Moments And Tail Behavior Of Matrix-T Distributions

Why Degrees of Freedom Matter

A matrix-t model can look similar to a matrix-normal model near its center while assigning much more probability to extreme values. Its degrees of freedom control that difference: they determine how slowly the tails decay and, consequently, whether means, variances, and higher moments are finite. These distinctions matter when interpreting a fitted scale matrix or deciding whether a covariance summary is mathematically defined.

This tutorial fixes a particular matrix-t scale convention, then derives its moment thresholds and covariance. The scale matrices in the density are not generally the covariance matrices; the conversion depends on the degrees of freedom.

Definition and Parameter Convention

Let $X$ be a $p\times q$ random matrix, let $M$ be a $p\times q$ location matrix, and let $U$ and $V$ be positive-definite $p\times p$ and $q\times q$ matrices, respectively. Let $\nu>0$ be the degrees of freedom, and write $d=pq$. We use the convention below throughout.

Definition: $X$ has a matrix-t distribution with location $M$, row scale $U$, column scale $V$, and $\nu$ degrees of freedom if its density, with respect to Lebesgue measure on the $pq$ entries, is $$ f(X)= \frac{\Gamma\!\left((\nu+d)/2\right)} {\Gamma(\nu/2)(\nu\pi)^{d/2}|U|^{q/2}|V|^{p/2}} \left[ 1+\frac{1}{\nu}\operatorname{tr}\!\left\{ U^{-1}(X-M)V^{-1}(X-M)^{T} \right\} \right]^{-(\nu+d)/2}. $$

Here $\Gamma$ is the gamma function. Under column-wise vectorization, the scale matrix associated with the quadratic form is $V\otimes U$. In particular, $U$ and $V$ are scale parameters, not the row and column covariance matrices of $X$. We will derive that covariance exists only for $\nu>2$, and then includes the multiplier $\nu/(\nu-2)$.

Tail Decay and Moment Existence

Whiten the centered matrix by setting $Y=U^{-1/2}(X-M)V^{-1/2}$, using the symmetric positive-definite square roots. The quadratic form in the density becomes $\|Y\|_F^2$, where $\|Y\|_F$ is the Frobenius norm. Apart from its normalizing constant, the density of the $d$-dimensional vector of entries of $Y$ is therefore $(1+\|Y\|_F^2/\nu)^{-(\nu+d)/2}$.

Theorem: For any real $s\geq 0$, the absolute radial moment $E[\|X-M\|_F^s]$ is finite if and only if $s<\nu$. In particular, the matrix mean exists entrywise if and only if $\nu>1$; second moments exist if and only if $\nu>2$. The whitened radial tail has order $r^{-\nu}$: $P(\|Y\|_F>r)$ is asymptotic to a positive constant times $r^{-\nu}$.

Proof. For $Y$, integration in $d$-dimensional polar coordinates reduces the finiteness of $E[\|Y\|_F^s]$ to the finiteness of

\int_0^\infty r^{d-1+s} \left(1+\frac{r^2}{\nu}\right)^{-(\nu+d)/2}\,dr.

Near zero, the integrand is of order $r^{d-1+s}$, which is integrable because $d\geq1$ and $s\geq0$. As $r$ tends to infinity, it is of order $r^{s-\nu-1}$. The integral over large $r$ is finite exactly when $s-\nu-1<-1$, or $s<\nu$. The linear transformation between $Y$ and $X-M$ is invertible. Positive definiteness of $U,V$ gives positive constants $a,b$ such that $a\|Y\|_F\leq\|X-M\|_F\leq b\|Y\|_F$. Thus the two matrices have the same moment-existence thresholds.

For the tail statement, the radial density of $\|Y\|_F$ is proportional to $r^{d-1}(1+r^2/\nu)^{-(\nu+d)/2}$. Its large-$r$ order is $r^{-\nu-1}$. Integrating this tail from $r$ to infinity gives order $r^{-\nu}$, with a positive finite leading constant. Finally, each entry of $X-M$ is a nonzero linear functional of the whitened vector. Its absolute moments have the same threshold; equivalently, the entrywise mean requires $\nu>1$, and entrywise second moments require $\nu>2$. At equality, the relevant absolute-moment integral diverges. $\square$

The strict inequalities are important. Symmetry of the density does not create an expectation when the absolute first moment is infinite. For $\nu\leq1$, a symmetric principal-value calculation may suggest the location $M$, but the ordinary mean is undefined.

Worked Example: A Mean Without a Covariance

Let $X$ be a $2\times1$ matrix-t variable with $M=0$, $U=I_2$, $V=[1]$, and $\nu=1.5$. Which moments exist?

Here $d=2$. For the first absolute radial moment, the threshold is $1<1.5$, so $E[\|X\|_F]$ is finite. Symmetry about zero then gives $E[X]=0$.

For a second moment, the required condition would be $2<1.5$, which is false. Thus the entries have no finite variances, and a covariance matrix for $X$ does not exist. The density is centered at zero, but zero is its mean only because the first moment exists; it does not imply finite variance.

Deriving the Mean and Covariance

Theorem: Under the stated matrix-t convention, if $\nu>1$, then $E[X]=M$. If $\nu>2$, then $$ \operatorname{Cov}(\operatorname{vec}X) =\frac{\nu}{\nu-2}(V\otimes U), \qquad \operatorname{Cov}(X_{ij},X_{kl}) =\frac{\nu}{\nu-2}U_{ik}V_{jl}. $$ For $\nu\leq1$, the mean is undefined; for $1<\nu\leq2$, the mean exists but the covariance is undefined.

