int_0^\infty 1_{\{X>x\}}\,dx\right] =\int_0^\infty E[1_{\{X>x\}}]\,dx =\int_0^\infty P(X>x)\,dx =\int_0^\infty S_X(x)\,dx.

No density is required for this argument. It also does not require $E[X]$ to be finite: if the survival curve has infinite area, then the expected severity is infinite. Whether the survival function is defined using $P(X>x)$ or, in some conventions, $P(X\ge x)$ does not change this integral. The two versions can differ at points where $X$ has probability mass, but changing a function at isolated points does not change its area.

Calculating Area for Different Survival Curves

For a survival curve given in pieces, integrate each piece over its own interval and add the results. For a step function, each interval contributes a rectangle: interval width multiplied by the survival probability on that interval. The endpoints of the intervals do not affect the integral.

Worked Example: A Discrete Claim Severity

A hypothetical equipment policy has claim severity $X$, in dollars, with $P(X=0)=0.20$, $P(X=1{,}000)=0.50$, and $P(X=3{,}000)=0.30$. Use the survival curve to calculate $E[X]$, and check the result directly from the probability distribution.

Method. First determine the probability that a claim exceeds a threshold in each interval. Then sum the rectangular areas under the step-shaped survival curve.

For $0\le x<1{,}000$, the claim exceeds $x$ when its value is either $1{,}000$ or $3{,}000$. Thus $S_X(x)=0.50+0.30=0.80$. For $1{,}000\le x<3{,}000$, only a claim of $3{,}000$ exceeds $x$, so $S_X(x)=0.30$. For $x\ge 3{,}000$, the survival probability is 0. The value at an interval endpoint has no effect on the integral.

E[X] =\int_0^{1{,}000}0.80\,dx +\int_{1{,}000}^{3{,}000}0.30\,dx +\int_{3{,}000}^{\infty}0\,dx

The first area is $1{,}000(0.80)=800$ dollars. The second area is $(3{,}000-1{,}000)(0.30)=600$ dollars. Their sum is $800+600=1{,}400$ dollars.

Check from the probability distribution. Directly, $E[X]=0(0.20)+1{,}000(0.50)+3{,}000(0.30)=0+500+900=1{,}400$ dollars. The two calculations agree. This method avoids the common mistake of using the probability at each claim value as though it were the height of the survival curve across an interval.

Worked Example: A Piecewise Linear and Exponential Tail

For a hypothetical property severity $X$, measured in dollars, suppose the survival curve is $S_X(x)=1-x/4{,}000$ for $0\le x\le2{,}000$, and $S_X(x)=0.50e^{-(x-2{,}000)/2{,}000}$ for $x>2{,}000$. The curve starts at 1, equals 0.50 at $2{,}000$, decreases thereafter, and tends to 0. Calculate the mean severity.

Method. Split the tail integral at $2{,}000$, where the formula changes. Integrate the linear section and the exponential tail separately.

E[X] =\int_0^{2{,}000}\left(1-\frac{x}{4{,}000}\right)dx +\int_{2{,}000}^{\infty}0.50e^{-(x-2{,}000)/2{,}000}\,dx.

For the first section, an antiderivative is $x-x^2/8{,}000$. Its value at $2{,}000$ is $2{,}000-4{,}000{,}000/8{,}000=2{,}000-500=1{,}500$, and its value at 0 is 0. The first area is therefore $1{,}500$ dollars.

For the tail, set $u=(x-2{,}000)/2{,}000$, so $dx=2{,}000\,du$. The lower limit becomes 0 and the upper limit becomes infinity:

\int_{2{,}000}^{\infty}0.50e^{-(x-2{,}000)/2{,}000}\,dx =1{,}000\int_0^\infty e^{-u}\,du =1{,}000[-e^{-u}]_0^\infty =1{,}000.

Adding the areas gives $E[X]=1{,}500+1{,}000=2{,}500$ dollars. The calculation includes the entire tail beyond $2{,}000$; stopping the integral at that point would omit $1{,}000$ dollars of area.

