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Let $\{X_i\}_{i=1}^{n}$ be a sequence of i.i.d. random variables with finite mean $\mu$, and finite variance.
Show $$ \dbinom{n}{2}^{-1} \sum_{1\le i < j\le n} X_i X_j \xrightarrow{p} \mu^2. $$

This is a question in Grimmet, where it is stated that there are various way of doing this.
One way he shows, is to write \begin{align} \dbinom{n}{2}^{-1} \sum_{1\le i < j\le n} X_i X_j &= \frac{n}{n-1}\left( \frac{1}{n} \sum_{i=1}^n X_i \right) ^ 2 - \frac{1}{n-1}\left( \frac{1}{n} \sum_{i=1}^n X_i^2 \right) \\ &= A - B \end{align} then $A \xrightarrow{p} \mu^2$ and $B\xrightarrow{p} 0$ and the result follows.

What are the other ways of solving this problem?

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up vote 2 down vote accepted

What are the other ways of solving this problem?

Show that the variance goes to zero. To do that, consider $Y_{ij}=X_iX_j-\mu^2$ and $$ S_n=\sum_{1\leqslant i\ne j\leqslant n}Y_{ij}=2\sum_{1\leqslant i\lt j\leqslant n}(X_iX_j-\mu^2). $$ Then $E[S_n]=0$ and $S_n^2$ is the sum of the random variables $Y_{ij}Y_{km}$ for $i\ne j$, $k\ne m$, $\{i,j,k,m\}$ subset of $\{1,2,\ldots,n\}$ of size $2$ or $3$ or $4$.

  • If the size of $\{i,j,k,m\}$ is $2$, then $E[Y_{ij}Y_{km}]=E[Y_{12}^2]=a$ with $a=\sigma^2(\sigma^2+2\mu^2)$. There are $2n(n-1)$ such terms.
  • If the size of $\{i,j,k,m\}$ is $3$, then $E[Y_{ij}Y_{km}]=E[Y_{12}Y_{13}]=b$ with $b=\sigma^2\mu^2$. There are $4n(n-1)(n-2)$ such terms.
  • If the size of $\{i,j,k,m\}$ is $4$, then $E[Y_{ij}Y_{km}]=E[Y_{12}Y_{34}]=0$. There are $n(n-1)(n-2)(n-3)$ such terms.

Thus, $E[S_n^2]=2n(n-1)a+4n(n-1)(n-2)b=\Theta(n^3)$ and $$ \frac2{n(n-1)}\sum_{1\leqslant i\lt j\leqslant n}X_iX_j=\mu^2+\frac1{n(n-1)}S_n\stackrel{L^2}{\to}\mu^2. $$

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