Change of Variables in Second Moment

Question

$X$ is a non-negative continuous random variable with pdf $f(x)$. $G(t) = \int_{t}^\infty f(x) dx$. Show that $$E[X^2] = 2 \int_{0}^\infty tG(t) dt$$.

I tried to write out E[X^2] (second moment of X) and see if that corresponds to the G(t) equation. I wasn't able to do that well because I wasn't sure of how the change of variables works.

Answer 1

I'm assuming $E[X^2] < \infty$. Integration by parts gives

$$2\int_0^b t\, G(t)\, dt = b^2 G(b) - \int_0^b t^2 G'(t)\, dt = b^2 G(b) + \int_0^b t^2 f(t)\, dt.$$

Taking the limit as $b \to \infty$ results in

$$2\int_0^\infty t G(t)\, dt = \int_0^\infty t^2 f(t)\, dt = E[X^2].$$

To see that $\lim\limits_{b\to \infty} b^2 G(b) = 0$, note that for $b > 0$,

$$E[X^2] = \int_0^b x^2 f(x)\, dx + \int_b^\infty x^2 f(x)\, dx \ge \int_0^b x^2 f(x)\, dx + b^2 G(b).$$

So

$$0 \le b^2 G(b) \le E[X^2] - \int_0^b x^2 f(x)\, dx.$$

The right-hand side tends to $0$ as $b\to \infty$, so $\lim\limits_{b\to \infty} b^2 G(b) = 0$.

Answer 2

Write $x^2=\int_{t=0}^x 2 t\,dt$ and then interchange the order of integration: $$ \begin{align} E(X^2)&= \int_{x=0}^\infty \left(\int_{t=0}^x 2 t\,dt \right)f(x)\,dx\\ &=\int_{x=0}^\infty \left(\int_{t=0}^x 2 tf(x)\,dt\right)\,dx\\ &=\int_{t=0}^\infty \left(\int_{x=t}^\infty 2 tf(x)\,dx\right)\,dt\\ &=\int_{t=0}^\infty 2t\left(\int_{x=t}^\infty f(x)\,dx\right)\,dt \end{align} $$ Another example of this approach can be seen in this answer.