$\newcommand{\etr}{\textrm{etr}}$

Two definitions of the matrix variate Beta type I distribution were proposed. We will denote them by $\mathcal{B}I_p^{(1)}(a,b,\Theta_1,\Theta_2)$ and $\mathcal{B}I_p^{(2)}(a,b,\Theta_1,\Theta_2)$, where $\Theta_1$ and $\Theta_2$ are the noncentrality parameters. Take two independent Wishart random matrices $W_1 \sim \mathcal{W}_p(2a, I_p, \Theta_1)$ and $W_2 \sim \mathcal{W}_p(2b, I_p, \Theta_2)$. Then $\mathcal{B}I_p^{(1)}(a,b,\Theta_1,\Theta_2)$ is the distribution of $$ U_1 = {(W_1+W_2)}^{-\frac12}W_1{(W_1+W_2)}^{-\frac12}, $$ while $\mathcal{B}I_p^{(2)}(a,b,\Theta_1,\Theta_2)$ is the distribution of $$ U_2 = W_1^\frac12{(W_1+W_2)}^{-1}W_1^\frac12. $$ The condition $a+b > \frac12(p-1)$ is required in order for $W_1+W_2$ to be invertible.

In the central case, i.e. when both $\Theta_1$ and $\Theta_2$ are the null matrices, these two distributions are the same. More generally, as we will see, they are the same when $\Theta_1$ and $\Theta_2$ are scalar.

Similarly, two definitions of the matrix variate Beta type II distribution were proposed. We will denote them by $\mathcal{B}II_p^{(1)}(a,b,\Theta_1,\Theta_2)$ and $\mathcal{B}II_p^{(2)}(a,b,\Theta_1,\Theta_2)$. The first one is the distribution of $$ V_1 = W_2^{-\frac12} W_1 W_2^{-\frac12}, $$ while the second one is the distribution of $$ V_2 = W_1^\frac12 {W_2}^{-1} W_1^\frac12. $$ The condition $b > \frac12(p-1)$ is required in order for $W_2$ to be invertible.

Similarly to the type I, these two distributions are the same in the central case, and more generally when $\Theta_1$ and $\Theta_2$ are scalar.

Under the second definition, the Beta type I distribution is related to the Beta type II distribution by $U_2 \sim V_2{(I_p+V_2)}^{-1}$.

Hypergeometric matrix function

The densities of the matrix Beta distributions involve the hypergeometric function of matrix argument ${}_0!F_1$. We will use the property ${}_0!F_1(\alpha, AB)={}_0!F_1(\alpha, BA)$ (to simplify the densities when $\Theta_1$ or $\Theta_2$ are scalar).

Results

$\mathcal{B}I_p^{(1)}(a, b, \Theta_1, \Theta_2)$

Recall that $$ U_1 = {(W_1+W_2)}^{-\frac12} W_1 {(W_1+W_2)}^{-\frac12}. $$ It is clear from this definition that $$ I_p - U_1 \sim \mathcal{B}I_p^{(1)}(b, a, \Theta_2, \Theta_1). $$ The density of $U_1$ is $$ \begin{aligned} \mathcal{B}I_p^{(1)}(U \mid a, b, \Theta_1, \Theta_2) \propto \, & {\det(U)}^{a-\frac12(p+1)} {\det(I_p-U)}^{b-\frac12(p+1)} \ & \int_{S>0} \etr\left(-S\right) {\det(S)}^{a+b-\frac12(p+1)} {}_0!F_1\left(a, \frac{1}{2}\Theta_1S^{\frac12} U S^\frac12\right) {}_0!F_1\left(b, \frac{1}{2}\Theta_2S^{\frac12}(I_p-U)S^\frac12\right) \mathrm{d}S. \end{aligned} $$

