I know how to show $\mathbb{E}(X) = \int _0 ^{\infty} \mathbb{P} (X>t) dt$, when $X$ is a nonnegative random variable. I'm struggling to prove the following:
$\mathbb{E}(X | \mathcal{F}) = \int _0 ^{\infty} \mathbb{P} (X>t | \mathcal{F}) dt$
I appreciate any hint. Thanks.
Observe that the right-hand side is $\mathcal F$-measurable, so it suffices to show that it satisfies the universal property of the left-hand side. To that end, let $A\in \mathcal F$. Then by the Fubini-Tonelli theorem, \begin{align*} \int_A \int_0^\infty\mathbb P(X>t\mid\mathcal F)\,dt\,d\mathbb P &= \int_0^\infty\int_A\mathbb P(X>t\mid\mathcal F)\,d\mathbb P\,dt\\ &= \int_0^\infty\mathbb P(X>t,A)\,dt \\ &= \int_0^\infty\mathbb P(\mathbf 1_AX>t) \,dt =\int_AX\,d\mathbb P. \end{align*} I leave it to you to justify the first equality by verifying the hypotheses of the Fubini-Tonelli theorem are met. The second equality holds by the definition of conditional probability, and the last by the equality $\mathbb E(X) = \int_0^\infty\mathbb P(X>t)\,dt$ you already know.