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\} $.

Question

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.

Answer 1

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 $\{ \lvert X\rvert > y\} \subseteq \{ \lvert X\rvert > x\}$.