Matrix Beta Type I From Independent Wishart Matrices

From Wishart Matrices to a Normalized Component

A useful way to obtain a matrix beta type I variable is to split a positive-definite matrix into two independent Wishart contributions, then compare one contribution with their sum. The matrix ratio is not formed by ordinary division: matrix multiplication is not commutative, and the normalized component must be defined so that it is symmetric and positive definite. A Cholesky factor of the sum provides a precise construction.

The shared scale matrix is essential to the clean result. When both Wishart matrices have the same scale, their exponential density terms combine into a term involving only the sum. The remaining part of the transformed density then depends only on the normalized component. We derive that separation, including the Jacobian, rather than assuming it.

Definition: Let \(A\) and \(B\) be independent \(p\times p\) Wishart random matrices with \(A\sim W_p(r,\Sigma)\) and \(B\sim W_p(s,\Sigma)\), where \(\Sigma\) is positive definite and \(r,s>p-1\). Set \(X=A+B\), and write the Cholesky decomposition \(X=LL^T\), with \(L\) lower triangular and having positive diagonal entries. Define the normalized component \(U=L^{-1}A L^{-T}\), where \(L^{-T}=(L^{-1})^T\).

The construction is invertible: \(A=LUL^T\) and \(B=L(I-U)L^T\). Since \(A\) and \(B\) are positive definite, the resulting component satisfies \(0<U<I\), meaning both \(U\) and \(I-U\) are positive definite. We use this domain in the derivation below.

The Change of Variables

The densities of the two independent Wishart matrices, with respect to Lebesgue measure on the distinct entries of a symmetric matrix, are

f_A(A)=\frac{|A|^{(r-p-1)/2}\exp\{-\tfrac12\operatorname{tr}(\Sigma^{-1}A)\}} {2^{rp/2}|\Sigma|^{r/2}\Gamma_p(r/2)},\qquad A>0,

and the same expression for \(B\) with \(r\) replaced by \(s\). Here \(|A|\) denotes determinant, and the multivariate gamma function is

\Gamma_p(c)=\pi^{p(p-1)/4}\prod_{j=1}^{p}\Gamma\!\left(c-\frac{j-1}{2}\right), \qquad c>\frac{p-1}{2}.

To calculate the Jacobian, first use \((A,B)\mapsto(X,A)\), where \(X=A+B\). This linear transformation has absolute Jacobian one: for each independent matrix entry it replaces \((A_{ij},B_{ij})\) by \((X_{ij},A_{ij})\), with determinant of absolute value one. Next, at fixed \(X\), the transformation from \(U\) to \(A=LUL^T\) is a congruence transformation on symmetric matrices.

Jacobian fact: For a nonsingular \(p\times p\) matrix \(C\), the map \(H\mapsto CHC^T\) on symmetric matrices has absolute Jacobian \(|C|^{p+1}\).

For completeness, when \(C\) is diagonal with diagonal entries \(c_1,\ldots,c_p\), a diagonal coordinate \(H_{ii}\) is multiplied by \(c_i^2\), and an off-diagonal coordinate \(H_{ij}\), \(i<j\), by \(c_ic_j\). Their product is \(\prod_i |c_i|^{2+(p-1)}=|C|^{p+1}\). Orthogonal congruences have absolute Jacobian one, so a singular-value decomposition extends the result to every nonsingular \(C\). Applying the fact with \(C=L\), and using \(|L|^2=|X|\), gives the Jacobian for \((X,U)\mapsto(X,A)\):

\left|\frac{\partial(A,B)}{\partial(X,U)}\right| =|X|^{(p+1)/2}.

The dependence of \(L\) on \(X\) does not add another factor: the derivative in the \((X,U)\mapsto(X,A)\) calculation is block triangular, with the identity on the \(X\) coordinates and the fixed-\(X\) congruence derivative on the \(U\) coordinates. The Jacobian of the preceding \((A,B)\mapsto(X,A)\) step is one.

Factorization and the Matrix Beta Type I Law

Under the inverse transformation, \(|A|=|X||U|\) and \(|B|=|X||I-U|\). The two trace terms also combine: \(\operatorname{tr}(\Sigma^{-1}A)+\operatorname{tr}(\Sigma^{-1}B)=\operatorname{tr}(\Sigma^{-1}X)\). Multiplying the Wishart densities and the Jacobian therefore gives the joint density of \(X\) and \(U\):

\begin{aligned} f_{X,U}(X,U) ={}& \frac{|X|^{(r+s-p-1)/2}\exp\{-\tfrac12\operatorname{tr}(\Sigma^{-1}X)\}} {2^{(r+s)p/2}|\Sigma|^{(r+s)/2}\Gamma_p(r/2)\Gamma_p(s/2)} \\ &\quad{}\times |U|^{(r-p-1)/2}|I-U|^{(s-p-1)/2}. \end{aligned}

The power of \(|X|\) includes the Jacobian contribution: the two original determinant powers contribute \((r+s-2p-2)/2\), and the Jacobian contributes \((p+1)/2\). Their sum is \((r+s-p-1)/2\), the Wishart determinant power for \(r+s\) degrees of freedom.

Theorem: Under the conditions above, \(X=A+B\sim W_p(r+s,\Sigma)\), and \(X\) is independent of \(U=L^{-1}AL^{-T}\). The density of \(U\) is \[ f_U(U)=\frac{\Gamma_p((r+s)/2)}{\Gamma_p(r/2)\Gamma_p(s/2)} |U|^{(r-p-1)/2}|I-U|^{(s-p-1)/2},\qquad 0<U<I. \] This distribution is the matrix beta type I distribution with parameters \(r/2\) and \(s/2\).

