Matrix Beta Type II And The Matrix Odds Transform

From a Matrix Interval to the Positive-Definite Cone

For a scalar probability $y$ between zero and one, the odds $y/(1-y)$ convert the interval $(0,1)$ into the positive half-line. The matrix counterpart does the same for eigenvalues: it carries a symmetric matrix whose eigenvalues lie between zero and one to a positive-definite matrix. This change of variables turns the bounded support of a matrix beta type I law into the entire positive-definite cone.

The transformation is not entrywise. For a symmetric matrix $Y$ satisfying $0<Y<I$, define $X=Y(I-Y)^{-1}$. Since $Y$ commutes with $I-Y$ and its inverse, $X$ is symmetric. Its eigenvalues are the scalar odds of the eigenvalues of $Y$.

Definition: A symmetric positive-definite $p\times p$ random matrix $X$ has a matrix beta type II distribution with parameters $a,b>(p-1)/2$, written here $X\sim\operatorname{BetaII}_p(a,b)$, if its density with respect to Lebesgue measure on the $p(p+1)/2$ distinct entries is $$ f_X(X)=\frac{1}{B_p(a,b)}|X|^{a-(p+1)/2}|I+X|^{-(a+b)},\qquad X>0. $$ Here $|A|$ denotes the determinant of $A$, $I$ is the $p\times p$ identity matrix, and $B_p(a,b)=\Gamma_p(a)\Gamma_p(b)/\Gamma_p(a+b)$, where $$ \Gamma_p(c)=\pi^{p(p-1)/4}\prod_{j=1}^{p}\Gamma\!\left(c-\frac{j-1}{2}\right),\qquad c>(p-1)/2. $$

The parameter conditions ensure that the multivariate gamma factors are finite and the density is integrable. The matrix beta type I density, used below, is supported on $0<Y<I$ and is proportional to $|Y|^{a-(p+1)/2}|I-Y|^{b-(p+1)/2}$, with normalizing constant $B_p(a,b)$.

The Odds Map and Its Support

For a symmetric matrix, $0<Y<I$ means that both $Y$ and $I-Y$ are positive definite. Diagonalize $Y=Q\operatorname{diag}(y_1,\ldots,y_p)Q^\mathsf{T}$, where $Q$ is orthogonal and $0<y_i<1$. Then $$ Y(I-Y)^{-1}=Q\operatorname{diag}\!\left(\frac{y_1}{1-y_1},\ldots,\frac{y_p}{1-y_p}\right)Q^\mathsf{T}. $$ Every transformed eigenvalue is positive, so $X>0$. Conversely, for any $X>0$, define $$ Y=X(I+X)^{-1}. $$ If $X$ has eigenvalues $x_i>0$, then $Y$ has eigenvalues $x_i/(1+x_i)$, all strictly between zero and one. Substitution gives $Y(I-Y)^{-1}=X$. Thus the maps are inverses and establish a one-to-one correspondence between the two supports.

Key result: The matrix odds transform $X=Y(I-Y)^{-1}$ is a bijection from $0<Y<I$ onto $X>0$. Its inverse is $Y=X(I+X)^{-1}$. The transformed support is the full cone of symmetric positive-definite matrices.

Jacobian on Symmetric Matrices

The Jacobian must be calculated on the space of symmetric matrices, not on all $p\times p$ matrices. For an invertible matrix $A$, the linear map $H\mapsto AHA^\mathsf{T}$ on symmetric matrices has absolute Jacobian $|A|^{p+1}$. This fact gives the Jacobian of the inverse odds map directly.

Jacobian lemma: With Lebesgue measure on the distinct entries of a symmetric matrix, the absolute determinant of the linear transformation $H\mapsto AHA^\mathsf{T}$, for invertible $A$, is $|A|^{p+1}$.

