Density Generators for Elliptical Laws

From Ellipsoidal Contours to a Density Generator

For an elliptical density, the value of the density depends on the location of a point only through its squared distance from the center, measured in the geometry determined by a positive-definite matrix. This makes the density contours ellipsoids, but it also gives a compact way to describe many different laws: specify a single nonnegative function of that quadratic distance.

The function is called a density generator. It is not, by itself, generally a probability density on the real line. Its normalization must account for the volume of spherical shells in the dimension under consideration. That dimension-dependent condition is essential when comparing, for example, Gaussian, multivariate t, and bounded uniform laws.

Definition: Let \(p\geq 1\), let \(\mu\in\mathbb{R}^p\), and let \(\Sigma\) be a positive-definite \(p\times p\) matrix. Define the squared Mahalanobis distance \(q(x)=(x-\mu)^{\mathsf T}\Sigma^{-1}(x-\mu)\). A density of the form \(f(x)=|\Sigma|^{-1/2}g(q(x))\), \(x\in\mathbb{R}^p\), is an elliptical density with generator \(g\), provided \(g:[0,\infty)\to[0,\infty]\) is measurable and satisfies \(\int_0^\infty t^{p/2-1}g(t)\,dt=\Gamma(p/2)/\pi^{p/2}\). Here \(|\Sigma|\) denotes the determinant, and \(\Gamma\) is the gamma function.

This definition assumes an ordinary density with respect to \(p\)-dimensional Lebesgue measure and a positive-definite \(\Sigma\). Singular laws supported on lower-dimensional subspaces require a different treatment. A generator is defined only up to changes on sets of one-dimensional Lebesgue measure zero: such changes do not alter the resulting density as a probability law.

Why the Normalizing Integral Has This Form

The radial integral follows by transforming the ellipsoidal distance into an ordinary Euclidean radius. Choose an invertible matrix \(A\) with \(AA^{\mathsf T}=\Sigma\), and write \(x=\mu+Az\). Then \(q(x)=\lVert z\rVert^2\), and the Jacobian determinant is \(|A|=|\Sigma|^{1/2}\). Thus the determinant factor in the density cancels the Jacobian factor in integration.

In \(\mathbb{R}^p\), the surface area of the unit sphere is \(2\pi^{p/2}/\Gamma(p/2)\). In polar coordinates, \(z=ru\), where \(r\geq 0\) and \(u\) is a unit direction, the volume element has radial factor \(r^{p-1}dr\). Substituting \(t=r^2\) gives \(r^{p-1}dr=\tfrac12 t^{p/2-1}dt\). Together, these factors yield the normalization below.

Theorem: For measurable \(g\geq 0\) and positive-definite \(\Sigma\), the function \(f(x)=|\Sigma|^{-1/2}g(q(x))\) integrates to one on \(\mathbb{R}^p\) if and only if \(\int_0^\infty t^{p/2-1}g(t)\,dt=\Gamma(p/2)/\pi^{p/2}\). If this condition holds and \(X\) has density \(f\), then \(Q=q(X)\) has density \(f_Q(t)=\{\pi^{p/2}/\Gamma(p/2)\}t^{p/2-1}g(t)\) for \(t>0\), and zero for \(t<0\).

Proof. Apply \(x=\mu+Az\) to the integral of \(f\). Since \(dx=|\Sigma|^{1/2}dz\), the determinant factors cancel:

\int_{\mathbb{R}^p}f(x)\,dx =\int_{\mathbb{R}^p}g(\lVert z\rVert^2)\,dz =\frac{2\pi^{p/2}}{\Gamma(p/2)} \int_0^\infty r^{p-1}g(r^2)\,dr =\frac{\pi^{p/2}}{\Gamma(p/2)} \int_0^\infty t^{p/2-1}g(t)\,dt.

The change of variable in the last equality is \(t=r^2\). Consequently, the integral of \(f\) equals one exactly when the stated generator condition holds. For the distribution of \(Q\), the same polar-coordinate calculation restricted to radii with \(r^2\) in a measurable set \(B\subseteq[0,\infty)\) gives

P(Q\in B) =\frac{\pi^{p/2}}{\Gamma(p/2)} \int_{B\cap(0,\infty)}t^{p/2-1}g(t)\,dt.

