It is known that $$ E[\:\{\:(X+Y) - E(X+Y)\:\}^2\:] = E[\:\{\:X - E(X)\:\}^2\:] + E[\:\{Y - E(Y)\:\}^2\:] $$ (that it is the variance of the sum is the sum of variances given that the r.v's are independent. But I have what I think it is an interesting question:
\begin{align*} E[\:\{\:(X+Y) - E(X+Y)\:\}^n\:] = E[\:\{X - E(X)\:\}^n\:] + E[\:\{\:Y - E(Y)\:\}^n\:] \end{align*}
is it true?
I think it is. However, I have not been able to prove it yet. Can someone give a hint or an idea?
Thanks.
This is false.
Take $X,Y$ independent, both mean $0$. Then you ask whether, for all $n\geq 0$, $$ \mathbb{E}[(X+Y)^n] = \mathbb{E}[X^n]+\mathbb{E}[Y^n] $$ Clearly, this "feels" false, so let's try the simplest examples: $X,Y$ both uniform on $\{-1,1\}$ (i.e, Rademacher random variables), and $n=4$. Then $$ \mathbb{E}[(X+Y)^4] = \mathbb{E}[X^4] + 4\mathbb{E}[X^3]\mathbb{E}[Y] + 6\mathbb{E}[X^2]\mathbb{E}[Y^2] + 4\mathbb{E}[X]\mathbb{E}[Y^3] + \mathbb{E}[Y^4] = 8 $$ using independence, the fact that $X^{2k} = 1$ and $X^{2k+1} = X$, and the fact that both r.v.'s have mean zero. But obviously, $$ \mathbb{E}[X^4] + \mathbb{E}[Y^4] = 2 \neq 8\,. $$
I used $n=4$ since, as you point out, it is true for $n=2$, and it happens to be true as well for $n=3$ for independent Rademacher r.v.'s, as one can check.