Choosing Between Matrix-Normal, Matrix-T, And Matrix-Beta Models

Three Families, Three Generating Mechanisms

A matrix of measurements may look Gaussian, heavy-tailed, or constrained to lie between zero and the identity, depending on how it was generated. The family name should describe that mechanism, not merely the matrix's dimensions or the fact that its entries are related. The key distinction here is between fixed Gaussian scales, random Gaussian scales, and a normalized ratio of Wishart matrices.

Throughout, matrices have $p$ rows and $q$ columns. The row scale $U$ is a $p\times p$ symmetric positive-definite matrix, and the column scale $V$ is a $q\times q$ symmetric positive-definite matrix. Write $\Gamma_p(a)$ for the multivariate gamma function, defined for $a>(p-1)/2$ by

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

We first state the model conventions that will be used to make the comparison precise.

Definition: A $p\times q$ random matrix $X$ has a matrix-normal distribution $MN_{p,q}(M,U,V)$ if its density is $$ f(X)=\frac{\exp\!\left\{-\tfrac12\operatorname{tr}\!\left[U^{-1}(X-M)V^{-1}(X-M)^{T}\right]\right\}}{(2\pi)^{pq/2}|U|^{q/2}|V|^{p/2}}, \qquad X\in\mathbb R^{p\times q}. $$ Here $M$ is fixed, and $U,V$ are fixed positive-definite scale matrices. Equivalently, $\operatorname{vec}(X)$ is multivariate normal with mean $\operatorname{vec}(M)$ and covariance $V\otimes U$.

In the matrix-normal model, the scales are fixed parameters: there is no additional random matrix changing the spread from realization to realization. To define the matrix-t convention used here, let $W\sim W_p(\nu,I_p)$, where $\nu>p-1$, and, conditionally on $W$, let

X\mid W\sim MN_{p,q}\!\left(M,\ U^{1/2}W^{-1}U^{1/2},\ V\right).

The symmetric positive-definite square root is used for $U^{1/2}$. This convention makes the row and column scales explicit; other matrix-t conventions exist, so specifying the mixture matters.

Definition: Under this Gaussian–Wishart mixture, $X$ has a matrix-t distribution with degrees of freedom $\nu$, location $M$, and scales $U,V$, written $MT_{p,q}(\nu,M,U,V)$. Its density is $$ f(X)= \frac{\Gamma_p((\nu+q)/2)} {\Gamma_p(\nu/2)\pi^{pq/2}|U|^{q/2}|V|^{p/2}} \left|I_p+U^{-1/2}(X-M)V^{-1}(X-M)^TU^{-1/2}\right|^{-(\nu+q)/2}. $$

For the third family, the object is not an unconstrained data matrix. It is a symmetric matrix formed by normalizing one Wishart matrix against another. Define $W_p(a,I_p)$ to have density proportional to $|A|^{(a-p-1)/2}\exp\{-\operatorname{tr}(A)/2\}$ on positive-definite $A$.

Definition: Let $A\sim W_p(a,I_p)$ and $B\sim W_p(b,I_p)$ be independent, with $a,b>p-1$. Set $S=A+B$, let $L$ be the lower-triangular Cholesky factor satisfying $S=LL^T$, and define $$ C=L^{-1}A L^{-T}. $$ Then $C$ has a matrix-beta type I distribution, denoted $Beta_p(a/2,b/2)$. Its support is $C>0$ and $I_p-C>0$, and its density is $$ f(C)=\frac{|C|^{(a-p-1)/2}|I_p-C|^{(b-p-1)/2}}{B_p(a/2,b/2)}, \qquad B_p(x,y)=\frac{\Gamma_p(x)\Gamma_p(y)}{\Gamma_p(x+y)}. $$

Worked Examples: Classify by the Construction

Worked Example: Fixed Gaussian Scales Give a Matrix-Normal Law

Suppose $X$ is a $2\times3$ Gaussian matrix with fixed mean $M$, row scale $U=\operatorname{diag}(4,1)$, and column scale $V=\operatorname{diag}(1,2,1)$. There is no random scale in its construction. Which family applies?

Both scale matrices are positive definite, since all their diagonal entries are positive. The stated density is therefore $MN_{2,3}(M,U,V)$, not a matrix-t density. For instance, if $M=0$, the density at $X=0$ is $$ f(0)=\frac{1}{(2\pi)^3 |U|^{3/2}|V|^{2/2}} =\frac{1}{(2\pi)^3(4)^{3/2}(2)} =\frac{1}{128\pi^3}. $$ The determinant powers use the dimensions: $|U|=4$ is raised to $q/2=3/2$, while $|V|=2$ is raised to $p/2=1$. Fixed scales, rather than the numerical values of the scales, determine the family.

