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Let $M \in \mathbb{R}^{m,n}$. The entries are in {$0, 1$} (with the value $1$ having probability $p$, and the value $0$ having probability $(1-p)$).

What is the expected rank of $M$?

Follow-up: how does this change if the entries are in {$-1, 1$} instead of {$0, 1$}?

There is already a post on a similar question here: Expected rank of a random binary matrix?

But that post assumes $p=0.5$, and I am more curious about non-uniform random entries, e.g. $p=0.75$.

I dived into the paper the post linked, which listed many other formulas, but none of them assumed the entries are drawn from a Bernoulli distribution where $p \neq 0.5$.

user3667125
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  • Have you attempted to do computer simulations ? – Jean Marie May 02 '24 at 11:59
  • I have not, I was hoping there was a way to determine this analytically. Do you think simulations are the only tractable way to find out? – user3667125 May 02 '24 at 12:04
  • Yes, they are a "tractable" way 1) to build conjectures 2) to "verify" them... They have become of common use among probabilists. – Jean Marie May 02 '24 at 12:06
  • Intuitively, the average rank should drop to $1$ when $p \to 1$ or $p \to 0$. – Jean Marie May 02 '24 at 12:12
  • I just saw in the referenced Stack Exchange question that Fields medallist Jean Bourgain has a 50 pages paper on the issue $p=q=1/2$. Therefore, I conclude (hastily ?) that your question hasn't an easy-to-establish "analytical" solution. – Jean Marie May 02 '24 at 14:59
  • Thank you @JeanMarie that is a very good point you raised. I will simply run the simulations myself, post the conjectures, and probably accept my own results as the answer to this post. – user3667125 May 02 '24 at 20:58

1 Answers1

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Upon reading the paper I linked in the original post a lot closer, I realize that the author does in fact go over the general case where the matrix entries draw from a Bernoulli distribution with probability $p$.

According to the paper, the probability that the matrix $M$ will be singular will have the upper bound:

$$P(\text{M is singular}) \le \left(\sqrt{\hat{p}} + o(1) \right)^n$$

where $\hat{p} = \max(p, 1-p)$ and $o(1)$ is some negligible constant.

As a result, the probability that $M$ is non-singular is

$$P(\text{M is non-singular}) \ge 1 - \left(\sqrt{\hat{p}} + o(1) \right)^n$$

Finding the exact expected rank seems like it will be complex and require approximation via simulations. However, for $\hat{p}$ close enough to 0.5 (e.g. $0.25 \le \hat{p} \le 0.75$), and for large enough $n$ (e.g. $n \ge 100$), it seems like the expected rank of $M$ will be very close to $\min(m,n)$ as $M$ is highly likely ($ > 0.95$) to be non-singular. And as $p$ approaches closer to $0$ or $1$ (which implies $\hat{p}$ approaches closer to $1$), the probability and therefore expected rank drops off drastically.

Note that when $p=0$, $rank(M) = 0$, and when $p=1$, $rank(M)=1$.

user3667125
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