I have two random vectors $x,y \in \mathbb{R}^n$, with joint probability law $P(x,y)$, and each has zero mean, $\mathbb{E} x = \mathbb{E} y = 0$. I build the rank-one matrix $M = xy^\intercal$. My goal is to show that: $$|\text{Tr}[(\mathbb{E} M)^2]| \leq (\mathbb{E} \text{Tr} M)^2$$
Or as a function of $x,y$, $$|\text{Tr}[(\mathbb{E} xy^\intercal)^2]| \leq (\mathbb{E} x^\intercal y)^2$$
I tried to use Cauchy-Schwartz but couldn't get the result. If you have any leads please let me know :)
This is not true. Let $u,v\stackrel{\text{i.i.d.}}{\sim}\operatorname{Unif}(\{-1,1\}),\ x=(u,v)^T$ and $y=(u,-v)^T$. That is, suppose the $xy^T$ is uniformly distributed over the sample space \begin{aligned} &\left\{\pmatrix{1\\ 1}\pmatrix{1&-1}, \ \pmatrix{1\\ -1}\pmatrix{1&1}, \ \pmatrix{-1\\ 1}\pmatrix{-1&-1}, \ \pmatrix{-1\\ -1}\pmatrix{-1&1}\right\}\\ =\, &\left\{\pmatrix{1&-1\\ 1&-1}, \ \pmatrix{1&1\\ -1&-1}, \ \pmatrix{1&1\\ -1&-1}, \ \pmatrix{1&-1\\ 1&-1}\right\}. \end{aligned} Then $E(xy^T)=\pmatrix{1\\ &-1}$ and $\operatorname{tr}\left((E(xy^T))^2\right)=2>0=\left(\operatorname{tr}(E(xy^T))\right)^2=\left(E(\operatorname{tr}(xy^T))\right)^2$.