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Is the below result true for any random variable $X$?

$|\mathbb{E}(X)|\leq \mathbb{E}(|X|)$

Below is my attempt so far..

For any $X$, we have $X\leq|X|$ and taking expectation on both sides yields, $\mathbb{E}(X)\leq \mathbb{E}(|X|)$ , the quantity on the right is always positive but the quantity on the left could be both positive or negative.

The main result comes from taking absolute values on both sides of the expression $\mathbb{E}(X)\leq \mathbb{E}(|X|)$ which might not always be true (e.g., $-5<2$ but $|-5|>2$).

Can anyone help me with the correct reasoning behind the inequality $|\mathbb{E}(X)|\leq \mathbb{E}(|X|)$?

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2 Answers 2

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You can use the triangle inequality. Sketch:

$$\vert E(X)\vert = \vert \sum k \Pr(X=k) \vert \leq \sum \vert k \vert \Pr(X=k) =E(\vert X\vert)$$

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  • $\begingroup$ Any random variable is discrete? $\endgroup$ Commented Sep 2, 2020 at 8:58
  • $\begingroup$ It works the same with any random variable. That's why this is only a sketch, to show the general idea and intuition. $\endgroup$
    – YJT
    Commented Sep 2, 2020 at 9:00
  • $\begingroup$ thanks, interesting proof for the discrete case as well. $\endgroup$
    – Smokey
    Commented Sep 2, 2020 at 9:53
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$|EX|=EX$ or $|EX|=-EX$. In the first case use the fact that $X \leq |X|$ and in the second case use the fact that $-X \leq |X|$.

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