Stochastic Radial Representations

Why Separate Direction and Radius?

An elliptical density describes contours through a quadratic form: points with the same value of that form lie on the same ellipsoid. A complementary description separates a draw into its direction and its distance from the center. That decomposition is useful because it distinguishes the geometry of the contours from how probability is distributed across them. In particular, the direction can be uniform after adjusting for the scatter matrix, while all variation in distance is captured by a single nonnegative random variable.

This tutorial derives that decomposition and the precise connection between the radial variable and the density generator. Throughout, let $p\geq 2$, let $\mu\in\mathbb R^p$, and let $\Sigma$ be a positive-definite $p\times p$ matrix. Consider an absolutely continuous elliptical distribution with density

f_X(x)=|\Sigma|^{-1/2}g\!\left((x-\mu)^T\Sigma^{-1}(x-\mu)\right),

where $g:[0,\infty)\to[0,\infty)$ is a measurable density generator and the displayed density integrates to one. Here $|\Sigma|$ denotes the determinant. The matrix $\Sigma$ specifies the ellipsoidal geometry; it is not assumed here to be the covariance matrix.

Definition: Let $S^{p-1}=\{u\in\mathbb R^p:u^Tu=1\}$, and let $U$ be uniform on this unit sphere, meaning that its distribution is normalized surface-area measure. A stochastic radial representation of $X$ is $$ X=\mu+RAU, $$ where $R\geq 0$ is a random variable independent of $U$, and $A$ is a nonsingular matrix satisfying $AA^T=\Sigma$. The variable $R$ is the radius in the coordinates transformed by $A^{-1}$.

From the Density Generator to the Radius

Choose any nonsingular factor $A$ such that $AA^T=\Sigma$, for example a Cholesky factor, and set $Y=A^{-1}(X-\mu)$. Since $\Sigma^{-1}=A^{-T}A^{-1}$, the quadratic form becomes $Y^TY$. The change of variables has Jacobian factor $|A|=|\Sigma|^{1/2}$, so the density of $Y$ is

f_Y(y)=g(y^Ty)=g(\|y\|^2).

Thus $Y$ is spherically symmetric: its density depends on $y$ only through its distance from the origin. Let $s_{p-1}$ denote the surface area of the unit sphere in $\mathbb R^p$:

s_{p-1}=\frac{2\pi^{p/2}}{\Gamma(p/2)}.

Polar integration expresses volume as $r^{p-1}\,dr$ times surface area on the sphere. Consequently, the density of $R=\|Y\|$ is

h_R(r)=s_{p-1}r^{p-1}g(r^2),\qquad r>0.

The factor $r^{p-1}$ is essential: a thin spherical shell grows in volume as its radius increases. The generator $g(r^2)$ gives the density at a point on that shell, while the shell's surface area supplies the additional factor. The resulting radial density satisfies

\int_0^\infty s_{p-1}r^{p-1}g(r^2)\,dr=1.

Theorem: If $X$ has the absolutely continuous elliptical density above, then $$ X\overset{d}{=}\mu+RAU, $$ where $A A^T=\Sigma$, $U$ is uniform on $S^{p-1}$, and $R$ is independent of $U$ with density $h_R(r)=s_{p-1}r^{p-1}g(r^2)$ for $r>0$.

Proof. The transformed vector $Y=A^{-1}(X-\mu)$ has density $g(\|y\|^2)$. In polar coordinates, write $y=ru$, where $r>0$ and $u\in S^{p-1}$. For a radial density, the joint probability element for radius and direction is

g(r^2)r^{p-1}\,dr\,dS(u),

where $dS$ is surface-area measure. Since uniform probability measure on the sphere is $dS(u)/s_{p-1}$, the same element can be written

\left[s_{p-1}r^{p-1}g(r^2)\,dr\right]\frac{dS(u)}{s_{p-1}}.

