I was wondering (on a setting where $\vec X_i \sim \mathcal{N}(\vec\mu, \mathbb{I})$ are $n$ random $d$-dimensional multivariate normal vectors with unknown mean $\vec\mu$) how I could obtain a lower bound, with high probability, on the spectral norm:

$$ \left\| \frac{1}{n} \sum_{i=1}^n \vec X_i \vec X_i^T - \hat{\vec\mu} \hat{\vec\mu}^T - \mathbb{I} \right\|$$

without having to depend my bound on $\|\vec\mu\|$.

In the above quantity, $\hat{\vec\mu}$ denotes an estimate of the mean value, for which I have that, with high probability and $n=O(\text{something})$ it holds $\| \hat{\vec\mu} - \vec\mu \| \leq \epsilon$ i.e. I can get a "good" approximation to $\vec\mu$.

Given that the covariance matrix of $\vec X_i$ is merely $\mathbb{I}$, this should be enough for me to lower bound the above quantity without $\|\vec\mu\|$.

It seems to me intuitive that I should be able to get the above quantity a lower bound (with high probability) without relying on some bound of $\|\vec\mu\|$ but I'm not sure how to obtain that. Trying to break the $\frac{1}{n} \sum_{i=1}^n \vec X_i \vec X_i^T$ into $\frac{1}{n} \sum_{i=1}^n (\vec X_i - \vec\mu) (\vec X_i - \vec\mu)^T$ hasn't led me somewhere (?), because then I need to additionally bound terms like $$\left\|\vec\mu \left( \frac{1}{n} \sum_{i=1}^n \vec X_i - \vec\mu \right)^T \right\|$$ which would introduce into my bound the terms $\|\vec\mu\|$ that I do not desire.

What could I do to lower bound the above quantity and avoid any dependence on $\|\vec\mu\|$?

In the above, $\|\cdot\|$ denotes the Euclidean norm for vectors and the spectral 2-norm for matrices.

  • $\begingroup$ Is each $X_i$ a column vector? $\endgroup$
    – Neal Young
    May 15, 2021 at 18:55
  • $\begingroup$ @NealYoung yes, it is a $d$-dimensional column vector. $\endgroup$
    – Jay
    May 15, 2021 at 22:21


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