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.


1 Answer 1


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

  • $\begingroup$ 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 $(p+s)/n$? $\endgroup$
    – BongJung
    Jul 27, 2021 at 8:24
  • $\begingroup$ I would have guessed $O(p+p^2s)$, since there are $p(p-1)$ off-diagonal entries $\endgroup$ Jul 28, 2021 at 1:57

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