# 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 $$p, it is known that \begin{align} ||\mathbf{S}-\mathbf{\Sigma}||_{\text{F}}=O_p(\frac{1}{\sqrt{n}}). \end{align} My question is that if $$p\asymp n^{\beta}$$ for some constant $$0<\beta<1$$, does this still hold? Though $$p 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
• Thank you for your answer! I understood your answer. Then if $\mathbf{\Sigma}$ has non-zero off diagonal entries $s=O(p)$, then would it be reasonable to guess the frobenius norm grows as $(p+s)/n$? Jul 27, 2021 at 8:24
• I would have guessed $O(p+p^2s)$, since there are $p(p-1)$ off-diagonal entries Jul 28, 2021 at 1:57