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Suppose $(X_n)_n\geq 1$ are i.i.d. $\mathbb E(X)=\mu$, $\sigma^2= \operatorname{Var}(X)<\infty$. I want to show $$\frac 1 {n(n-1)} \sum_{1\,\leq\, i,j\,\leq\, n,\,\, i\,\neq\, j} X_iX_j\to \mu^2$$ in probability.

Here is what I tried.

Applying Strong law of large number we have $$\frac 1 {n^2} \left( \sum_{1\leq i\leq n} X_i\right)^2\to \mu^2$$ in probability. For all $\epsilon>0$, $\mathbb P(|1/n^2(\sum_{1\leq i\leq n} X_i)^2-\mu^2|\geq\epsilon)\to 0 $. Expand this we have $\mathbb P(|(\sum_{1\leq i,j\leq n, i\neq j} X_iX_j-n(n-1)\mu^2+\sum_{1\leq i\leq n}X_i^2-n\mu^2|\geq n^2\epsilon)\to 0 $. I don't know how to proceed from here. In particular , I'm wondering how to use Var$(X)<\infty$.

Michael Hardy
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Eddie
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1 Answers1

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Write

$$\sum_{1\leq i,j\leq n,i\neq j}X_{i}X_j=(\sum_{i=1}^nX_i)^2-\sum_{i=1}^nX_i^2.$$

Divide both sides by $n^2$. The first term on the right goes to $\mu^2$ almost surely. For the second term, $E[X_i^2]=\mbox{Var}(X_i)+E[X_i]^2=<\infty$. By SLLN the second term will go to 0 since there's a factor of $n^2$. Now use the linearity of almost-sure limits, to conclude the entire right side goes to $\mu$. Finally observe that $n(n-1)/n^2\rightarrow 1$ to conclude that you can replace $n^2$ with $n(n-1)$.

Alex R.
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