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Actually this question is similar to the question here

The following is from Durrett's Probability Theory and Examples question 2.2.1:

Let $X_{1}, X_{2}, \ldots$ be uncorrelated with $\mathbb{E} X_{i} = \mu_{i}$ and $\frac{Var X_{i}}{i} \to 0$ as $i\to \infty$. Let $S_{n} = X_{1}+X_{2} + \ldots + X_{n}$ and $\nu_{n} = \mathbb{E} S_{n}/n$. Then as $n\to \infty$, $S_{n}/n - \nu_{n} \to 0$ in $L^{2}$ and in probability.

It can be shown that \begin{align*} \mathbb{E}\left[\left(\frac{S_{n}}{n} -\nu_{n} \right)^{2} \right] = \frac{1}{n^{2}} \sum_{i=1}^{n} Var(X_{i}) \end{align*} Now I want to show that $\frac{1}{n^{2}} \sum_{i=1}^{n} Var(X_{i}) \to 0$. My approach to this is for any fixed $\varepsilon>0$ I can find $N$ such that for all $i\geq N$ I have $Var(X_{i})/i < \varepsilon$. Now, if I knew $Var (X_{i})/i \leq M<\infty$ for $i=1, \ldots N$ then I can have \begin{align*} \frac{1}{n^{2}} \sum_{i=1}^{n} Var(X_{i}) &\leq \frac{1}{n} \sum_{i=1}^{n} \frac{Var(X_{i})}{i}\\ &=\frac{1}{n} \sum_{i=1}^{N} \frac{Var(X_{i})}{i} + \frac{1}{n} \sum_{i=N+1}^{n} \frac{Var(X_{i})}{i}\\ &\leq \frac{NM}{n} + \frac{(n-N-1)\varepsilon}{n} \end{align*} so that the result follows by taking $n\to \infty$ and noting that $\varepsilon$ is arbitrary. The issue is I don't know that there exists an $M$ such that $Var (X_{i})/i \leq M<\infty$ for $i=1, \ldots N$...

Presumably we can have $Var (X_{i})/i =\infty$? Is there anyway to prove that this is not the case? Or is my approach to the proof incorrect? Any help is appreciated.

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  • $\begingroup$ Since every convergent sequence is bounded, you can find an $M>0$ such that $Var(X_i)/i \leq M$ for all $i$. Why do you think $Var(X_i)$ can be $\infty$? $\endgroup$ – Jeff Nov 10 '16 at 4:18
  • $\begingroup$ @Jeff I agree that every convergent sequence of real numbers is bounded. However the proof that every convergent sequence of real numbers is bounded requires that the first $N$ terms in the sequence are finite (which is fine, since they are assumed to be real numbers). I do not know if $\mathbb{E}X_{i}^{2} < \infty$ so I do not know if $Var(X_{i})/i<\infty$. For example, why can't I have a sequence in this case that goes $\{\infty, \infty, 1, 1/2, 1/4, 1/16, \ldots\}$? $\endgroup$ – möbius Nov 10 '16 at 12:40
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    $\begingroup$ I think implicit in the theorem statement is an assumption that each $X_i$ has a finite second moment. So you can assume $\mathbb{E}X_i^2 < \infty$. $\endgroup$ – Jeff Nov 10 '16 at 20:59
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Suppose $E(X_N^2)=\infty$ for some positive integer $N$. Then

\begin{align*} \mathbb{E}\left[\left(\frac{S_{n}}{n} -\nu_{n} \right)^{2} \right] = \frac{1}{n^{2}} \sum_{i=1}^{n} Var(X_{i})=\infty \end{align*} for $n>N$ and thus there is no way to get the $L^2$ convergence.

If you go back to the beginning of section 2.2.1 in Durrett's book, you can see that he does assume finite second moment when he defines what are uncorrelated random variables:

enter image description here

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Typically the assumption of uncorrelatedness implicitly means that the random variables are square-integrable. Otherwise the expectation $E[X_i X_j]$ might be ill-defined.

Also, if at least one of the random variables were not square-integrable, you would have $\operatorname{Var}(S_n/n) = \infty$ for all $n$, which means that the sequence is in fact not $L^2$-convergent.

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