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Let $X$, $Y$ be independent random variables, $E\left(\left|X\right|^p\right)<+\infty$ where $p\geq 1$ and $E(Y)=0$. Show that $E\left(\left|X+Y\right|^p\right)\geq E(\left|X\right|^p)$, where $E\left(\cdot\right)$ stands for expectation.

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@Srivatsan: you're right, there wasn't that typo before I edited it for the first time. – Davide Giraudo Jul 27 '11 at 8:30
Thank you for your hint.I didn`t think of jensen inequality before. – cheng Jul 27 '11 at 9:20
@cheng: If you have a proof, you could write it yourself as an answer, wait a reasonable amount of time to see whether people agree with it, then accept your answer. – Did Jul 28 '11 at 15:41

Hint: for every fixed $x$ and every random variable $Y$, $E(|x+Y|^p)≥|x+E(Y)|^p$.

Jensen's inequality seems to be the way to prove this, hence the first goal is to find a convex function somewhere... Once you know this, the rest should be easy.

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Do you mean integrate with respect to x?but we won`t get E|X+Y|^p on the left side ,but an iterated integration. – cheng Aug 21 '11 at 1:48
Sorry? Call $u(x)=E(|x+Y|^p)$, then $E(u(X))=E(|X+Y|^p)$. This is a consequence of the independence of $X$ and $Y$ (hence the distribution of $(X,Y)$ is a product distribution) and of Fubini's theorem. – Did Aug 21 '11 at 2:03
@cheng, any further question on this solution? – Did Aug 27 '11 at 18:59

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