Finding Severity Moments from the CDF

Why Find Moments From a CDF?

A severity model is often specified by its cumulative distribution function rather than by a convenient density. This is common when the model has probability masses, a piecewise definition, or a mixture of continuous and discrete components. In those cases, a direct density calculation may be awkward or incomplete. For a nonnegative loss, the CDF still provides a systematic way to calculate raw moments.

The key is to integrate the complement of the CDF, $1-F_X(x)$, with an appropriate weight. This method works for discrete, continuous, and mixed distributions. It also gives a useful check on direct calculations using probability tables or a density.

Definition: The CDF of a loss $X$ is $F_X(x)=P(X\le x)$. Its complement at $x$, $1-F_X(x)=P(X>x)$, is the probability that the loss exceeds $x$. For a nonnegative loss and any real $r>0$, the $r$th raw moment is $$ E[X^r]=r\int_0^\infty x^{r-1}\bigl(1-F_X(x)\bigr)\,dx, $$ where the integral is allowed to be infinite. In particular, $E[X]$ uses weight 1, and $E[X^2]$ uses weight $2x$. If both are finite, $\operatorname{Var}(X)=E[X^2]-(E[X])^2$.

Why the CDF Formula Holds

For each fixed nonnegative value of $X$, the area under $r x^{r-1}$ from 0 to $X$ equals $X^r$. Taking expectations converts that area into a probability at each integration point. The derivation is valid even when $X$ has atoms: it does not assume a density.

X^r=\int_0^\infty r x^{r-1}\mathbf{1}_{\{X>x\}}\,dx \quad\Longrightarrow\quad E[X^r]=\int_0^\infty r x^{r-1}P(X>x)\,dx =r\int_0^\infty x^{r-1}\bigl(1-F_X(x)\bigr)\,dx.

The first equality holds because the indicator $\mathbf{1}_{\{X>x\}}$ is 1 exactly for $0\le x<X$, and is 0 otherwise. The exchange of expectation and integration is justified for this nonnegative integrand by Tonelli's theorem. Finally, $P(X>x)=1-P(X\le x)=1-F_X(x)$, including when the distribution has a jump at $x$.

The formula applies to nonnegative losses. If a variable can be negative, this one-sided integral does not directly calculate its full raw moment; positive and negative parts must be handled separately. Also, if the displayed integral diverges, the corresponding moment is infinite rather than a finite value that can be used in a variance calculation.

Using the Formula With Different CDFs

For the mean, integrate $1-F_X(x)$ over the nonnegative range. For the second raw moment, multiply that complement by $2x$ before integrating. For integer $r\ge 1$, the general formula weights it by $r x^{r-1}$. The CDF's jumps do not require separate jump terms in these integrals; each jump changes the complement over an interval, while a value at a single endpoint does not change an ordinary integral.

The support and every interval of the CDF must still be read correctly. In a discrete distribution, for example, $F_X(x)$ is constant between mass points, and the constant is the cumulative probability already reached—not the probability at the next mass point.

Worked Example: Moments From a Discrete CDF

A hypothetical portfolio's claim severity $X$, in thousands of dollars, has probabilities $P(X=0)=0.20$, $P(X=2)=0.50$, and $P(X=5)=0.30$. Use its CDF to find $E[X]$, $E[X^2]$, and $\operatorname{Var}(X)$.

Method. Use the CDF moment formula. The CDF is 0 below 0, equals 0.20 on $0\le x<2$, equals 0.70 on $2\le x<5$, and equals 1 for $x\ge5$. Thus, its complement is 0.80 on $[0,2)$, 0.30 on $[2,5)$, and 0 beyond 5.

For the mean, integrate the complement on each interval:

E[X]=\int_0^2 0.80\,dx+\int_2^5 0.30\,dx =0.80(2)+0.30(3)=1.60+0.90=2.50.

For the second raw moment, include the weight $2x$:

\begin{aligned} E[X^2] &=2\int_0^2 0.80x\,dx+2\int_2^5 0.30x\,dx\\ &=1.60\left[\frac{x^2}{2}\right]_0^2 +0.60\left[\frac{x^2}{2}\right]_2^5\\ &=1.60(2)+0.60\left(\frac{25-4}{2}\right) =3.20+6.30=9.50. \end{aligned}

Therefore,

\operatorname{Var}(X)=9.50-(2.50)^2=9.50-6.25=3.25.

Check. Directly from the probability table, $E[X]=0(0.20)+2(0.50)+5(0.30)=2.50$, and $E[X^2]=0^2(0.20)+2^2(0.50)+5^2(0.30)=9.50$. Both agree. The common mistake this setup avoids is using 0.50 instead of 0.70 for the CDF on $[2,5)$: the CDF there includes the mass at both 0 and 2.

Worked Example: A Continuous CDF Without Integrating Its Density

A hypothetical loss has CDF $F_X(x)=x^2/16$ for $0\le x\le4$, with $F_X(x)=0$ below 0 and $F_X(x)=1$ above 4. Find its mean, second raw moment, and variance using the CDF.

Method. On the support, the complement is $1-x^2/16$. Outside the support, it is 1 below 0 and 0 above 4; only the interval from 0 to 4 contributes to the nonnegative-loss formula. For the mean,

E[X]=\int_0^4\left(1-\frac{x^2}{16}\right)dx =\left[x-\frac{x^3}{48}\right]_0^4 =4-\frac{64}{48} =\frac{8}{3}.