Using the Formula for a Heavy-Tailed Severity

A survival function can also make the existence of the mean visible. Consider a Pareto Type II (Lomax) severity with scale parameter $\theta>0$, shape parameter $\alpha>0$, support $x\ge0$, and survival function $S_X(x)=(1+x/\theta)^{-\alpha}$. The curve extends over an unbounded range, so its tail behavior matters to whether its area is finite.

Worked Example: Mean of a Pareto Type II Severity

A hypothetical portfolio uses a Pareto Type II severity model with $\theta=1{,}200$ dollars and $\alpha=2.5$. Its survival function is $S_X(x)=(1+x/1{,}200)^{-2.5}$ for $x\ge0$. Calculate the mean severity using the survival function and verify that it is finite.

Method. Integrate the survival curve over all $x\ge0$, using a change of variable to reduce the integral to a power integral.

E[X]=\int_0^\infty \left(1+\frac{x}{1{,}200}\right)^{-2.5}dx.

Set $u=1+x/1{,}200$, so $dx=1{,}200\,du$. As $x$ runs from 0 to infinity, $u$ runs from 1 to infinity. Therefore,

E[X] =1{,}200\int_1^\infty u^{-2.5}\,du =1{,}200\left[\frac{u^{-1.5}}{-1.5}\right]_1^\infty =1{,}200\left(0-\left(-\frac{1}{1.5}\right)\right) =800.

The mean severity is $800$ dollars, which is less than the scale parameter $1{,}200$ dollars. The integral converges because the power exponent is $2.5>1$; for this survival form, the mean is finite when $\alpha>1$. The result checks both the calculation and the required tail condition.

Common Mistake: Integrating the CDF Instead

The mean is the area under the survival curve, not generally the area under the CDF. Since $S_X(x)=1-F_X(x)$ for a continuous severity, replacing $S_X(x)$ by $F_X(x)$ changes the integral. In fact, for an unbounded nonnegative severity, the CDF usually approaches 1, so its area over the entire positive axis is infinite.

For example, let $X$ have an exponential distribution with mean $1{,}000$ dollars. Its survival function is $S_X(x)=e^{-x/1{,}000}$, so the correct calculation is $\int_0^\infty e^{-x/1{,}000}dx=1{,}000$ dollars. The wrong calculation $\int_0^\infty F_X(x)dx$ integrates $1-e^{-x/1{,}000}$, which tends to 1 and has infinite area. The survival function records exceedance probabilities, which is why its area—not the CDF's area—gives the mean.

Key takeaway: For any nonnegative severity, compute $E[X]$ by integrating $S_X(x)=P(X>x)$ from zero to infinity. For a piecewise curve, add the areas of its pieces; for a discrete severity, these are rectangles. Check that the full tail is included and that the resulting area is finite if a finite mean is required.

Check Your Understanding

For each problem, use the survival-function area and check that the stated curve or distribution has the required support and tail behavior.

A hypothetical severity has $S_X(x)=0.90$ for $0\le x<500$, $S_X(x)=0.30$ for $500\le x<2{,}500$, and $S_X(x)=0$ for $x\ge2{,}500$, with amounts in dollars. Find $E[X]$. (Answer: $\$1{,}050$)
A hypothetical claim severity has $S_X(x)=1-x/6{,}000$ for $0\le x\le3{,}000$, then $S_X(x)=0.50e^{-(x-3{,}000)/4{,}000}$ for $x>3{,}000$. Find its mean in dollars. (Answer: $\$4{,}250$)
A discrete hypothetical severity is $0$ with probability 0.10, $\$500$ with probability 0.60, and $\$2{,}500$ with probability 0.30. Find its mean using the survival curve. (Answer: $\$1{,}050$)
A hypothetical Pareto Type II severity has survival $S_X(x)=(1+x/1{,}500)^{-3}$ for $x\ge0$, with amounts in dollars. Calculate its mean. (Answer: $\$750$)
A hypothetical severity is a mixture: with probability 0.40 it is exponential with mean $\$500$, and with probability 0.60 it is exponential with mean $\$1{,}500$. Use the mixture survival curve to find its mean. (Answer: $\$1{,}100$)

What You'll Learn