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Let $(J_{ij})$ be an $n \times n$ random matrix with i.i.d Gaussian centered coefficients with $\displaystyle \mathbb{E}[J_{ij}^2] = \frac{\sigma^2}{n}$.

Let the random variable $A_n(\sigma)$ defined as the number of real solutions in $\mathbb{R}^n$ of : $$-x_i + \sum_{j=1}^n J_{ij} \phi(x_j) = 0\mbox{ for all }1\leq i \leq n$$ where $\phi(x) = \arctan(x)$.

The question is : what is the law of $A_n(\sigma)$ ? In particular, its expectation ?

I know how to solve "by hand" the case n=1, and n=2, but then it becomes really painful.

Any idea?

Thank you!

Edit: I have a conjecture but I do not know if it is true:

$$\lim_{n \to \infty} \frac{1}{n}\log \mathbb{E}[A_n(\sigma)] = C(\sigma)$$ with $C(\sigma)=0$ for $\sigma <1$ and $C(\sigma)=O((\sigma-1)^2)$ for $\sigma \to 1^+$.

What do you think of this?

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Have you really solved it for $n=1$ ? What is your result ? –  leonbloy Jul 8 '11 at 0:43
    
In your equation, must this hold for all $i$ independently, or are implicitly summing over all $i$? –  A Walker Nov 18 '11 at 15:14
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@AWalker: I assumed that it must hold for each $i$. The equation could be written in vectorial form ${\bf x} = {\bf J} atan({\bf x})$ where $atan(\cdot)$ is applyied to each element. Seems difficult. –  leonbloy Nov 18 '11 at 16:23

1 Answer 1

I'll make an attempt. I think this reduces your problem to an integral, but I don't know if it can be found in closed form.

Let $d\mu(J)$ be the probability density on the matrices $J$; the expected value for the number of solutions to the equation will be given by $$ \int d\mu(J) \int d\vec{x}\ [\vec{x}=J\vec{\phi}], $$ where $\vec{\phi} = \arctan\vec{x}$. Change the order of integration to get $$ \int d\vec{x}\int d\mu(J)\ [\vec{x}=J\vec{\phi}], $$ then we can find the inner integral. The outer integral is taken over all real vectors $\vec{x}\in\mathbb{R}^n$.

The condition $\vec{x}=J\vec{\phi}$ consists of $n$ independent conditions, each over $n$ entries of $J$. Consider one row of that equation: $$ \int d\mu(\vec{J}_i) [x_i = \vec{J}_i \cdot\vec{\phi}]. $$ Because $J_{ij}$ are i.i.d. Gaussian, with mean 0, we can assume without loss of generality that $\vec{\phi}$ is pointing along, say, the axis of $J_{i1}$, which removes $J_{i2},\ldots,J_{in}$ from the integrand, and implies that $$ \int d\mu(\vec{J}_i) \left[x_i = J_{i1} \|\vec{\phi}\|\right] = (2\pi\sigma^2)^{-1/2}\exp\left (-\frac{x_i^2/\|\vec{\phi}\|^2}{2\sigma^2} \right). $$

Now, $$\int d\mu(J) [1_{\vec{x}=J\vec{\phi}}] = (2\pi\sigma^2)^{-n/2} \exp\left(-\frac{\|\vec{x}\|^2/\|\vec{\phi}\|^2}{2\sigma^2}\right), $$ and so the answer is $$ I = (2\pi\sigma^2)^{-n/2} \int d\vec{x}\ \exp\left(-\frac1{2\sigma^2}\frac{\|\vec{x}\|^2}{\|\arctan \vec{x}\|^2} \right). $$

Now, take the fact that $|\arctan x|\leq |x|$, and approximate the integral by $$ I \leq (2\pi\sigma^2)^{-n/2}e^{-1/(2\sigma^2)}\pi^n + \cdots, $$ where $\pi^n$ is the volume of the cube $[-\pi/2,\pi/2]^n$, and I'm assuming (though this seems a reasonable guess) that the contribution from outside the cube is asymptotically negligible. Then $$ \lim_{n\to\infty} \frac1n \log I < \frac12\log(\pi/2) - \log\sigma. $$ Unfortunately, I'm not at all certain that this particular bound is either tight or even correct.

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Dear Kirill, thank you a lot for your attempt. It looks like you are trying to apply Kac-Rice formula. But unfortunately I think your formula is missing the absolute value of the Jacobian determinant. Do you agree, or am I missing something ? –  mellow Feb 7 '12 at 15:26

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