Multivariate Student Distributions as Elliptical Examples

Why the Multivariate Student Distribution Matters

The multivariate Student distribution is an important elliptical model for data that may produce more extreme observations than a multivariate normal model. Its density has ellipsoidal contours, so location and scatter can be interpreted geometrically, but its tails decay at a polynomial rather than an exponential rate. The degrees of freedom parameter controls how heavy those tails are and whether moments such as the variance exist.

We derive its density from a normal vector scaled by an independent chi-square variable. The derivation gives the normalizing constant as well as the form of the density generator. We then use the same construction to establish the moment conditions, rather than inferring them from the visual appearance of the density.

Definition and Density

Let $p$ be a positive integer, $\nu>0$, $\mu\in\mathbb{R}^p$, and let $\Sigma$ be a symmetric positive-definite $p\times p$ matrix. Write $|\Sigma|$ for its determinant. For $x\in\mathbb{R}^p$, define the squared Mahalanobis distance from $\mu$ using $\Sigma$ by

\delta(x)=(x-\mu)^{\mathsf T}\Sigma^{-1}(x-\mu).

Definition: The $p$-dimensional Student distribution with degrees of freedom $\nu$, location $\mu$, and scatter matrix $\Sigma$, written $X\sim t_p(\nu,\mu,\Sigma)$, has density $$ f(x)=\frac{\Gamma((\nu+p)/2)}{\Gamma(\nu/2)(\nu\pi)^{p/2}|\Sigma|^{1/2}} \left(1+\frac{\delta(x)}{\nu}\right)^{-(\nu+p)/2},\qquad x\in\mathbb{R}^p. $$ Here $\Gamma$ is the gamma function. The matrix $\Sigma$ is called the scatter matrix; it is not generally the covariance matrix.

This density is elliptical: apart from its normalizing constant, it is a function of $\delta(x)$ alone. The set of points having a fixed positive value $c$ of $\delta(x)$ is an ellipsoid centered at $\mu$. The eigenvectors of $\Sigma$ give its principal directions, and the eigenvalues determine the squared lengths along those directions, up to the common contour factor $c$. The density is largest at $x=\mu$, so $\mu$ is its unique mode and its center of symmetry.

The scatter matrix describes the shape and scale in the density, whether or not a covariance exists. When covariance does exist, it is a degrees-of-freedom-dependent multiple of $\Sigma$, as we establish below.

Deriving the Density

Let $Z\sim N_p(0,I_p)$ and $U\sim\chi^2_\nu$ be independent, where $I_p$ is the $p\times p$ identity matrix. Choose a matrix $A$ with $AA^{\mathsf T}=\Sigma$, and define

X=\mu+A\sqrt{\frac{\nu}{U}}\,Z.

Because $\Sigma$ is positive definite, such an $A$ exists and is nonsingular. For each fixed $u>0$, the conditional density of $X$ given $U=u$ is multivariate normal with mean $\mu$ and covariance matrix $(\nu/u)\Sigma$. Its density at $x$ is

f_{X\mid U=u}(x)= (2\pi)^{-p/2}|\Sigma|^{-1/2} \left(\frac{u}{\nu}\right)^{p/2} \exp\left(-\frac{u\,\delta(x)}{2\nu}\right).

The density of $U$ is $u^{\nu/2-1}e^{-u/2}/(2^{\nu/2}\Gamma(\nu/2))$ for $u>0$. Integrating the conditional density against this density gives the marginal density of $X$. All integrands are nonnegative, so this conditioning and integration is justified by Tonelli's theorem.

\begin{aligned} f(x) &=\frac{(2\pi)^{-p/2}|\Sigma|^{-1/2}\nu^{-p/2}} {2^{\nu/2}\Gamma(\nu/2)} \int_0^\infty u^{(\nu+p)/2-1} \exp\left\{-\frac{u}{2}\left(1+\frac{\delta(x)}{\nu}\right)\right\}\,du\\ &=\frac{(2\pi)^{-p/2}|\Sigma|^{-1/2}\nu^{-p/2}} {2^{\nu/2}\Gamma(\nu/2)} \frac{\Gamma((\nu+p)/2)} {\left[\frac12\left(1+\frac{\delta(x)}{\nu}\right)\right]^{(\nu+p)/2}}\\ &=\frac{\Gamma((\nu+p)/2)} {\Gamma(\nu/2)(\nu\pi)^{p/2}|\Sigma|^{1/2}} \left(1+\frac{\delta(x)}{\nu}\right)^{-(\nu+p)/2}. \end{aligned}

The second equality uses the gamma integral $\int_0^\infty u^{a-1}e^{-bu}\,du=\Gamma(a)b^{-a}$, valid for $a,b>0$. Here $a=(\nu+p)/2>0$ and $b=(1+\delta(x)/\nu)/2>0$. The resulting expression is the stated multivariate Student density.

