Affine Closure of Elliptical Families

Why Affine Closure Matters

Elliptical distributions extend familiar features of multivariate normal distributions to laws whose contours and dependence structure are organized by a location and a scatter matrix. A linear map can rotate, stretch, shear, project, or embed a random vector. Knowing that these operations preserve ellipticity lets us describe the transformed law without finding its density or constructing a new random vector from scratch.

The key rule is concise: under $Y=b+BX$, the location becomes $b+B\mu$, and a scatter matrix $S$ becomes $BSB^\top$. The rule applies even when $B$ is rectangular or singular. Singularity can reduce the dimension of the set on which the transformed vector lies; it does not, by itself, destroy ellipticity.

Definition and Transformation Rule

Definition: A random vector $X$ in $\mathbb{R}^p$ has an elliptical distribution with location $\mu$ and scatter matrix $S$ if its characteristic function has the form $$ \varphi_X(t)=E[\exp(i t^\top X)] =\exp(i t^\top\mu)\,\psi(t^\top S t), \qquad t\in\mathbb{R}^p, $$ where $S$ is a symmetric positive semidefinite $p\times p$ matrix and $\psi$ is a scalar characteristic generator for which the displayed function is a characteristic function. The law may be singular; $S$ is called nonsingular when it is positive definite, but this alone does not imply that the law has a Lebesgue density. The location and scatter are parameters in this representation; location need not equal the mean, and scatter need not equal the covariance.

This definition is consistent with the affine-image description of elliptical laws: if $X=\mu+AZ$, where $Z$ is spherical and $A$ is a matrix, then $S=AA^\top$ is a scatter matrix. The generator and scatter parameterization need not be unique, since a rescaling of $S$ can be offset by a rescaling of the argument of $\psi$. In what follows, $S$ means the scatter matrix chosen in the stated representation.

Theorem (Affine closure): Suppose $X$ in $\mathbb{R}^p$ is elliptical with location $\mu$, scatter matrix $S$, and characteristic generator $\psi$. For any $q\times p$ real matrix $B$ and any $b\in\mathbb{R}^q$, the vector $Y=b+BX$ is elliptical with location $$ \mu_Y=b+B\mu $$ and scatter matrix $$ S_Y=BSB^\top. $$ The transformed generator is $\psi$. No full-rank assumption on $B$ is required.

Proof of Affine Closure

For any $t\in\mathbb{R}^q$, the characteristic function of $Y$ is obtained by substituting $b+BX$ into its definition. The vector $B^\top t$ belongs to $\mathbb{R}^p$, so the characteristic function formula for $X$ applies at that argument:

\begin{aligned} \varphi_Y(t) &=E[\exp(i t^\top(b+BX))]\\ &=\exp(i t^\top b)\,E[\exp(i(B^\top t)^\top X)]\\ &=\exp(i t^\top b)\exp(i (B^\top t)^\top\mu) \,\psi\!\left((B^\top t)^\top S(B^\top t)\right)\\ &=\exp(i t^\top(b+B\mu))\,\psi(t^\top BSB^\top t). \end{aligned}

The last line has the defining elliptical form in dimension $q$, with location $b+B\mu$ and scatter $BSB^\top$. This matrix is symmetric because $S$ is symmetric. It is positive semidefinite because, for every $t\in\mathbb{R}^q$,

t^\top BSB^\top t=(B^\top t)^\top S(B^\top t)\geq 0.

Thus the transformed characteristic function is valid, and the transformed vector is elliptical, including when $BSB^\top$ is singular. $\square$

Interpreting the Transformed Scatter

The matrix product $BSB^\top$ records how the linear map acts on the original scatter. It is not necessary to find a new density, nor to invert $B$. If $B$ is square and invertible, the transformed scatter remains positive definite whenever $S$ is positive definite. If $B$ is rank-deficient, the new scatter can be singular.

