Wishart 행렬의 로그 결정자의 예상 값

하자 , 즉,이에 따라 배포 차원 Wishart의 평균과 분산 자유와도 . 대한 표현식을 원합니다. 여기서 결정자입니다. $\Lambda \sim \mathcal W_D(\nu, \Psi)$ $D \times D$ $\nu \Psi$ $\nu$ $E(\log |\Lambda|)$ $|\Lambda|$

나는 이것에 대한 답변을 약간 봤으며 충돌하는 정보를 얻었습니다. 이 백서 에서는 여기서디 감마 함수를 나타내고,

E (\log | Λ |) = D \log 2 + \log | Ψ | + \sum_{i = 1}^{D} ψ (\frac{ν - i + 1}{2})

$E(\log|\Lambda|) = D \log 2 + \log |\Psi| + \sum_{i = 1} ^ D \psi\left(\frac{\nu - i + 1} 2\right)$

ψ (\cdot)

$\psi(\cdot)$

\frac{d}{d x} \log Γ (x)

$\frac d {dx} \log \Gamma(x)$ ; 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

ν^{D} \frac{| Λ |}{| Ψ |} \sim χ_{ν}^{2} χ_{ν - 1}^{2} \dots χ_{ν - D + 1}^{2} . (†)

$\nu^D \frac{|\Lambda|}{|\Psi|} \sim \chi^2_\nu \chi^2_{\nu - 1} \cdots\chi^2_{\nu - D + 1}. \qquad (\dagger)$

E (\log | Λ |) = D \log 2 - D \log ν + \log | Ψ | + \sum_{i = 1}^{D} ψ (\frac{ν - i + 1}{2})

$E(\log | \Lambda|) = D \log 2 - D \log \nu + \log |\Psi| + \sum_{i = 1} ^ D \psi\left(\frac{\nu - i + 1} 2\right)$ which is derived using the fact that

E (\log χ_{ν}^{2}) = \log (2) + ψ (ν / 2)

$E(\log \chi^2_\nu) = \log(2) + \psi(\nu /2)$ . I checked this calculation starting from

(†)

$(\dagger)$ and it seems okay, but we have an extra

- D \log ν

$-D\log\nu$ .

distributions wishart

— guy
소스

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{| Λ |}{| Ψ |} \sim χ_{ν}^{2} χ_{ν - 1}^{2} \dots χ_{ν - D + 1}^{2} . († †)

$\frac{|\Lambda|}{|\Psi|} \sim \chi^2_\nu \chi^2_{\nu - 1} \cdots\chi^2_{\nu - D + 1}. \qquad (\dagger\dagger)$ 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.

— guy
소스

I like the Cholesky version better - you have square root of chi-square on the diagonal and standard normal on the lower triangle.

— probabilityislogic

@probabilityislogic Thanks for the tip! Remembering it like that seems easier and more useful.

— guy

Hey I'm trying to derive the expectation of the log Wishart (stated in Bishop's book), that looks complicated, did you find any source to derive the result?

— avocado