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Suppose $\{X_1, X_2, X_3, \ldots\}$ is an infinite sequence of random variables such that $E[X_i]=0$ for all $i$, and $E[X_iX_j]=0$ whenever $i \neq j$. Further suppose the variances $\sigma_i^2 = E[X_i^2]$ are finite and satisfy $\sum_{i=1}^{\infty} \sigma_i^2 < \infty$.

Define $S_n = \sum_{i=1}^nX_i$. When can one conclude that $S_n$ converges almost surely as $n\rightarrow\infty$?

I can show that, with prob 1, the limit exists (and is a real number) over a particular subsequence $n[m]$, so that $\lim_{m\rightarrow\infty}S_{n[m]}$ exists as a (random) real number. I suspect that, in general, we do not have convergence. However, counter-examples seem hard.

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    $\begingroup$ Related: en.wikipedia.org/wiki/Kolmogorov%27s_three-series_theorem $\endgroup$ – dsaxton Jul 27 '15 at 19:41
  • $\begingroup$ I suppose $S_n$ can be viewed as a Cauchy sequence in a metric space with inner product $<X,Y> = E[XY]$, in which case a counter-example might be inspired by examples of incomplete metric spaces. $\endgroup$ – Michael Jul 27 '15 at 19:50
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    $\begingroup$ Kolmogorov 3 series criterion is for independent seq. of r.v.'s $\endgroup$ – Saty Jul 27 '15 at 19:56
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    $\begingroup$ If one replaces the requirement $E[X_iX_j]=0$ for all $i \neq j$ with the more stringent requirement that $E[X_n|X_1, \ldots, X_{n-1}]=0$ for all $n\geq 2$, then almost sure convergence follows from the Doob martingale convergence theorem, since $S_n$ is then a martingale with $$\sup_n E[|S_n|] \leq \sup_n \sqrt{E[S_n^2]} < \infty$$. $\endgroup$ – Michael Jul 27 '15 at 22:14
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    $\begingroup$ @Michael: No, your claim is fine. :) I meant to say that if the question posed by the op turned out to be true in general, this would imply Carlesons theorem. The correct link I meant is math.stackexchange.com/questions/1376110/… $\endgroup$ – PhoemueX Jul 28 '15 at 6:30
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This answer is first adapted from comments in: does convergence in $L^p$ imply convergence almost everywhere. Consider a probability space $[0, 1]$ with uniform measure. Define the random variable $X_{2^i + k}$ by $$X_{2^i + k} = 2^{i/2}\chi_{\left[\frac{k}{2^i}, \frac{k+1/2}{2^i}\right]} - 2^{i/2}\chi_{\left[\frac{k + 1/2}{2^i}, \frac{k+1}{2^i}\right]}$$ for $i \geq 1, 0 \leq k < 2^i$. These random variables (known as Haar Functions) have mean zero. Furthermore, they are uncorrelated. However, the sum of these random variable do not satisfy the requirements of your theorem because $Var(\sum_{n=1}^\infty X_n) = \infty$.

Here we change our point of view to restate the question as follows: Does there exist an orthonormal basis of $L^2([0,1])$ such that an element written in this basis converges in $L^2$ but does not converge pointwise almost everywhere. I asked this question in this thread and David Mitra was kind enough to point me towards the paper: Topics in Orthogonal Functions - Price. On pg. 598 the author states the Haar functions defined above form a complete orthonormal basis. Then pg. 603 states

For every complete orthonormal system $\Phi$ there is an $L^2$ function $Y$ whose $\Phi$-Fourier series can be rearranged to diverge almost everywhere.

So we can finish as follows. Take the Haar functions as above (a complete orthonormal set) and the function $Y$ defined above. We write this function's $\Phi$-Fourier series as $$\sum_{n=1}^\infty \langle Y, X_n \rangle X_n$$ Now there exists a rearrangement $\sigma$ of the indices of sum such that it diverges almost everywhere. Now let $Y_n = \langle Y, X_{\sigma(n)} \rangle X_{\sigma(n)}$ be this rearrangement of terms. We can see $$\sum_{n=1}^\infty Y_n$$ satisfies the requirements of the problem. Indeed $$\sum_{n=1}^\infty\operatorname{Var} (Y_n) = \operatorname{Var}(Y) < \infty.$$ Furthermore, $\sum Y_n$ diverges almost everywhere.

Note: A fun corollary is that if you have a sequence $X_n$ that does converge almost everywhere (and satisfies the questions requirements), and it forms a complete basis (its already practically orthonormal), then a rearrangement of itself diverges almost everywhere.

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  • $\begingroup$ @Michael - good point - that is less ambiguous. Thanks. $\endgroup$ – muaddib Jul 28 '15 at 22:32
  • $\begingroup$ Theorem 14 of that paper (page 608) suggests that a randomized ordering of the $Y_n$ variables gets us back to probability 1 convergence. $\endgroup$ – Michael Jul 28 '15 at 22:41
  • $\begingroup$ I think by "diverges almost everywhere" they just mean the limit does not exist (not that it goes to infinity). According to my comment in the question, the limit should indeed exist as a real number almost everywhere over a subsequence, even for those crazily arranged variables. $\endgroup$ – Michael Jul 28 '15 at 22:45
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    $\begingroup$ @Michael - Not sure if if you are referring to my last Note. There I meant there exists a specific reordering that causes it to diverge. But, yes, I see by Thm 14 that the "generic" rearrangement does converge. Neat, and odd stuff. $\endgroup$ – muaddib Jul 28 '15 at 22:45

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