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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 $\langle y,v \rangle >0.$

enter image description here

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  • $\begingroup$ $A = -\int_{y\in R^2} \langle y,v\rangle (y\otimes y) d\mu(y+\bar{x}) $ $= \int_{y\in R^2, \langle y,v\rangle > 0} \langle y,v\rangle (y\otimes y) (d\mu(-y+\bar{x})-d\mu(y+\bar{x}))$. I suspect that $d\mu(-y+\bar{x})-d\mu(y+\bar{x}) > 0$ when $\langle y,v\rangle > 0$. Is that true? $\endgroup$
    – Yoav Kallus
    Commented Jan 30, 2020 at 14:56
  • $\begingroup$ @YoavKallus what you write could be true and is interesting, however it is difficult to say since $\overline{x}$ is not explicit. I will try some numerics. $\endgroup$
    – Kung Yao
    Commented Jan 30, 2020 at 15:07
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    $\begingroup$ @YoavKallus so it is mostly true but not everywhere $\endgroup$
    – Kung Yao
    Commented Jan 30, 2020 at 15:25
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    $\begingroup$ @KungYao: You shouldn't call $A$ a random matrix because it is not random. $\endgroup$
    – Abdelmalek Abdesselam
    Commented Jan 30, 2020 at 17:57
  • $\begingroup$ @AbdelmalekAbdesselam you are right of course. $\endgroup$
    – Kung Yao
    Commented Jan 30, 2020 at 20:56

1 Answer 1

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Something is strange (about the question) because, due to symmetry, the expectation of $X$ is just $(c,c)$ where $c$ is the expectation of the random variable with the density $p(x)$ proportional to $e^{-x^2-x^4+2x}$. Thus, the off-diagonal matrix elements are merely $$ -\iint (x_1-c)(x_2-c)[(x_1-c)+(x_2-c)]p(x_1)p(x_2)dx_1dx_2=0 $$ because each of two terms splits into a product in which one of the factors is $\int (x-c)p(x)dx=0$. So why to ask about eigenvalues if the matrix is diagonal?

Similarly, the diagonal entries are $$ -\iint (x_1-c)^2[(x_1-c)+(x_2-c)]p(x_1)p(x_2)dx_1dx_2 \\ =-\int(x-c)^3p(x)dx, $$ which is shown to be non-negative in this post

Am I missing anything?

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  • $\begingroup$ wow, yes you seem to be right. Let me digest this post you are mentioning first, though. But just for my own understanding, would this other post also explain why it is true if you do not go into $(1,1)$ but any other non-zero direction, if it is even true?- Going through it once, it seems so and we would just have some $c_1,c_2$, would you agree? $\endgroup$
    – Kung Yao
    Commented Jan 31, 2020 at 23:26
  • $\begingroup$ @KungYao Yes, you'll still have the product structure though it will be $(c_1,c_2)$ and the diagonal elements will be different and have $v_1$ and $v_2$ in them, so the signs of the third central moments will flip together with the signs of $v_1,v_2$, i.e., with the sign of the $y$-parameter in that other post. $\endgroup$
    – fedja
    Commented Jan 31, 2020 at 23:31
  • $\begingroup$ I see, so what made my question trivial, as you say, is that the individual components factorize. So let us consider the measure with density $p(x) = e^{-\langle x-v, \Sigma^{-1} x-v \rangle - \vert x_1 \vert^4-\vert x_2 \vert^4}$ where $\Sigma$ is positive-definite. Then your approach suggests to study for every $z$: $\int \langle x-c, z \rangle^2 \langle x-c, v \rangle p(x) \ dx.$ Proceeding as in your other answer leads the condition that for all $a>0$ we have $\int_{\vert \langle x,z \rangle \vert \ge a} \langle x,v \rangle p(x+c) \ dx \ge 0.$ $\endgroup$
    – Kung Yao
    Commented Feb 1, 2020 at 1:39
  • $\begingroup$ So we can differentiate this thing with respect to $a$ and co-area formula yields a surface integral proportional to $-\int_{\vert \langle x,z \rangle \vert = \pm a} \langle x,v \rangle p(x+c) \ dx. $ This is, as far as I can tell very similar to what you were doing in the other answer. However, in the multi-dimensional case I don't think I can close the argument. Do you see how to do it?-Of course since you showed more than necessary in the other answer, this might be too much to ask for. Maybe the result is also wrong in the multi-dimensional case. Do you have any intution about this? $\endgroup$
    – Kung Yao
    Commented Feb 1, 2020 at 1:43
  • $\begingroup$ given your answer to the other post, I guess that is the problem you have been looking for in my post ;) I would be very curious to hear about the answer $\endgroup$
    – Kung Yao
    Commented Feb 1, 2020 at 1:48

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