The first bracket is a probability element for $R$, and the second factor is uniform probability measure for $U$. Their product shows that $R$ and $U$ are independent, with the stated radial density and uniform direction. The density is absolutely continuous, so $P(Y=0)=0$; the direction is therefore defined with probability one. Finally, $Y=RU$, and hence $X=\mu+AY=\mu+RAU$. $\square$

Recovering the Generator from a Radial Law

The relationship also works in the opposite direction. Suppose a nonnegative random variable $R$ has a density $h_R$, $U$ is independent of $R$ and uniform on $S^{p-1}$, and $Y=RU$. Away from the origin, polar coordinates give the density of $Y$ as the radial density divided by the shell's surface-area factor:

f_Y(y)=\frac{h_R(\|y\|)}{s_{p-1}\|y\|^{p-1}},\qquad y\ne 0.

It follows that the associated generator, defined up to values at individual points that do not affect a density, is

g(t)=\frac{h_R(\sqrt{t})}{s_{p-1}t^{(p-1)/2}},\qquad t>0.

Theorem: If $h_R$ is a probability density on $(0,\infty)$, then $Y=RU$, with $R$ independent of a uniform $U$, has density $g(\|y\|^2)$ for $y\ne 0$, where $$ g(t)=\frac{h_R(\sqrt{t})}{s_{p-1}t^{(p-1)/2}}. $$ For any nonsingular $A$ and location $\mu$, $X=\mu+AY$ then has an elliptical density with $\Sigma=AA^T$.

Proof. Let $B$ be a measurable region in $\mathbb R^p$. Independence and uniformity give the probability of landing in a polar-coordinate element $dr\,dS(u)$ as $h_R(r)\,dr\,dS(u)/s_{p-1}$. On the other hand, a density $f_Y(y)$ assigns probability $f_Y(ru)r^{p-1}\,dr\,dS(u)$ to that element. Equating these expressions gives

f_Y(ru)r^{p-1}=\frac{h_R(r)}{s_{p-1}}, \qquad f_Y(ru)=\frac{h_R(r)}{s_{p-1}r^{p-1}}.

Substituting $r=\|y\|$ and then $t=r^2$ yields the stated generator. Its integral over $\mathbb R^p$ is one, since polar integration gives

\int_{\mathbb R^p}g(\|y\|^2)\,dy =\int_0^\infty s_{p-1}r^{p-1} \frac{h_R(r)}{s_{p-1}r^{p-1}}\,dr =\int_0^\infty h_R(r)\,dr =1.

A random variable with a density has no point mass at zero, so this construction has no additional atom at $Y=0$. Applying the nonsingular affine transformation $X=\mu+AY$ gives the elliptical density with quadratic form $(x-\mu)^T(AA^T)^{-1}(x-\mu)$. $\square$

Worked Examples

Worked Example: The Bivariate Normal Radius

Let $p=2$, $\mu=0$, and $\Sigma=I_2$, with the standard bivariate normal density. Its generator is $g(t)=(2\pi)^{-1}e^{-t/2}$. Find the radial density and identify the distribution of $R^2$.

The unit circle has circumference $s_1=2\pi$. Therefore

h_R(r)=2\pi r\frac{1}{2\pi}e^{-r^2/2} =re^{-r^2/2},\qquad r>0.

To obtain the density of $Q=R^2$, use $r=\sqrt q$ and $dr/dq=1/(2\sqrt q)$:

h_Q(q)=h_R(\sqrt q)\frac{1}{2\sqrt q} =\sqrt q\,e^{-q/2}\frac{1}{2\sqrt q} =\frac12e^{-q/2},\qquad q>0.

This is the chi-square density with two degrees of freedom. As a check, $\int_0^\infty re^{-r^2/2}\,dr=1$, by the substitution $v=r^2/2$. Thus the bivariate normal can be represented as $X=RU$, with a uniform direction and an independent radius having the density above.

Worked Example: A Uniform Distribution on a Disk

Let $Y$ be uniform on the unit disk in $\mathbb R^2$. Find its generator and the distribution of its radius.

The disk has area $\pi$, so its density is $1/\pi$ when $\|y\|\leq 1$, and zero otherwise. Therefore a generator is

g(t)=\frac{1}{\pi}\mathbf{1}_{[0,1]}(t).

Using $s_1=2\pi$, the radius density is

h_R(r)=2\pi r\frac{1}{\pi}\mathbf{1}_{[0,1]}(r^2) =2r\mathbf{1}_{(0,1)}(r).

