Uniform Laws on Ellipsoids

From a Ball to an Ellipsoid

Elliptical distributions often have density contours that are ellipsoids, but not every distribution supported on an ellipsoid has the same radial behavior. A particularly concrete model begins with a point drawn uniformly from a ball and maps that point to an ellipsoid by a linear transformation. The resulting distribution has bounded support, constant density inside that support, and a radius distribution determined directly by dimension.

Let $p$ be a positive integer, and write $\|u\|$ for the Euclidean length of $u\in\mathbb R^p$. The unit ball is $B_p=\{u:\|u\|\leq 1\}$. Let $U$ be uniform on this ball, let $\mu\in\mathbb R^p$, and let $A$ be a nonsingular $p\times p$ matrix. Define $X=\mu+AU$. The support of $X$ is the ellipsoid centered at $\mu$ with shape matrix $\Sigma=AA^\mathsf{T}$.

Definition: A random vector $X$ is uniform on the ellipsoid with center $\mu$ and positive-definite shape matrix $\Sigma$ if it has constant density on $E=\{x:(x-\mu)^\mathsf{T}\Sigma^{-1}(x-\mu)\leq 1\}$ and zero density outside $E$. Equivalently, $X=\mu+AU$, where $U$ is uniform on $B_p$, $A$ is nonsingular, and $AA^\mathsf{T}=\Sigma$. The boundary has probability zero.

Here $\Sigma$ describes the ellipsoid’s shape and orientation; it is not the covariance matrix of $X$. We derive that covariance below. The quadratic form $(x-\mu)^\mathsf{T}\Sigma^{-1}(x-\mu)$ measures squared distance from the center in coordinates scaled to the ellipsoid.

Worked Example: A Uniform Point on a Two-Dimensional Ellipse

Suppose $U$ is uniform on the unit disk and $X=\mu+AU$, where $A=\begin{pmatrix}3&0\\0&2\end{pmatrix}$ and $\mu=(1,-1)^\mathsf{T}$. Find the shape matrix and the probability that $X$ lies within half the ellipse’s radial distance from its center.

The shape matrix is $\Sigma=AA^\mathsf{T}=\begin{pmatrix}9&0\\0&4\end{pmatrix}$. Thus the support is $(x_1-1)^2/9+(x_2+1)^2/4\leq1$, an ellipse with semiaxes $3$ and $2$.

Being within half the radial distance means $\|U\|\leq 1/2$, or equivalently $(X-\mu)^\mathsf{T}\Sigma^{-1}(X-\mu)\leq1/4$. In two dimensions, the area of a disk of radius $r$ is proportional to $r^2$, so this probability is $(1/2)^2=1/4$. The full radial derivation appears below.

The Radius of a Uniform Point

Set $R=\|U\|$. Uniformity means that probability is proportional to volume. A ball of radius $r$ has volume $r^p$ times the volume of the unit ball, so the probability of falling inside that smaller ball is $r^p$, for $0\leq r\leq1$.

Theorem: If $U$ is uniform on $B_p$, then $R=\|U\|$ has distribution function $$ P(R\leq r)= \begin{cases} 0,&r<0,\\ r^p,&0\leq r\leq1,\\ 1,&r>1, \end{cases} $$ and density $f_R(r)=p r^{p-1}$ for $0<r<1$, and zero otherwise. The direction $U/\|U\|$ is uniform on the unit sphere and is independent of $R$; the event $U=0$ has probability zero.

To derive the distribution function, compare volumes: for $0\leq r\leq1$, the ball of radius $r$ has volume $r^p\operatorname{Vol}(B_p)$, while the full ball has volume $\operatorname{Vol}(B_p)$. Their ratio is $r^p$. Differentiating $r^p$ on $(0,1)$ gives the stated density, whose integral is $\int_0^1p r^{p-1}\,dr=1$.

The independence assertion follows from polar coordinates. In those coordinates, volume has the form $r^{p-1}\,dr\,d\omega$, where $d\omega$ is surface measure for direction. Since the density of $U$ is constant on the ball, the joint probability element factors into a function of $r$ and a uniform measure over directions. Thus radius and direction are independent.

Worked Example: Radius and Mean Squared Radius in Four Dimensions

Let $U$ be uniform on the unit ball in four dimensions. Find $P(R\leq0.7)$, the density of $R$, and $E[R^2]$.

The radius distribution gives $P(R\leq0.7)=0.7^4=0.2401$. The density on $0<r<1$ is $f_R(r)=4r^3$.

Using this density, $$ E[R^2]=\int_0^1 r^2\,4r^3\,dr =4\int_0^1r^5\,dr =\frac{4}{6} =\frac{2}{3}. $$ As a check, the general integral is $E[R^2]=p/(p+2)$; for $p=4$, this is $4/6=2/3$.

Covariance and the Shape Matrix

Symmetry about the origin gives $E[U]=0$. Sign symmetry also makes every cross-moment $E[U_iU_j]$ with $i\ne j$ equal to zero. Coordinate-permutation symmetry makes the diagonal moments equal. Their common value can be found by summing them: $\sum_{i=1}^p E[U_i^2]=E[\|U\|^2]=E[R^2]$.

