Radial-Angular Representations of Elliptical Vectors

Why Separate Radius and Direction?

An elliptical vector is an affine image of a spherically symmetric vector. The spherical part has no preferred direction: rotating it does not change its distribution. This symmetry allows a useful separation into two ingredients—a nonnegative radius and a random direction. The radius determines how far the spherical vector lies from the origin, while the direction is uniform over all orientations and independent of that radius.

This decomposition is useful both for understanding and for constructing elliptical laws. It does not require a density, and the radius need not follow a familiar distribution. Throughout, $p$ is a positive integer, vectors are in $\mathbb{R}^p$, and $\|z\|$ denotes the Euclidean norm.

Definition: A random vector $Z$ in $\mathbb{R}^p$ is spherically symmetric if $QZ$ has the same distribution as $Z$ for every $p\times p$ orthogonal matrix $Q$, that is, every $Q$ satisfying $Q^TQ=I_p$. The unit sphere is $S^{p-1}=\{u\in\mathbb{R}^p:\|u\|=1\}$. A random vector $U$ is uniform on the unit sphere if its distribution is the rotation-invariant probability measure on $S^{p-1}$.

The uniform distribution on $S^{p-1}$ is a probability distribution on directions, not a distribution over the ball. For $p=2$, it means that the angle is uniform on $[0,2\pi)$. For $p=1$, the sphere consists of the two points $-1$ and $1$, each with probability $1/2$.

Definition: A vector $X$ has a radial-angular representation with location $\mu$ and matrix $A$ if its distribution can be written $$ X \overset{d}{=} \mu+R A U, $$ where $R\geq 0$, $U$ is uniform on $S^{p-1}$, and $R$ and $U$ are independent. Here $A$ is a $p\times p$ matrix; it need not be invertible.

The equality is in distribution: the representation specifies the law of $X$, and does not require the variables on the right to be the original random variables on the same probability space. In the affine-image definition of an elliptical vector, $X=\mu+AZ$ with $Z$ spherical. The key step is to show that $Z$ can be represented in distribution as $RU$.

Deriving the Representation

For a spherical vector $Z$, define its radius $R=\|Z\|$. When $R>0$, the direction of $Z$ is $Z/\|Z\|$. Rotational invariance says that no direction is preferred at any fixed radius. The following result makes that statement precise, including when $Z$ has an atom at zero.

Theorem: If $Z$ is spherically symmetric in $\mathbb{R}^p$, then there exist independent random variables $R\geq 0$ and $U$, with $U$ uniform on $S^{p-1}$, such that $Z\overset{d}{=}RU$. The distribution of $R$ is the distribution of $\|Z\|$.

Proof. Let $Q$ be an orthogonal matrix drawn from the rotation-invariant probability measure on the orthogonal group, independently of $Z$. For any fixed unit vector $u$, the vector $Qu$ is uniform on $S^{p-1}$: its distribution is unchanged by any fixed rotation. Consequently, for each fixed nonzero $z$, $Qz$ has the distribution of $\|z\|U$, where $U$ is uniform on the sphere.

Spherical symmetry implies $QZ\overset{d}{=}Z$. To identify this distribution, let $h$ be any bounded measurable function on $\mathbb{R}^p$. Conditioning on $Z$ and using the distribution of $Qz$ for each fixed $z\neq 0$ gives

E[h(Z)] =E[h(QZ)] =E\left[\int_{S^{p-1}}h(\|Z\|u)\,\sigma(du)\right],

where $\sigma$ denotes the uniform probability measure on the sphere. If $Z=0$, the integrand is $h(0)$ for every $u$, so the same identity holds there as well. Now take $R$ with the distribution of $\|Z\|$ and take $U\sim\sigma$ independently of $R$. The final expectation is $E[h(RU)]$. Since the equality holds for every bounded measurable $h$, $Z\overset{d}{=}RU$. By construction, $R$ has the required distribution and is independent of $U$. $\square$

The proof uses the uniform direction as a distributional construction. If $P(Z=0)>0$, the direction $Z/\|Z\|$ is undefined on that event; there is no need to assign it an artificial value. Instead, take a fresh independent uniform $U$, as in the theorem.

