show $\mathbb{E} \vert X \vert = \int_0^\infty \mathbb{P}(\{\vert X \vert>y\})dy \leq \sum_{n=0}^\infty\mathbb P \{\vert X \vert>y\}$.

Looking for a hint to show $\mathbb{E} \vert X \vert = \int_0^\infty \mathbb{P}(\{\vert X \vert>y\})dy \leq \sum_{n=0}^\infty\mathbb P \{\vert X \vert>y\}$.

This is from Theorem 2.3.7 in Durrett (Probability: Theory and examples)

The first equality makes sense by Fubini and the definition of expectation (Durrett Lemma 2.2.8). I'm having a hard time showing the second, though. My gut intuition makes me feel like it should be the other direction.

• Isn't there a symbol missing in the RHS? $\{\lvert X\rvert > y\}$ is a set, assuming usual notations; and there is no $n$ in the summand, only an unbound "$y$." – Clement C. Oct 20 '15 at 13:35
• Sum should start from 0, otherwise take any $X$, s.t. $|X|\leq$ and $RHS=0$. – A.S. Oct 20 '15 at 13:38
• Even though this post is slightly different, I’d like to link it to the current choice of mother post. Also see the meta post for (abstract) duplicates. – Lee David Chung Lin Nov 13 '18 at 13:38
• A proof using Fubini (and integrating $dP$): math.stackexchange.com/questions/536442/… – D.R. Nov 20 '19 at 8:29

As you mention, the first equation is essentially Fubini, along with rewriting $\mathbb{P}\{\lvert X\rvert > y\}$ as an integral.
For the inequality on the right: observe that \begin{align} \int_0^\infty \mathbb{P}\{ \lvert X\rvert > y\} dy &= \sum_{n=0}^\infty \int_n^{n+1} \mathbb{P}\{ \lvert X\rvert > y\} dy \leq \sum_{n=0}^\infty \int_n^{n+1} \mathbb{P}\{ \lvert X\rvert > n\} dy \\ &= \sum_{n=0}^\infty \mathbb{P}\{ \lvert X\rvert > n\} \end{align} using the fact that for $y \geq x$, $\mathbb{P}\{ \lvert X\rvert > y\} \leq \mathbb{P}\{ \lvert X\rvert > x\}$ since $\{ \lvert X\rvert > y\} \subseteq \{ \lvert X\rvert > x\}$.