In Gupta & Nagar's book *Matrix variate distributions*, the noncentral Beta type I(B) distribution with parameters $a$, $b$ and noncentrality parameter $\Theta$ is defined by $U={(S\_1+S\_2)}^{-\frac12}S\_1{(S\_1+S\_2)}^{-\frac12}$ where $S\_1 \sim W(2a, I, \Theta)$ is independent of $S\_2\sim W(2b, I)$ (Wishart distributions), and the noncentral Beta type II(B) distribution is defined by $V={(S\_2)}^{-\frac12}S\_1{(S\_2)}^{-\frac12}$ where $S\_1$ and $S\_2$ as above.


Then it is claimed that $V \sim U{(I-U)}^{-1}$. However my simulations show that this distributional equality does not hold. I've found that it holds when $\Theta = \textrm{diag}(\theta, \ldots, \theta)$. Otherwise I get $\det(V) \sim \det\bigl(U{(I-U)}^{-1}\bigr)$ and $\textrm{tr}(V) \sim \textrm{tr}\bigl(U{(I-U)}^{-1}\bigr)$ but not $V \sim U{(I-U)}^{-1}$.


Is it true that that $V \sim U{(I-U)}^{-1}$? That would mean I do something wrong in my simulations but I'm pretty sure my simulations are correct, and I'm rather wondering why the equality holds when $\Theta = \textrm{diag}(\theta, \ldots, \theta)$.


### Note


With my simulations again, I've found that $V \sim U{(I-U)}^{-1}$ if we define $U = S\_1^\frac12{(S\_1+S\_2)}^{-1}S\_1^\frac12$ and $V = S\_1^\frac12 S\_2^{-1}S\_1^\frac12$. These are other proposed definitions of the matrix Beta distributions.


Update
======


I think I misread. In fact the book is a bit confusing regarding this point. There are two Beta type I(B) distributions in the book: the one defined above and another one which was originally defined by Asoo. The book claims that $V \sim U{(I-U)}^{-1}$ when $U$ has the Asoo distribution, there's no such claim for the $U$ defined as above. So the real question is: why this equality holds when $\Theta = \textrm{diag}(\theta, \ldots, \theta)$? 



Noncentral matrix beta distributions of type I and II
The density of $U$ is (see Gupta & Nagar page 189)
$$
f(U) = C\_f {\det(U)}^{a - \frac12(p+1)} {\det(I\_p-U)}^{b - \frac12(p+1)} \\
\int\_{S>0} {\det(S)}^{a+b-\frac12(p+1)} 
\exp\left(-\mathrm{tr}\left(\frac{S}{2}\right)\right)
{}\_0\!F\_1\left(a, \frac{1}{4}\Theta\_1 S^\frac12 US^\frac12\right) \mathrm{d}S \\ 
= K\_f {\det(U)}^{a - \frac12(p+1)} {\det(I\_p-U)}^{b - \frac12(p+1)} \\
\int\_{S>0} {\det(S)}^{a+b-\frac12(p+1)} 
\exp\bigl(-\mathrm{tr}(S)\bigr)
{}\_0\!F\_1\left(a, \frac{1}{2}\Theta\_1 S^\frac12 US^\frac12\right) \mathrm{d}S
$$
(where $C\_f$ and $K\_f$ are constants).


The density of $V$ is (see Gupta & Nagar page 192)
$$
g(V) = C\_g {\det(V)}^{a - \frac12(p+1)} {\det(I+V)}^{\frac12 (a+b)} 
{}\_1\!F\_1\left(a+b, a, \frac12 \Theta\_1 V{(I+V)}^{-1}\right).
$$
Make the change of variables $U = V{(I\_p+V)}^{-1}$, so $V = U{(I\_p-U)}^{-1}$ and 
$I\_p + V = {(I\_p-U)}^{-1}$. 
The Jacobian is ${\det(I\_p -U)}^{-(p+1)}$. 
We find that the density of $U$ is 
$$
h(U) = C\_g {\det(U)}^{a - \frac12(p+1)} {\det(I\_p-U)}^{b - \frac12(p+1)} 
{}\_1\!F\_1\left(a+b, a, \frac12 \Theta\_1 U\right).
$$


The integral in $f(U)$ does not simplify to something nice. Hence the density $f(U)$ and $h(U)$ are not the same.


But when $\Theta\_1$ is scalar, by [integral involving hypergeometric function of matrix argument](https://mathoverflow.net/questions/287866/integral-involving-hypergeometric-function-of-matrix-argument), 
$$
f(U) = 
K\_f {\det(U)}^{a - \frac12(p+1)} {\det(I\_p-U)}^{b - \frac12(p+1)} \\
\int\_{S>0} {\det(S)}^{a+b-\frac12(p+1)} 
\exp\bigl(-\mathrm{tr}(S)\bigr)
{}\_0\!F\_1\left(a, \frac{1}{2}\Theta\_1 S U\right) \mathrm{d}S \\
 = K\_f\Gamma\_p(a+b) {\det(U)}^{a - \frac12(p+1)} {\det(I\_p-U)}^{b - \frac12(p+1)} {}\_1\!F\_1\left(a+b, a, \frac12 \Theta\_1 U\right),
$$
the same as $h(U)$.


