Tail Integral Formulas For Nonnegative Losses

Why Calculate Moments From the Tail?

A loss distribution can be described by its probabilities near each possible loss amount or by the probability that the loss exceeds each threshold. The second description is the survival function. Integrating that tail gives a direct route to the mean and higher raw moments, including for distributions that combine a point mass with a continuous range of losses.

This method is useful when the survival function is simpler than the density or probability mass function. It also handles discrete, continuous, and mixed distributions in one formula. The loss $X$ is assumed to be nonnegative throughout; the formula does not apply unchanged to a loss variable that can take negative values.

Definition: For a nonnegative loss $X$, define its survival function by $S_X(x)=P(X>x)$ for $x\geq 0$. For any real $k>0$, the $k$th raw moment, if finite, is $$ E[X^k]=k\int_0^\infty x^{k-1}S_X(x)\,dx. $$ The integral is allowed to be infinite: if it diverges, then $E[X^k]=\infty$.

Why the Tail-Integral Formula Holds

The key observation is that a realized loss $X$ can be represented by accumulating a quantity over every threshold below it. For any fixed nonnegative value $z$,

z^k=\int_0^\infty kx^{k-1}\mathbf{1}_{\{z>x\}}\,dx,

because the indicator is 1 from $x=0$ up to $x=z$, and 0 beyond $z$. Thus the integral is $\int_0^z kx^{k-1}dx=z^k$. Substituting the random loss $X$ for $z$ and taking expectations gives the formula: the expectation of the indicator $\mathbf{1}_{\{X>x\}}$ is $P(X>x)=S_X(x)$. Since the integrand is nonnegative, the order of expectation and integration can be interchanged even when the result is infinite.

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

For $k=1$, this becomes $E[X]=\int_0^\infty S_X(x)\,dx$. For $k=2$, it becomes $E[X^2]=2\int_0^\infty xS_X(x)\,dx$. The survival function uses the strict inequality $X>x$. Atoms can make its value jump at particular thresholds, but changing the value of an integrand at isolated points does not change these integrals.

Worked Example: A Mixed Loss With an Atom at Zero

A hypothetical loss $X$, measured in thousands of dollars, has survival function $S_X(x)=0.40(1-x/2)$ for $0\leq x<2$, and $S_X(x)=0$ for $x\geq 2$. Find $E[X]$ and $E[X^2]$. The specified survival function implies $P(X>0)=0.40$, so the remaining probability, 0.60, is at zero.

Method. Apply the tail-integral formula. The survival function vanishes at and above 2, so both integrals stop at 2.

For the mean,

\begin{aligned} E[X] &=\int_0^2 0.40(1-x/2)\,dx\\ &=0.40\left[x-\frac{x^2}{4}\right]_0^2\\ &=0.40(2-1)\\ &=0.40. \end{aligned}

For the second raw moment,

\begin{aligned} E[X^2] &=2\int_0^2 x\,0.40(1-x/2)\,dx\\ &=0.80\int_0^2 (x-x^2/2)\,dx\\ &=0.80\left[\frac{x^2}{2}-\frac{x^3}{6}\right]_0^2\\ &=0.80\left(2-\frac{8}{6}\right)\\ &=0.80\left(\frac{2}{3}\right) =\frac{8}{15}\approx 0.5333. \end{aligned}

The units are thousands of dollars for the mean and squared thousands of dollars for the second raw moment. As a check, the survival curve is nonnegative, decreases from 0.40 to 0, and has support only below 2, so the calculated moments are finite and nonnegative.

Common mistake avoided. An incorrect antiderivative for $x-x^2/2$ is $x^2/2-x^3/12$. Its derivative is $x-x^2/4$, not $x-x^2/2$. Using it gives $1.0667$, which is not the second moment for the stated survival function. The correct antiderivative is $x^2/2-x^3/6$, giving $8/15$.

Step Functions: Discrete and Mixed Losses

When $X$ is discrete, $S_X(x)$ is typically a step function. The tail-integral formula still applies: split the integral at each point where the survival function changes, and integrate over each interval. A jump at a loss amount reflects probability at that amount; it does not require a separate correction to the integral.

Worked Example: Moments From a Discrete Severity Distribution

A hypothetical claim amount $X$, in thousands of dollars, equals 0 with probability 0.20, 1 with probability 0.50, and 3 with probability 0.30. Find its mean and second raw moment using its survival function.

Method. First identify the survival probability on each interval. For $0\leq x<1$, losses of 1 or 3 exceed $x$, so $S_X(x)=0.50+0.30=0.80$. For $1\leq x<3$, only the loss of 3 exceeds $x$, so $S_X(x)=0.30$. For $x\geq 3$, $S_X(x)=0$.

The mean is

\begin{aligned} E[X] &=\int_0^1 0.80\,dx+\int_1^3 0.30\,dx\\ &=0.80(1-0)+0.30(3-1)\\ &=0.80+0.60=1.40. \end{aligned}

The second raw moment is

\begin{aligned} E[X^2] &=2\left(\int_0^1 0.80x\,dx+\int_1^3 0.30x\,dx\right)\\ &=2\left(0.80\left[\frac{x^2}{2}\right]_0^1 +0.30\left[\frac{x^2}{2}\right]_1^3\right)\\ &=2\left(0.80\cdot\frac{1}{2} +0.30\cdot\frac{9-1}{2}\right)\\ &=2(0.40+1.20)=3.20. \end{aligned}

Check directly from the probabilities: $E[X]=0(0.20)+1(0.50)+3(0.30)=1.40$, and $E[X^2]=0^2(0.20)+1^2(0.50)+3^2(0.30)=3.20$. The two methods agree. A common mistake is to use $S_X(x)=P(X\geq x)$ without respecting the interval boundaries; here the correct strict-tail values apply on the intervals shown, and endpoint choices do not affect the integrals.

