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