$E[x_i^2 x_j^2]$ for white Gaussian noise

Question

If $x_n$ is a discrete time random signal and is white Gaussian noise (ergodic and WSS) so

$$E[x_n x_{n+l}]=\sigma ^2 \delta (l)$$

and

$$E[x_n]=0$$

Where $n \in \mathbb{R}$ and $l\in\mathbb{R}$

then what is:

$$\sum_{i=0}^{N-1}\sum_{j=0}^{N-1} E[x_i^2 x_j^2]$$

The problem I'm having is that I don't know if $E[x_i^2 x_j^2]=E[x_i^2]E[j_i^2]\;\forall \;i \ne j$

The expectaton values of the summands factorize for i and j different from each other, so you are left with the expectation value of the fourth powers, this is easy to evaluate. — Count Iblis, Mar 14 '15 at 16:35
Actually I remembered I do know how to prove $E[x_n^4]=3\sigma ^4$. But why does the expectation value factorise for $i\ne j$? — texasflood, Mar 14 '15 at 17:00
It factorizes because the joint probability density p(xi, xj) factorizes into a product of Gaussians for xi and xj. — Count Iblis, Mar 14 '15 at 17:55
Ok that should have been obvious, I think I need a coffee. Problem solved. Thanks — texasflood, Mar 14 '15 at 17:58
For reference the answer is $3\sigma^4 N +\sigma^4N(N-1)=\sigma^4N(N+2)$ — texasflood, Mar 14 '15 at 18:02

score 0 · Accepted Answer · edited Apr 13 '17 at 12:21

The expectation values of the summands factorize for $i\ne j$ because the joint probability density $p(x_i, x_j)$ factorizes into a product of Gaussians for $x_i$ and $x_j$. So you are left with the expectation value of the fourth powers. -- Count Iblis

I do know how to prove $E[x_n^4]=3\sigma ^4$. For reference the answer is $$3\sigma^4 N +\sigma^4N(N-1)=\sigma^4N(N+2)$$ -- texasflood

A reference for $E[x_n^4]=3\sigma ^4$: Calculation of the n-th central moment of the normal distribution $\mathcal{N}(\mu,\sigma^2)$

$E[x_i^2 x_j^2]$ for white Gaussian noise

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