# proving that $\mathbb{E}[X] = \int_{0}^{\infty}\mathbb{P}(X \geq t) dt$ in the case of non-negative random variables.

say we're in a probability space $$(\Omega, \Sigma, \mathbb{P})$$ and $$X : (\Omega, \Sigma) \to (\mathbb{R}, \mathscr{B}_{\mathbb{R}})$$ is random variable

I resulted in the right result but I feel like the math is bad, please check it :

\begin{align} & \int_{0}^{\infty}\mathbb{P}(X \geq t) dt = \int_0^{\infty}\int_{X \geq t}d\mathbb{P}\,dt \,\,\,\,\\ = & \int_{X \geq 0}\int_{0}^{X}dt\,d\mathbb{P} \,\,(\star) = \int_{X \geq 0}X\,d\mathbb{P} \\ = & \int_{\Omega}X\,d\mathbb{P} \,\,\,\, \text{due to non-negativity} = \mathbb{E}[X] \end{align}

where I'm not sure I'm doing legal stuff is $$(\star)$$, I used Fubini-Tonelli theorem to change order of integration, but when I changed bounds of integration I reasoned the following way :

we have : $$t\geq 0$$ and $$X\geq t$$ meaning on the one hand we have $$X \geq 0$$ and $$0 \leq t \leq X$$

am I doing this right ?

• Looks fine to me. – drhab Nov 16 '18 at 14:17

The integral before $$(\star)$$ can be written as $$\int_\Omega\int_{\mathbb R}f(t,\omega)\mathrm dt\mathrm dP(\omega)$$, where $$f(t,\omega)=1$$ if $$X(\omega)\geqslant t$$ and $$0$$ otherwise. Then Fubini's theorem can be used in this setting.