This identifies the density of \(Q\) on \((0,\infty)\). The event \(Q=0\) is the single point \(X=\mu\), which has probability zero under a Lebesgue density. Hence there is no additional atom at zero. \(\square\)

Worked Examples: Three Generators

Worked Example: The Multivariate Gaussian Generator

Suppose \(X\) has the \(p\)-dimensional Gaussian density with mean \(\mu\) and positive-definite covariance matrix \(\Sigma\). Its density can be written \(f(x)=(2\pi)^{-p/2}|\Sigma|^{-1/2}\exp\{-q(x)/2\}\). Thus the generator is \(g(t)=(2\pi)^{-p/2}e^{-t/2}\), for \(t\geq 0\).

To check the required normalization, put \(a=p/2\). Using the gamma integral and then \(u=t/2\),

\int_0^\infty t^{p/2-1}g(t)\,dt =(2\pi)^{-p/2}\int_0^\infty t^{a-1}e^{-t/2}\,dt =(2\pi)^{-p/2}2^a\Gamma(a) =\frac{\Gamma(p/2)}{\pi^{p/2}}.

The generator therefore gives a normalized density in every dimension \(p\). It is positive for all \(t\geq0\), and the induced squared distance has density \(f_Q(t)=t^{p/2-1}e^{-t/2}/\{2^{p/2}\Gamma(p/2)\}\) for \(t>0\). In particular, its essential support is unbounded.

Worked Example: The Multivariate t Generator

Let \(\nu>0\) and \(\Sigma\) be positive definite. The multivariate t density with \(\nu\) degrees of freedom, center \(\mu\), and scale matrix \(\Sigma\) has generator

g(t)= \frac{\Gamma((\nu+p)/2)} {\Gamma(\nu/2)(\nu\pi)^{p/2}} \left(1+\frac{t}{\nu}\right)^{-(\nu+p)/2}, \qquad t\geq0.

To verify its normalization, set \(a=p/2\), \(b=\nu/2\), and substitute \(u=t/\nu\). The beta integral \(\int_0^\infty u^{a-1}(1+u)^{-(a+b)}du=\Gamma(a)\Gamma(b)/\Gamma(a+b)\), valid for \(a,b>0\), gives

\begin{aligned} \int_0^\infty t^{a-1}g(t)\,dt &=\frac{\Gamma(a+b)}{\Gamma(b)(\nu\pi)^a} \nu^a\int_0^\infty u^{a-1}(1+u)^{-(a+b)}\,du\\ &=\frac{\Gamma(a+b)}{\Gamma(b)\pi^a} \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)} =\frac{\Gamma(p/2)}{\pi^{p/2}}. \end{aligned}

The conditions \(a>0\) and \(b>0\) hold because \(p\geq1\) and \(\nu>0\). This generator is positive at every finite \(t\geq0\), so the resulting squared distance has unbounded support. Compared with the Gaussian generator, its tail decreases as a power rather than exponentially; no claim about moments is needed to identify this difference in density shape.

Worked Example: Uniform Distribution on an Ellipsoid

Fix \(R>0\), and consider the uniform distribution on the ellipsoid \(\{x:q(x)\leq R^2\}\). Its density is constant inside that ellipsoid and zero outside, so its generator must be \(g(t)=c\) for \(0\leq t\leq R^2\) and \(g(t)=0\) for \(t>R^2\).

The normalizing condition determines \(c\). Integrating the power of \(t\) yields

c\int_0^{R^2}t^{p/2-1}\,dt =c\,\frac{(R^2)^{p/2}}{p/2} =c\,\frac{2R^p}{p} =\frac{\Gamma(p/2)}{\pi^{p/2}}.

Solving gives \(c=p\Gamma(p/2)/(2\pi^{p/2}R^p)=\Gamma(p/2+1)/(\pi^{p/2}R^p)\). The induced density of \(Q\), for \(0<t<R^2\), is therefore

f_Q(t)=\frac{\pi^{p/2}}{\Gamma(p/2)} t^{p/2-1}\frac{\Gamma(p/2+1)}{\pi^{p/2}R^p} =\frac{p}{2R^p}t^{p/2-1}.

