Let $S_n$ be the sample standard deviation of $(X_i)_{1\le i\le n}$ for a sequence of iid variables $X_i$. It is easily proved (as below) that it underestimates $\sigma$, where $\sigma^2 ={\rm Var} X_i$. In common examples such as the normal distribution it is the case that ${\rm E}S_n\rightarrow \sigma$ as $n\to \infty$. Is it always the case whenever $\sigma$ is finite?
My starting point is the usual proof of the fact that ${\rm E\,}S_n < \sigma$ (in non degenerate cases) $$ 0<{\rm Var\,} S_n ={\rm E\,}S_n^2 - ({\rm E\,}S_n)^2 = \sigma^2 - ({\rm E\,}S_n)^2\,. $$ and then the answer is affirmative if ${\rm Var\,} S_n \to 0$. I know that ${\rm Var\,} S_n^2 \to 0$ in the case that the fourth moments of $X_i$ are finite. If we show that $$ {\rm Var\,} S_n^2\to 0 \implies {\rm Var\,}S_n \to 0 $$ then the question would be settled in the case ${\rm E\,}X_i^4<\infty$, but I don't know even how to show this (I'm posting a general question about this as well).