If $\Theta_1$ and $\Theta_2$ are scalar: $$ \begin{aligned} \mathcal{B}I_p^{(1)}(U \mid a, b, \theta_1 I_p, \theta_2 I_p) \propto\, & {\det(U)}^{a-\frac12(p+1)} {\det(I_p-U)}^{b-\frac12(p+1)} \ & \int_{S>0} \etr\left(-S\right) {\det(S)}^{a+b-\frac12(p+1)} {}_0!F_1\left(a, \frac{\theta_1}{2}SU\right) {}_0!F_1\left(b, \frac{\theta_2}{2}S(I_p-U)\right) \mathrm{d}S. \end{aligned} $$

$\mathcal{B}I_p^{(2)}(a, b, \Theta_1, \Theta_2)$

Recall that $$ U_2 = W_1^\frac12{(W_1+W_2)}^{-1}W_1^\frac12. $$

The density of $U_2$ is $$ \begin{aligned} \mathcal{B}I_p^{(2)}(U \mid a, b, \Theta_1, \Theta_2) \propto \, & {\det(U)}^{-b-\frac12(p+1)} {\det(I_p-U)}^{b-\frac12(p+1)} \ & \int_{S>0} \etr\left(-S U^{-1}\right) {\det(S)}^{a+b-\frac12(p+1)} {}_0!F_1\left(a, \frac{1}{2}\Theta_1S\right) {}_0!F_1\left(b, \frac{1}{2}\Theta_2 S^\frac12 U^{-1}(I_p-U)S^{\frac12}\right)\mathrm{d}S. \end{aligned} $$

If $\Theta_1$ and $\Theta_2$ are scalar, it is equal to the density of $\mathcal{B}I_p^{(1)}(a, b, \Theta_1, \Theta_2)$.

$\mathcal{B}II_p^{(1)}(a,b,\Theta_1,\Theta_2)$

More generally, we give the density of $$ V_1 = {(W_2^{-\frac12})}' W_1 W_2^{-\frac12} $$ where $W_2^{\frac12}{(W_2^{\frac12})}' = W_2.$

This density is $$ \begin{aligned} \mathcal{B}II_p^{(1)}(V \mid a, b, \Theta_1, \Theta_2) \propto \, & {\det(V)}^{a-\frac12(p+1)} \ & \int_{S>0} \etr\bigl(-(I_p+V)S\bigr) {\det(S)}^{a+b-\frac12(p+1)} {}_0!F_1\left(a, \frac{1}{2}\Theta_1{(S^{\frac12})}' V S^{\frac12}\right) {}_0!F_1\left(b, \frac{1}{2}\Theta_2S\right) \mathrm{d}S. \end{aligned} $$

If $\Theta_1$ is scalar, the distribution does not depend on the choice of $W_2^\frac12$.

$\mathcal{B}II_p^{(2)}(a,b,\Theta_1,\Theta_2)$

Recall that $$ V_2 = W_1^{\frac12} W_2^{-1} {(W_1^{\frac12})}'. $$ It is clear from the definitions that $$ V_2^{-1} \sim \mathcal{B}II_p^{(1)}(b,a,\Theta_2,\Theta_1). $$

$$ \begin{aligned} \mathcal{B}II_p^{(2)}(V \mid a, b, \Theta_1, \Theta_2) \propto \, & {\det(V)}^{-b-\frac12(p+1)} \ & \int_{S >0} \etr\left(-S(I_p+V^{-1})\right) {\det(S)}^{a+b-\frac12(p+1)}
{}_0!F_1\left(a, \frac{1}{2}\Theta_1S\right) {}_0!F_1\left(b, \frac{1}{2}\Theta_2{(S^\frac12)}' V^{-1} S^\frac12\right) \mathrm{d}S. \end{aligned} $$

If $\Theta_2$ is scalar, the distribution does not depend on the choice of $W_1^\frac12$.

If $\Theta_1$ and $\Theta_2$ are scalar, this is the same distribution as $\mathcal{B}II_p^{(1)}(a,b,\Theta_1,\Theta_2)$.

If we take $W_1^{\frac12}$ the symmetric square root of $W_1$, then $V_2{(I_p+V_2)}^{-1} \sim \mathcal{B}I_2(a,b,\Theta_1,\Theta_2)$.

