Let $X$ be a $\sigma$-subGaussian random vector on $\mathbb R^n$ (for large $n \ge 3$), meaning that the random variable $X^Tv$ is $\sigma$-subGaussian for every unit vector $v \in \mathbb R^n$. Consider the $n$-by-$n$ psd matrix $\Sigma := \mathbb E[UU^T]$, where $U := X/\|X\|_2$. It is clear that every eigenvalue of $\Sigma$ lies in the interval $[0, 1]$. In fact, $\text{tr}\Sigma \le 1$.

Question 1. What is a good estimate for the largest eigenvalue of $\Sigma $ ?

Question 2. Same question without the sub-Gaussianity assumption.

My Current approach

Let $\lambda$ be an eigenvalue of $\Sigma$ and $z$ be a unit vector in the corresponding eigenspace.

For any $\delta \in [0,1]$, let $G_\delta := \{x \in \mathbb R^n \mid |x^Tz| > \delta\}$. then

$$ \begin{split} \lambda &= \lambda\|z\|^2 = z^T(\lambda z) = z^T\Sigma z = z^T E[UU^T]z = E[z^TUU^Tz] = E|U^Tz|^2\\ &= E[|U^Tz|^2 \mid U \in G_\delta]P(U \in G_\delta) + E[|U^Tz|^2 \mid U \in G^c_\delta)P(U \in G^c_\delta)\\ &\le P(U \in G_\delta) + \delta^2P(U \in G_\delta^c) = (1 - \delta^2)P(U \in G_\delta) + \delta^2. \end{split} $$

That is,

$$ \lambda \le (1-\delta^2)P(|U^Tz| > \delta) + \delta^2,\; \forall \delta \in [0, 1]. \tag{1} $$

Thus, if I had a bound on $P(|U^Tz| > \delta)$, I could plug it in (1) and then minimize over $\delta \in [0, 1]$ to get (a perhaps good) upper bound on $\lambda$.

For simplificy, suppose $X \sim \mathcal N(0,\sigma^2 I_n)$. A little computation (e.g, see https://mathoverflow.net/a/227237/78539) reveals that

$$ \begin{split} P(|U^Tz| > \delta) &= 2\omega_{n-2}\int_{\sqrt{\delta}}^\infty (1-t^2)^{(n-3)/2}dt=\omega_{n-2}\int_\delta^\infty s^{-1/2}(1-s)^{(n-3)/2}ds\\ &=\omega_{n-2}\left(1-I\left(\delta;\frac{1}{2},\frac{n-1}{2}\right)\right) = \omega_{n-2}I\left(1-\delta;\frac{n-1}{2},\frac{1}{2}\right), \end{split} \tag{2} $$

where $\omega_{n-2}=\dfrac{2\pi^{\frac{n-1}{2}}}{\Gamma(\frac{n-1}{2})}$ is the surface area of the unit sphere in $\mathbb R^{n-2}$, and $I(t; a, b) := \int_{0}^t s^{a-1}(1-s)^{b-1}ds$ is the incomplete beta function.

Edit: Bounding $P(|U^Tz| > \delta)$

Theorem ($U^Tz$ is sub-exponential! ). For every $\delta \in [0, 1]$, it holds that $$ P(|U^Tz| > \delta) \le e^{-\frac{n-1}{4}\delta}. \tag{3} $$

Proof. Let $p = I(1-\delta; 1/2, (n-1)/2)$. It is known since [Temme (1992)] that for $p \in (0, 1)$ and large $a > 0$, the solution of the equation $p = I(t; a,b)$ is given (approximately) by

$$ t=t_p(a, b) \approx e^{-(1/a)Q_{1-p}(\Gamma(b,1))}, \tag{4} $$

where $Q_{1-p}(\Gamma(b,1))$ is the $1-p$ quantile of the unit-scale gamma distribution with shape parameter $b$. Now by standard concentration results (e.g see Boucheron et al. textbook),

$$ Q_{1-p}(\Gamma(b,1)) \le \log(1/p) + \sqrt{2b\log(1/p)}. \tag{5} $$

In particular, for $b=1/2$ we get

$$ Q_{1-p}(\Gamma(1/2,1)) \le \log(1/p) + \sqrt{\log(1/p)} \le 2\log(1/p). \tag{6} $$

Putting (2), (4), and (6) together and using the basic inequality $e^{-t} \ge 1-t\;\forall t > -1$, we see that $$ \begin{split} 1-\delta &\ge t_{2p}\left((n-1)/2,1/2\right) \ge e^{-\frac{2Q_{1-2p}(\Gamma(1/2,1))}{n-1}} \ge e^{-\frac{2}{n-1}\left(\log\left(\frac{1}{2p}\right) + \sqrt{\log\left(\frac{1}{2p}\right)}\right)}\\ & \ge 1 - \frac{2\left(\log\left(\frac{1}{2p}\right) + \sqrt{\log\left(\frac{1}{2p}\right)}\right)}{n-1} \ge 1-\frac{4\log\left(\frac{1}{2p}\right)}{n-1}, \end{split} $$

from which (3) follows upon combining with (2). $\quad\quad\Box$