If $A$ is chosen uniformly at random over all possible $m \times n$ (0,1)-matrices, what is the expected size of the absolute value of the determinant of $AA^T$. We can assume $m < n$ and all arithmetic is over the reals.


In [Expected determinant of a random NxN matrix](https://mathoverflow.net/questions/13008/expected-determinant-of-a-random-nxn-matrix) the expected size of the determinant for random (0,1) matrices was given as $\sqrt{(n+1)!}/ 2^n$ by David Speyer.




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As pointed out by Richard Stanley, my question is incorrect in its reference to the related problem solved by David Speyer (and others). I should have said that when $A$ is a random $n$ by $n$ (0,1)-matrix then $\sqrt{E((\det A)^2)} = \sqrt{(n+1)!}/ 2^n$



Expected size of determinant of $AA^T$ for non-square random $A$
By the Cauchy-Binet theorem, $\det AA^T=\sum (\det B)^2$, where $B$
ranges over all $m\times m$ submatrices of $A$. The expected value of
$(\det B)^2$ is $(m+1)!/4^m$, so the expected value of $\det AA^T$ is
${n\choose m}(m+1)!/4^m$. This is also the expected value of $|\det
AA^T|$ since $\det AA^T\geq 0$.


Incidentally, it is misleading to say that the expected size of the
determinant of a random $n\times n$ $(0,1)$-matrix $A$ is
$\sqrt{(n+1)!}/2^n$. It is true that $(n+1)!/4^n$ is the expected
value of $(\det A)^2$, but this doesn't mean that $\sqrt{(n+1)!}/2^n$
is the expected value of $|\det A|$.


