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.

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    $\begingroup$ 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 "$y$." $\endgroup$ – Clement C. Oct 20 '15 at 13:35
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    $\begingroup$ Sum should start from 0, otherwise take any $X$, s.t. $|X|\leq$ and $RHS=0$. $\endgroup$ – A.S. Oct 20 '15 at 13:38
  • $\begingroup$ 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. $\endgroup$ – Lee David Chung Lin Nov 13 '18 at 13:38
  • $\begingroup$ A proof using Fubini (and integrating $dP$): math.stackexchange.com/questions/536442/… $\endgroup$ – 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 $\{ \lvert X\rvert > y\} \subseteq \{ \lvert X\rvert > x\}$.

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  • $\begingroup$ Very cute proof! $\endgroup$ – karmanaut Oct 21 '15 at 1:19

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