Support And Density Of Matrix Beta Type I

Support Is a Matrix Constraint

A scalar beta variable takes values between zero and one. Its matrix analogue must satisfy two simultaneous positivity conditions: the matrix itself is positive definite, and its difference from the identity is also positive definite. These conditions describe an open region in the space of symmetric matrices, rather than a collection of unrelated entrywise bounds.

For a concrete connection to the earlier material on independent Wishart matrices, let $A\sim W_p(\nu_1,I_p)$ and $B\sim W_p(\nu_2,I_p)$ be independent, with $\nu_1,\nu_2>p-1$. The symmetric matrix $U=(A+B)^{-1/2}A(A+B)^{-1/2}$ has a matrix beta type I distribution. The symmetric form matters: it gives a random variable in the space of symmetric matrices with a directly interpretable positive-definite support.

Definition: For $p\geq 1$ and $a,b>(p-1)/2$, a symmetric $p\times p$ random matrix $U$ has a matrix beta type I distribution with parameters $a,b$ if its density, with respect to Lebesgue measure on the $p(p+1)/2$ distinct entries of a symmetric matrix, is $$ f(U)=\frac{\Gamma_p(a+b)}{\Gamma_p(a)\Gamma_p(b)} |U|^{a-(p+1)/2}|I_p-U|^{b-(p+1)/2}, \qquad 0\prec U\prec I_p, $$ and is zero outside this support. Here $0\prec U\prec I_p$ means that both $U$ and $I_p-U$ are positive definite, $|\cdot|$ denotes determinant, and $$ \Gamma_p(c)=\pi^{p(p-1)/4}\prod_{j=1}^{p}\Gamma\!\left(c-\frac{j-1}{2}\right), \qquad c>(p-1)/2, $$ is the multivariate gamma function.

The Wishart construction gives $a=\nu_1/2$ and $b=\nu_2/2$. Thus the requirement $\nu_i>p-1$ is exactly the parameter requirement $a,b>(p-1)/2$. The support is written with strict inequalities because a positive-definite Wishart matrix is nonsingular with probability one under these degrees-of-freedom conditions.

Theorem: For a symmetric matrix $U$, the condition $0\prec U\prec I_p$ holds if and only if every eigenvalue of $U$ is strictly between zero and one.

Proof. By the spectral theorem, $U=Q\Lambda Q^\mathsf{T}$ for an orthogonal $Q$ and a diagonal matrix $\Lambda$ of eigenvalues $\lambda_1,\ldots,\lambda_p$. The eigenvalues of $U$ are positive exactly when $U$ is positive definite. Also, $I_p-U=Q(I_p-\Lambda)Q^\mathsf{T}$, whose eigenvalues are $1-\lambda_1,\ldots,1-\lambda_p$. Therefore $I_p-U$ is positive definite exactly when every $\lambda_i<1$. Combining the two conditions proves the equivalence. $\square$

Equivalently, the support is the set of symmetric matrices whose eigenvalues all lie in $(0,1)$. It is not enough for the diagonal entries to lie in that interval: off-diagonal entries also affect the eigenvalues and positive definiteness.

Worked Example: Checking Membership in the Support

Consider $$ U=\begin{pmatrix}0.6&0.2\\0.2&0.5\end{pmatrix}. $$ Check whether $U$ is in the support.

A symmetric $2\times2$ matrix is positive definite if its top-left entry and determinant are positive. Here $0.6>0$, and $$ |U|=(0.6)(0.5)-(0.2)^2=0.30-0.04=0.26>0. $$ For the second condition, $$ I_2-U=\begin{pmatrix}0.4&-0.2\\-0.2&0.5\end{pmatrix}. $$ Its top-left entry is $0.4>0$, and its determinant is $$ |I_2-U|=(0.4)(0.5)-(-0.2)^2=0.20-0.04=0.16>0. $$ Thus both matrices are positive definite, so $U$ is in the support. Its eigenvalues are $(11\pm\sqrt{17})/20$, approximately $0.756$ and $0.344$, which also verifies that both lie between zero and one.