To prove the lemma, use a singular value decomposition $A=U D V^\mathsf{T}$, where $U,V$ are orthogonal and $D=\operatorname{diag}(d_1,\ldots,d_p)$ with each $d_i>0$. Congruence by an orthogonal matrix preserves Euclidean volume on the symmetric-matrix space, so the absolute Jacobian is the same as for $H\mapsto DHD$. The diagonal coordinate $h_{ii}$ is multiplied by $d_i^2$, and the off-diagonal coordinate $h_{ij}$, for $i<j$, is multiplied by $d_i d_j$. The product of all these factors is $$ \left(\prod_{i=1}^{p}d_i^2\right)\left(\prod_{i<j}d_i d_j\right) =\prod_{i=1}^{p}d_i^{p+1} =|A|^{p+1}. $$ This proves the lemma. $\square$

For $Y=X(I+X)^{-1}=I-(I+X)^{-1}$, its differential is $$ dY=(I+X)^{-1}(dX)(I+X)^{-1}. $$ Indeed, differentiating the inverse gives $d[(I+X)^{-1}]=-(I+X)^{-1}(dX)(I+X)^{-1}$. The Jacobian lemma, with $A=(I+X)^{-1}$, therefore gives the absolute Jacobian of $X\mapsto Y$: $$ \left|\frac{\partial Y}{\partial X}\right|=|I+X|^{-(p+1)}. $$ Equivalently, the forward transformation has Jacobian $|I-Y|^{-(p+1)}$, since $I-Y=(I+X)^{-1}$. Keeping track of which direction the Jacobian refers to is essential in the density calculation.

Deriving the Type II Density

Suppose $Y\sim\operatorname{BetaI}_p(a,b)$, with $a,b>(p-1)/2$, and set $X=Y(I-Y)^{-1}$. The inverse map gives the determinant identities $$ |Y|=\frac{|X|}{|I+X|},\qquad |I-Y|=\frac{1}{|I+X|}. $$ The change-of-variables formula multiplies the type I density evaluated at $Y=X(I+X)^{-1}$ by the Jacobian $|I+X|^{-(p+1)}$. Combining these factors yields the result below.

Theorem: If $Y\sim\operatorname{BetaI}_p(a,b)$ and $X=Y(I-Y)^{-1}$, for $a,b>(p-1)/2$, then $X\sim\operatorname{BetaII}_p(a,b)$.

To verify the density, begin with the type I kernel and substitute the determinant identities and Jacobian:

\begin{aligned} f_X(X) &=\frac{1}{B_p(a,b)} \left(\frac{|X|}{|I+X|}\right)^{a-(p+1)/2} \left(\frac{1}{|I+X|}\right)^{b-(p+1)/2} |I+X|^{-(p+1)}\\ &=\frac{1}{B_p(a,b)}|X|^{a-(p+1)/2} |I+X|^{-\{a-(p+1)/2+b-(p+1)/2+(p+1)\}}\\ &=\frac{1}{B_p(a,b)}|X|^{a-(p+1)/2}|I+X|^{-(a+b)},\qquad X>0. \end{aligned}

The exponents combine because $-a+(p+1)/2-b+(p+1)/2-(p+1)=-(a+b)$. The support is $X>0$, as established by the bijection. Since this is a one-to-one change of variables from a normalized type I density, the resulting type II density integrates to one. This completes the derivation. $\square$

Worked Examples

Worked Example: Scalar Odds and a Beta Type II Density

Let $Y\sim\operatorname{Beta}(2,3)$, with density $f_Y(y)=12y(1-y)^2$ for $0<y<1$. Set $X=Y/(1-Y)$. Find the support and density of $X$.

The inverse is $y=x/(1+x)$, so $x>0$ and $dy/dx=(1+x)^{-2}$. Substitution gives $$ f_X(x)=12\frac{x}{1+x}\left(\frac{1}{1+x}\right)^2\frac{1}{(1+x)^2} =\frac{12x}{(1+x)^5},\qquad x>0. $$ For $p=1$, the matrix beta type II formula has determinant exponents $a-1=1$ and $-(a+b)=-5$; it gives exactly this density. Its integral is one: with $y=x/(1+x)$, the integral becomes $\int_0^1 12y(1-y)^2\,dy=1$.

