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I'm trying to prove that:

Given a sequence $(X_n)_{n \geq 1}$ of independent and identically distributed random variables, $E(X_i^2) < +\infty$ for all $i \geq 1$, then $$\frac{2}{n(n-1)}\sum\limits_{1 \leq i < j \leq n} X_i X_j \xrightarrow[n\to +\infty]{} E(X)^2 \;\; \text{a.s}$$

My trick is to write $$\sum\limits_{1 \leq i < j \leq n} X_i X_j = \frac{1}{2}\left(\left(\sum\limits_{k=1}^{n} X_k\right)^2 - \sum\limits_{k=1}^{n}X_k^2\right)$$ then using the strong law of large numbers:$$\frac{1}{n}\sum\limits_{k=1}^{n} X_k \xrightarrow[n \to +\infty]{} E(X) \;\; \text{a.s}$$so$$\left(\frac{1}{n}\sum\limits_{k=1}^{n} X_k\right)^2 \xrightarrow[n \to +\infty]{} E(X)^2 \;\; \text{a.s}$$and$$\frac{1}{n}\sum\limits_{k=1}^{n} X_k^2 \xrightarrow[n \to +\infty]{} E(X^2) \;\; \text{a.s}$$Then$$\frac{2}{n(n-1)}\sum\limits_{1 \leq i < j \leq n} X_i X_j = \frac{n^2}{n(n-1)}\left(\frac{1}{n}\sum\limits_{k=1}^{n} X_k\right)^2 - \frac{1}{n-1}\left(\frac{1}{n}\sum\limits_{k=1}^{n}X_k^2\right) \xrightarrow[n \to +\infty]{} E(X)^2 \;\; \text{a.s}$$

But I find this solution is not natural since it strictly depends on the identity $$\sum\limits_{1 \leq i < j \leq n} X_i X_j = \frac{1}{2}\left(\left(\sum\limits_{k=1}^{n} X_k\right)^2 - \sum\limits_{k=1}^{n}X_k^2\right)$$then it would not work for another kind of sum.

I'm thinking of using another fact that $$\sum\limits_{k=1}^{n}\frac{1}{n}f\left(X_k,\dots,X_{k+m-1}\right) \xrightarrow[n \to +\infty]{} E(f\left(X_1,\dots,X_{m-1}\right)) \;\; \text{a.s}$$ for any $m > 1$ and integrable function $f$ (this technique is used for this question).

Ta Thanh Dinh
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  • I agree that the technique and identity seem restrictive, but I'm not sure what exactly you want here. For example, do you want a proof that uses the last identity that you wrote? The problem is that when you try and fit this into the appropriate situation, $m$ won't be fixed in what you're considering. Also, what "other kind of sum" are you considering and why would you expect this identity to hold for that kind of sum? – Sarvesh Ravichandran Iyer Feb 28 '24 at 06:53
  • Sure, the problem is $m$ is not fixed so the direct application of this technique would not work. I'm trying to find if there is another trick to get around this limit. – Ta Thanh Dinh Feb 28 '24 at 06:59

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