Check that $E[(X - Y)^3]=0$ for every i.i.d. $(X,Y)$

Question

If $X$, $Y$ are two independent and identically distributed random variables of mean $0$, it is true that $\mathbb{E}(X - Y)^3 = 0$?

Answer 1

Let me get you started: if $(X,Y)$ is i.i.d., $(Y,X)$ is distributed like $(X,Y)$ hence $Y-X$ is distributed like $X-Y$ hence $E(u(Y-X))=E(u(X-Y))$ for every $u$ such that the expectations exist. For $u(t)=t^3$ (or any odd function), this yields $E((X-Y)^3)=E((Y-X)^3)=-E((X-Y)^3)$ hence...

Thus, the answer is "Yes, provided $E(|X|^3)$ is finite" (otherwise $E((Y-X)^3)$ does not exist), and the condition that the mean is zero is irrelevant (both conclusions are stated by @drhab in a comment).

Answer 2

I'll get you started: $$ E(X-Y)^3 = E(X^3)-E(3X^2Y)+E(3XY^2)-E(Y^3) = E(X^3)-3E(X^2)E(Y)+3E(X)E(Y^2)-E(Y^3) $$ The last step followed since $X^2$ is independent of $Y$, and the same for $X$ and $Y^2$. Now, how can you simplify this further?