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Does anyone know anything about the determinant of a random $n\times n$ row stochastic matrix? What I have in mind is that the rows are independently selected from the uniform distribution on the unit $(n-1)$-dimensional simplex: $x_1+\cdots+x_n=1$. I'm interested in upper (and lower) bounds on the expected absolute value of the determinant as a function of $n$.

Thanks for any references! I found something due to Nguyen for the random doubly stochastic matrices, but didn't see anything for the easier (?) singly stochastic case.

EDIT thanks to Igor's answer below, I have an answer to the original question that seems likely. The paper referenced by Igor gives the empirical spectral distribution of $A_n$ to be uniform on the disk of radius $1/\sqrt n$. This suggests that the determinant should be something like $ (ne)^{-n/2}$. A result somewhat like this was proved for matrices with iid entries by Nguyen and Vu.

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  • $\begingroup$ Did you find any other paper or write one yourself on distribution of such determinants? $\endgroup$ – Maesumi Nov 20 '17 at 14:17
  • $\begingroup$ Could you please list the title of the paper by Nguyen and Vu you referenced? $\endgroup$ – Hans Jan 30 '18 at 22:28
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The paper to look at is Bordenave, Caputo, Chafai, which is cited in Nguen's paper.

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  • $\begingroup$ Thanks Igor - It seems that that model is a bit different to the one I'm considering: they are taking rows formed by taking iid non-negative entries and then normalizing (so that the $i$th entry is $X_i/(X_1+\ldots+X_n)$), which seems to be somewhat different (even if the $X$'s are uniform) than uniformly sampling over the simplex (e.g. conditional on $X_1+\ldots+X_n=n-0.5$, all of the $X_i$'s are above 0.5). From the introduction to the paper, the spirit seems to be that small changes to the distribution make small changes to the spectrum so presumably this paper gives the right distribution. $\endgroup$ – Anthony Quas Dec 2 '16 at 1:00
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    $\begingroup$ One obtains the uniform distribution on the simplex by taking $X_i$ to be iid exponential variables. $\endgroup$ – Nick Cook Dec 2 '16 at 1:52
  • $\begingroup$ Oh yes. I see that. $\endgroup$ – Anthony Quas Dec 2 '16 at 2:20

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