Proofs

$\mathcal{B}I_p^{(1)}(a,b,\Theta_1,\Theta_2)$

Recall that $\mathcal{B}I_p^{(1)}(a,b,\Theta_1,\Theta_2)$ is the distribution of $$ U = {(W_1+W_2)}^{-\frac12}W_1{(W_1+W_2)}^{-\frac12}. $$

The joint density of $S_1$ and $S_2$ is $$ C \, \etr\left(-\frac12 S_1\right)\etr\left(-\frac12 S_2\right) {\det(S_1)}^{a-\frac12(p+1)} {\det(S_2)}^{b-\frac12(p+1)} {}0!F_1\left(a, \frac{1}{4}\Theta_1S_1\right) {}_0!F_1\left(b, \frac{1}{4}\Theta_2S_2\right). $$ Using the transformation $S_1+S_2=S$ and $S_1 = S^{\frac12} U S^\frac12$, with Jacobian $J(S_1, S_2 \rightarrow U, S) = {\det(S)}^{\frac12(p+1)}$, we get the pdf of $(U,S)$ as $$ \begin{aligned} C\, & {\det(U)}^{a-\frac12(p+1)} {\det(I_p-U)}^{b-\frac12(p+1)} \ & \etr\left(-\frac12 S\right) {\det(S)}^{a+b-\frac12(p+1)} {}_0!F_1\left(a, \frac{1}{4}\Theta_1S^{\frac12} U S^\frac12\right) {}_0!F_1\left(b, \frac{1}{4}\Theta_2S^{\frac12}(I_p-U) S^\frac12\right). \end{aligned} $$ Thus the density of $U_1$ is $$ \begin{aligned} C\, & {\det(U)}^{a-\frac12(p+1)} {\det(I_p-U)}^{b-\frac12(p+1)} \ & \int{S>0} \etr\left(-\frac12 S\right) {\det(S)}^{a+b-\frac12(p+1)} {}0!F_1\left(a, \frac{1}{4}\Theta_1S^{\frac12}U{(S^\frac12)}'\right) {}_0!F_1\left(b, \frac{1}{4}\Theta_2S^{\frac12}(I_p-U)S^\frac12\right) \mathrm{d}S \ = C\,& {\det(U)}^{a-\frac12(p+1)} {\det(I_p-U)}^{b-\frac12(p+1)} \ & \int{S>0} \etr(-S) {\det(S)}^{a+b-\frac12(p+1)} {}_0!F_1\left(a, \frac{1}{2}\Theta_1S^{\frac12} U S^\frac12\right) {}_0!F_1\left(b, \frac{1}{2}\Theta_2S^{\frac12}(I_p-U)S^\frac12\right) \mathrm{d}S. \end{aligned} $$

$\mathcal{B}I_p^{(2)}(a,b,\Theta_1,\Theta_2)$

Recall that $\mathcal{B}I_p^{(2)}(a,b,\Theta_1,\Theta_2)$ is the distribution of $$ U = W_1^\frac12{(W_1+W_2)}^{-1}W_1^\frac12. $$

