Conditional Laws Within Elliptical Families

Why Conditioning Changes an Elliptical Law

For a Gaussian vector, conditioning shifts the center and reduces the scatter in a precise way. In a general elliptical family, the same partitioned-matrix geometry applies, but the radial generator also changes: the observed coordinates add a fixed amount to the quadratic distance in the joint density. Thus the conditional law is elliptical, but it need not retain the original generator.

This distinction matters in multivariate modeling. The conditional location describes where the unobserved block is centered after observing the other block; the Schur complement describes its contour shape; and the generator determines how probability is distributed across those contours. For the multivariate t distribution, conditioning updates the location, rescales the scatter, and increases the degrees of freedom.

Partitioning the Density

Let $X=(X_1',X_2')'$, where $X_1$ has dimension $p$ and $X_2$ has dimension $q$. Suppose $X$ has an elliptical density with location $\mu=(\mu_1',\mu_2')'$, positive-definite scatter matrix $\Sigma$, and generator $g_{p+q}$:

f(x_1,x_2)=|\Sigma|^{-1/2}g_{p+q}\!\left((x-\mu)'\Sigma^{-1}(x-\mu)\right).

Assume this is a proper density, so its integral over $\mathbb{R}^{p+q}$ is one. Partition the scatter matrix conformably:

\Sigma= \begin{pmatrix} \Sigma_{11}&\Sigma_{12}\\ \Sigma_{21}&\Sigma_{22} \end{pmatrix}, \qquad \Sigma_{21}=\Sigma_{12}'.

Because $\Sigma$ is positive definite, $\Sigma_{22}$ is positive definite. Define the Schur complement $S=\Sigma_{11.2}$, the conditional location $m(x_2)$, and the observed quadratic distance $\delta_2$ by

S=\Sigma_{11}-\Sigma_{12}\Sigma_{22}^{-1}\Sigma_{21},\qquad m(x_2)=\mu_1+\Sigma_{12}\Sigma_{22}^{-1}(x_2-\mu_2),\qquad \delta_2=(x_2-\mu_2)'\Sigma_{22}^{-1}(x_2-\mu_2).

The Schur complement $S$ is positive definite. It gives the shape of the conditional elliptical contours, while $m(x_2)$ is their center. The quadratic distance $\delta_2$ records how far the observed block lies from its location in the $\Sigma_{22}$ metric.

Definition: For $t\geq 0$, define the integrated generator $H_q(t)=\int_{\mathbb{R}^p}g_{p+q}(t+u'u)\,du$, whenever this integral is finite. It integrates out the $p$-dimensional block while holding the other block's quadratic distance fixed.

The Conditional Density Formula

The block inverse identity gives the joint quadratic form as the sum of the observed block's distance and the residual distance from the conditional center:

(x-\mu)'\Sigma^{-1}(x-\mu) = \delta_2+ \{x_1-m(x_2)\}'S^{-1}\{x_1-m(x_2)\}.

Also, the block determinant identity gives $|\Sigma|=|\Sigma_{22}||S|$. These two identities isolate the role of each block in the joint density.

Theorem: Suppose the joint density above is proper, and for a given $x_2$, $0<H_q(\delta_2)<\infty$. Then the marginal density of $X_2$ at $x_2$ is $f_2(x_2)=|\Sigma_{22}|^{-1/2}H_q(\delta_2)$. A conditional density of $X_1$ given $X_2=x_2$ is $$ f_{1\mid 2}(x_1\mid x_2) = |S|^{-1/2} \frac{g_{p+q}\!\left(\delta_2+r\right)}{H_q(\delta_2)}, \qquad r=\{x_1-m(x_2)\}'S^{-1}\{x_1-m(x_2)\}. $$ This formula applies at values $x_2$ where the marginal density is positive and finite; conditional densities are determined only up to sets of $x_2$ having marginal probability zero.

To prove the result, fix $x_2$ and write $b=x_2-\mu_2$. In the marginal integral over $x_1$, make the change of variables $u=S^{-1/2}(x_1-m(x_2))$, where $S^{-1/2}$ is the symmetric positive-definite inverse square root. The Jacobian gives $dx_1=|S|^{1/2}du$, and the quadratic form becomes $\delta_2+u'u$. Therefore

\begin{aligned} f_2(x_2) &=\int_{\mathbb{R}^p}|\Sigma|^{-1/2} g_{p+q}\!\left(\delta_2+\{x_1-m(x_2)\}'S^{-1}\{x_1-m(x_2)\}\right)\,dx_1\\ &=|\Sigma_{22}|^{-1/2}\int_{\mathbb{R}^p}g_{p+q}(\delta_2+u'u)\,du\\ &=|\Sigma_{22}|^{-1/2}H_q(\delta_2). \end{aligned}

Dividing the joint density by this positive marginal density and using the determinant identity yields the displayed conditional formula. It integrates to one: applying the same change of variables to its integral over $x_1$ gives $H_q(\delta_2)/H_q(\delta_2)=1$. This proves the theorem. $\square$

The formula shows exactly what is and is not preserved. The conditional contours are centered at $m(x_2)$ and shaped by $S$, but their radial density is proportional to $g_{p+q}(\delta_2+r)$. The observed distance shifts the argument of the generator. In general, this shifted generator does not belong to the same family as the original one.

