Normalizing Generators and Moment Conditions

Why the Generator Needs Normalizing

An elliptical density separates its shape across directions from its behavior as distance from the center increases. That radial behavior is encoded by a nonnegative function called a generator. Choosing a generator does not automatically produce a probability density: its radial mass must be integrable, and the density needs a dimension-dependent constant that makes its total integral equal to one.

The same generator can behave differently as the dimension changes. A tail that is integrable in one dimension may fail to be integrable in a higher dimension, because the volume of a shell grows with its radius. The radial integral makes both normalization and moment existence explicit.

Definition: Let $p\geq 1$, let $\mu\in\mathbb{R}^p$, and let $\Sigma$ be a symmetric positive-definite $p\times p$ matrix. For a measurable generator $g:[0,\infty)\to[0,\infty]$, an elliptical density has the form $$ f_X(x)=c_p|\Sigma|^{-1/2}g\!\left((x-\mu)'\Sigma^{-1}(x-\mu)\right),\qquad x\in\mathbb{R}^p, $$ where $c_p$ is chosen so that the density integrates to one. The matrix $\Sigma$ describes the quadratic-form contours; this definition alone does not assert that it is the covariance matrix.

Write $J_{p}(g)=\int_0^\infty t^{p/2-1}g(t)\,dt$. The necessary and sufficient condition for a finite, positive normalizing constant is $0<J_p(g)<\infty$. The condition includes integrability both near zero and in the tail. A generator that is zero almost everywhere, or one whose integral diverges, cannot be turned into a probability density by multiplying by a finite positive constant.

Deriving the Normalizing Constant

First change coordinates to $z=\Sigma^{-1/2}(x-\mu)$, where $\Sigma^{-1/2}$ is the symmetric positive-definite inverse square root. The Jacobian gives $dx=|\Sigma|^{1/2}dz$, and the quadratic form becomes $z'z$. Thus the integral of the unnormalized part of the density is reduced to a radial integral in ordinary Euclidean space.

Theorem: If $g$ is measurable, nonnegative, and $0<J_p(g)<\infty$, then the unique positive constant that normalizes the elliptical density is $$ c_p=\frac{\Gamma(p/2)}{\pi^{p/2}J_p(g)} =\frac{\Gamma(p/2)}{\pi^{p/2}\displaystyle\int_0^\infty t^{p/2-1}g(t)\,dt}. $$ If $J_p(g)$ is infinite, no finite positive $c_p$ normalizes this form to a probability density.

Proof. In the transformed coordinates, the integral before multiplication by $c_p$ is $\int_{\mathbb{R}^p}g(z'z)\,dz$. Polar coordinates $z=ru$, with $r\geq0$ and $u$ on the unit sphere $S^{p-1}$, give $$ \int_{\mathbb{R}^p}g(z'z)\,dz =|S^{p-1}|\int_0^\infty r^{p-1}g(r^2)\,dr. $$ The sphere’s surface area is $|S^{p-1}|=2\pi^{p/2}/\Gamma(p/2)$. With the substitution $t=r^2$, $dr=\tfrac12t^{-1/2}dt$, so $$ \int_0^\infty r^{p-1}g(r^2)\,dr =\frac12\int_0^\infty t^{p/2-1}g(t)\,dt =\frac12J_p(g). $$ The determinant factors from the change of coordinates cancel the factor $|\Sigma|^{-1/2}$ in the density. Therefore the full integral is $c_p\,\pi^{p/2}J_p(g)/\Gamma(p/2)$. It equals one exactly for the displayed value of $c_p$. If $J_p(g)=\infty$, every positive finite multiplier leaves an infinite integral; if $J_p(g)=0$, the unnormalized function has zero integral. $\square$

The density is defined up to changes on sets of Lebesgue measure zero, as usual for probability densities. The integral condition is often easier to inspect after writing it in the radius $r=\sqrt{t}$: the relevant integral becomes $\int_0^\infty r^{p-1}g(r^2)\,dr$. The factor $r^{p-1}$ is the contribution of the expanding spherical shells.

Worked Example: Normalizing a Power-Exponential Generator

Take $p=2$ and $g(t)=\exp(-t^2)$. In the defining integral, $p/2=1$, so $$ J_2(g)=\int_0^\infty e^{-t^2}\,dt =\frac12\Gamma\!\left(\frac12\right) =\frac{\sqrt{\pi}}{2}. $$ The normalizer is therefore $$ c_2=\frac{\Gamma(1)}{\pi J_2(g)} =\frac{1}{\pi(\sqrt{\pi}/2)} =\frac{2}{\pi^{3/2}}. $$ Thus the density is $f_X(x)=\frac{2}{\pi^{3/2}}|\Sigma|^{-1/2}\exp(-q^2)$, where $q=(x-\mu)'\Sigma^{-1}(x-\mu)$. Its radial integral is finite and positive, and the theorem shows that this coefficient makes the density integrate to one.

