Positive definiteness of bounded distribution

Question

Let $x\in \mathbb{R}^n$ be a random vector distributed on the compact set $\mathcal{D} \subset \mathbb{R}^n$ according to a certain distribution. Let $||x||_2\leq X$ for every $x \in \mathcal{D}$. Let $\sigma$ be the minimum singular value of $\mathbb{E}[xx^T]$. How can one prove that $$\frac{\mathbb{E}[xx^T]}{\sigma}- \frac{yy^T}{X^2} \succeq 0,\quad \forall y\in \mathcal{D}\,,$$ where $\succeq 0$ denotes positive-semidefiniteness?

Answer 1

I do not think that you can ask for positive-definiteness. Indeed, if you take $n=1$ and $\mathcal{D} = \{1\}$ then $x$ is constant equal to $1$ and $\sigma = X = 1$ so that $$ \frac{\mathbb{E}[xx^{T}]}{\sigma} - \frac{y y^{T}}{X^2} = 0 $$ for $y = 1$ the unique element of $\mathcal{D}$.

However, it is true that the matrix that you consider is always positive. Indeed, if $u \in \mathbb{R}^n$ then we have $$ \langle \mathbb{E}[x x^{T}] u, u \rangle \geq \sigma \|u\|_2^2 $$ since $\mathbb{E}[x x^{T}]$ is positive, and hence its minimum singular value is just its smallest eigenvalue. Moreover, we have $$ \langle yy^{T} u ,u \rangle = (y y^{T} u)^{T} u = u^{T} y y^{T} u = \langle u,y \rangle^2 \leq \|u\|_2^2 \|y\|_2^2 \leq X^2 \|u\|_2^2, $$ where we applied Cauchy--Schwarz inequality and the definition of $X$. It follows from the inequalities above that $$ \langle \left( \frac{\mathbb{E}[xx^{T}]}{\sigma} - \frac{y y^{T}}{X^2} \right) u , u \rangle \geq 0, $$ and thus the symmetric matrix $\frac{\mathbb{E}[xx^{T}]}{\sigma} - \frac{y y^{T}}{X^2}$ is positive.

Notice however that you need to make some assumption of non-degeneracy on $x$ in order to have $\sigma \neq 0$.

I do not know in which particular context you are working, but notice for instance that if $y$ is not an eigenvector of $\mathbb{E}[x x^{T}]$ associated to its smallest eigenvalue, then it follows from the analysis above that your matrix is definite positive (because you know the case of equality in Cauchy--Schwarz).