Elliptical Densities and Ellipsoidal Contours

From a Radial Density to Ellipsoidal Geometry

A multivariate density can have a central location and spread that vary by direction. In an elliptical model, the density is organized by a single nonnegative radial quantity: a squared distance measured relative to a positive-definite scatter matrix. This structure makes its contours tractable. The matrix determines their shape and orientation, while the radial generator determines which of those contours occur at a given density level.

The word “radial” here does not mean distance from the origin in ordinary Euclidean coordinates. It means distance after accounting for the location and directional scaling encoded by the scatter matrix. Keeping that distinction clear is essential when interpreting an elliptical contour.

Definition: Let $X$ take values in $\mathbb{R}^p$. An elliptical density with location vector $\mu\in\mathbb{R}^p$, positive-definite scatter matrix $S$, and radial generator $g$ has the form $$ f(x)=c_g |S|^{-1/2}g\!\left((x-\mu)^{\mathsf T}S^{-1}(x-\mu)\right),\qquad x\in\mathbb{R}^p. $$ Here $S$ is symmetric and positive definite, $|S|$ is its determinant, $g:[0,\infty)\to[0,\infty)$ is measurable, and $c_g>0$ is a normalizing constant. We assume $g$ is not zero almost everywhere and that $\int_{\mathbb{R}^p}g(z^{\mathsf T}z)\,dz$ is finite, so the constant can be chosen to make $f$ integrate to one. The associated quadratic form is $$ q(x)=(x-\mu)^{\mathsf T}S^{-1}(x-\mu). $$

Because $S$ is positive definite, $q(x)\geq 0$, with equality only at $x=\mu$. The density depends on $x$ only through $q(x)$. Thus points with the same value of $q$ have the same density, provided the generator assigns the same value to that radial argument. The determinant factor adjusts the density for the scale and volume change represented by $S$; the quadratic form identifies the contour geometry.

Constant-Density Sets and Their Eigenstructure

A constant-density contour at height $a>0$ is the set of points satisfying $f(x)=a$. Substitution into the density formula shows that its possible radial values are precisely the solutions of $$ g(q)=\frac{a|S|^{1/2}}{c_g}. $$ If $g$ is strictly decreasing, there can be at most one such value $q=r^2$. When that value exists and is positive, the contour is $q(x)=r^2$, the boundary of a filled ellipsoid $q(x)\leq r^2$. If $g$ is not one-to-one, a single density height can instead correspond to several radial values and therefore several ellipsoidal surfaces.

Theorem: Let $S$ be symmetric positive definite and let $r>0$. The set $q(x)=r^2$ is an ellipsoid centered at $\mu$. If $\lambda_1,\ldots,\lambda_p$ are the eigenvalues of $S$ and $v_1,\ldots,v_p$ are corresponding orthonormal eigenvectors, its principal axes point along the $v_i$, and its semiaxis length in direction $v_i$ is $r\sqrt{\lambda_i}$.

Proof. By the spectral theorem, $S=U\Lambda U^{\mathsf T}$, where $U$ has the orthonormal eigenvectors as columns and $\Lambda=\operatorname{diag}(\lambda_1,\ldots,\lambda_p)$. Positive definiteness gives $\lambda_i>0$ for every $i$, so $S^{-1}=U\Lambda^{-1}U^{\mathsf T}$. Write the coordinates of $x-\mu$ in this eigenvector basis as $z=U^{\mathsf T}(x-\mu)$. Then $$ q(x)=z^{\mathsf T}\Lambda^{-1}z=\sum_{i=1}^p\frac{z_i^2}{\lambda_i}. $$ Consequently $q(x)=r^2$ is equivalent to $$ \sum_{i=1}^p\frac{z_i^2}{r^2\lambda_i}=1. $$ This is the standard equation of an ellipsoid in the orthonormal coordinates $z_i$, with semiaxis $r\sqrt{\lambda_i}$ along coordinate direction $i$, which is $v_i$ in the original coordinates. $\square$

The eigenvectors give orientation; the eigenvalues give relative axis lengths. A large eigenvalue means a longer semiaxis in that eigenvector’s direction. The common factor $r$ changes all axis lengths together as one moves to a different radial contour. If all eigenvalues are equal, every contour is a sphere; otherwise the contours are stretched differently across directions.