Theorem: For $U$ uniform on $B_p$, $E[U]=0$ and $\operatorname{Cov}(U)=I_p/(p+2)$. For $X=\mu+AU$, with $A$ nonsingular and $\Sigma=AA^\mathsf{T}$, $E[X]=\mu$ and $$ \operatorname{Cov}(X)=\frac{\Sigma}{p+2}. $$

For the first claim, the radius density gives $$ E[R^2]=\int_0^1r^2p r^{p-1}\,dr =p\int_0^1r^{p+1}\,dr =\frac{p}{p+2}. $$ Since the diagonal moments are equal, each is $E[U_i^2]=E[R^2]/p=1/(p+2)$. Together with the zero cross-moments, this yields $\operatorname{Cov}(U)=I_p/(p+2)$. The affine transformation then gives $E[X]=\mu+A E[U]=\mu$, and $$ \operatorname{Cov}(X)=A\operatorname{Cov}(U)A^\mathsf{T} =\frac{AA^\mathsf{T}}{p+2} =\frac{\Sigma}{p+2}. $$ This also verifies why the shape matrix is not the covariance: the covariance is smaller by the factor $p+2$. $\square$

Worked Example: Covariance After a Nondiagonal Transformation

In two dimensions, let $A=\begin{pmatrix}2&1\\0&1\end{pmatrix}$, and let $X=\mu+AU$ for $U$ uniform on the unit disk. Find the shape matrix, covariance, and density inside the support.

First, $$ \Sigma=AA^\mathsf{T} =\begin{pmatrix}2&1\\0&1\end{pmatrix} \begin{pmatrix}2&0\\1&1\end{pmatrix} =\begin{pmatrix}5&1\\1&1\end{pmatrix}. $$ The determinant is $5(1)-1(1)=4$, so $\det(\Sigma)=4$. Since $p=2$, the covariance is $$ \operatorname{Cov}(X)=\frac{\Sigma}{4} =\begin{pmatrix}5/4&1/4\\1/4&1/4\end{pmatrix}. $$

The unit disk has area $\pi$. The linear transformation multiplies area by $|\det A|=2$, so the ellipsoid has area $2\pi$. A uniform distribution on this ellipsoid therefore has density $1/(2\pi)$ at every interior point, and zero outside. This agrees with the general density $1/(\pi\sqrt{\det\Sigma})=1/(2\pi)$ for $p=2$.

Probabilities for Concentric Ellipsoidal Regions

For $X=\mu+AU$, define the squared Mahalanobis distance $Q=(X-\mu)^\mathsf{T}\Sigma^{-1}(X-\mu)$. Because $\Sigma=AA^\mathsf{T}$, the inverse-matrix identity gives $Q=U^\mathsf{T}U=R^2$. Thus a region defined by $Q\leq q$ is the image of a ball of radius $\sqrt q$. This gives a direct formula for its probability.

Result: For $0\leq q\leq1$, $$ P\big((X-\mu)^\mathsf{T}\Sigma^{-1}(X-\mu)\leq q\big)=q^{p/2}. $$ Consequently, if $0\leq q_1\leq q_2\leq1$, the probability of the concentric shell $q_1<(X-\mu)^\mathsf{T}\Sigma^{-1}(X-\mu)\leq q_2$ is $q_2^{p/2}-q_1^{p/2}$.

Indeed, $Q=R^2$, so $Q\leq q$ is exactly $R\leq\sqrt q$. Substituting into the radius distribution gives $P(Q\leq q)=(\sqrt q)^p=q^{p/2}$. The shell probability is the difference between the probabilities at its outer and inner boundaries. Boundaries have probability zero because the distribution has a density.

Worked Example: A Shell Probability in Three Dimensions

Let $X=\mu+AU$, where $U$ is uniform on the three-dimensional unit ball and $A$ is nonsingular. Find the probability that $X$ lies between radial fractions $0.5$ and $0.8$ of the ellipsoid’s maximum radius.

The radial condition is $0.5<R\leq0.8$, where $R=\|U\|$. In three dimensions, $P(R\leq r)=r^3$, so $$ P(0.5<R\leq0.8)=0.8^3-0.5^3=0.512-0.125=0.387. $$ Equivalently, the squared-distance thresholds are $q_1=0.25$ and $q_2=0.64$, and the shell probability is $0.64^{3/2}-0.25^{3/2}=0.512-0.125=0.387$. The answer does not depend on the ellipsoid’s orientation or semiaxis lengths, provided distance is measured using its own shape matrix.

Interpretation and a Common Pitfall

The radius distribution has an important dimensional effect. Although the density is constant throughout the ball, the volume of successive radial layers grows with radius. In dimension $p$, the density of $R$ is proportional to $r^{p-1}$; as dimension increases, a larger share of the probability lies near the boundary. This is a volume effect, not a rise in the pointwise density near the boundary.

A common mistake is to assume that uniformity in a ball makes the radius uniform on $[0,1]$. That would give $P(R\leq r)=r$, but it ignores the fact that the sub-ball of radius $r$ has only $r^p$ of the full ball’s volume. In two dimensions, for instance, the probability of radius at most $1/2$ is $1/4$, not $1/2$. The correction is to calculate probabilities by volume ratios, or use the derived formula $P(R\leq r)=r^p$.

The ellipsoid’s shape affects the directions and scales of variation, while the dimension determines the normalized radius law. To calculate a probability for a concentric ellipsoidal region, first express its boundary using the shape matrix, convert the squared-distance cutoff to $q$, and then use $q^{p/2}$. This clean separation is useful when comparing ellipsoids with different orientations or axis lengths.

Check Your Understanding

Use the radius and ellipsoid formulas to answer the following questions.

For a point uniform on the unit ball in five dimensions, what is $P(R\leq0.6)$?
If $X=\mu+AU$ in dimension $p$ and $AA^\mathsf{T}=\Sigma$, what is $\operatorname{Cov}(X)$?
In dimension four, what is the probability that squared Mahalanobis distance is at most $0.25$?
Why is the shape matrix of a uniform ellipsoid generally not its covariance matrix?
For radial cutoffs $a$ and $b$, with $0\leq a\leq b\leq1$, write the probability of the shell $a<R\leq b$ in dimension $p$.

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