This integrates to one: \(\int_0^{R^2}\frac{p}{2R^p}t^{p/2-1}dt =\frac{p}{2R^p}\frac{2R^p}{p}=1\). The law is concentrated on the closed ellipsoid \(q(x)\leq R^2\); the boundary itself has probability zero under this density.

Support: Use the Generator Almost Everywhere

The examples illustrate two support patterns. The uniform generator vanishes beyond a finite squared distance, so the distribution is confined to a bounded ellipsoid. The Gaussian and t generators have positive tails on intervals arbitrarily far out, so their squared distances are not bounded above. A rigorous support statement must, however, depend on where the generator is nonzero on sets of positive measure, not on isolated point values.

Support rule: The distribution of \(Q\) has unbounded essential support if and only if, for every \(M>0\), \(g\) is nonzero on a set of positive Lebesgue measure in \((M,\infty)\). It is concentrated on \(Q\leq C\) for some finite \(C\) if and only if \(g(t)=0\) for almost every \(t>C\).

To see the rule, the density of \(Q\) is a positive constant times \(t^{p/2-1}g(t)\) for \(t>0\). The factor \(t^{p/2-1}\) is strictly positive there. Thus \(P(Q>M)>0\) exactly when \(g\) is positive on a subset of \((M,\infty)\) with positive measure. If instead \(g=0\) almost everywhere beyond \(C\), then \(P(Q>C)=0\). These statements concern the probability law and are unchanged by modifying \(g\) on a null set.

Worked Example: Why Isolated Positive Values Do Not Give Unbounded Support

In dimension \(p=2\), let \(g(t)=1/\pi\) on \([0,1]\), set \(g(n)=1\) for integers \(n\geq2\), and let \(g(t)=0\) at all other points. The generator is positive at arbitrarily large arguments, but those extra positive values occur only on a countable set, which has Lebesgue measure zero.

For \(p=2\), the normalization condition is \(\int_0^\infty g(t)\,dt=1/\pi\). The interval \([0,1]\) contributes \(1/\pi\), while the countable set of integer points contributes zero to the integral. Hence the generator is normalized. Also \(Q\) has density \(f_Q(t)=\pi g(t)\) for \(t>0\), so its law is uniform on \([0,1]\) and \(Q\leq1\) almost surely.

The correct conclusion is not that pointwise positivity arbitrarily far out implies unbounded support. Here \(g\) is nonzero beyond 1 only on a null set, so the distribution has bounded support. This is also why generators that agree almost everywhere describe the same probability law.

What the Generator Representation Provides

The generator separates the radial shape of an elliptical density from its center and matrix of ellipsoidal distances. Once the dimension and generator are specified, the normalization condition fixes the overall scale of the density, and the density of \(Q\) follows directly. This is useful when checking whether a proposed formula is a probability density and when comparing how different elliptical laws allocate probability across radial distances.

A frequent mistake is to treat \(g\) as an ordinary one-dimensional density and require \(\int_0^\infty g(t)\,dt=1\). That omits the shell-volume factor \(t^{p/2-1}\) and the dimension-dependent constant. The required integral is instead \(\int_0^\infty t^{p/2-1}g(t)\,dt=\Gamma(p/2)/\pi^{p/2}\). A second pitfall is to infer support from isolated nonzero values of \(g\); support is a property of the probability law and is therefore determined up to null sets.

Key takeaway: An elliptical density has the form \(f(x)=|\Sigma|^{-1/2}g(q(x))\), but a valid generator must satisfy a dimension-specific weighted integral condition. Its radial law and support are governed by \(t^{p/2-1}g(t)\), interpreted almost everywhere.

Check Your Understanding

Use the generator form and support rules to answer the following questions.

For dimension \(p\), write the normalization condition that a generator \(g\) must satisfy.
Why does the determinant factor in an elliptical density cancel when the change of variables \(x=\mu+Az\), \(AA^{\mathsf T}=\Sigma\), is made?
For the uniform ellipsoid example, what is the density of \(Q\) on \(0<t<R^2\)?
A generator is positive at an unbounded sequence of isolated points and zero elsewhere beyond a finite interval. Does this alone imply unbounded support for \(Q\)? Explain.
Which part of the multivariate t generator indicates that it has a power-decreasing rather than exponentially decreasing tail?