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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][1] 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][2]

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.

Please let me know about any thoughts you may have on this. [1]: https://mathoverflow.net/posts/351509/edit [2]: https://gofile.io/?c=YnM19e

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.

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][1] 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][2]

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. [1]: https://mathoverflow.net/posts/351509/edit [2]: https://gofile.io/?c=YnM19e

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][1] 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][2]

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.

Please let me know about any thoughts you may have on this. [1]: https://mathoverflow.net/posts/351509/edit [2]: https://gofile.io/?c=YnM19e

This question is about the positive definiteness of a (non-random) matrix that is defined using random variables as follows:

We fix the vector $v=(1,1)$ (I am mostly interested in this choice, but 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 in [this thread][1] 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 $$d\mu(x) \propto e^{-\langle (x-v), \Sigma^{-1} (x-v) \rangle-\vert x_1 \vert^4-\vert x_2 \vert^4} \ dx.$$ factorite.

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][2]

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. [1]: https://mathoverflow.net/posts/351509/edit [2]: https://gofile.io/?c=YnM19e

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][1] 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][2]

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. [1]: https://mathoverflow.net/posts/351509/edit [2]: https://gofile.io/?c=YnM19e