Proof of "Prove that a sub-gaussian and isotropic random vector over a finite set T implies that the set is exponentially large"

Question

Here the original question was asked and answered. However I have a question to the solution. If I get it right they try to show $\frac 12 I_n \leq \mathbf{E} YY^T \leq I_n$ by proving $$ \mathbf{E} \left[ \langle X,\theta \rangle^2 {\bf 1}_{\{||X|| > 4C \sqrt{n}\}} \right] \leq \left( \mathbf{E} \left[ \langle X,\theta \rangle^4 \right] \cdot \mathbf{P}(||X||^2 > 16C^2n)\right)^{1/2} \leq \sqrt{\frac{4C^2}{16C^2}} = \frac{1}{2}$$ for all unit vectors $\theta$. However I don't see how you get $\mathbf{E} Y_i Y_j=0$ for $i\neq j$ from that. I hope someone can help me here.

Answer 1

In the linked answer, the inequality sign $\le$ in $\frac12\,I_n\le EYY^T\le I_n$ is not meant in the sense of the entrywise comparison. Rather, it is meant in this sense: for any two symmetric matrices $A$ and $B$, we write $A\le B$ if $B-A$ is positive semidefinite.

In this case, we have $Y=X1_{\|X\|\le4C\sqrt{n}}$, where $EXX^T=I_n$ and $$E\langle X,\theta \rangle^2 1_{\|X\|>4C\sqrt{n}}\le\tfrac12=\tfrac12\, E\langle X,\theta \rangle^2$$ for any unit vector $\theta$. So, $$\theta^T EYY^T\theta=E\langle Y,\theta \rangle^2 =E\langle X,\theta \rangle^2 1_{\|X\|\le4C\sqrt{n}}\le E\langle X,\theta \rangle^2=\theta^T EXX^T\theta=\theta^T I_n\theta,$$ which means $EYY^T\le I_n$, and $$E\langle Y,\theta \rangle^2=E\langle X,\theta \rangle^2-E\langle X,\theta \rangle^2 1_{\|X\|>4C\sqrt{n}}\ge \tfrac12\,E\langle X,\theta \rangle^2,$$ which similarly means $EYY^T\ge\tfrac12\,I_n$.