Talk:Wishart distribution
From Wikipedia, the free encyclopedia
 |
This article is within the scope of WikiProject Statistics, which collaborates to improve Wikipedia's coverage of statistics. If you would like to participate, please visit the project page. |
![f_{\mathbf W}(w)=
\frac{
\left|w\right|^{(n-p-1)/2}
\exp\left[ - {\rm trace}({\mathbf V}^{-1}w/2 )\right]
}{
2^{np/2}\left|{\mathbf V}\right|^{n/2}\Gamma_p(n/2)
}](../../../../math/2/2/8/2282948f1300222d990f8258ed687f34.png)
shoud be
![f_{\mathbf W}(w)=
\frac{
\left|w\right|^{(n-p-1)/2}
\exp\left[ - \frac{1}{2}{\rm trace}({\mathbf V}^{-1}w )\right]
}{
2^{np/2}\left|{\mathbf V}\right|^{n/2}\Gamma_p(n/2)
}](../../../../math/1/2/8/128f9fbca0e3f0deb8524d87a633dd76.png)
?
- Well, the two say exactly the same thing. Perhaps there is some reason why one way of saying it is preferable in a given context, but certainly if either is correct then so is the other. Michael Hardy 22:37, 18 March 2007 (UTC)
- Thanks for comments. I confused that trace(A / x) = 1 / xptrace(A). —The preceding unsigned comment was added by 150.29.217.146 (talk) 04:09, 19 March 2007 (UTC).
I agree, but if one wants to be mathematically strict V/2 is not defined, where V is matrix. Matric algebra defines the product of a non-zero scalar with a matrix and thus (1/2) * V is well defined, but V/2 is not. In this sense V/2 is rather a convenient convention. So from the above equations, the second (using 1/2 outside of the trace) seems more mathematically sound. having said this I have used many times the first equation, simply because of convenience. KT
