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.

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

  • 1
    $\begingroup$ so you're asking for the distribution of $S=\sum_n\log w_n$, with $w_n$, $n=1,2,\ldots N$, an eigenvalue of $W$; the entire probability distribution $P(\{w_n\})$ is known, which allows you to calculate $P(S)$ for small $N$ by integration; asymptotic results for $N\gg 1$ can be readily obtained, but exact closed form expressions for $P(S)$ for arbitrary $N$ are not forthcoming; for the expectation value of $S$, see stats.stackexchange.com/questions/34165/… $\endgroup$ Commented Sep 28, 2016 at 13:41
  • 2
    $\begingroup$ One can factor $\log \det W$ as $\sum_{j=1}^n \log \mathrm{dist}(X_j, V_j)^2$, where $X_j$ is the $j^{th}$ row of $A$ and $V_j$ is the space spanned by the previous $j-1$ rows. Each of the distances squared is a chi-square random variable with n-j complex degrees of freedom (or 2(n-j) real degrees of freedom, if you prefer), and are jointly independent, so this in principle gives an exact formula. $\endgroup$
    – Terry Tao
    Commented Sep 28, 2016 at 17:03
  • $\begingroup$ There are also central limit theorems for this distribution, see e.g. stat.yale.edu/~hz68/Covariance-Determinant.pdf $\endgroup$
    – Terry Tao
    Commented Sep 28, 2016 at 17:05

1 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}];
 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.