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Let $X \in \mathbb{R}^{m \times n}$ and $Y \in \mathbb{R}^{n \times k} $ be two independent Gaussian random matrices, i.e., with entries independently sampled from $\mathcal{N}(0,1)$ (a normal distribution with zero mean and unit variance). We define the positive orthant probability $$p_+(m,n,k) \triangleq P(\forall i,j: (XY)_{ij} >0).$$ Question: How does $\log_2 p_+(m,n,k)$ behave asymptotically, to leading order, in the limit $ n \rightarrow \infty$ and $\alpha m = \beta k = n $, for some positive constants $\alpha$ and $\beta$?

Note that if $n\rightarrow \infty$ but $m$ and $k$ remain fixed, then from the central limit theorem $\frac{1}{\sqrt{n}}XY$ becomes a Gaussian matrix with independent entries, and then we get $\log_2 p_+(m,n,k) = -mk$. However, I wish to know if how this asymptotic behavior persists when $m$ and $k$ are of similar magnitude as $n$. Specifically, what is the largest positive constant $r>1$ for which we have

$$\lim_{\underset{\alpha m = \beta k = n}{n\rightarrow\infty}} \frac{1}{n^{r}} \log_2 p_+(m,n,k) < 0 \,.$$

I guess the answer should depend on $\alpha$ and $\beta$. Note I relaxed the conditions in this question from the previous version, since I didn't get an answer so far.

Thanks in advance!

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    $\begingroup$ It certainly gets off quite a bit even if $k=2$ and $m$ is comparable to $n$. Indeed, in that case we just need to take the expectation of the $m$-th power of the area of the intersection of two random half-spheres on a sphere of total area $1$. We have a chance like $e^{-nc^2}$ to have it like $(1+c)/4$, which gives an extra exponential factor $e^{-nc^2+cm}$, so $c$ can be of constant size. This effect only magnifies when $k$ goes up. So, how precise do you want to be (the exact asymptotics may be quite a headache, so you'd better let us know what is the minimum you'll agree to settle at). $\endgroup$
    – fedja
    Sep 16, 2016 at 2:05
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    $\begingroup$ I didn't follow your specific example (e.g., what does $c$ represent?), but I think I understand the general issue. I've added more information on the required precision of the solution. I hope this is sufficient. $\endgroup$
    – Daniel Soudry
    Sep 16, 2016 at 8:08
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    $\begingroup$ There is some invariance in the problem, e.g. if I apply the kxk matrix EI which is the identity except EI_{11} = -1 I flip the signs in the first row, however EI X has the same distribution as X and therefore the prob of being in the positive quadrant is the same as that only the first row is negative. $\endgroup$
    – user83457
    Sep 16, 2016 at 11:23
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    $\begingroup$ Yes, you can flip the sign of any row or column. However, this gives only $2^{k+m}$ options --- this allows us to bound $p_+$ above by $2^{-k-m}$, which is much higher then $2^{-km}$, the behavior I'm interested in. $\endgroup$
    – Daniel Soudry
    Sep 16, 2016 at 11:31
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    $\begingroup$ $2^{-km}$ is certainly out of question. You can only hope for $2^{-n^r}$ with some $3/2\le r\le 5/3$ but the precision you requested is much higher than given by these trivial bounds so I'm not posting them yet... $\endgroup$
    – fedja
    Sep 16, 2016 at 13:19

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