Expected value of the log-determinant of a Wishart matrix

Question

Let $\Lambda \sim \mathcal W_D(\nu, \Psi)$, i.e. distributed according to a $D \times D$ dimensional Wishart distribution with mean $\nu \Psi$ and degrees of freedom $\nu$. I would like an expression for $E(\log |\Lambda|)$ where $|\Lambda|$ is the determinant.

I've google'd a bit for the answer to this and have gotten some conflicting information. This paper explicitly states that $$ E(\log|\Lambda|) = D \log 2 + \log |\Psi| + \sum_{i = 1} ^ D \psi\left(\frac{\nu - i + 1} 2\right) $$ where $\psi(\cdot)$ denotes the digamma function $\frac d {dx} \log \Gamma(x)$; the paper does not give a source for this fact as far as I can tell. This is also the formula used on the wikipedia page for the Wishart, which sites Bishop's Pattern Recognition text.

On the other hand, google turned up this discussion with a linked paper that states that $$ \nu^D \frac{|\Lambda|}{|\Psi|} \sim \chi^2_\nu \chi^2_{\nu - 1} \cdots\chi^2_{\nu - D + 1}. \qquad (\dagger) $$ They conclude by stating that $$ E(\log | \Lambda|) = D \log 2 - D \log \nu + \log |\Psi| + \sum_{i = 1} ^ D \psi\left(\frac{\nu - i + 1} 2\right) $$ which is derived using the fact that $E(\log \chi^2_\nu) = \log(2) + \psi(\nu /2)$. I checked this calculation starting from $(\dagger)$ and it seems okay, but we have an extra $-D\log\nu$.

Answer 1

As I was getting ready to post this, I was able to answer my own question. In accordance with general StackExchange etiquette I've decided to post it anyways in hopes that someone else who runs into this problem might find this in the future, possibly after running into the same issues with sources that I did. I've decided to answer it immediately so that no one wastes time on it since the solution isn't interesting.

$(\dagger)$ is wrong, because the paper linked to in the discussion was using a different parametrization of the Wishart; this wasn't noticed by the discussants. What we should actually have is $$ \frac{|\Lambda|}{|\Psi|} \sim \chi^2_\nu \chi^2_{\nu - 1} \cdots\chi^2_{\nu - D + 1}. \qquad (\dagger\dagger) $$ After this correction, the two formulas lead to the same answer.

At any rate, I think think $(\dagger\dagger)$ is an interesting relationship.

EDIT:

Following probabilisticlogic's advice we can write $\Lambda \stackrel d = \Psi^{1/2} L L^T \Psi^{1/2}$ where lower triangular $L$ has $N(0, 1)$ elements off the diagonal and $\sqrt{\chi^2_{\nu - i + 1}}, (i = 1, ..., D)$ elements on the diagonal. Taking the determinent of both sides gives $(\dagger\dagger)$ immediately.