# Upper-bound for eigenvalues of $E [UU^T]$, where $U$ is uniformly distributed on the unit $n$-sphere

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 simplicity, suppose $$X \sim \mathcal N(0,\sigma^2 I_n)$$. Since $$U$$ is uniformly distributed on the unit $$n$$-sphere, it follows by symmetry that for every unit vector $$z \in \mathbb R^d$$, the random variable $$U^Tz$$ has the same distribution as $$U_1$$ (the first coordinate of the random vector $$U$$), which in turn (by the Archimedean projection property) has the same distribution as the first coordinate of a point draw uniformly in the unit ball in $$\mathbb R^{n-2}$$. Thus, $$P(U_1 > \delta)$$ is the probability that a random point in the unit ball in $$\mathbb R^{n-2}$$ lies in on given side of an equatorial hyperplane, we have

$$\begin{split} P(|U^Tz| > \delta) &= P(|U_1| > \delta)= 2P(U_1 > \delta) = 1-I\left(\delta;\frac{1}{2}, \frac{n-1}{2}\right)\\ &= I\left(1-\delta;\frac{n-1}{2},\frac{1}{2}\right), \end{split} \tag{2}$$ $$\begin{split} P(|U^Tz| > \delta) &= 1-I\left(\delta;\frac{1}{2}, \frac{n-1}{2}\right) = I\left(1-\delta;\frac{n-1}{2},\frac{1}{2}\right), \end{split} \tag{2}$$

where $$I(t; a, b)$$ is the normalized incomplete beta function, defined by $$I_t(t; a, b) := B(t;a,b) / B(1; a, b)$$, with $$B(t; a, b):= \int_{0}^t s^{a-1}(1-s)^{b-1}ds$$.

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

Theorem ($$U^Tz$$ is sub-exponential! ). Let $$U$$ be uniformly distributed on the unit $$n$$-sphere and let $$z$$ be a fixed vector on this sphere. If $$n$$ is large enough, then for every $$\delta \in [0, 1]$$, it holds that $$P(|U^Tz| > \delta) \le e^{-\frac{n-1}{4}\delta}. \tag{3}$$ Consequently, we have the spectral bound $$\lambda_{\max}(\Sigma) \le \min_{0 \le \delta \le 1}(1-\delta^2)e^{-\frac{n-1}{4}\delta} + \delta^2 \sim \frac{C\log\log n}{n^2}, \tag{*}$$ for some positive absolute constant independent of $$n$$.

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 $$a=(n-1)/2$$ and $$b=1/2$$ 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). Finally, (*) follows from (1) and (3) and the estimate obtained here (the constant $$C$$ can be made explicit). $$\quad\quad\Box$$

Below is a graphical visualization of the bound obtained with the above ingredients.

• Oops, seems I went through all the trouble of establishing the bound (*) for nothing. Indeed, by symmetry of the sphere, for every unit vector $z \in \mathbb R^d$, the random variable $U^Tz$ has the same distribution as $U_1$ (the first coordinate of the random $n$-dimensional vector $U$). Thus, we can use the argument in mathoverflow.net/a/315232/78539 to get the (slightly more precise) bound $P(|U^Tz| > \delta) = P(|U_1| > \delta) = P(|\mathcal N(0, 1)| > \sqrt{n}\delta) + \mathcal O(1/n)$, and there are well-known tail bounds for the standard normal distribution $\mathcal N(0,1)$. – dohmatob Apr 19 at 16:54