Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
@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

1 Answer 1

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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.