Worked Example: A Wishart Scale Mixture Gives a Matrix-T Law

Take $p=q=2$, $\nu=5$, $M=0$, $U=\operatorname{diag}(1,4)$, and $V=I_2$. Generate $W\sim W_2(5,I_2)$, then draw $X\mid W\sim MN_{2,2}(0,U^{1/2}W^{-1}U^{1/2},I_2)$. Evaluate the marginal density at $X=\operatorname{diag}(1,2)$.

Because $W$ is random, this is the matrix-t construction. For the specified $X$, $$ U^{-1/2}X V^{-1}X^T U^{-1/2} =\operatorname{diag}(1,1), \qquad \left|I_2+U^{-1/2}X V^{-1}X^TU^{-1/2}\right|=4. $$ The determinant exponent is $-(\nu+q)/2=-7/2$, so this factor contributes $4^{-7/2}=1/128$. Also $|U|^{q/2}=4$ and $|V|^{p/2}=1$. Since $\Gamma_2(z)=\pi^{1/2}\Gamma(z)\Gamma(z-1/2)$, $$ \frac{\Gamma_2(7/2)}{\Gamma_2(5/2)} =\frac{\Gamma(7/2)\Gamma(3)}{\Gamma(5/2)\Gamma(2)} =\frac{5}{2}\cdot 2=5. $$ Thus the marginal density is $$ f(X)=\frac{5}{4\pi^2}\cdot\frac{1}{128} =\frac{5}{512\pi^2}. $$ Integrating out the random precision, rather than simply asserting that the conditional distribution is Gaussian, is what changes the marginal family to matrix-t.

Worked Example: A Normalized Wishart Ratio Gives a Matrix-Beta Law

Let $A\sim W_2(5,I_2)$ and $B\sim W_2(6,I_2)$ be independent. For a possible realization, suppose $S=A+B$ and the Cholesky-normalized matrix is $C=L^{-1}AL^{-T}=\operatorname{diag}(0.25,0.60)$, where $S=LL^T$. Identify its distribution and check the density factors.

The degrees of freedom satisfy $5>1$ and $6>1$, as required for $p=2$. Both eigenvalues of $C$ are positive and below one, so $C>0$ and $I_2-C>0$. The law is $Beta_2(5/2,3)$. The determinant factors are $$ |C|=0.25(0.60)=0.15,\qquad |I_2-C|=0.75(0.40)=0.30. $$ Their exponents are $(5-2-1)/2=1$ and $(6-2-1)/2=3/2$. Consequently, the density at this matrix is $$ f(C)=\frac{(0.15)^1(0.30)^{3/2}}{B_2(5/2,3)} =\frac{0.15(0.30)^{3/2}}{\Gamma_2(5/2)\Gamma_2(3)/\Gamma_2(11/2)}. $$ This is a density on symmetric matrices in the beta support, not a density for an unrestricted rectangular Gaussian matrix.

Why the Wishart Ratio Is Matrix-Beta

The beta density follows from a change of variables that separates the total $S=A+B$ from the normalized component $C$. The Wishart density for $A\sim W_p(a,I_p)$ is $$ f_A(A)=\frac{|A|^{(a-p-1)/2}\exp\{-\operatorname{tr}(A)/2\}}{2^{ap/2}\Gamma_p(a/2)}. $$ The corresponding formula for $B$ replaces $a$ by $b$. The condition $a,b>p-1$ ensures the stated Wishart distributions are proper.

For fixed $S$, the transformation $C\mapsto A=LCL^T$ has Jacobian $|L|^{p+1}=|S|^{(p+1)/2}$ on the space of symmetric matrices. This congruence Jacobian is the standard one for $H\mapsto LHL^T$. In the joint transformation, first replace $(A,B)$ by $(S,A)$, whose absolute Jacobian is one; then replace $A$ by $C$ holding $S$ fixed. The resulting joint Jacobian is therefore $|S|^{(p+1)/2}$.

Result: For independent $A\sim W_p(a,I_p)$, $B\sim W_p(b,I_p)$, with $a,b>p-1$, the Cholesky-normalized ratio $C=L^{-1}AL^{-T}$, $S=A+B=LL^T$, has density $Beta_p(a/2,b/2)$ and is independent of $S$.