Worked Example: Conditional Geometry with a Partitioned Scatter Matrix

Let $X_1$ and $X_2$ be scalar components with location $\mu=(0,0)'$, scatter matrix $\Sigma=\begin{pmatrix}5&2\\2&3\end{pmatrix}$, and any proper elliptical generator $g_2$. Find the conditional location, scatter, observed distance, and generator form when $X_2=x_2$.

The block for the observed component is $\Sigma_{22}=3$, and the cross-block is $\Sigma_{12}=2$. Hence $S=5-2(1/3)2=11/3$, while $m(x_2)=0+(2/3)x_2$ and $\delta_2=x_2^2/3$.

The residual quadratic distance is $r=(x_1-2x_2/3)^2/(11/3)$. Thus, wherever $H_1(x_2^2/3)$ is positive and finite,

f_{1\mid 2}(x_1\mid x_2) = \sqrt{\frac{3}{11}}\, \frac{g_2\!\left(x_2^2/3+(x_1-2x_2/3)^2/(11/3)\right)} {H_1(x_2^2/3)}.

The conditional center changes linearly with $x_2$, and the conditional contour scale is $11/3$. The generator also depends on $x_2$ through its shifted argument and normalizing integral; without specifying $g_2$, one cannot claim that the conditional law retains a particular named family.

A Concrete Non-t Example

A compactly supported elliptical distribution makes the generator shift especially visible. Let $(X_1,X_2)$ be uniform on the unit disk, with density $1/\pi$ when $x_1^2+x_2^2<1$ and zero otherwise. This is an elliptical density with location zero, identity scatter, and a generator that is constant inside the unit-radius support and zero outside.

Worked Example: Conditioning a Uniform Distribution on a Disk

Find the conditional distribution of $X_1$ given $X_2=0.6$. Here $S=1$, $m(0.6)=0$, and $\delta_2=0.6^2=0.36$. The conditional support requires $\delta_2+x_1^2<1$, or $|x_1|<\sqrt{0.64}=0.8$.

The marginal density of $X_2$ at $0.6$ is the joint density integrated across the permitted horizontal chord:

f_2(0.6)=\int_{-0.8}^{0.8}\frac{1}{\pi}\,dx_1=\frac{1.6}{\pi}.

Consequently the conditional density is $(1/\pi)/(1.6/\pi)=1/1.6=0.625$ on $(-0.8,0.8)$, and zero outside. Its integral is $0.625(1.6)=1$. Observing $X_2=0.6$ has shortened the conditional support; the conditional law is uniform on that shorter interval, not uniform on the original interval $(-1,1)$.

The Multivariate t Update

For a $d$-dimensional multivariate t distribution with degrees of freedom $\nu>0$, location $\mu$, and positive-definite scatter $\Sigma$, use the density

f(x)= \frac{\Gamma((\nu+d)/2)} {\Gamma(\nu/2)(\nu\pi)^{d/2}|\Sigma|^{1/2}} \left(1+\frac{(x-\mu)'\Sigma^{-1}(x-\mu)}{\nu}\right)^{-(\nu+d)/2}.

Here $\Sigma$ is the t scatter parameter, not generally the covariance matrix. The density has a generator proportional to $(1+q/\nu)^{-(\nu+d)/2}$. For the partition above, write $d=p+q$ and let $S$, $m(x_2)$, and $\delta_2$ have their previous meanings.

Theorem: If $X$ has this $d$-dimensional multivariate t distribution, then, at every $x_2$, the conditional distribution is $$ X_1\mid X_2=x_2 \sim t_p\!\left( \nu+q,\; m(x_2),\; \frac{\nu+\delta_2}{\nu+q}S \right). $$ The three entries denote degrees of freedom, location, and scatter, respectively.

To derive the result, substitute the t generator into the conditional density formula. Apart from factors that do not depend on $x_1$, the conditional kernel is

\left(1+\frac{\delta_2+r}{\nu}\right)^{-(\nu+p+q)/2} = \left(1+\frac{\delta_2}{\nu}\right)^{-(\nu+p+q)/2} \left(1+\frac{r}{\nu+\delta_2}\right)^{-(\nu+p+q)/2}.

