Does the kurtosis need to be finite for the sample variance to be consistent?

Question

It is known (see this answer) that if $\mu_4$ is the fourth central moment of a distribution and $\sigma$ is the standard deviation, then we can write $$\operatorname{Var}(S^2_n)=\frac{1}{n}\left[\mu_4-\frac{n-3}{n-1}\sigma^4\right].$$ This goes to $0$ as $n\to\infty$, hence if the kurtosis (and therefore the fourth moment) is finite, then the sample variance $S^2$ is consistent for the population variance (since $\mathbb{E}(S^2_n)=\sigma^2)$.

Does the converse hold? Does the kurtosis need to be finite for $S^2_n$ to converge in probability to $\sigma^2$?

I have tried simulating the sampling distribution of $S^2$ for various distributions; I tried looking at the sampling distribution for a $t_4$ population (which only has up to the third moment) as well as a $t_5$ population (which has finite kurtosis), and this seems to support the idea that the kurtosis does in fact need to exist (but obviously there could be a counterexample with another distribution).

Answer 1

We do not need the Kurtosis to be finite for so called weak consistency (convergence in probability), but the so called strong consistency (convergence in the 2nd moment) is equivalent to that the Kurtosis is finite.

In fact, the sample variance $S^2$ almost surely converges to population variance $\sigma^2$ whenever $\sigma^2$ is finite. Indeed, you can write $$S^2=\frac{n}{n-1} \left ( \frac{1}{n} \sum_{i=1}^n (X_i-\mu)^2 - (\bar{X}-\mu)^2 \right ).$$

Then, by twice using the strong law of large numbers and properties of almost sure convergence:

$$(\bar{X}-\mu)^2 \to 0 \quad \text{a.s.}$$ $$\frac{1}{n} \sum_{i=1}^n (X_i-\mu)^2 \to \mathbb E ((X-\mu)^2)=\sigma^2 \quad \text{a.s.},$$

for $\color{blue}{\sigma^2<\infty}$ we obtain

$$\color{blue}{S^2\to \sigma^2 \quad \text{a.s.}},$$

which also implies

$$\color{blue}{S^2\to \sigma^2 \quad \text{in probability}}. $$

However, considering that $S^2$ is unbiased estimator of $\sigma^2$, $\color{blue}{\kappa<\infty}$ if and only if

$$ \color{blue}{S^2\to \sigma^2 \quad \text{in 2nd moment}}. $$