More precisely, $BSB^\top$ is positive semidefinite by the proof above. If $S$ is positive definite, then $\operatorname{rank}(BSB^\top)=\operatorname{rank}(B)$. To see this, write $S=S^{1/2}S^{1/2}$. Then $BSB^\top=(BS^{1/2})(BS^{1/2})^\top$, whose rank equals the rank of $BS^{1/2}$; multiplication by the invertible $S^{1/2}$ does not change the rank of $B$. Consequently a rank-$r$ transformation produces rank-$r$ scatter in this case.

A null direction of $BSB^\top$ has zero quadratic scatter in that direction. In a singular normal example, this corresponds to a deterministic linear constraint on the transformed coordinates. For general elliptical laws, the scatter describes the shape parameterization, and the actual support may depend on the distribution as well. Singularity always signals that the scatter is not full rank, but support claims should be made with the law in view.

Worked Example: A Shear of a Two-Dimensional Elliptical Vector

Let $Z$ be a spherical vector in $\mathbb{R}^2$, and set $X=\mu+AZ$, where $\mu=(1,-2)^\top$ and $A=\begin{pmatrix}2&0\\1&1\end{pmatrix}$. Apply the affine transformation $Y=b+BX$, with $b=(3,0)^\top$ and $B=\begin{pmatrix}1&1\\0&1\end{pmatrix}$. Find the transformed location and scatter.

The original scatter is $S=AA^\top$. Multiplying the matrices gives

S= \begin{pmatrix}2&0\\1&1\end{pmatrix} \begin{pmatrix}2&1\\0&1\end{pmatrix} = \begin{pmatrix}4&2\\2&2\end{pmatrix}.

The transformed location is

b+B\mu = \begin{pmatrix}3\\0\end{pmatrix} + \begin{pmatrix}1&1\\0&1\end{pmatrix} \begin{pmatrix}1\\-2\end{pmatrix} = \begin{pmatrix}3\\0\end{pmatrix} + \begin{pmatrix}-1\\-2\end{pmatrix} = \begin{pmatrix}2\\-2\end{pmatrix}.

For the scatter, first calculate $BS$, then multiply by $B^\top$:

BS= \begin{pmatrix}1&1\\0&1\end{pmatrix} \begin{pmatrix}4&2\\2&2\end{pmatrix} = \begin{pmatrix}6&4\\2&2\end{pmatrix}, \qquad BSB^\top= \begin{pmatrix}6&4\\2&2\end{pmatrix} \begin{pmatrix}1&0\\1&1\end{pmatrix} = \begin{pmatrix}10&4\\4&2\end{pmatrix}.

As a check, $BA=\begin{pmatrix}3&1\\1&1\end{pmatrix}$, and $(BA)(BA)^\top=\begin{pmatrix}10&4\\4&2\end{pmatrix}$, the same result. The shear is invertible, so the transformed scatter remains full rank.

Moments When They Exist

Affine closure concerns the distributional form and does not require moments. When first and second moments do exist, the transformed mean and covariance obey corresponding rules. These are separate from the location-and-scatter rule: only under additional conditions can location be identified with mean and scatter with covariance.

Moment transformation: If $E[\|X\|^2]<\infty$, then for $Y=b+BX$, $$ E[Y]=b+B E[X], \qquad \operatorname{Var}(Y)=B\operatorname{Var}(X)B^\top. $$

For completeness, let $m=E[X]$. Linearity of expectation gives $E[Y]=b+Bm$, and hence $Y-E[Y]=B(X-m)$. Therefore

\begin{aligned} \operatorname{Var}(Y) &=E[(Y-E[Y])(Y-E[Y])^\top]\\ &=E[B(X-m)(X-m)^\top B^\top]\\ &=B\,E[(X-m)(X-m)^\top]\,B^\top\\ &=B\operatorname{Var}(X)B^\top. \end{aligned}

If $X$ has the affine representation $X=\mu+AZ$, where $E[Z]=0$ and $\operatorname{Var}(Z)=\kappa I$ for some finite $\kappa$, then $E[X]=\mu$ and $\operatorname{Var}(X)=\kappa AA^\top=\kappa S$. The same scalar factor applies after transformation: $\operatorname{Var}(Y)=\kappa BSB^\top$. Without finite second moments, this covariance calculation is unavailable, while the affine-closure theorem still applies.