Using the Formula for Continuous Tails

For a continuous loss, a smooth survival function often makes the integrals straightforward. The formula can also reveal whether a moment exists: a tail that decreases slowly enough can make a high-order moment infinite even when the mean is finite. The convergence decision must be made for the particular power $k$; existence of the mean alone does not guarantee that the second or a higher moment is finite.

Worked Example: First and Second Moments From a Smooth Tail

A hypothetical nonnegative loss, measured in dollars, has survival function $S_X(x)=e^{-x/2}$ for $x\geq0$. Find its mean and second raw moment. The parameter in the exponent is $1/2$ per dollar, and the tail tends to zero as $x$ tends to infinity.

Method. Substitute the given survival function into each tail integral. For the mean,

\begin{aligned} E[X] &=\int_0^\infty e^{-x/2}\,dx\\ &=\left[-2e^{-x/2}\right]_0^\infty\\ &=0-(-2)=2. \end{aligned}

For the second raw moment,

E[X^2]=2\int_0^\infty xe^{-x/2}\,dx.

To evaluate the integral, integrate by parts with $u=x$ and $dv=e^{-x/2}dx$, so $du=dx$ and $v=-2e^{-x/2}$:

\begin{aligned} \int_0^\infty xe^{-x/2}\,dx &=\left[-2xe^{-x/2}\right]_0^\infty +2\int_0^\infty e^{-x/2}\,dx\\ &=0+2(2)=4. \end{aligned}

The boundary term is zero: at $x=0$ it is zero, and as $x\to\infty$, $xe^{-x/2}\to0$. Therefore $E[X^2]=2(4)=8$. Both integrals converge, which checks that these moments are finite. The common error to avoid is omitting the leading factor $k=2$ in the second-moment formula; the integral alone is 4, but the second raw moment is 8.

Moment Existence and Practical Checks

For a nonnegative loss, the tail-integral formula is valid whether the distribution is continuous, discrete, or mixed. Its main practical requirements are to use the correct survival function, integrate over the full range where it is positive, and include the factor $k$. The moment may be infinite, so evaluating an improper integral includes checking its behavior as $x$ grows without bound.

Confirm the variable is nonnegative.
The stated formula applies to a nonnegative loss $X$. A variable that can be negative needs a different treatment.

Write the strict survival function.
Use $S_X(x)=P(X>x)$. For a step function, find its value on each interval between jumps.

Choose the moment order.
Use $k\int_0^\infty x^{k-1}S_X(x)\,dx$, retaining the leading $k$.

Evaluate and check.
Split at jumps or support boundaries, check convergence if the range is unbounded, and verify that the result is nonnegative and has the appropriate units.

These formulas calculate raw moments, not central moments. In particular, the second raw moment is $E[X^2]$, not the variance. When the first and second moments are finite, the familiar relation $\operatorname{Var}(X)=E[X^2]-(E[X])^2$ can then be used. Keeping the moment definition clear prevents confusion between the integral for $E[X^2]$ and the subsequent calculation of variance.

A useful final check is to compare the scale of the result with the loss range. If $X\leq b$, then $X^k\leq b^k$, so $E[X^k]\leq b^k$. For a loss with an unbounded range, this bound is unavailable; convergence of the tail integral must be checked directly. This is especially important for higher moments, since the factor $x^{k-1}$ gives more weight to large losses as $k$ increases.

Key takeaway: For $k>0$ and nonnegative $X$, $E[X^k]=k\int_0^\infty x^{k-1}S_X(x)\,dx$. The formula includes discrete jumps and atoms automatically. Correct integration, the factor $k$, and convergence at infinity are essential checks.

Check Your Understanding

For each problem, use the survival function and tail-integral formula. Loss amounts are in the units stated.

A hypothetical loss has $S_X(x)=0.25$ for $0\leq x<4$ and $S_X(x)=0$ for $x\geq4$. Find $E[X]$ and $E[X^2]$. (Answer: $E[X]=1$, $E[X^2]=4$.)
A loss is 0, 1, or 3 thousand dollars with probabilities 0.25, 0.50, and 0.25, respectively. Write its survival function by intervals and calculate its mean and second raw moment. (Answer: $S_X(x)=0.75$ on $0\leq x<1$, $0.25$ on $1\leq x<3$, and 0 on $x\geq3$; $E[X]=1.25$ thousand dollars, $E[X^2]=2.75$ thousand-dollars squared.)
A hypothetical loss has $S_X(x)=e^{-x/4}$ for $x\geq0$, with $x$ measured in dollars. Use the tail integrals to find its mean and second raw moment. (Answer: $E[X]=4$ dollars, $E[X^2]=32$ dollars squared.)
A loss has $S_X(x)=(1-x/5)^2$ for $0\leq x<5$ and 0 for $x\geq5$. Calculate its first and second raw moments. (Answer: $E[X]=5/3$, $E[X^2]=25/6$.)
A hypothetical loss has $S_X(x)=0.60$ for $0\leq x<2$, $S_X(x)=0.20$ for $2\leq x<5$, and 0 for $x\geq5$. Find the mean and second raw moment, and verify that the result is consistent with the maximum loss. (Answer: $E[X]=1.8$, $E[X^2]=6.6$; $X\leq5$, so $E[X^2]\leq25$, as required.)

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