Density and Determinant Powers

The two determinants measure how close the matrix is to the two boundaries of the support. In eigenvalue terms, $|U|=\prod_{i=1}^p\lambda_i$ and $|I_p-U|=\prod_{i=1}^p(1-\lambda_i)$. The first determinant becomes small when eigenvalues approach zero; the second becomes small when eigenvalues approach one. Their powers in the density therefore control how the density changes near those boundaries.

The exponents are $a-(p+1)/2$ and $b-(p+1)/2$, not $a$ and $b$ themselves. Either exponent may be negative even though the density is integrable. For example, the allowed condition $a>(p-1)/2$ implies $a-(p+1)/2>-1$; it does not require that exponent to be positive. A negative exponent means the density can increase as the corresponding determinant approaches zero. This behavior does not put probability mass on the boundary: the distribution has a density on the open support, and the boundary is excluded.

Key interpretation: The factor $|U|^{a-(p+1)/2}$ controls behavior as eigenvalues approach zero, while $|I_p-U|^{b-(p+1)/2}$ controls behavior as eigenvalues approach one. The multivariate beta constant makes the density integrate to one over $0\prec U\prec I_p$.

The density can be obtained from the independent-Wishart construction. The derivation also explains why the support has these two positivity constraints.

Result: If $A$ and $B$ are independent $W_p(\nu_1,I_p)$ and $W_p(\nu_2,I_p)$ matrices, respectively, then $U=(A+B)^{-1/2}A(A+B)^{-1/2}$ has the density in the definition with $a=\nu_1/2$ and $b=\nu_2/2$.

Derivation. Set $S=A+B$, so that $A=S^{1/2}US^{1/2}$ and $B=S^{1/2}(I_p-U)S^{1/2}$. The transformation maps positive-definite $A,B$ to $S\succ0$ and $0\prec U\prec I_p$. At fixed $S$, the linear map $H\mapsto S^{1/2}HS^{1/2}$ on symmetric matrices has Jacobian $|S|^{(p+1)/2}$. To see this, diagonalize $S$, with eigenvalues $s_1,\ldots,s_p$. The map scales diagonal coordinates by $s_i$ and off-diagonal coordinates $H_{ij}$, $i<j$, by $\sqrt{s_i s_j}$. Its determinant is $\prod_i s_i\prod_{i<j}\sqrt{s_i s_j}=|S|^{(p+1)/2}$.

The joint Wishart density is proportional to $|A|^{(\nu_1-p-1)/2}|B|^{(\nu_2-p-1)/2}\exp[-\operatorname{tr}(S)/2]$. Congruence gives $|A|=|S||U|$ and $|B|=|S||I_p-U|$. Multiplying by the Jacobian separates the factors in $U$ from those in $S$: the $U$-dependent terms are $$ |U|^{(\nu_1-p-1)/2}|I_p-U|^{(\nu_2-p-1)/2}. $$ The remaining $S$-dependent terms are proportional to $|S|^{(\nu_1+\nu_2-p-1)/2}\exp[-\operatorname{tr}(S)/2]$, the kernel of a $W_p(\nu_1+\nu_2,I_p)$ density. Integrating over $S$ leaves the normalized matrix beta type I density. The Wishart normalizing constants give its constant as $\Gamma_p((\nu_1+\nu_2)/2)/[\Gamma_p(\nu_1/2)\Gamma_p(\nu_2/2)]$. Substituting $a=\nu_1/2$ and $b=\nu_2/2$ gives the stated result. $\square$

Evaluating the Density

A density value is a height relative to the chosen coordinate measure; it is not, by itself, the probability of observing a particular matrix. For $p=2$, the measure is over the three entries $u_{11},u_{12},u_{22}$. The determinant calculations and the support check must both be done before evaluating the formula.

Worked Example: Recovering the Scalar Beta Density

Let $p=1$, $a=2$, and $b=3$. The multivariate gamma function reduces to the ordinary gamma function, and the support is $0<u<1$.