To verify this result, use $A=LCL^T$ and $B=L(I_p-C)L^T$. Then $|A|=|S||C|$, $|B|=|S||I_p-C|$, and $\operatorname{tr}(A+B)=\operatorname{tr}(S)$. Multiplying the independent Wishart densities and the Jacobian gives a joint density proportional to $$ |S|^{(a+b-p-1)/2}e^{-\operatorname{tr}(S)/2} |C|^{(a-p-1)/2}|I_p-C|^{(b-p-1)/2}. $$ The first factor is the kernel of $W_p(a+b,I_p)$; the second is the matrix-beta kernel. The constants also separate: the product of the Wishart constants is $$ \frac{1}{2^{(a+b)p/2}\Gamma_p(a/2)\Gamma_p(b/2)}. $$ The Wishart density for $S$ contributes $1/[2^{(a+b)p/2}\Gamma_p((a+b)/2)]$, leaving the beta normalizer $1/B_p(a/2,b/2)=\Gamma_p((a+b)/2)/[\Gamma_p(a/2)\Gamma_p(b/2)]$. Thus the joint density factors into a density in $S$ and a density in $C$, proving both the claimed beta law and independence. $\square$

Deriving the Matrix-T Density From Mixing

The matrix-t density can likewise be checked from its construction, which prevents confusion between a fixed-scale normal and a scale mixture. Write $E=X-M$. The conditional matrix-normal density contributes the determinant factor $|W|^{q/2}$ and exponential term involving $W$. Multiplying it by the $W_p(\nu,I_p)$ density gives a kernel in $W$ proportional to $$ |W|^{(\nu+q-p-1)/2} \exp\!\left\{-\tfrac12\operatorname{tr}(WK)\right\}, \quad K=I_p+U^{-1/2}EV^{-1}E^TU^{-1/2}. $$ For positive-definite $K$ and $r>p-1$, the Wishart integral is $$ \int_{W>0}|W|^{(r-p-1)/2}e^{-\operatorname{tr}(WK)/2}\,dW =2^{rp/2}\Gamma_p(r/2)|K|^{-r/2}. $$ Here $r=\nu+q>p-1$, and $K$ is positive definite because it is the identity plus a positive-semidefinite matrix. Applying the integral and collecting the conditional and prior constants yields $$ f(X)= \frac{\Gamma_p((\nu+q)/2)} {\Gamma_p(\nu/2)\pi^{pq/2}|U|^{q/2}|V|^{p/2}} |K|^{-(\nu+q)/2}. $$ This is the matrix-t density stated above. The derivation shows precisely where the heavier-tailed determinant factor comes from: it is the result of integrating over the random precision $W$, not an extra term in a fixed matrix-normal model. $\square$

A Practical Choice and a Common Pitfall

When choosing among these families, start with the random object and its generating steps. A rectangular matrix with a fixed Gaussian mean and fixed row and column scales is matrix-normal. A rectangular matrix drawn conditionally as Gaussian but with a Wishart-distributed precision integrated out is matrix-t under the convention defined here. A symmetric matrix made by normalizing one of two independent Wishart components by their sum is matrix-beta type I, when the Wishart factors share the same scale after whitening.

Construction	Typical object	Key diagnostic
Gaussian with fixed $U,V$	Unrestricted $p\times q$ matrix	Scales are parameters, not random variables
Gaussian conditional on Wishart precision	Unrestricted $p\times q$ matrix	Integrating out the random precision gives a matrix-t law
Normalized independent Wishart ratio	Symmetric $p\times p$ matrix	Support is $C>0$ and $I_p-C>0$

Common mistake: “The observations are Gaussian, so the marginal matrix must be matrix-normal.” This reasoning ignores what is conditioned on. In a scale mixture, the conditional distribution is Gaussian, but its precision varies randomly; integrating out that precision produces the matrix-t density. The correction is to identify whether the scales are fixed or random before naming the marginal law.

A second tempting shortcut is to call any matrix formed from covariances or cross-products matrix-beta. That is not enough. The beta result above requires independent Wishart factors with compatible scale matrices and a specified normalization. Check the construction and the support; an arbitrary ratio, especially one involving different unadjusted scales, need not have the stated beta law.

Key takeaway: Classify the matrix by its construction, then verify dimensions, parameter conditions, and support. Fixed Gaussian scales point to matrix-normal; Wishart-mixed Gaussian precision points to matrix-t; a properly normalized independent Wishart ratio points to matrix-beta type I.

Check Your Understanding

For each question, focus on the generating mechanism as well as the matrix's shape.

A $3\times2$ matrix is Gaussian with fixed row and column scales. Which family applies, and what feature rules out a matrix-t mixture?
A Gaussian matrix is generated conditionally on a Wishart-distributed precision. What changes when the precision is integrated out?
For $p=3$, what inequalities on $a$ and $b$ are required for the two Wishart factors in the matrix-beta construction?
Why must a matrix-beta type I realization satisfy both $C>0$ and $I_p-C>0$?
Two independent Wishart matrices have different unadjusted scale matrices. Is the beta result above automatic? Explain which assumption needs checking.

Construction	Typical object	Key diagnostic
Gaussian with fixed \(U,V\)	Unrestricted \(p\times q\) matrix	Scales are parameters, not random variables
Gaussian conditional on Wishart precision	Unrestricted \(p\times q\) matrix	Integrating out the random precision gives a matrix-t law
Normalized independent Wishart ratio	Symmetric \(p\times p\) matrix	Support is \(C>0\) and \(I_p-C>0\)