Naive question about expectation: how do we prove that $\textbf{E}(X) = \int_{0}^{\infty}\textbf{P}(X > x)\mathrm{d}x$?

Question

As the title says, let us consider a non-negative continuous random variable. Prove the following statement is true \begin{align*} \textbf{E}(X) = \int_{0}^{\infty}\textbf{P}(X>x)\mathrm{d}x \end{align*}

Any help is appreciated. Thanks in advance!

Answer 1

$$\int_0^\infty \mathbf{P}(X > x)\, dx = \int_0^\infty \int_x^\infty f_X(y)\, dy\, dx =^* \int_0^\infty \int_0^y f_X(y)\, dx\, dy = \int_0^\infty y f_X(y)\, dy = \mathbf{E}(X)$$

You should convince yourself by sketching the domain of integration that the change of integration limits and order at the starred equals sign is licit.

Answer 2

Hint

Let $f_X$ the density function of $X$.

$$\mathbb P\{X>x\}=\int_{x}^\infty f_X(u)\,\mathrm d u.$$