The normalizing constant is $$ \frac{\Gamma(5)}{\Gamma(2)\Gamma(3)} =\frac{24}{1\cdot 2}=12. $$ Since $(p+1)/2=1$, the determinant powers are $a-1=1$ and $b-1=2$. Thus $$ f(u)=12u(1-u)^2,\qquad 0<u<1. $$ At $u=1/2$, $$ f(1/2)=12(1/2)(1/2)^2 =12\cdot\frac18=\frac32. $$ This is the ordinary $\operatorname{Beta}(2,3)$ density, showing that the matrix formula includes the scalar beta distribution as the $p=1$ case.

Worked Example: A Two-Dimensional Density Value

Take $p=2$ and $a=b=2$. The condition $a,b>(p-1)/2=1/2$ holds. First calculate $$ \Gamma_2(2)=\sqrt{\pi}\,\Gamma(2)\Gamma(3/2) =\sqrt{\pi}(1)\frac{\sqrt{\pi}}{2}=\frac{\pi}{2}, $$ and $$ \Gamma_2(4)=\sqrt{\pi}\,\Gamma(4)\Gamma(7/2) =\sqrt{\pi}(6)\frac{15\sqrt{\pi}}{8}=\frac{45\pi}{4}. $$ Therefore the density constant is $$ \frac{\Gamma_2(4)}{\Gamma_2(2)^2} =\frac{45\pi/4}{(\pi/2)^2}=\frac{45}{\pi}. $$ At $U=\operatorname{diag}(1/4,3/4)$, both $|U|$ and $|I_2-U|$ equal $3/16$. Each exponent is $2-3/2=1/2$, so $$ f(U)=\frac{45}{\pi}\left(\frac{3}{16}\right)^{1/2} \left(\frac{3}{16}\right)^{1/2} =\frac{135}{16\pi}. $$ The eigenvalues $1/4$ and $3/4$ confirm that this matrix is in the support.

Worked Example: A Nondiagonal Density Value

Keep $p=2$ and $a=b=2$, but take $$ U=\begin{pmatrix}1/2&1/6\\1/6&1/2\end{pmatrix}. $$ Its determinant is $$ |U|=\frac14-\frac1{36}=\frac8{36}=\frac29. $$ Because $I_2-U$ has diagonal entries $1/2$ and off-diagonal entries $-1/6$, its determinant is also $2/9$. The eigenvalues of $U$ are $1/2+1/6=2/3$ and $1/2-1/6=1/3$, so the matrix lies in the support. Using the constant $45/\pi$ and the two exponents $1/2$, $$ f(U)=\frac{45}{\pi}\left(\frac29\right)^{1/2} \left(\frac29\right)^{1/2} =\frac{45}{\pi}\frac29=\frac{10}{\pi}. $$ The off-diagonal entries contribute to both determinants; using only the diagonal entries would give the wrong density value.

Common Pitfall and Practical Use

A frequent error is to treat $0\prec U\prec I_p$ as an entrywise condition, checking only that each displayed entry is between zero and one. For instance, $$ V=\begin{pmatrix}1/2&3/4\\3/4&1/2\end{pmatrix} $$ has diagonal entries in $(0,1)$, but $|V|=1/4-9/16=-5/16<0$, so it is not positive definite and is outside the support. The correction is to check positive definiteness of both $U$ and $I_p-U$, or equivalently to check that all eigenvalues of $U$ lie strictly between zero and one.

This distinction matters in applications because the support is part of the model, not a cosmetic restriction on the formula. A matrix outside the support has density zero, even if the determinant powers can be written algebraically. For a valid matrix, the determinants compactly summarize the eigenvalue contributions, while the multivariate gamma ratio supplies the scale required for a probability density.

Check Your Understanding

Use the support and density definitions to answer these questions.

State the two positive-definiteness conditions that define the support of a matrix beta type I variable.
Why is it insufficient to check that every entry of a symmetric matrix lies between zero and one?
For $p=3$, what are the determinant exponents in terms of $a$ and $b$?
What does a negative exponent on $|U|$ indicate about density behavior as an eigenvalue approaches zero?
For $p=1$, identify the familiar distribution represented by the matrix beta type I law.