Which Moments Exist?

Let $R=\{\,(X-\mu)'\Sigma^{-1}(X-\mu)\,\}^{1/2}$, the radius measured in the quadratic-form geometry. Once the density is normalized, its radial moments are determined by the same integral as its normalizer, with an extra power of the radius.

Theorem: Suppose $0<J_p(g)<\infty$, and let $s\geq0$. Then $E[R^s]$ is finite if and only if $$ \int_0^\infty t^{(p+s)/2-1}g(t)\,dt<\infty. $$ When this integral is finite, $$ E[R^s] =\frac{\displaystyle\int_0^\infty t^{(p+s)/2-1}g(t)\,dt} {\displaystyle\int_0^\infty t^{p/2-1}g(t)\,dt}. $$

Proof. In the transformed coordinates $z=\Sigma^{-1/2}(X-\mu)$, the radius is $\|z\|=R$. The density of $z$ is $c_pg(z'z)$. Applying polar coordinates to the nonnegative integrand $\|z\|^s g(z'z)$ gives $$ E[R^s] =c_p|S^{p-1}|\int_0^\infty r^{p+s-1}g(r^2)\,dr =c_p|S^{p-1}|\frac12\int_0^\infty t^{(p+s)/2-1}g(t)\,dt. $$ These equalities hold for nonnegative integrands even if the integral is infinite. Applying the same calculation with $s=0$, and using that the density integrates to one, shows that the ratio of the order-$s$ radial integral to the order-zero integral is $E[R^s]$, including the possibility of an infinite value. Hence the stated finiteness condition is both necessary and sufficient. $\square$

For a fixed positive-definite $\Sigma$, Euclidean distance $\|X-\mu\|$ and $R$ bound each other up to finite positive constants. In particular, if $\lambda_{\min}$ and $\lambda_{\max}$ are the smallest and largest eigenvalues of $\Sigma$, then $$ \lambda_{\min}R^2\leq \|X-\mu\|^2\leq\lambda_{\max}R^2. $$ Consequently, finiteness of a nonnegative moment of the radius is equivalent to finiteness of the corresponding moment of $\|X-\mu\|$. This condition is also what permits the usual absolute moments of the coordinates and their products up to that order. For example, a finite second radial moment is required for a finite covariance matrix.

Power-Exponential Generators

A useful family is $g(t)=\exp(-\lambda t^\beta)$, with $\lambda>0$ and $\beta>0$. The parameter $\beta$ controls the radial tail, while $\lambda$ sets the rate of decay. To evaluate the normalizing integral, substitute $v=\lambda t^\beta$. Then $$ J_p(g)=\frac{1}{\beta}\lambda^{-p/(2\beta)} \Gamma\!\left(\frac{p}{2\beta}\right), \qquad c_p=\frac{\beta\lambda^{p/(2\beta)}\Gamma(p/2)} {\pi^{p/2}\Gamma(p/(2\beta))}. $$ The integral is finite for every $s\geq0$, so all nonnegative radial moments exist. The same substitution in the moment formula yields $$ E[R^s]=\lambda^{-s/(2\beta)} \frac{\Gamma((p+s)/(2\beta))}{\Gamma(p/(2\beta))}. $$

Worked Example: Checking the Gaussian Special Case

Set $\beta=1$ and $\lambda=1/2$, so $g(t)=e^{-t/2}$. The formula gives $$ c_p=\frac{(1/2)^{p/2}\Gamma(p/2)} {\pi^{p/2}\Gamma(p/2)} =(2\pi)^{-p/2}. $$ This is the usual Gaussian density coefficient when $\Sigma$ is the covariance matrix. As a check on the radial moment calculation, $$ E[R^2]=2\frac{\Gamma((p+2)/2)}{\Gamma(p/2)} =2\frac{\Gamma(p/2+1)}{\Gamma(p/2)} =p, $$ using $\Gamma(a+1)=a\Gamma(a)$. The result is consistent with the sum of $p$ standardized squared Gaussian coordinates.