Worked Example: Evaluating a Bivariate Student Density

Suppose $X\sim t_2(5,\mu,\Sigma)$, where $\mu=(0,0)^{\mathsf T}$ and $\Sigma=\operatorname{diag}(4,1)$. Find the density at $x=(2,1)^{\mathsf T}$.

The determinant is $|\Sigma|=4$, and $\Sigma^{-1}=\operatorname{diag}(1/4,1)$. Thus

\delta(x)= \begin{pmatrix}2&1\end{pmatrix} \begin{pmatrix}1/4&0\\0&1\end{pmatrix} \begin{pmatrix}2\\1\end{pmatrix} =\frac{2^2}{4}+1^2=2.

Substitute $p=2$, $\nu=5$, and $|\Sigma|^{1/2}=2$ into the density. Since $\Gamma(7/2)/\Gamma(5/2)=5/2$, its constant is $(5/2)/(5\pi\cdot2)=1/(4\pi)$. Therefore

f(2,1)=\frac{1}{4\pi}\left(1+\frac{2}{5}\right)^{-7/2} =\frac{1}{4\pi}\left(\frac{7}{5}\right)^{-7/2}.

The larger first diagonal entry of $\Sigma$ represents greater scatter in the first coordinate direction. The calculation also shows why the distance uses $\Sigma^{-1}$: a displacement of 2 in the first coordinate contributes $2^2/4=1$, rather than $2^2$, to the squared Mahalanobis distance.

Degrees of Freedom and Tail Decay

The generator in the density is $g(s)=(1+s/\nu)^{-(\nu+p)/2}$, apart from the normalizing constant. To see its tail behavior, fix a nonzero direction $v$ and consider points $x=\mu+rv$ as $r$ grows. Then $\delta(\mu+rv)=r^2v^{\mathsf T}\Sigma^{-1}v$, with $v^{\mathsf T}\Sigma^{-1}v>0$. Consequently, along this direction the density decays at rate $r^{-(\nu+p)}$. Larger $\nu$ gives faster decay; smaller $\nu$ gives heavier tails.

A useful way to establish exactly which moments are finite is to examine the construction. Let $R=\|A^{-1}(X-\mu)\|$, where $\|\cdot\|$ is Euclidean length. Then $R=\sqrt{\nu/U}\,\|Z\|$. The squared length $V=\|Z\|^2$ has a $\chi^2_p$ distribution and is independent of $U$.

Theorem: For $q>0$, $E[\|A^{-1}(X-\mu)\|^q]$ is finite if and only if $q<\nu$. In particular, the mean of $X$ exists for $\nu>1$; its covariance exists for $\nu>2$ and is $$ \operatorname{Cov}(X)=\frac{\nu}{\nu-2}\Sigma. $$

Proof. Since $R^q=\nu^{q/2}V^{q/2}U^{-q/2}$ and $V,U$ are independent, its expectation factors into a finite positive constant $ \nu^{q/2}E[V^{q/2}] $ times $E[U^{-q/2}]$, allowing the latter expectation to be infinite. The first expectation is finite for every $q>0$, because the chi-square density has an exponentially decaying tail and its power near zero is integrable here. For the second, the chi-square density shows that the integrand near zero is proportional to $u^{(\nu-q)/2-1}$. Its integral at zero is finite exactly when $(\nu-q)/2-1>-1$, or $q<\nu$; at infinity, the exponential factor makes the integral finite. This proves the moment criterion.

