Suppose $A$ is a real $N \times P$ matrix, $P \geq N$, with entries drawn independently according to $A_{ij} \sim \mathcal{N}(0,1)$. Then $W = A \, A^\top$ is a member of the real Wishart ensemble. What is the distribution of $\det W$? I'm particularly interested in the large $N$ behaviour. There might be a qualitative distinction between cases when $N / P(N) \rightarrow c$, with $c = 1$ or $c < 1$. In that case I'd be more interested in the former. Thanks!
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2 Answers
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The Distribution of the Determinant of a Complex Wishart Distributed Matrix proves that the determinant is distributed as the product of independent random variables with a chi-squared distribution, $${\rm det}\,W\simeq\chi^2_P\chi^2_{P-1}\cdots\chi^2_{P-N+1}.$$ (The title of the paper refers to the complex case, but the real ensemble is also considered towards the end.)
Results for the large-$N$ asymptotics can be found here. If we take $N,P\rightarrow\infty$ at fixed ratio $c=N/P<1$, the asymptotics is $$\frac{\log{\rm det}\,(W/P)-\sum_{k=1}^N\log(1-k/P)}{\sqrt{-2\log(1-c)}}\rightarrow z,$$ with $z$ normally distributed. For $c\rightarrow 0$ this means that $(2c)^{-1/2}\log{\rm det}\,(W/P)$ tends to a normal distribution. At the other extreme, for $c\rightarrow 1$ one has $$\frac{\log{\rm det}\,(W/P)+P+\tfrac{1}{2}\log(P/2\pi)}{\sqrt{2\log P}}\rightarrow z,$$
answered Apr 25, 2017 at 14:36
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$\begingroup$ Thanks for your answer. I'm having trouble understanding the large $N$,$P$ behaviour, though... What do you expect as a typical value for $\det W$ as $N,P \rightarrow \infty$? When I do some numerics (and take $P = N +$few), I seem to find that $N!$ is in the ball park of the mean of $\det W$. However if I draw random numbers according to your asymptotic formula, I seem to get much smaller numbers... Is it correct to assume $\det W$ is supposed to be drawn from $\exp [\mathcal{N}(0,\sqrt{2 \log P})]$ (where the second argument of $\mathcal N$ is the standard deviation)? $\endgroup$– LatraceCommented Apr 25, 2017 at 15:28
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$\begingroup$ I'm guessing there's some $N!$ or $P!$ missing to make this formula correct. $\endgroup$– LatraceCommented Apr 25, 2017 at 15:36
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$\begingroup$ Thanks again for your answer! I saw I made a crucial mistake in the formulation of the problem... The case of most interest is $c = 1$, not $c = 0$. Sorry about that. $\endgroup$– LatraceCommented Apr 25, 2017 at 16:24
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$\begingroup$ I'm sorry but I still cannot understand the result you write. Suppose $P = 2N$. Then your formula tells us we should expect $\det W$ to be of order 1 (or less), right? This doesn't make sense at all... Formula (11) in the paper you cite seems closer to the truth, but there $P = N$. $\endgroup$– LatraceCommented Apr 25, 2017 at 16:36
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$\begingroup$ By the way, from basic expectations of the eigenvalues of real Wishart matrices one expects an $N!$ growth of $\det W$. The case $c<1$ is clearest, because the smallest eigenvalue of $W$ has an expected value of order $N$. For $P = N + \nu$ the situation may be different, but I still expect (super) exponential growth of $\det W$ with $N$. $\endgroup$– LatraceCommented Apr 25, 2017 at 16:54
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I've been looking at this question for a while. The accepted answer is correct but I recommend An Introduction to Multivariate Statistical Analysis by Theodore W. Anderson pages 250-254. This has an alternate proof which I think is more intuitive.