Student t Generators and Tail Thresholds

For $\nu>0$, the multivariate Student t density uses the generator $g(t)=(1+t/\nu)^{-(\nu+p)/2}$. The exponent depends on the dimension as well as the degrees of freedom. Its normalizer follows directly from the radial integral; no construction of the distribution is needed here.

For $a>0$ and $b>0$, the beta integral identity is $\int_0^\infty u^{a-1}(1+u)^{-(a+b)}du=B(a,b)=\Gamma(a)\Gamma(b)/\Gamma(a+b)$. Substitute $u=t/\nu$ into $J_p(g)$, with $a=p/2$ and $b=\nu/2$, to obtain $$ J_p(g)=\frac{\nu^{p/2}}{2} \frac{\Gamma(p/2)\Gamma(\nu/2)}{\Gamma((p+\nu)/2)}. $$ The normalizer is consequently $$ c_p=\frac{\Gamma((p+\nu)/2)} {\Gamma(\nu/2)(\nu\pi)^{p/2}}. $$

Worked Example: A Two-Dimensional Student t Density

Let $p=2$ and $\nu=4$. The generator is $g(t)=(1+t/4)^{-3}$. Its radial integral is $$ J_2(g)=\int_0^\infty (1+t/4)^{-3}\,dt =4\int_1^\infty v^{-3}\,dv =4\left[\frac{1}{2}\right] =2, $$ where $v=1+t/4$. Therefore $$ c_2=\frac{\Gamma(1)}{\pi(2)} =\frac{1}{2\pi}. $$ The radial second moment is finite, since its numerator integral is $$ \int_0^\infty t(1+t/4)^{-3}\,dt =16\int_1^\infty (v-1)v^{-3}\,dv =16\left(\int_1^\infty v^{-2}\,dv-\int_1^\infty v^{-3}\,dv\right) =16\left(1-\frac12\right)=8. $$ Thus $E[R^2]=8/2=4$. This calculation checks both the normalizing constant and the existence of the second radial moment for these parameters.

More generally, the order-$s$ radial integral for the Student t generator is a beta integral with parameters $(p+s)/2$ and $(\nu-s)/2$. The first parameter is positive for $p\geq1$ and $s\geq0$; the second must be positive. Therefore $$ E[R^s]<\infty\quad\Longleftrightarrow\quad 0\leq s<\nu. $$ When $s<\nu$, evaluating the ratio gives $$ E[R^s]=\nu^{s/2} \frac{\Gamma((p+s)/2)\Gamma((\nu-s)/2)} {\Gamma(p/2)\Gamma(\nu/2)}. $$ At $s=\nu$, the radial integral diverges logarithmically at infinity; for $s>\nu$, its tail also diverges. Thus the boundary is strict: $\nu=2$ is not enough for a finite second moment, whereas $\nu>2$ is.

Common Pitfall: Confusing the Radial Integral and the Normalizer

The integral $J_p(g)$ is not itself the normalizing constant. It is the generator’s radial mass after the power of $t$ appropriate to dimension $p$ is included. The constant $c_p$ is its reciprocal multiplied by $\Gamma(p/2)/\pi^{p/2}$. Forgetting either factor can produce a density whose integral is not one.

A second error is to test tail decay using $g(t)$ alone. The density’s total mass involves $t^{p/2-1}g(t)$, and its order-$s$ radial moment involves $t^{(p+s)/2-1}g(t)$. In the Student t family, this distinction is why the density exists for every $\nu>0$, while its second moment exists only for $\nu>2$. Finally, the matrix $\Sigma$ in the density’s quadratic form should not automatically be called a covariance matrix; a covariance requires the relevant second moments to exist.

Key takeaway: Normalize an elliptical generator by evaluating $J_p(g)=\int_0^\infty t^{p/2-1}g(t)\,dt$. Test a radial moment of order $s$ by replacing $p/2$ with $(p+s)/2$. For power-exponential generators every nonnegative radial moment exists; for the Student t generator, order $s$ exists exactly when $s<\nu$.

Check Your Understanding

Use the radial integral and moment conditions to answer the following.

For $p=1$, what integral must be finite for $g$ to define a normalized elliptical density?
For $g(t)=e^{-\lambda t^\beta}$, how does the normalizer change as $\lambda$ changes while $p$ and $\beta$ remain fixed?
For a Student t generator with $\nu=3$, which of the radial moments of orders $1$, $2$, and $3$ are finite?
Why can a generator have a finite normalizing integral but an infinite second radial moment?
For $p=2$ and $g(t)=e^{-t^2}$, verify that the normalizer is $2/\pi^{3/2}$ using the formula for $J_p(g)$.