<p>I was wondering (on a setting where <span class="math-container">$\vec X_i \sim \mathcal{N}(\vec\mu, \mathbb{I})$</span> are <span class="math-container">$n$</span> random <span class="math-container">$d$</span>-dimensional multivariate normal vectors with <strong>unknown</strong> mean <span class="math-container">$\vec\mu$</span>) how I could obtain a lower bound, with high probability, on the spectral norm:</p>
<p><span class="math-container">$$ \left\| \frac{1}{n} \sum_{i=1}^n \vec X_i \vec X_i^T - \hat{\vec\mu} \hat{\vec\mu}^T - \mathbb{I} \right\|$$</span></p>
<p>without having to depend my bound on <span class="math-container">$\|\vec\mu\|$</span>.</p>
<p>In the above quantity, <span class="math-container">$\hat{\vec\mu}$</span> denotes an estimate of the mean value, for which I have that, with high probability and <span class="math-container">$n=O(\text{something})$</span> it holds <span class="math-container">$\| \hat{\vec\mu} - \vec\mu \| \leq \epsilon$</span> i.e. I can get a &quot;good&quot; approximation to <span class="math-container">$\vec\mu$</span>.</p>
<p>Given that the covariance matrix of <span class="math-container">$\vec X_i$</span> is merely <span class="math-container">$\mathbb{I}$</span>, this should be enough for me to lower bound the above quantity without <span class="math-container">$\|\vec\mu\|$</span>.</p>
<p>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 <span class="math-container">$\|\vec\mu\|$</span> but I'm not sure how to obtain that. Trying to break the <span class="math-container">$\frac{1}{n} \sum_{i=1}^n \vec X_i \vec X_i^T$</span> into <span class="math-container">$\frac{1}{n} \sum_{i=1}^n (\vec X_i - \vec\mu) (\vec X_i - \vec\mu)^T$</span> hasn't led me somewhere (?), because then I need to additionally bound terms like <span class="math-container">$$\left\|\vec\mu \left( \frac{1}{n} \sum_{i=1}^n \vec X_i - \vec\mu \right)^T \right\|$$</span> which would introduce into my bound the terms <span class="math-container">$\|\vec\mu\|$</span> that I do not desire.</p>
<p>What could I do to lower bound the above quantity and avoid any dependence on <span class="math-container">$\|\vec\mu\|$</span>?</p>
<p>In the above, <span class="math-container">$\|\cdot\|$</span> denotes the Euclidean norm for vectors and the spectral 2-norm for matrices.</p>

CC BY-SA 4.0
2021-05-15T18:55:40.603Z
Is each $X_i$ a column vector?
Neal Young
CC BY-SA 4.0
2021-05-15T22:21:37.337Z
@NealYoung yes, it is a $d$-dimensional column vector.
Jay
<lower-bounds><matrices><norms>
Obtaining a lower bound of a matrix norm
Jay
# Obtaining a lower bound of a matrix norm

(Asked by: username_0 on 2021-05-14)

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.

**Tags:** lower-bounds, matrices, norms
> **Comments:**
> **username_1**: Is each $X_i$ a column vector?  
> **username_0**: @NealYoung yes, it is a $d$-dimensional column vector.  

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## 0 answers