The first factor is constant in $x_1$ and is absorbed into the conditional normalizing constant. Put $\nu_*=\nu+q$ and $c=(\nu+\delta_2)/(\nu+q)$. Since $c\nu_*=\nu+\delta_2$, the remaining kernel is

\left(1+\frac{r}{\nu+\delta_2}\right)^{-(\nu_*+p)/2} = \left(1+\frac{\{x_1-m(x_2)\}'S^{-1}\{x_1-m(x_2)\}}{c\nu_*}\right)^{-(\nu_*+p)/2}.

This is the kernel of a $p$-dimensional t density with degrees of freedom $\nu_*$, location $m(x_2)$, and scatter $cS$: its inverse scatter in the quadratic form is $(cS)^{-1}=c^{-1}S^{-1}$. The conditional density formula has already established that this kernel is normalized to integrate to one. The t density with these parameters is also normalized, so the two normalized densities agree. This proves the update. $\square$

The degrees of freedom increase by $q$, the dimension of the observed block. The conditional scatter is multiplied by $c=(\nu+\delta_2)/(\nu+q)$, so it depends both on how far the observation lies from $\mu_2$ and on the dimension conditioned upon. It is not automatically the conditional covariance: that covariance exists only when $\nu+q>2$, and then equals $\nu_*cS/(\nu_*-2)$.

Worked Example: Conditioning a Three-Dimensional t Distribution

Let $X$ have a three-dimensional t distribution with $\nu=5$, zero location, and scatter $\Sigma=\begin{pmatrix}4&1&0.5\\1&2&0\\0.5&0&1\end{pmatrix}$. Take $X_1$ to be the first coordinate and observe $X_2=(X_2^{(1)},X_2^{(2)})'=(2,1)'$.

The observed-block scatter is $\Sigma_{22}=\operatorname{diag}(2,1)$, and the cross-block is $\Sigma_{12}=(1,0.5)$. Thus $S=4-(1,0.5)\operatorname{diag}(1/2,1)(1,0.5)'=4-(0.5+0.25)=13/4$. The conditional location is $(1,0.5)\operatorname{diag}(1/2,1)(2,1)'=1+0.5=1.5$. The observed distance is $\delta_2=(2,1)\operatorname{diag}(1/2,1)(2,1)'=2+1=3$.

There are $q=2$ observed coordinates, so the conditional degrees of freedom are $5+2=7$, and the conditional scatter is $(5+3)/(5+2)\times(13/4)=26/7$. Therefore $X_1\mid X_2=(2,1)'$ is a univariate t distribution with 7 degrees of freedom, location $1.5$, and scatter $26/7$. Its density is

f_{1\mid 2}(x_1\mid (2,1)') = \frac{\Gamma(4)}{\Gamma(7/2)\sqrt{26\pi}} \left(1+\frac{(x_1-1.5)^2}{26}\right)^{-4}.

The denominator inside the power is $7(26/7)=26$, as required by the t density parameterization. The location, scale, and degrees of freedom each reflect a different part of the conditioning update.

Common Pitfall and Practical Meaning

A common mistake is to use the Gaussian conditional formula and conclude that every elliptical conditional law has the same generator as its joint law. The matrix calculation for the conditional center and Schur complement is useful, but it does not determine the radial density. The correct conditional generator is proportional to $g_{p+q}(\delta_2+r)$, normalized by $H_q(\delta_2)$. Only after examining that expression can one identify the conditional family.

For the t case, another error is to leave the degrees of freedom unchanged or to use $\delta_2$ as though it were a scalar residual without reference to $\Sigma_{22}$. The degrees of freedom increase by the number of conditioned coordinates, and $\delta_2$ must be computed with $\Sigma_{22}^{-1}$. These details affect both the conditional scatter and tail behavior.

Key takeaway: In an elliptical partition, the Schur complement determines conditional contour shape, the cross-block determines conditional location, and the observed quadratic distance shifts the generator. For a multivariate t law, this yields degrees of freedom $\nu+q$ and scatter $(\nu+\delta_2)S/(\nu+q)$.

Check Your Understanding

Use the definitions and derivations above to answer the following questions.

In the conditional density formula, what does $H_q(\delta_2)$ integrate out, and why must it be positive and finite?
For a partitioned scatter matrix, identify which matrix determines the conditional contour shape and which terms determine the conditional location.
For a multivariate t vector with degrees of freedom $\nu$, what are the conditional degrees of freedom after observing $q$ coordinates?
In the three-dimensional t example, recompute the conditional scatter if the observed quadratic distance were $7$ rather than $3$, leaving the other parameters unchanged.
Why does the conditional density of the uniform disk example have a shorter support after observing $X_2=0.6$?