Using the transformation $S_1+S_2 = S_1^{\frac12}U^{-1}S_1^{\frac12}$, with Jacobian $J(S_1, S_2 \rightarrow U, S_1) = {\det(S_1)}^{\frac12(p+1)} {\det(U)}^{-(p+1)}$, we get the pdf of $(U,W_1)$ as $$ \begin{aligned} C \, & {\det(S_1)}^{a+b-\frac12(p+1)} {\det(U)}^{-(p+1)} {\det(U^{-1}-I_p)}^{b-\frac12(p+1)} \ & \etr\left(-\frac12 S_1U^{-1}\right) {}0!F_1\left(a, \frac{1}{4}\Theta_1S_1\right) {}_0!F_1\left(b, \frac{1}{4}\Theta_2S_1^{\frac12}(U^{-1}-I_p)S_1^\frac12\right) \ = C \, & {\det(S_1)}^{a+b-\frac12(p+1)} {\det(U)}^{-b-\frac12(p+1)} {\det(I_p-U)}^{b-\frac12(p+1)} \ & \etr\left(-\frac12 S_1U^{-1}\right) {}_0!F_1\left(a, \frac{1}{4}\Theta_1S_1\right) {}_0!F_1\left(b, \frac{1}{4}\Theta_2S_1^{\frac12}U^{-1}(I_p-U)S_1^\frac12\right). \end{aligned} $$ Thus the density of $U$ is $$ \begin{aligned} C\, & {\det(U)}^{-b-\frac12(p+1)} {\det(I_p-U)}^{b-\frac12(p+1)} \ & \int{S_1>0} \etr\left(-\frac12 S_1U^{-1}\right) {\det(S_1)}^{a+b-\frac12(p+1)} {}0!F_1\left(a, \frac{1}{4}\Theta_1S_1\right) {}_0!F_1\left(b, \frac{1}{4}\Theta_2S_1^{\frac12}U^{-1}(I_p-U)S_1^\frac12\right) \mathrm{d}S_1\ = C\, &
{\det\bigl(U{(I_p-U)}^{-1}\bigr)}^{-b-\frac12(p+1)} {\det(I_p-U)}^{-(p+1)} \ & \int
{S>0} \etr\left(-S U^{-1}\right) {\det(S)}^{a+b-\frac12(p+1)} {}0!F_1\left(a, \frac{1}{2}\Theta_1S\right) {}_0!F_1\left(b, \frac{1}{2}\Theta_2S^{\frac12}U^{-1}(I_p-U)S^\frac12\right) \mathrm{d}S. \end{aligned} $$ Let's derive the density of $U{(I_p-U)}^{-1}$. Using the transformation $U = V{(I_p+V)}^{-1}$ with Jacobian $J(U \rightarrow V) = {\det(I_p+V)}^{-(p+1)}$, we get the density of $V$ as $$ \begin{aligned} C\, & {\det(V)}^{-b-\frac12(p+1)} \ & \int{S>0} \etr\bigl(-S (I_p+V^{-1})\bigr) {\det(S)}^{a+b-\frac12(p+1)} {}_0!F_1\left(a, \frac{1}{2}\Theta_1S\right) {}_0!F_1\left(b, \frac{1}{2}\Theta_2S^{\frac12}V^{-1}S^\frac12\right) \mathrm{d}S. \end{aligned} $$ This is the density of $\mathcal{B}II_p^{(2)}(a,b,\Theta_1,\Theta_2)$.

$\mathcal{B}II_p^{(1)}(a,b,\Theta_1,\Theta_2)$

Recall that $\mathcal{B}II_p^{(1)}(a,b,\Theta_1,\Theta_2)$ is the distribution of $$ V = {(W_2^{-\frac12})}' W_1 W_2^{-\frac12}. $$

Transforming $S_1 = {(S_2^{\frac12})}' V S_2^{-\frac12}$ with Jacobian $J(S_1, S_2 \rightarrow V, S_2) = {\det(S_2)}^{\frac12(p+1)}$, we get the density of $(V,S_2)$ as $$ \begin{aligned} C\, & {\det(V)}^{a-\frac12(p+1)} {\det(S_2)}^{a+b-\frac12(p+1)} \ & \etr\left(-\frac12 (I_p+V)S_2\right) {}0!F_1\left(a, \frac{1}{4}\Theta_1 {(S_2^{\frac12})}' V S_2^{-\frac12}\right) {}_0!F_1\left(b, \frac{1}{4}\Theta_2S_2\right). \end{aligned} $$ Thus, the density of $V$ is $$ \begin{aligned} C\, & {\det(V)}^{a-\frac12(p+1)} \ & \int{S>0} \etr\bigl(-(I_p+V)S\bigr) {\det(S)}^{a+b-\frac12(p+1)} {}_0!F_1\left(a, \frac{1}{2}\Theta_1{(S^{\frac12})}' V S^{-\frac12}\right) {}_0!F_1\left(b, \frac{1}{2}\Theta_2S\right) \mathrm{d}S. \end{aligned} $$

