Infinite variance in the Law of large numbers

Question

Let $(X_{n})_{n>0}$ be independent random variables, with $E(X_{n})=\mu$ and $V(X_{n})=\sigma_{i}^{2}$.

If $\sigma_{i}^{2}$. are not bounded, is it true that $\frac{\sum_{n=1}^{N} X_{n}}{N}$ converges in probability to $\mu$ when $N\to+\infty$?

If the answer is No, what would be a counter-example? If it is Yes, how can I prove it?

Answer 1

If the $\sigma_i^2$ increase extremely quickly, then the sample means can have large variances and prevent convergence in probability.

Concrete counterexample:

Let $X_n \sim N(0, n^2)$. Then $\bar{X}_N := \frac{1}{N} \sum_{n=1}^N X_n \sim N(0, \frac{(N+1)(2N+1)}{6N})$, so $P(|\bar{X}_N| > \epsilon)$ is strictly increasing as $N \to \infty$.

Response to comment:

Since $\bar{X}_N / \sqrt{\frac{(N+1)(2N+1)}{6N}} \sim N(0,1)$ we have $P(|\bar{X}_N| > \epsilon) = P\left(|Z| > \epsilon/\sqrt{\frac{(N+1)(2N+1)}{6N}}\right)$. As $N \to \infty$, the quantity $\epsilon/\sqrt{\frac{(N+1)(2N+1)}{6N}}$ decreases to zero, making the probability increase.