$\mathbb{E}[X^2]= 2 \int _0 ^{\infty} x P(X>x) dx$

Question

I found the equation mentioned in the title while reading through a proof, and was wondering if there is a way to prove it without using measure theoretic induction, I think I remember having seeing it in a stochastic 1 script before, but don't know where exactly. Any help would be appreciated.

Answer 1

For non-negative $X$ you can prove it easily using Fubini's Theorem: $2\int_0^{\infty} xP\{X>x\}\, dx=2\int_0^{\infty} x \int_{\{X>x\}} \, dP\, dx=2 \int \int_0^{X}x\, dx dP=\int X^{2} \, dP$.

Answer 2

Assuming $X$ is nonnegative, one has:

$$\int^\infty_0 2x\mathbb{I}_{X>x} \,dx = X^2$$ where $\mathbb I$ is the indicator function. Taking expectation on both sides using Fubini gets you pretty much fast to the result.

Answer 3

If $Y$ is non-negative then $$\mathbb{E}[Y]= \int\limits_{y=0} ^{\infty} P(Y>y)\, dy$$

Now suppose $X$ is also non-negative and let $y=x^2$ (strictly increasing) so $\frac{dy}{dx}=2x$ and $P(Y > x^2) = P(X^2 > x^2) = P(X>x)$. Then by substitution $$\mathbb{E}[X^2]= \int\limits_{x=0} ^{\infty} P(X>x) \,2x \,dx$$

This will therefore not be true if $P(X <0) > 0$ but you could say more generally $$\mathbb{E}[X^2]= 2\int\limits_{x=0} ^{\infty} x\,P(|X|>x) \,dx$$