It integrates to one because $\int_0^1 2r\,dr=1$. Equivalently, $P(R\leq r)=r^2$ for $0\leq r\leq1$: the probability is the ratio of the disk of radius $r$, with area $\pi r^2$, to the unit disk's area $\pi$. The direction is uniform and independent of this radius. This example also shows that a constant generator inside its support does not imply a uniform radius; the shell-area factor makes larger radii more probable.

Worked Example: Recovering a Generator from a Chosen Radius

In two dimensions, suppose $R$ has density $h_R(r)=2r/(1+r^2)^2$ for $r>0$, and $U$ is an independent uniform direction. Find the density generator and verify normalization.

First check that the proposed radial density integrates to one. With $v=1+r^2$ and $dv=2r\,dr$,

\int_0^\infty\frac{2r}{(1+r^2)^2}\,dr =\int_1^\infty v^{-2}\,dv =1.

For $p=2$, the sphere-area factor is $s_1=2\pi$. The generator recovered from the radial density is

g(t)=\frac{h_R(\sqrt t)}{2\pi\sqrt t} =\frac{2\sqrt t}{2\pi\sqrt t(1+t)^2} =\frac{1}{\pi(1+t)^2},\qquad t>0.

Thus $Y=RU$ has density $f_Y(y)=1/[\pi(1+\|y\|^2)^2]$. Directly checking in polar coordinates gives

\int_{\mathbb R^2}f_Y(y)\,dy =2\pi\int_0^\infty\frac{r}{\pi(1+r^2)^2}\,dr =\int_0^\infty\frac{2r}{(1+r^2)^2}\,dr =1.

For a nonzero location and a nonsingular matrix $A$, the vector $X=\mu+ARU$ has the corresponding elliptical density, with $\Sigma=AA^T$. The radius law itself is unchanged by this affine transformation; $A$ determines how directions and distances are mapped into the observed coordinates.

Interpretation and Common Pitfalls

The representation separates two roles. The uniform variable $U$ supplies a direction in transformed coordinates, while $R$ determines how far a draw lies along that direction. The matrix $A$ then maps those spherical coordinates into ellipsoidal coordinates. Different radial densities can therefore yield different distributions with the same ellipsoidal geometry. Conversely, specifying a radial density determines the generator through the formula above.

A frequent error is to identify the radial density with $g(r^2)$ alone. That omits the shell-volume factor and generally gives an unnormalized or incorrect distribution. Another error is to confuse $R$ with $Q=R^2$. If the squared radius has density $h_Q$, the change of variables $q=r^2$ gives

h_R(r)=2r\,h_Q(r^2),\qquad r>0,

not $h_R(r)=h_Q(r^2)$. Combining this change of variables with the radial-generator identity gives $g(q)=2h_Q(q)/(s_{p-1}q^{p/2-1})$ for $q>0$.

Finally, an arbitrary radial random variable need not produce a density. For example, if $R$ is fixed at a positive constant, then $RU$ lies on a sphere and has probability one on a set of $p$-dimensional volume zero. It has a stochastic radial representation, but no density generator of the absolutely continuous form used here. The density formulas require the stated density assumption on $R$, or equivalently the absolutely continuous elliptical model.

Key takeaway: For an absolutely continuous elliptical density, $R=\|A^{-1}(X-\mu)\|$ has density $s_{p-1}r^{p-1}g(r^2)$. In the reverse direction, an independent uniform direction and radial density $h_R$ produce generator $g(t)=h_R(\sqrt t)/(s_{p-1}t^{(p-1)/2})$. The surface-area factor and the distinction between $R$ and $R^2$ are essential.

Check Your Understanding

Use the radial and generator relationships to answer these questions.

For $p=3$, what geometric factor multiplies $g(r^2)$ to obtain the radius density?
Why are the direction $U$ and radius $R$ independent for an absolutely continuous spherical density?
If a two-dimensional radius has density $h_R(r)=2r$ on $0<r<1$, what is the associated generator?
Explain why omitting $r^{p-1}$ from the radial density formula leads to an error.
Does a fixed positive radius yield an absolutely continuous density in $\mathbb R^p$? Explain.