We construct a non-random matrix using random variables as follows:

We fix the vector $v=(1,1)$ and choose $\Sigma \in \mathbb R^{2 \times 2}$ positive definite.

Let $X$ be a $\mathbb R^2$-valued random variable such that $X$ is distributed according to

$$d\mu(x) \propto e^{-\langle (x-v), \Sigma^{-1} (x-v) \rangle-\vert x_1 \vert^4-\vert x_2 \vert^4} \ dx.$$

We then define the following deterministic matrix using centred random variables $Y=X-\mathbb E(X)$

$$ \langle x,Ay\rangle = -\mathbb E\left(\langle Y,x \rangle \langle Y,\Sigma^{-1} v \rangle \langle Y, y\rangle\right).$$

Fedja proved in this thread that for any $v$ and $\Sigma$ diagonal, the matrix $A$ is positive definite. He understood that in this case the matrix is essentially diagonal, as components $x_1,x_2$ are uncorrelated. His proof also shows that for $x=\Sigma^{-1}v$ we have also for non-diagonal $\Sigma $

$$\langle x,Ax\rangle \ge 0$$

We then discussed in the comments whether this would also be true in the correlated case but did not get anywhere so far.

Numerically it seems that for all choices of $\Sigma$ I made so far, the matrix $A$ is positive definite.

Here you can find the Mathematica file I was using to verify this

Click me

My question therefore is: How can I show the eigenvalues of $A$ are non-negative or is this wrong once $\Sigma$ is not assumed to be diagonal?

Please let me know if you have any questions.