Proof. The displayed joint density separates into a function of \(X\) and a function of \(U\). The \(X\)-factor, including its constant, is exactly the Wishart density with \(r+s\) degrees of freedom and scale \(\Sigma\). The remaining factor has constant \(\Gamma_p((r+s)/2)/[\Gamma_p(r/2)\Gamma_p(s/2)]\), as seen by dividing the joint constant by the \(W_p(r+s,\Sigma)\) constant. Both factors are normalized densities: the first is a Wishart density, and the second must integrate to one because the joint density is normalized and factors into a normalized \(X\)-density times a nonnegative function of \(U\). Thus the factorization proves both the stated marginal law and independence. \(\square\)

Worked Constructions and Checks

Worked Example: The Scalar Case Gives an Ordinary Beta Variable

Let \(p=1\), \(A\sim W_1(4,1)\), and \(B\sim W_1(6,1)\) independently. In one dimension, these are chi-square variables with 4 and 6 degrees of freedom, and the Cholesky factor of \(X=A+B\) is \(\sqrt{X}\). Thus \(U=A/X=A/(A+B)\).

The matrix beta formula reduces to the ordinary beta density with parameters \(r/2=2\) and \(s/2=3\). Since \(\Gamma(2)=1\), \(\Gamma(3)=2\), and \(\Gamma(5)=24\), its coefficient is \(24/(1\cdot2)=12\). Therefore

f_U(u)=12u^{2-1}(1-u)^{3-1}=12u(1-u)^2,\qquad 0<u<1.

For a normalization check, \(\int_0^1 12u(1-u)^2\,du=12(1/2-2/3+1/4)=12(1/12)=1\). This is the scalar special case of the Wishart-sum construction, not a separate assumption about a matrix ratio.

Worked Example: Evaluating a Two-Dimensional Beta Density

Take \(p=2\), \(r=5\), \(s=7\), and \(\Sigma=I\). The matrix beta exponents are \((5-2-1)/2=1\) and \((7-2-1)/2=2\). The normalizing coefficient is

\frac{\Gamma_2(6)}{\Gamma_2(5/2)\Gamma_2(7/2)} =\frac{1260}{\pi}.

Indeed, \(\Gamma_2(6)=3543.75\pi\), \(\Gamma_2(5/2)=3\pi/4\), and \(\Gamma_2(7/2)=15\pi/4\); their ratio is \(3543.75\pi/(45\pi^2/16)=1260/\pi\). For \(U=\operatorname{diag}(1/4,1/2)\), \(|U|=1/8\), while \(|I-U|=(3/4)(1/2)=3/8\). Hence

f_U(U)=\frac{1260}{\pi}\left(\frac18\right)\left(\frac38\right)^2 =\frac{1260\cdot9}{512\pi} =\frac{2835}{128\pi} \approx 7.05.

The density is with respect to the three distinct entries of a symmetric \(2\times2\) matrix. Its value at a point is a density value, not the probability of observing that exact matrix.

Worked Example: Applying the Normalization to Specific Matrices

Consider positive-definite matrices \(A=\operatorname{diag}(2,1)\) and \(B=\operatorname{diag}(1,3)\). Their sum is \(X=\operatorname{diag}(3,4)\), with Cholesky factor \(L=\operatorname{diag}(\sqrt{3},2)\). Directly applying the definition gives

U=L^{-1}AL^{-T} =\operatorname{diag}\left(\frac{2}{3},\frac14\right).

The complementary component is \(I-U=\operatorname{diag}(1/3,3/4)\), which is also positive definite. Reconstructing the original matrices checks the transformation: \(LUL^T=\operatorname{diag}(2,1)=A\) and \(L(I-U)L^T=\operatorname{diag}(1,3)=B\). This calculation illustrates the coordinates used in the density derivation; a random pair of independent Wishart matrices with a common scale produces the beta law for these normalized coordinates.

Why the Common Scale Matters

The cancellation in the trace term is what allows the joint density to factor. Both matrices must have the same scale \(\Sigma\): then the exponent is a function of \(A+B\) alone. If the scales differ, the trace contribution is instead \(\operatorname{tr}(\Sigma_A^{-1}A)+\operatorname{tr}(\Sigma_B^{-1}B)\), which generally retains dependence on both the sum and the normalized component after transformation. The factorization above—and therefore this derivation of the matrix beta type I law—does not follow.

A common mistake is to write \(A(A+B)^{-1}\) and treat it as an ordinary scalar fraction. That product need not be symmetric, so it is not the positive-definite component used here. The Cholesky normalization \(L^{-1}AL^{-T}\) is symmetric by construction, and its complement is \(L^{-1}BL^{-T}=I-U\). Another error is to omit the Jacobian: doing so leaves the wrong power of \(|X|\), so the sum no longer has the \(W_p(r+s,\Sigma)\) density. The factor \(|X|^{(p+1)/2}\) is indispensable.

Key takeaway: Two independent Wishart matrices with a common scale split into an independent Wishart sum and a Cholesky-normalized matrix beta type I component. The determinant powers come from the original Wishart densities; the correct power for the sum requires the congruence Jacobian.

Check Your Understanding

Use the transformation and factorization above to answer these questions.

Why does \(L^{-1}AL^{-T}\) have a symmetric matrix form, and why is its complement positive definite?
For \(p=3\), what is the absolute Jacobian of the congruence map \(H\mapsto LHL^T\) in terms of \(|X|\), where \(X=LL^T\)?
What degrees of freedom and scale does the sum \(A+B\) have under the theorem?
Identify the step in the density derivation that fails if \(A\) and \(B\) have different scale matrices.
In the scalar example, explain why the normalized component has beta parameters \(r/2\) and \(s/2\), rather than \(r\) and \(s\).