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

We fix the vector $v=(1,1).$

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

$$d\mu(x) \propto e^{-\Vert x-v \Vert^2-\vert x_1 \vert^4-\vert x_2 \vert^4} \ dx.$$

We then define the following matrix for $Y=X-\mathbb E(X)$

$$ \langle x,Ay\rangle = -\mathbb E\left(\langle Y,x \rangle \langle Y,v \rangle \langle Y, y\rangle\right).$$

Using wolframalpha I find that the normalization constant of this measure is $0.62,$ the expectation value of $X$ is $\mathbb E(X)=0.27 v$ and the eigenvalues of the matrix $A$ are non-negative.

Observations:

It seems like any non-zero $v$ does the job, i.e. $A$ will be positive semi-definite but I am only interested in this particular choice.

What seems to be important for this to be true is that we have the power $4$ appearing in the probability measure, as the expectation would vanish in the purely Gaussian case by symmetry.

My question therefore is: How can I show the eigenvalues of $A$ are non-negative?

Please let me know if you got any questions, I am happy to investigate further ideas. At the moment, I am a bit puzzled by this problem.

ADDENDUM to the comments: To respond to a conjecture made in a comment:

I used the following Mathematica code

```
Plot3D[If[Sign[x1*1 + x2*1] >= 0, (Exp[-((-x1 + 0.27) - 1)^2 + ((-x2 + 0.27) - 1)^2 - (-x1 +
0.27)^4 - (-x2 + 0.27)^4] -
Exp[-((x1 + 0.27) - 1)^2 + ((x2 + 0.27) - 1)^2 - (x1 +
0.27)^4 - (x2 + 0.27)^4])/0.62, 0], {x1, -2, 2}, {x2, -2,2}, PlotRange -> All]
```

This produces the following plot for $\mu(-y+\mathbb E(X)) -\mu(y+\mathbb E(X))$ which shows that it is almost everywhere (up to some very small region) true that $\mu(-y+\mathbb E(X)) >\mu(y+\mathbb E(X))$ if $\langle y,v \rangle >0.$