Suppose $X$ is a nonnegative and continuous random variable with a survival function $S(x)$. Show that

$$\mathsf{Var}(X)=2\int_0^\infty tS(t) \, dt-\left( \int_0^\infty S(t) \, dt\right)^2$$

Using the fact that

$$f(x) =−\frac{dS(x)}{dx}$$

we have that

$$\begin{align*} \mathsf{Var}(X) &=\mathsf{E}(X^2)-\mathsf{E}(X)^2\\\\ &=\int_0^\infty t^2 f(t)\,dt-\left(\int_0^ \infty tf(t) \, dt \right)^2 \\\\ &=-\int_0^\infty t^2 \, dS(t)-\left(\int_0^\infty t \, dS(t) \right)^2 \\\\ &=-\left(t^2 S(t)\bigg\rvert_0^\infty-\int_0^\infty 2tS(t) \, dt \right) - \left(tS(t)\bigg\rvert_0^\infty-\int_0^\infty S(t) \, dt \right)^2 \end{align*}$$

I could get the desired result by showing

$$t^2 S(t) \bigg\rvert_0^\infty = tS(t) \bigg\rvert_0^\infty = 0 $$

but how do we know if $t$ and $t^2$ go to $\infty$ slower than $S(t)$?

  • 2
    $\begingroup$ It may be worth considering the finiteness of the variance in the case of a distibution where $S(t)$ goes to 0 slowly enough that the integrals you look at don't have a finite value (you may want to write improper integrals out as a limit). You may need to make some assumptions about $S$ sufficient to guarantee it has a variance (otherwise the whole exercise is kind of moot), and then see whether under those assumptions the integral converges.. $\endgroup$
    – Glen_b
    Sep 9, 2019 at 2:48

1 Answer 1


Your attempt to prove this using integration-by-parts is the wrong approach, and that is what is leading you to require unnecessary conditions. The simpler approach is to express the moments using double-integration, and then use interchange of integrals to get the survival function. For any non-negative continuous random variable $X$ you can use the interchange of integrals to write the $k$th raw moment as:

$$\begin{equation} \begin{aligned} \mathbb{E}(X^k) = \int \limits_0^\infty t^k f(t) \ dt &= \int \limits_0^\infty \Bigg[ r^k f(t) \Bigg]_{r=0}^{r=t} \ dt \\[6pt] &= \int \limits_0^\infty \int \limits_0^t k r^{k-1} f(t) \ dr \ dt \\[6pt] &= k \int \limits_0^\infty r^{k-1} \int \limits_r^\infty f(t) \ dt \ dr \\[6pt] &= k \int \limits_0^\infty r^{k-1} S(r) \ dr. \\[6pt] \end{aligned} \end{equation}$$

Substitution of this general result will give you the required expressions for the first and second moments, which then gives you the desired variance expression.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.