Uncorrelated Identically Distributed Random Variables Let $X_n$ be a sequence of uncorrelated identically distributed random variables with $\mathbb{E}(X_n) = 0$ and $\operatorname{Var}(X_n) = \sigma^2$
Prove that $\mathbb{E}$$(X_n + X_m)^2 = 2\sigma^2$ for any $n \neq m$
So here is what I have so far:
I can rewrite $\mathbb{E}(X_n + X_m)^2$ to give $\mathbb{E}(X_n^2 + 2X_n X_m + X_m^2)$
Am I allowed to then write $\mathbb{E}(X_n^2) +2\mathbb{E}(X_n)\mathbb{E}(X_n) + \mathbb{E}(X_m^2)$?
Because this would allow me to say that (because $\mathbb{E}$($X_n$) = $0$) we have:
$\mathbb{E}$($X_n^2$) + $\mathbb{E}$($X_m^2$)
We also know that $\mathbb{E}$($X_n^2$) = $\sigma^2$ + ($\mathbb{E}$($X_n$))^2
And hence $\mathbb{E}$($X_n^2$) = $\sigma^2$  (because $\mathbb{E}$($X_n$) = $0$))
Therefore $\mathbb{E}$($X_n^2$) + $\mathbb{E}$($X_m^2$) = $\sigma^2$ + $\sigma^2$ = 2$\sigma^2$
And so $\mathbb{E}$$(X_n + X_m)^2$ = 2$\sigma^2$ for any n $\neq$ m
Is this right? Or am I somehow assuming independence?
 A: Yes, your calculations are correct! As confirmation:
Since the variables $(X_n)_{n \in \mathbb N}$ are uncorrelated you know that $$E[X_nX_m]=E[X_n]E[X_m] \tag 1$$ for all $n\neq m \in \mathbb N$. Thus, as you already have $$E[(X_n+X_m)^2]=E[X_n^2+2X_nX_m+X_m^2]$$ where the RHS by the linearity of the expectation becomes $$E[X_n^2]+2E[X_nX_m]+E[X_m^2]$$ Since, by the formula of the variance $$Var(X_j)=E[X_j^2]-E[X_j]^2$$ wed have that $$E[X_j^2]=Var(X_j)+E[X_j]^2=σ^2+μ^2 \tag2$$ and due to the given condition (1), we have that $$\begin{align*}E[(X_n+X_m)^2]&=[X_n^2]+2E[X_nX_m]+E[X_m^2]=σ^2+μ^2+2(μμ)+σ^2+μ^2=\\\\&=2σ^2+4μ^2\end{align*}$$ Since $μ=0$ the result follows.
A: $$ E((X_n+X_m)^2) = E(X_n^2) + E(X_m^2) + 2E(X_n X_m) = E((X_n - \mu_{X_n})^2) + E((X_m - \mu_{X_m})^2) + E((X_n - \mu_{X_n}) (X_m - \mu_{X_m})) = var(X_n) + var(X_m) + cor(X_n,X_m) = \sigma ^2 + \sigma ^2 + 0 = 2 \sigma ^2 .$$
Repeatedly above I use the (given) fact that $\mu_{X_n} = 0 \quad \forall n \in \mathbb{N}$.
A: Let me use $E,V,C$ for expectation, variance, and covariance for ease of typing. Also, let $X,Y$ be your $X_m,X_n$.
Since $V[X+Y] = E[(X+Y)^2]-E[X+Y]^2$, we have
$$E[(X+Y)^2]=V[X+Y] +E[X+Y]^2$$
Now we are given $E[X]=E[Y]=0$, so $E[X+Y]^2 =(E[X]+E[Y])^2=(0+0)^2=0$, so
$$E[(X+Y)^2]=V[X+Y]+0=V[X+Y].$$
But
$$V[X+Y]=C[X+Y,X+Y]=C[X,X]+C[Y,Y]+2C[X,Y]$$
$$=V[X] +V[Y]+0= 2\sigma^2$$
with $C[X,Y ]=0$ following from the fact that $X $ and $Y$ are uncorrelated