Corollary: If $Z$ is spherical and $X\overset{d}{=}\mu+AZ$, then $$ X\overset{d}{=}\mu+RAU, $$ where $R$ has the distribution of $\|Z\|$, $U$ is uniform on $S^{p-1}$, and $R$ and $U$ are independent.

This follows by substituting the theorem’s representation $Z\overset{d}{=}RU$ into the affine-image expression. The same argument applies if $A$ is singular: the random direction is first chosen in $\mathbb{R}^p$, then mapped by $A$.

Constructing Spherical and Elliptical Laws

The representation also works in reverse. Start with any nonnegative random variable $R$, take an independent uniform direction $U$, and set $Z=RU$. For every orthogonal $Q$, $QU$ is uniform on the sphere and remains independent of $R$. Hence $QZ=R(QU)\overset{d}{=}RU=Z$, so $Z$ is spherical. Applying a fixed affine transformation then gives an elliptical vector.

Result: A vector $Z=RU$, with $R\geq 0$ independent of a uniform $U$ on $S^{p-1}$, is spherically symmetric. Therefore every vector $X=\mu+RAU$ constructed with these conditions is an affine image of a spherical vector.

This gives a flexible construction recipe: choose the radius law, choose the location and matrix, and use an independent uniform direction. Different choices of the radial distribution can produce different elliptical laws even when $\mu$ and $A$ stay fixed.

Worked Example: A Bivariate Normal Vector in Polar Form

Let $G=(G_1,G_2)^T$, where $G_1$ and $G_2$ are independent $N(0,1)$ random variables. Show its radial-angular representation, then construct an affine image with $\mu=(1,-1)^T$ and $A=\begin{pmatrix}2&0\\0&1\end{pmatrix}$.

The joint density of $G$ at $g=(g_1,g_2)^T$ is $(2\pi)^{-1}\exp\{-(g_1^2+g_2^2)/2\}$. Write $g=(r\cos\theta,r\sin\theta)^T$, where $r>0$ and $0\leq\theta<2\pi$. The polar-coordinate Jacobian is $r$, so the joint density of radius and angle is

f_{R,\Theta}(r,\theta) =\frac{r}{2\pi}e^{-r^2/2} =\left(r e^{-r^2/2}\right)\left(\frac{1}{2\pi}\right), \qquad r>0,\quad 0\leq\theta<2\pi.

The factorization shows that $R$ and $\Theta$ are independent. The radial density integrates to one because $\int_0^\infty r e^{-r^2/2}\,dr=1$; the angle is uniform on $[0,2\pi)$. Thus $U=(\cos\Theta,\sin\Theta)^T$ is uniform on the unit circle and $G\overset{d}{=}RU$.

With the specified $A$ and $\mu$, the elliptical vector has the representation $$ X\overset{d}{=} \begin{pmatrix}1\\-1\end{pmatrix} +R \begin{pmatrix}2&0\\0&1\end{pmatrix} \begin{pmatrix}\cos\Theta\\\sin\Theta\end{pmatrix} = \begin{pmatrix}1+2R\cos\Theta\\-1+R\sin\Theta\end{pmatrix}. $$ The same independent $R$ and uniform $\Theta$ generate the standard bivariate normal before the affine transformation. Here the matrix changes how each direction is mapped, while the radial law remains the one derived above.

Worked Example: A Spherical Law With Only Two Possible Radii

Let $\Theta$ be uniform on $[0,2\pi)$. Independently, let $R=1$ with probability $0.7$ and $R=3$ with probability $0.3$. Define $U=(\cos\Theta,\sin\Theta)^T$ and $Z=RU$. Identify the distributional structure and give an elliptical transform.

