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

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

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 $\lambda$.

A little computation (e.g, see https://mathoverflow.net/a/227237/78539) reveals that

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

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 $I(t; a, b) := \int_{0}^t s^a(1-s)^bds$ is the incomplete beta function.