# Product of two random Gaussian matrices - orthant probability

In my studies of random matrices, I recently came across this challenge:

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 this asymptotic behavior persists when $m$ and $k$ are of similar magnitude as $n$. In other words, for what positive constants $a, b$ and $c$, do we have

$$\lim_{\underset{\alpha m = \beta k = n}{n\rightarrow\infty}} \frac{1}{m ^{a }k^{b}} \log_2 p_+(m,n,k) = -c \,.$$

Is it $a=b=c=1$, like in the case of fixed $m$ and $k$, or something else? I guess the answer should depend on $\alpha$ and $\beta$.

If no exact answer will be given to my question, I'll award the answer to whoever supplies the tightest upper bound (non-trivial, in which $a+b>1$).