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?"
 A: 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 $n$.
A: 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.
