I know from this question and from other sources that the standard deviation of the sample variance $S^2=\frac{1}{n-1}\sum\left(x_i-\bar{x}\right)^2$ is,
$$ \sigma_{S^2}=\sqrt{\frac{\mu_4}{n}-\frac{n-3}{n(n-1)}\mu_2^2} $$
where $\mu_4$ and $\mu_2$ are respectively the fourth and second central moments. What I'm wondering is if I could estimate $\sigma_{S^2}$ from a sample by using the above formula and replacing $\mu_4$ and $\mu_2$ by the fourth and second sample central moments i.e.
$$ m_4=\frac{1}{n}\sum\left(x_i-\bar{x}\right)^4 \ , \ m_2=\frac{1}{n}\sum\left(x_i-\bar{x}\right)^2 $$
It seems to me it would be biased since $m_2$ is already a biased estimator of $\mu_2$. Is it at least a consistent estimator of $\sigma_{S^2}$? Do you know any other better way of estimating it?
