convergence of sample covariance matrix in case sample size depends on dimesion Let $X_1,X_2,\dots,X_n$ be random sample from $\mathcal{N}_p(\mathbf{0},\mathbf{\Sigma})$ and put $\mathbf{S}=\frac{1}{n}\sum_{i=1}^nX_iX_i^t$, which is sample covariance matrix. If $p<n$, it is known that
\begin{align}
||\mathbf{S}-\mathbf{\Sigma}||_{\text{F}}=O_p(\frac{1}{\sqrt{n}}).
\end{align}
My question is that if $p\asymp n^{\beta}$ for some constant $0<\beta<1$, does this still hold? Though $p<n$ is being held for all sufficiently large $n$, I guess the given convergence rate does not hold because $n$ depends on $p$. I've tried to prove this but failed eventually. Is my guess wrong? Otherwise, can anyone give me a hint? Thanks in advance.

The variance of the elements of $nS$ is $n(\sigma^2_{ij}+\sigma^2_{ii}\sigma^2{jj})$,
so the variance of the elements of $S$ is $n^{-1}(\sigma^2_{ij}+\sigma^2_{ii}\sigma^2{jj})$.
The expected value of the square of the Frobenius norm of $S-\Sigma$ is the sum of these. The value will depend on how $\Sigma$ changes as $n\to\infty$, but it's reasonably going to increase at a rate somewhere between $p$ (eg for the identity matrix) and $p^2$ (for symmetric positive correlation). So, the expectation of the Frobenius norm grows as something between $p/n$ and $p^2/n$, which is not $O(1/\sqrt{n})$ in your case. And the Wishart distribution is sufficiently well behaved in the tails that it won't be $O_p(1/\sqrt{n})$ either
