This question is about the positive definiteness of a (non-random) matrix that is defined using random variables as follows:
We fix the vector $v=(1,1)$ (yet, it seems the final result does not depend on it) and choose a positive definite covariance matrix $\Sigma \in \mathbb R^{2 \times 2}$.
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 already 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 $x_1,x_2$ in the measure $$d\mu(x) \propto e^{-\langle (x-v), \Sigma^{-1} (x-v) \rangle-\vert x_1 \vert^4-\vert x_2 \vert^4} \ dx.$$ factorize.
His proof also shows that at least for the special choice $x=\Sigma^{-1}v$, we have also for non-diagonal $\Sigma $, that $\langle x,Ax\rangle \ge 0.$
We then discussed in the comments whether one can extend the proof to non-diagonal $\Sigma$, but did not succeed so far.
Numerically it seems that for all choices of $\Sigma$ and $v$ 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
Another way of expressing this matrix $A$ is to write $$A = \int_{\langle x,\Sigma^{-1} v \rangle>0} (x \otimes x) \langle x,\Sigma^{-1} v \rangle \ (d\mu(x+\mathbb E(X))-d\mu(x-\mathbb E(X))).$$
We observe that for $\langle x,\Sigma^{-1} v \rangle>0$ the integrand $(x \otimes x) \langle x,\Sigma^{-1} v \rangle $ is always positive. So if we could show that "most of the time" $$\mu(x+\mathbb E(X))\ge \mu(x-\mathbb E(X))$$
this would imply the result.
My question therefore is: How can I show the eigenvalues of $A$ are non-negative?
Numerically, I made the following observations:
The positive definiteness of $A$ holds independent of the choice of $v \neq 0$ and $\Sigma.$
- If we write $$d\mu(x) \propto e^{-\langle (x-v), \Sigma^{-1} (x-v) \rangle-\vert x_1 \vert^p-\vert x_2 \vert^p} \ dx.$$
then for $p<2$ the eigenvalues of $A$ become negative, zero for $p=2$ and greater than zero for $p>4.$
This seems to be consistent with what Fedja proved in the one-dimensional case.