For $q=1$, this criterion gives finite first absolute moments of the coordinates precisely when $\nu>1$, so the mean exists then. Symmetry of $Z$ around zero gives $E[X]=\mu$. For $\nu>2$, the gamma integral yields $E[U^{-1}]=1/(\nu-2)$, and $E[ZZ^{\mathsf T}]=I_p$. Independence therefore gives

\operatorname{Cov}(X) =A\,E\left[\frac{\nu}{U}ZZ^{\mathsf T}\right]A^{\mathsf T} =\frac{\nu}{\nu-2}AA^{\mathsf T} =\frac{\nu}{\nu-2}\Sigma.

If $\nu\leq2$, the second moment of the distance from $\mu$ is infinite, so a finite covariance matrix does not exist. If $\nu\leq1$, even the first absolute moments are infinite, and symmetry does not make an ordinary expectation exist. $\square$

Worked Example: Comparing Moment Existence

Consider two univariate Student distributions centered at zero with the same scatter $\sigma^2=4$, but with degrees of freedom $\nu=3$ and $\nu=5$. Compare their means, variances, and density decay.

For $\nu=3$, the mean exists because $3>1$, and symmetry makes it zero. The variance exists because $3>2$, and is

\operatorname{Var}(X)=\frac{3}{3-2}\,4=12.

For $\nu=5$, the mean also exists and is zero. Its variance is

\operatorname{Var}(X)=\frac{5}{5-2}\,4=\frac{20}{3}.

In dimension one, the density along either tail decays as $|x|^{-(\nu+1)}$. The $\nu=3$ density therefore has rate $|x|^{-4}$, while the $\nu=5$ density has rate $|x|^{-6}$. The smaller degrees of freedom produce heavier tails, even though both distributions have finite variances.

Worked Example: Reading Scatter and Covariance Separately

Let $X\sim t_2(4,\mu,\Sigma)$, with $\mu=(1,-1)^{\mathsf T}$ and $\Sigma=\operatorname{diag}(9,4)$. Find the squared Mahalanobis distance and density at $x=(4,3)^{\mathsf T}$, and identify the covariance matrix.

The displacement is $x-\mu=(3,4)^{\mathsf T}$, and $\Sigma^{-1}=\operatorname{diag}(1/9,1/4)$. Hence

\delta(x)=\frac{3^2}{9}+\frac{4^2}{4}=1+4=5.

Here $|\Sigma|^{1/2}=6$. The density constant is $\Gamma(3)/[\Gamma(2)(4\pi)6]=2/(24\pi)=1/(12\pi)$, so

f(x)=\frac{1}{12\pi}\left(1+\frac{5}{4}\right)^{-3} =\frac{1}{12\pi}\left(\frac{4}{9}\right)^3 =\frac{16}{2187\pi}.

Since $\nu=4>2$, the covariance exists and equals $\frac{4}{4-2}\Sigma=2\Sigma=\operatorname{diag}(18,8)$. Thus $\Sigma$ controls the elliptical shape in the density, whereas the covariance is twice that scatter matrix for these degrees of freedom.

Common Pitfall: Calling Scatter the Covariance

A common mistake is to read the matrix $\Sigma$ in the density as the covariance matrix, just as one might for a multivariate normal distribution. That conclusion is wrong for the Student distribution. The matrix is always the scatter parameter, but covariance exists only when $\nu>2$, and then it is $\nu\Sigma/(\nu-2)$. For example, at $\nu=3$, the covariance is $3\Sigma$, not $\Sigma$; at $\nu=2$, the scatter remains well-defined but the covariance does not.

The distinction is useful in practice: a Student model can retain a well-defined location and elliptical shape even when its covariance is infinite. Degrees of freedom are not merely a shape adjustment to the center of the density; they determine tail decay and which moment-based summaries are available.

Key takeaway: In $t_p(\nu,\mu,\Sigma)$, $\mu$ is the location and mode, while $\Sigma$ is the scatter matrix. The density decays polynomially, and covariance equals $\nu\Sigma/(\nu-2)$ only when $\nu>2$.

Check Your Understanding

Use the density and moment conditions developed above.

For $X\sim t_3(6,\mu,\Sigma)$, does the mean exist? Does the covariance exist?
In a $p$-dimensional Student density, what is the rate of decay along a fixed ray moving away from its location?
For $\nu=4$, express the covariance matrix in terms of the scatter matrix $\Sigma$.
Why does a symmetric Student distribution with $\nu\leq1$ not have an ordinary mean?
Describe how changing $\nu$ affects tail heaviness while leaving $\mu$ and $\Sigma$ fixed.