Distribution of log of determinant Wishart matrix

Question

Are there any exact or approximate distributions for the log of the determinant of Wishart matrix?

$W=A^\dagger A$

where $A$ is an i.i.d complex Gaussian random matrix with zero-mean and variance of 1.

$\mathbb{P}(\log\det W)$?

Answer 1

There is a neat closed form expression for the characteristic function of this quantity. Let me derive this. Again let me fix notation $C:= X \cdot X^T$ where $X$ is an $n\times T$ real random matrix with i.i.d standardized Gaussian distributed entries.

\begin{align} &\chi_{\log(\det(C))}(k) := E\left[ e^{\imath k \log\det(C)} \right] \tag1\\[6pt] = {} & N_{n,T} \int\limits_{{\mathbb R}_+^n} e^{\imath k \sum_{j=1}^n \log(\lambda_j)} \cdot \prod_{1 \le i < j \le n} | \lambda_i - \lambda_j| \cdot \prod_{j=1}^n \lambda_j^{\frac{T-n-1}{2}} \cdot \exp\left( -\frac{1}{2} \sum\limits_{j=1}^n \lambda_j \right) \cdot \prod_{j=1}^n d \lambda_j \tag2\\[6pt] = {} & N_{n,T} \int\limits_{{\mathbb R}_+^n} \prod_{1 \le i < j \le n} | \lambda_i - \lambda_j| \cdot \prod_{j=1}^n \lambda_j^{\frac{T-n-1}{2}+ \imath k} \cdot \exp\left( -\frac{1}{2} \sum_{j=1}^n \lambda_j \right) \cdot \prod\limits_{j=1}^n d \lambda_j \tag3\\[6pt] = {} & N_{n,T} \cdot \left( 2^{\frac{n T}{2}+\imath n k} \Gamma\left[\frac{n T}{2}+\imath n k \right]\right) \cdot \left( \frac{\pi ^{\frac{1-n}{2}} n! \left(\prod _{j=0}^{n-3} \Gamma \left(\frac{n-j}{2}\right)\right) \prod_{j=0}^{n-1} \Gamma \left(\frac{1}{2} (-j+2 i k+T)\right)}{\Gamma \left(\frac{1}{2} n (2 i k+T)\right)}\right) \tag4 \\[6pt] = {} &2^{\imath n k} \cdot \prod\limits_{j=0}^{n-1} \frac{\Gamma(\frac{T-j}{2} + \imath k)}{\Gamma(\frac{T-j}{2})} \tag5 \end{align}

In $(2)$ we used the expression for the joint distribution of the eigenvalues in the Wishart ensemble see here for example. Here the normalization constant reads $N_{n,T}:= \frac{(\sqrt{\pi})^{n-1} }{2^{(n T)/2} (\prod\limits_{j=0}^{n-1} \Gamma[\frac{T-j}{2}]) \cdot (\prod\limits_{j=0}^{n-3} \Gamma[\frac{n-j}{2}]) \cdot n! }$. In $(3)$ we simplified the expression, i.e. we absorbed absorbed the first exponential on the left into the product of the powers of the eigenvalues. In $(4)$ we used a parametrisation $(\lambda_i)_{i=1}^n \rightarrow (z(\lambda_i-\lambda_{i-1}))_{i=1}^n$ subject to $\lambda_0=$ and $\lambda_n= 1$ and we split the multivariate integral into integration over a product of two integrals the first one being $z \in {\mathbb R}_+$ and the second one being over a unit simplex. Here the first and the second expression in parentheses corresponds to the first and the second integral respectively. The first integral was trivial whereas the second one uses the result $(ii)$ from here with $T \rightarrow T + 2 \imath k$. As a sanity check we also verified this step numerically below:

In[2911]:= n = RandomInteger[{2, 3}]; T = 
 RandomInteger[{n + 1, 20}]; k = RandomReal[{0, 2}];
NIntegrate[
 Product[Abs[DD[l, xi1] - DD[l, xi2]], {xi1, 1, n}, {xi2, xi1 + 1, 
    n}] Product[(DD[l, xi1])^((T - n - 1)/2 + I k), {xi1, 1, n}], 
 Evaluate[Sequence @@ 
   Table[{l[xi1], If[xi1 == 1, 0, l[xi1 - 1]], 1}, {xi1, 1, n - 1}]]]
(n! (\!\(
\*UnderoverscriptBox[\(\[Product]\), \(j = 0\), \(\(-3\) + n\)]\(Gamma[
\*FractionBox[\(1\), \(2\)]\ \((\(-j\) + n)\)]\)\)) \!\(
\*UnderoverscriptBox[\(\[Product]\), \(j = 0\), \(\(-1\) + n\)]\(Gamma[
\*FractionBox[\(1\), \(2\)]\ \((\(-j\) + T + 2\ I\ k)\)]\)\))/(\[Pi]^(
 1/2 (-1 + n)) Gamma[(n (T + 2 I k))/2])

Out[2912]= -6.19729*10^-7 - 5.58825*10^-7 I

Out[2913]= -6.19729*10^-7 - 5.58825*10^-7 I

Finally in $(5)$ we just simplified the result. Now we check the normalization and the first moment:

\begin{eqnarray} \chi_{\log(\det(C))}(0) &=& 1 \\ \left.\frac{d}{d \imath k} \chi_{\log(\det(C))}(k)\right|_{k=0} &=& n \log(2) + \sum\limits_{j=0}^{n-1} \psi^{(0)} \left( \frac{T-j}{2}\right) \end{eqnarray} as expected.