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 matrix for $Y=X-\mathbb E(X)$

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

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. 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.

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?

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 matrix for $Y=X-\mathbb E(X)$

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

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. 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.

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.

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 matrix for $Y=X-\mathbb E(X)$

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

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

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

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?

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 matrix for $Y=X-\mathbb E(X)$

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

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. 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.

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?