Worked Example: A Rank-One Map Collapses a Normal Vector onto a Line

Let $X=(1,-1)^\top+Z$, where $Z\sim N(0,I_2)$. Set $B=\begin{pmatrix}1&1\\2&2\end{pmatrix}$ and $b=(0,1)^\top$. Compute the transformed distributional parameters and identify its constraint.

Here $\mu=(1,-1)^\top$, $S=I_2$, and $B\mu=(0,0)^\top$. Thus the transformed location is $b+B\mu=(0,1)^\top$. The transformed scatter is

BSB^\top=BB^\top = \begin{pmatrix}1&1\\2&2\end{pmatrix} \begin{pmatrix}1&2\\1&2\end{pmatrix} = \begin{pmatrix}2&4\\4&8\end{pmatrix}.

To verify the geometry directly, write $Z=(Z_1,Z_2)^\top$. Then $Y=b+BX=(U,\,2U+1)^\top$, where $U=Z_1+Z_2$. Independence and the standard normal variances give $U\sim N(0,2)$. Hence $Y$ lies on the line $y_2=2y_1+1$, and its covariance is $\operatorname{Var}(U)\begin{pmatrix}1&2\\2&4\end{pmatrix} =\begin{pmatrix}2&4\\4&8\end{pmatrix}$. The scatter has rank one, matching the rank of $B$.

Worked Example: Embedding a Vector into a Higher-Dimensional Space

Let $X\sim N((1,2)^\top,I_2)$, and define $Y=b+BX$, where $B=\begin{pmatrix}1&0\\0&1\\1&1\end{pmatrix}$ and $b=(0,0,4)^\top$. Find the location and scatter in $\mathbb{R}^3$, and state the support constraint.

Since $B\mu=(1,2,3)^\top$, the transformed location is $\mu_Y=(1,2,7)^\top$. With $S=I_2$, the scatter is

S_Y=BB^\top = \begin{pmatrix} 1&0&1\\ 0&1&1\\ 1&1&2 \end{pmatrix}.

The third coordinate satisfies $Y_3=X_1+X_2+4=Y_1+Y_2+4$, so the vector lies in a plane in $\mathbb{R}^3$. The matrix $B$ has rank two, and so does $S_Y$: this is an embedding of a two-dimensional random vector into three coordinates, not a collapse of its intrinsic dimension. The ambient three-dimensional scatter is nevertheless singular.

Common Mistake: Using a Density Change of Variables for a Singular Map

A tempting but incorrect argument is to say that every affine transformation changes a density by a Jacobian factor $1/|\det B|$. That formula requires a square, invertible transformation between spaces of the same dimension. In the rank-one example, $B$ is singular, so its determinant is zero and the transformed law is concentrated on a line; it has no ordinary two-dimensional density with respect to area. The correction is to use the characteristic-function transformation, or the affine-image representation, to obtain location and scatter. Neither method requires an inverse.

A related error is to call $BSB^\top$ the covariance without checking moments and the scale convention. It is always the transformed scatter for the selected elliptical representation. It is the covariance only when the covariance exists and its relationship to scatter has been established.

Key takeaway: For $Y=b+BX$, affine closure gives location $b+B\mu$ and scatter $BSB^\top$. When $S$ is positive definite, the rank of $B$ determines whether the transformed scatter stays full rank or becomes singular; for general positive semidefinite $S$, the transformed rank also depends on $S$. No inverse is needed.

Check Your Understanding

Use the affine-closure rule to answer each question.

If $X$ has location $\mu$ and scatter $S$, what are the location and scatter of $c+2X$, where $c$ is a fixed vector?
Why is $BSB^\top$ positive semidefinite even when $B$ is rectangular?
For a positive definite $S$, what is the rank of $BSB^\top$ in terms of the rank of $B$?
In the rank-one worked example, identify the linear constraint satisfied by every realization of $Y$.
Why can the affine-closure theorem be applied when $B$ is singular, even though an ordinary density change-of-variables formula cannot?