# Independent, unbounded variance, and almost sure convergence

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?"

• I don't see why anybody would downvote this. Jun 14, 2022 at 18:42
• @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 $n$. Jun 14, 2022 at 20:35
• 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.] Jun 14, 2022 at 20:38

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 $$n$$.

• 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 $X_n\neq 0$. Jun 14, 2022 at 20:52

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.

• Much appreciation for your integration skills :) Jun 14, 2022 at 18:27
• 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. Jun 14, 2022 at 18:32
• @irchans thank you for comments. I'll try to edit my answer Jun 14, 2022 at 18:40