The direction $U$ is uniform on the unit circle by the uniform distribution of $\Theta$, and it is independent of $R$ by construction. The reverse result therefore shows that $Z$ is spherical. Its radius is exactly $1$ with probability $0.7$ and $3$ with probability $0.3$; it is not necessary for the radius to have a density.

For example, choose $\mu=(0,2)^T$ and $A=\begin{pmatrix}1&1/2\\0&2\end{pmatrix}$. Then $$ X=\mu+RAU = \begin{pmatrix}0\\2\end{pmatrix} +R\begin{pmatrix}\cos\Theta+\frac12\sin\Theta\\2\sin\Theta\end{pmatrix} $$ is an affine image of a spherical vector, hence elliptical. The two possible values of $R$ give two distinct radial magnitudes before the transformation. In particular, this construction produces a law concentrated on transformed circular paths rather than a law with a density throughout the plane.

Worked Example: A Randomly Scaled Bivariate Normal Radius

Let $G$ be standard bivariate normal, and let $S$, independent of $G$, equal $1$ or $2$, each with probability $1/2$. Show that $Z=SG$ has a radial-angular representation and derive its radial density.

From the first example, write $G=R_0U$, where $R_0$ has density $f_{R_0}(r)=r e^{-r^2/2}$ for $r>0$, and $U$ is uniform on the unit circle independent of $R_0$. Since $S$ is independent of $G$, it is independent of the pair $(R_0,U)$. Thus $R=SR_0$ is independent of $U$, and $$ Z=SG=(SR_0)U=RU. $$ The reverse result verifies that $Z$ is spherical.

Conditionally on $S=1$, the radius has density $r e^{-r^2/2}$. If $S=2$, then $R=2R_0$, so the change-of-scale formula gives $$ f_{R\mid S=2}(r)=\frac12 f_{R_0}(r/2) =\frac{r}{4}e^{-r^2/8},\qquad r>0. $$ Weighting the two conditional densities by their probabilities yields $$ f_R(r)=\frac12 r e^{-r^2/2}+\frac{r}{8}e^{-r^2/8},\qquad r>0. $$ Each term integrates to $1/2$: the first is half a Rayleigh density, and the second is half the density of $2R_0$. Thus the displayed density integrates to one. Taking $X=\mu+AZ$ for any fixed $\mu$ and $A$ gives an elliptical vector with this non-Gaussian radial law.

Interpreting the Radius and Avoiding a Common Mistake

In $X=\mu+RAU$, $R$ is the radius of the spherical vector before applying $A$. It is generally not the Euclidean distance $\|X-\mu\|$. Indeed,

\|X-\mu\|=R\|AU\|.

Unless $A$ preserves lengths in every direction, $\|AU\|$ varies with $U$. For an invertible $A$, the representation instead gives $\|A^{-1}(X-\mu)\|=R$. If $A$ is singular, this inverse expression is unavailable, although the latent radius in the construction is still well defined.

Common mistake: Treating $R$ as the distance from $X$ to its location parameter leads to the incorrect identity $\|X-\mu\|=R$ for a general $A$. The correction is to retain the directional factor $\|AU\|$: the radius belongs to the spherical input, and the affine map can stretch different directions by different amounts.

The main practical advantage of the representation is that radial and directional choices can be made separately. It permits direct simulation, accommodates discrete or continuous radial laws, and makes clear why the affine image remains elliptical. It also prevents a further misconception: the location parameter $\mu$ is part of the representation, but no finite-mean assumption is needed to define it.

Check Your Understanding

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

What is the radial distribution associated with a spherical vector $Z$, and what is the distribution of its independent angular component?
Why does $Z=RU$ remain spherical when $R\geq 0$ is independent of a uniform $U$?
In the bivariate normal example, what factorization of the polar-coordinate joint density establishes independence of radius and angle?
For $X=\mu+RAU$, under what condition can the latent radius be recovered as $\|A^{-1}(X-\mu)\|$?
Why is $\|X-\mu\|=R$ generally incorrect after an affine transformation?

Radial-Angular Representations of Elliptical Vectors

What You'll Learn