Worked Example: Transforming a Two-Dimensional Matrix

Take $$ Y=\begin{pmatrix}1/3&0\\0&1/2\end{pmatrix}. $$ Its eigenvalues lie in $(0,1)$, so $0<Y<I$. Since $I-Y=\operatorname{diag}(2/3,1/2)$, its inverse is $\operatorname{diag}(3/2,2)$. Therefore $$ X=Y(I-Y)^{-1} =\begin{pmatrix}1/3&0\\0&1/2\end{pmatrix} \begin{pmatrix}3/2&0\\0&2\end{pmatrix} =\begin{pmatrix}1/2&0\\0&1\end{pmatrix}. $$ The result is positive definite. Applying the inverse verifies the calculation: $X(I+X)^{-1}=\operatorname{diag}((1/2)/(3/2),1/2)=Y$.

For a distributional check, let $Y\sim\operatorname{BetaI}_2(2,2)$. Here $B_2(2,2)=\pi/45$, obtained from $\Gamma_2(2)=\pi/2$ and $\Gamma_2(4)=45\pi/4$. At the transformed matrix, $|X|=1/2$ and $|I+X|=3$. The type II density is therefore $$ f_X(X)=\frac{45}{\pi}\left(\frac12\right)^{1/2}3^{-4} =\frac{5}{9\sqrt{2}\,\pi}. $$ This is the density value at that matrix, with respect to Lebesgue measure on its three distinct entries.

Worked Example: Checking a Two-Dimensional Jacobian

At $X=\operatorname{diag}(1,2)$, the inverse transformation has $A=(I+X)^{-1}=\operatorname{diag}(1/2,1/3)$. A symmetric perturbation has coordinates $$ H=\begin{pmatrix}h_{11}&h_{12}\\h_{12}&h_{22}\end{pmatrix}. $$ The differential $dY=AHA$ multiplies $h_{11}$ by $1/4$, $h_{12}$ by $1/6$, and $h_{22}$ by $1/9$. Thus its absolute Jacobian is $$ \frac14\cdot\frac16\cdot\frac19=\frac{1}{216}. $$ The general formula agrees: $|I+X|^{-3}=(2\cdot3)^{-3}=1/216$. The exponent $3=p+1$ reflects the three independent coordinates of a symmetric $2\times2$ matrix.

Interpretation and a Common Pitfall

The transformation acts on eigenvalues: each $y_i\in(0,1)$ becomes $x_i=y_i/(1-y_i)>0$, with inverse $y_i=x_i/(1+x_i)$. Eigenvalues near one map to large positive values, while eigenvalues near zero remain near zero. This gives the type II support its unbounded character. The determinant factor $|I+X|^{-(a+b)}$ in the density reflects both the transformed type I kernel and the volume adjustment from the Jacobian.

Common mistake: Applying the scalar odds formula separately to every matrix entry, or using the ordinary scalar derivative $(1-y)^{-2}$ as the matrix Jacobian. Neither is correct. The matrix map involves an inverse and matrix multiplication; it acts through the eigenvalues, and its Jacobian must account for all $p(p+1)/2$ independent symmetric coordinates. For the inverse map that Jacobian is $|I+X|^{-(p+1)}$, not a scalar derivative applied entrywise.

Key takeaway: The matrix odds transform maps $0<Y<I$ bijectively to $X>0$. Its inverse has Jacobian $|I+X|^{-(p+1)}$; applying it to a matrix beta type I density gives the type II kernel $|X|^{a-(p+1)/2}|I+X|^{-(a+b)}$.

Check Your Understanding

Use the support, inverse map, and Jacobian to answer these questions.

What is the support of $X=Y(I-Y)^{-1}$ when $0<Y<I$, and what is the inverse transformation?
For $p=3$, what exponent appears in the Jacobian of $X\mapsto X(I+X)^{-1}$, and why?
If $Y$ has eigenvalue $y=3/4$, what is the corresponding eigenvalue of $X$?
For a matrix beta type I law with parameters $a,b$, explain how the determinant exponents combine to give the power $-(a+b)$ on $|I+X|$.
Why is it incorrect to apply $y/(1-y)$ separately to the entries of a symmetric matrix?