Prove of $E\left(|X+Y|^a\right)\ge E\left(|Y|^a\right)$?

Question

Let $E(V)$ be the expectation of $V$. It is also known that $E(X)=0, a>1, E\left(|X|^a\right) < +\infty$ and $E\left(|Y|^a\right)< +\infty$. $X$ and $Y$ are independent. How can I prove that $E\left(|X+Y|^a\right)\ge E\left(|Y|^a\right)$?

Answer 1

Since $a>1$, the function $f: x\mapsto |x|^a$ is convex. (Of course this is true for $a\ge1$, too.) Your result follows from Jensen's inequality: conditional on $Y$ we have $E(f(X+Y)|Y) \ge f(E(X+Y|Y))=f(Y)$. And so on.