$\mathcal{B}II_p^{(2)}(a,b,\Theta_1,\Theta_2)$

Recall that $\mathcal{B}II_p^{(2)}(a,b,\Theta_1,\Theta_2)$ is the distribution of $$ V = W_1^{\frac12} W_2^{-1} {(W_1^{\frac12})}'. $$

Transforming $S_2 = {(S_1^\frac12)}' V^{-1} S_1^\frac12$ with Jacobian $J(S_1, S_2 \rightarrow V, S_1) = {\det(S_1)}^{\frac12(p+1)}{\det(V)}^{-(p+1)}$, we get the density of $(V,S_1)$ as $$ \begin{aligned} C\, & {\det(S_1)}^{a+b-\frac12(p+1)} {\det(V)}^{-b-\frac12(p+1)} \ & \etr\left(-\frac12 S_1\right)\etr\left(-\frac12 S_1V^{-1}\right) {}0!F_1\left(a, \frac{1}{4}\Theta_1S_1\right) {}_0!F_1\left(b, \frac{1}{4}\Theta_2{(S_1^\frac12)}' V^{-1} S_1^\frac12\right). \end{aligned} $$ Thus the density of $V$ is $$ \begin{aligned} C\, & {\det(V)}^{-b-\frac12(p+1)} \ & \int{S >0} \etr\bigl(-(I_p+V^{-1})S\bigr) {\det(S)}^{a+b-\frac12(p+1)}
{}0!F_1\left(a, \frac{1}{2}\Theta_1S\right) {}_0!F_1\left(b, \frac{1}{2}\Theta_2{(S^\frac12)}' V^{-1} S^\frac12\right) \mathrm{d}S. \end{aligned} $$ In the case when $\Theta_2$ is scalar, $$ {}_0!F_1\left(b, \frac{1}{2}\Theta_2{(S^\frac12)}' V^{-1} S^\frac12\right) = {}_0!F_1\left(b, \frac{1}{2}\Theta_2 S V^{-1}\right). $$ Doing the change of variables $R = S V^{-1}$ in the integral, we get the density of $V$ as $$ \begin{aligned} C\, & {\det(V)}^{a-\frac12(p+1)} \ & \int{R >0} \etr\bigl(-(I_p+V)R\bigr) {\det(R)}^{a+b-\frac12(p+1)}
{}_0!F_1\left(a, \frac{1}{2}\Theta_1 R V\right) {}_0!F_1\left(b, \frac{1}{2}\Theta_2 R\right) \mathrm{d}R. \end{aligned} $$ If in addition, $\Theta_1$ is scalar, this is the density of $\mathcal{B}II_p^{(1)}(a,b,\Theta_1,\Theta_2)$.

Simplification

One has ${}0!F_1(\alpha, \mathbf{0}) = 1$. Thus, when $\Theta_1$ or $\Theta_2$ is the null matrix, we get integrals like $$ \int{S>0} \etr(-ZS) {\det(S)}^{\alpha-\frac12(p+1)} {}_0!F_1\left(\beta, \frac{1}{2}\Theta S T \right) \mathrm{d}S, $$ for example in the density of $\mathcal{B}I_p^{(2)}(a,b,\Theta_1,\Theta_2)$ when $\Theta_2$ is the null matrix, or in the density of $\mathcal{B}I_p^{(1)}(a,b,\Theta_1,\Theta_2)$ when $\Theta_1$ is the null matrix and $\Theta_2$ is scalar.

This integral is equal to $$ \Gamma_p(\alpha){\det(Z)}^{-\alpha} {}_1!F_1\left(\alpha, \beta, \frac{1}{2}\Theta Z^{-1} T \right). $$



stla/matrixsampling documentation built on Aug. 27, 2019, 3:03 a.m.