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Consider a sequence of i.i.d. random variables, $(X_i)_{i=1}^n$, with zero mean and unit variance.

I want to calculate the limit (a.s.) of

$$ \frac{1}{n}\sum_{i\neq j}X_iX_j $$ as $n\to\infty$.

My initial guess was that this sum converge to $0$. But it can be seen that the variance is given by $$ \text{var}\left(\frac{1}{n}\sum_{i\neq j}X_iX_j \right) = \frac{n-1}{n}\mathbb{E}\left(X_1^2X_2^2\right) $$ which don't goes to zero, and so this guess is wrong.

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Let $$Y_n:=\sum_{\substack{i,j=1\\ i\neq j}}^nX_iX_j.$$ We can rewrite it as $$\left(\sum_{k=1}^nX_k\right)^2-\sum_{k=1}^nX_k^2.$$ This suggests the use of well known limit theorems.

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  • $\begingroup$ Yes, so, $1/n$ of the second term (with the minus) converges to the $E(X_1^2)$. But, the $1/n$ of the first term is essentially $(\frac{1}{\sqrt{n}}\sum X_i)^2$. Can we claim that this term goes to zero? Thus, "my" term converge to $-E(X_1^2)$? $\endgroup$
    – user91011
    Sep 17, 2013 at 12:03
  • $\begingroup$ It's not what central limit theorem says. $\endgroup$
    – Davide Giraudo
    Sep 17, 2013 at 15:38
  • $\begingroup$ The central limit theorem says that $(\frac{1}{\sqrt{n}}\sum X_i)$ goes in distribution to the Normal distribution with zero mean and variance $1$ (assuming that $EX^2 = 1$). So, we actually can say that $((\frac{1}{\sqrt{n}}\sum X_i))^2$ goes to $1$, and hence $Y_n$ goes to $0$. Right? $\endgroup$
    – user91011
    Sep 17, 2013 at 15:45

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