For the second raw moment,

\begin{aligned} E[X^2] &=2\int_0^4 x\left(1-\frac{x^2}{16}\right)dx\\ &=2\left[\frac{x^2}{2}-\frac{x^4}{64}\right]_0^4 =2(8-4)=8. \end{aligned}

The variance is consequently

\operatorname{Var}(X)=8-\left(\frac{8}{3}\right)^2 =8-\frac{64}{9} =\frac{8}{9}.

Check and trap. The calculated second moment exceeds the squared mean: $8>64/9$, as it must for a nonnegative variance. Integrating $F_X(x)$ instead of $1-F_X(x)$ for the mean would give $\int_0^4 x^2/16\,dx=4/3$, which is not the mean. The required integrand is the probability of exceeding the integration point.

Worked Example: A Mixed Severity With an Atom

A hypothetical payment severity is generated as follows: with probability 0.40, $X$ is uniform on $[0,2]$; with probability 0.60, $X=3$. Use the CDF to calculate $E[X]$, $E[X^2]$, and $\operatorname{Var}(X)$.

Method. For $0\le x<2$, the continuous component has accumulated probability $x/2$, weighted by 0.40, so $F_X(x)=0.20x$. For $2\le x<3$, the entire continuous component has accumulated probability 0.40, so $F_X(x)=0.40$. At $x=3$, the remaining 0.60 is added. Therefore, $1-F_X(x)$ equals $1-0.20x$ on $[0,2)$, 0.60 on $[2,3)$, and 0 above 3.

The mean is

\begin{aligned} E[X] &=\int_0^2(1-0.20x)\,dx+\int_2^3 0.60\,dx\\ &=\left[x-0.10x^2\right]_0^2+0.60(1) =(2-0.40)+0.60=2.20. \end{aligned}

The second raw moment is

\begin{aligned} E[X^2] &=2\int_0^2x(1-0.20x)\,dx+2\int_2^3 0.60x\,dx\\ &=2\left[\frac{x^2}{2}-\frac{0.20x^3}{3}\right]_0^2 +1.20\left[\frac{x^2}{2}\right]_2^3\\ &=2\left(2-\frac{8}{15}\right)+1.20\left(\frac{9-4}{2}\right)\\ &=\frac{44}{15}+3=\frac{89}{15}. \end{aligned}

Thus,

\operatorname{Var}(X)=\frac{89}{15}-(2.20)^2 =\frac{89}{15}-\frac{121}{25} =\frac{82}{75}\approx1.0933.

Check. From the mixture definition, the mean is $0.40(1)+0.60(3)=2.20$. The second moment is $0.40(4/3)+0.60(9)=89/15$, agreeing with the CDF calculation. The jump at 3 must not be omitted: its probability is represented by the complement remaining on the entire interval $[2,3)$.

Common Exam Traps

CDF-based moment calculations are often short, but a small setup error can affect every subsequent quantity. Check these points before evaluating an integral:

Use the complement. For a nonnegative severity, the integrand uses $1-F_X(x)$, not $F_X(x)$.
Track cumulative probability. Between two mass points, the CDF includes all probability at or below the lower point.
Use the right weight. The second raw moment requires $2x(1-F_X(x))$, not just $1-F_X(x)$.
Keep raw and central moments distinct. Calculate variance as $E[X^2]-(E[X])^2$; $E[X^2]$ alone is not the variance.
Check whether the integral is finite. An unbounded support does not by itself imply an infinite moment, but the integral must converge for that moment to exist.

A useful final check is to compare with the probability table, mixture definition, or density when one is available. A variance must be nonnegative, and a second raw moment must be at least the square of the mean. A mismatch usually signals an incorrect CDF interval, complement, or moment weight.

Key takeaway: For $X\ge0$, calculate raw moments from $r\int_0^\infty x^{r-1}(1-F_X(x))\,dx$. Read the CDF carefully across jumps and intervals, then obtain variance from the first and second raw moments when both are finite.

Check Your Understanding

For each problem, use the CDF or its complement as appropriate. Answers are given so you can check your calculations.

A hypothetical loss has $P(X=0)=0.10$, $P(X=3)=0.60$, and $P(X=6)=0.30$. Find $E[X]$, $E[X^2]$, and $\operatorname{Var}(X)$ using CDF intervals. (Answer: $E[X]=3.60$, $E[X^2]=16.20$, $\operatorname{Var}(X)=3.24$.)
A nonnegative severity has $F_X(x)=(x/5)^2$ for $0\le x\le5$, with the CDF equal to 0 below 0 and 1 above 5. Calculate the mean and variance from the CDF. (Answer: $E[X]=10/3$, $\operatorname{Var}(X)=25/18$.)
A hypothetical payment is uniform on $[0,2]$ with probability 0.50 and equals 4 with probability 0.50. Find its mean, second raw moment, and variance, accounting for the CDF jump at 4. (Answer: $E[X]=2.50$, $E[X^2]=26/3$, $\operatorname{Var}(X)=29/12$.)
A nonnegative loss has $F_X(x)=1-(1+x)^{-3}$ for $x\ge0$, and $F_X(x)=0$ for $x<0$. Use the CDF formula to find its first two raw moments and variance. (Answer: $E[X]=1/2$, $E[X^2]=1$, $\operatorname{Var}(X)=3/4$.)
A severity has $F_X(x)=1-(1+x)^{-1}$ for $x\ge0$, with $F_X(x)=0$ below 0. Determine whether the mean and second raw moment are finite by examining the corresponding CDF integrals. (Answer: $E[X]=\infty$ and $E[X^2]=\infty$; the variance is not finite.)

What You'll Learn