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Over at Reddit someone asked, "Is it possible for a sum of the sequence of independent random variables to converge in probability, even when their variance doesn't converge?"

I was wonder about a similar related question.

"If you have a sequence of independent real random variables $X_n$ and Var($X_n$) is unbounded, is it possible that the $X_n$ converge almost surely?"

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  • $\begingroup$ I don't see why anybody would downvote this. $\endgroup$ Jun 14, 2022 at 18:42
  • $\begingroup$ @Michael OK, so maybe we can combine Levon's idea with the law of large numbers. if we have i.i.d. random variables $Y_i$ such that for all positive integers $k$, $P(Y_i=2^k) = \pi^2/(12 k^2)$, $P(Y_i=-2^k) = \pi^2/(12 k^2)$, $P(Y=k)=0$ when $k$ is not a power of 2, and $X_n = \frac{1}{n} \sum_{i=1}^n Y_i$, then by SLLN the $X_n\rightarrow 0$ a.s., but I think $Var(X_n)=\infty$ for all $n$. $\endgroup$
    – irchans
    Jun 14, 2022 at 20:35
  • $\begingroup$ I deleted my comment since I didn't originally notice you were not talking about convergence of the sample means. I gave another answer below. [My original comment was that the standard SLLN works for iid random variables with finite mean but infinite variance.] $\endgroup$
    – Michael
    Jun 14, 2022 at 20:38

2 Answers 2

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This may be similar in spirit to the other answer but perhaps simpler. Let $\{Y_i\}_{i=1}^{\infty}$ be i.i.d. with zero mean and infinite variance, let $\{U_i\}_{i=1}^{\infty}$ be i.i.d. uniform $[0,1]$ and independent of $\{Y_i\}$. Define $$ X_n = Y_n1_{\{U_n\leq 1/n^2\}} \quad \forall n \in \{1, 2, 3, …\}$$ Then $\{X_n\}_{n=1}^{\infty}$ are mutually independent. Now $P[X_n\neq 0] \leq 1/n^2$ and Borel-Cantelli ensures $X_n\rightarrow 0$ almost surely. But $E[X_n]=0$ and $Var(X_n)=E[X_n^2] = E[Y_n^2](1/n^2) = \infty$ for all $n$.

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    $\begingroup$ Great. Thank you. The key seems to be that $P(\cap_{n=1}^\infty(\cup_{k=n}^\infty E_n))=0$ by Borel-Cantelli with $E_n$ being the event $X_n\neq 0$. $\endgroup$
    – irchans
    Jun 14, 2022 at 20:52
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Yes, such a sequence exists. Consider the following construction:

$$ X_m \sim p_m(x)=\sum_{k=1}^m \frac{m}{2} \cdot \mathbb{I}_{[k, k+1/m^2] \cup [-k-1/m^2, -k]} $$

By symmetry we have $\mathbb{E}[X_n]=0$. So, $$ \mathbb{D}X_m=\mathbb{E}[X_m^2] = \\ =\int_{\mathbb{R}} x^2 p_m(x)dx = 2 \cdot \sum_{k=1}^m 2m \cdot \int_k^{k+1/m^2} x^2dx $$ Now see my God-like integration technique: $$ \int_k^{k+1/m^2} x^2dx = \frac{1}{3} (k+1/m^2)^3 - \frac{1}{3} k = \\ =\frac{1}{3}(\frac{3k^2}{m^2}+\frac{3k}{m^4}+\frac{1}{m^6}) $$

The variance:

$$ \mathbb{D}X_m=4/3 \cdot \sum_{k=1}^m \frac{3k^2}{m}+\frac{3k}{m^3}+\frac{1}{m^5} = 4/3 \cdot 1 / m \cdot O(m^3) = O(m^2) \to \infty $$

is unbounded.

But the measure of a set where $p_m(x) \ne 0$

$$ 2m \cdot 1/m^2 = O(1 / m) \to 0 $$

So, $X_m$ almost surely converges to 0.

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  • $\begingroup$ Much appreciation for your integration skills :) $\endgroup$
    – irchans
    Jun 14, 2022 at 18:27
  • $\begingroup$ What random variable does the sequence of $(X_1, X_2, X_3, \ldots)$ converge almost surely to? Suppose $P(Y=0)=1$. The $X_m$ can't converge to $Y$ because $P(|X_m|<1/2)=0$ for all $m$. I think. $\endgroup$
    – irchans
    Jun 14, 2022 at 18:32
  • $\begingroup$ @irchans thank you for comments. I'll try to edit my answer $\endgroup$ Jun 14, 2022 at 18:40

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