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We have X and Y, independent, zero mean random variables. Show that:

$$E|X+Y| \geq E|X|$$

What I tried:

I treated them as continuous variables, though it should be true in general. But I could not derive the proof. I found that it should be proven:

$$\int_{-\infty}^{\infty}\int_{-x}^{\infty}(x+y)f(x)g(y)dydx \geq \int_0^{\infty} xf(x)dx$$

where we have the constraints:

$$\int f(x)dx=1$$

$$\int x\cdot f(x)dx = 0$$

$$\int g(y)dy=1$$

$$\int y\cdot g(y)dy = 0$$

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    $\begingroup$ Do you know Jensen's inequality? $\endgroup$ – kimchi lover Jul 20 '17 at 12:10
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    $\begingroup$ This was asked before. Use the fact that, by convexity, $$E(|x+Y|)\geqslant|x+E(Y)|=|x|$$ for every $x$, and integrate this inequality with respect to the distribution of $X$. $\endgroup$ – Did Jul 21 '17 at 12:58
  • $\begingroup$ @kimchilover I see what you are hinting at but, to show $E(|Z|)\geqslant|E(Z)|$, Jensen is hardly necessary. $\endgroup$ – Did Jul 21 '17 at 18:57
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    $\begingroup$ @Did could you show me please, where it was asked? Thank you! $\endgroup$ – AvanDavad Jul 27 '17 at 14:28
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    $\begingroup$ @Did you are totally right, I did the search before posting but I did not find it... $\endgroup$ – AvanDavad Jul 28 '17 at 16:37
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We have that

$$|EX| \leq E|X|$$

which is true in general.

$$E(abs(X+Y)| X) \geq |E(X+Y|X)| = |X + E(Y|X)| = |X|$$

take the expected value (and use the tower rule, aka. the law of total expectation):

$$EE(abs(X+Y)|X)\equiv E|X+Y| \geq E|X|$$

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