Proof. In whitened coordinates, the density of $Y$ depends only on $\|Y\|_F$, so it is invariant under orthogonal transformations of its $d$ entries. For $\nu>1$, the first absolute moment is finite by the preceding result. Symmetry under $Y\mapsto -Y$ therefore gives $E[Y]=0$, and the inverse whitening transformation gives $E[X]=M$.

For the covariance, suppose $\nu>2$. Orthogonal invariance implies that the covariance matrix of $\operatorname{vec}Y$ is $cI_d$ for some $c$, and hence $E[\|Y\|_F^2]=dc$. To calculate the radial second moment, define

I_k=\int_0^\infty r^{d-1+k} \left(1+\frac{r^2}{\nu}\right)^{-(\nu+d)/2}\,dr.

With $t=r^2/\nu$, the beta integral gives

I_k=\frac{\nu^{(d+k)/2}}{2} B\!\left(\frac{d+k}{2},\frac{\nu-k}{2}\right), \qquad k<\nu.

Consequently, for $\nu>2$,

E[\|Y\|_F^2] =\frac{I_2}{I_0} =\nu\, \frac{B((d+2)/2,(\nu-2)/2)} {B(d/2,\nu/2)} =\frac{\nu d}{\nu-2}.

The last equality follows from $B(a,b)=\Gamma(a)\Gamma(b)/\Gamma(a+b)$ and $\Gamma(z+1)=z\Gamma(z)$. Thus $c=E[\|Y\|_F^2]/d=\nu/(\nu-2)$, so $\operatorname{Cov}(\operatorname{vec}Y)=\nu I_d/(\nu-2)$. Since $\operatorname{vec}(X-M)=(V^{1/2}\otimes U^{1/2})\operatorname{vec}Y$, the covariance transformation rule gives $\operatorname{Cov}(\operatorname{vec}X)=\nu(V\otimes U)/(\nu-2)$. Reading its entries in column-wise vectorization yields $\operatorname{Cov}(X_{ij},X_{kl})=\nu U_{ik}V_{jl}/(\nu-2)$. For $\nu\leq2$, second moments do not exist by the moment threshold already proved. $\square$

Worked Example: Entrywise Covariances

Suppose $X$ is $2\times2$, with $\nu=3$, $M=0$, and $U=\begin{pmatrix}2&0.5\\0.5&1\end{pmatrix}$, $V=\begin{pmatrix}1&0.2\\0.2&3\end{pmatrix}$. Find $\operatorname{Var}(X_{12})$ and $\operatorname{Cov}(X_{11},X_{22})$.

Because $3>2$, the covariance exists, and its multiplier is $\nu/(\nu-2)=3/(3-2)=3$. For the variance, use $i=k=1$ and $j=l=2$:

\operatorname{Var}(X_{12}) =3U_{11}V_{22} =3(2)(3)=18.

For the covariance of $X_{11}$ and $X_{22}$, use $i=1,j=1,k=2,l=2$:

\operatorname{Cov}(X_{11},X_{22}) =3U_{12}V_{12} =3(0.5)(0.2)=0.3.

This calculation uses both off-diagonal scale entries. It also shows why it is not generally correct to treat the matrix entries as independent.

Worked Example: Which Norm Moments Are Finite?

Let $X$ be a $2\times3$ matrix-t variable with $M=0$, $U=I_2$, $V=I_3$, and $\nu=5$. Determine the tail order and whether its second, fourth, and sixth radial moments exist.

The dimension is $d=pq=6$. The whitened matrix is $Y=X$, and its radial tail has order $r^{-5}$. A radial moment of order $s$ is finite exactly when $s<5$. Therefore the second and fourth moments are finite, while the sixth moment is infinite.

In particular, the covariance exists and $\operatorname{Cov}(\operatorname{vec}X)=5I_6/(5-2)=5I_6/3$. The second radial moment can also be checked directly from the theorem:

E[\|X\|_F^2] =\frac{d\nu}{\nu-2} =\frac{6(5)}{3}=10.

The finite second moment does not imply that all higher moments exist: in this example, the fourth is finite but the sixth is not.

Common Pitfall: Calling the Scale a Covariance

A frequent error is to read $U$ and $V$ as covariance matrices and report $\operatorname{Cov}(\operatorname{vec}X)=V\otimes U$. Under the density convention used here, that omits the factor $\nu/(\nu-2)$, and the covariance formula is valid only when $\nu>2$. For $1<\nu\leq2$, the mean exists but there is no finite covariance to rescale. For $\nu\leq1$, even the mean is undefined.

The degrees of freedom also give a useful tail diagnostic. Increasing $\nu$ increases the tail exponent $\nu$, allowing more moments to exist; it does not change the moment thresholds into non-strict inequalities. A moment of order exactly $\nu$ still diverges.

Key takeaway: Under this matrix-t convention, absolute moments of order $s$ exist precisely when $s<\nu$. The mean is $M$ for $\nu>1$, and for $\nu>2$ the covariance is $\nu(V\otimes U)/(\nu-2)$, not merely $V\otimes U$.

Check Your Understanding

Use the density and scale convention stated above.

For a matrix-t variable with $\nu=2$, does its mean exist? Does its covariance exist?
What is the tail order of the whitened Frobenius norm when $\nu=7$?
If $\nu=4$, which absolute radial moments of orders one, three, and four are finite?
For $\nu>2$, explain why the row and column scale matrices alone do not give the covariance of $\operatorname{vec}X$.
For a $p\times q$ matrix-t variable, what is the dimension $d$ used in the radial-density argument, and does it change the moment threshold?