Worked Example: A Diagonal Scatter Matrix

Let $p=2$, $\mu=(1,-2)^{\mathsf T}$, and $S=\begin{pmatrix}4&0\\0&1\end{pmatrix}$. Take the Gaussian radial generator $g(q)=e^{-q/2}$. The matrix is positive definite, with determinant $4$ and inverse $\begin{pmatrix}1/4&0\\0&1\end{pmatrix}$. Thus $$ q(x)=\frac{(x_1-1)^2}{4}+(x_2+2)^2. $$

The contour $q(x)=1$ has equation $(x_1-1)^2/4+(x_2+2)^2=1$. Its center is $(1,-2)$; the eigenvectors are the coordinate directions and the eigenvalues are $4$ and $1$. The semiaxis lengths are therefore $1\sqrt{4}=2$ horizontally and $1\sqrt{1}=1$ vertically. Since $g(1)=e^{-1/2}$, the density is constant on this contour, at height $c_g|S|^{-1/2}e^{-1/2}$. For the bivariate Gaussian generator, $c_g=(2\pi)^{-1}$, so this height is $e^{-1/2}/(4\pi)$.

The filled region $q(x)\leq1$ is the interior of that ellipse. The level set $q(x)=1$ is only its boundary.

Worked Example: A Rotated Ellipse

Take $\mu=(0,0)^{\mathsf T}$ and $S=\begin{pmatrix}5&4\\4&5\end{pmatrix}$. The vectors $v_1=(1,1)^{\mathsf T}/\sqrt{2}$ and $v_2=(1,-1)^{\mathsf T}/\sqrt{2}$ are orthonormal eigenvectors. Direct multiplication gives $Sv_1=9v_1$ and $Sv_2=v_2$, so their eigenvalues are $9$ and $1$, respectively.

Let $u=v_1^{\mathsf T}x=(x_1+x_2)/\sqrt{2}$ and $v=v_2^{\mathsf T}x=(x_1-x_2)/\sqrt{2}$. The quadratic form is $q(x)=u^2/9+v^2$. At $q(x)=4$, divide by $4$ to obtain $$ \frac{u^2}{36}+\frac{v^2}{4}=1. $$ The semiaxis lengths are $2\sqrt{9}=6$ along $v_1$, the diagonal direction $x_1=x_2$, and $2\sqrt{1}=2$ along $v_2$, the direction $x_1=-x_2$. The off-diagonal entries of $S$ are reflected in the rotation of these axes away from the coordinate directions.

Affine Changes of Coordinates Preserve the Form

An invertible change of coordinates can rotate, stretch, or reflect an elliptical density without changing its radial structure. In particular, the transformed scatter matrix determines the transformed contour axes. This fact follows directly from the quadratic form and the density’s change-of-variables formula.

Theorem: Suppose $X$ has the elliptical density above, and let $Y=b+BX$, where $B$ is an invertible $p\times p$ matrix. Then $Y$ has location $b+B\mu$, scatter $BSB^{\mathsf T}$, and the same radial generator $g$.

Proof. Put $\mu_Y=b+B\mu$ and $S_Y=BSB^{\mathsf T}$. This scatter matrix is positive definite: for any nonzero $w$, $B^{\mathsf T}w\neq0$, and $$ w^{\mathsf T}S_Yw=(B^{\mathsf T}w)^{\mathsf T}S(B^{\mathsf T}w)>0. $$ Since $y-\mu_Y=B(x-\mu)$ and $S_Y^{-1}=B^{-\mathsf T}S^{-1}B^{-1}$, the transformed quadratic form satisfies $$ (y-\mu_Y)^{\mathsf T}S_Y^{-1}(y-\mu_Y) =(x-\mu)^{\mathsf T}S^{-1}(x-\mu)=q(x). $$ The change-of-variables formula gives $f_Y(y)=f_X(B^{-1}(y-b))/|\det B|$. Also, $|S_Y|^{1/2}=|\det B|\,|S|^{1/2}$. Substituting these identities into the density of $X$ yields $$ f_Y(y)=c_g|S_Y|^{-1/2}g\!\left((y-\mu_Y)^{\mathsf T}S_Y^{-1}(y-\mu_Y)\right), $$ as claimed. $\square$

Worked Example: A Nonmonotone Generator

Let $p=2$, $\mu=(0,0)^{\mathsf T}$, $S=I$, and $g(q)=\exp(-(q-2)^2)$. This generator is nonnegative and not monotone over $q\geq0$. A finite normalizing constant exists: in polar coordinates, $$ \int_{\mathbb{R}^2}g(x^{\mathsf T}x)\,dx =2\pi\int_0^\infty e^{-(r^2-2)^2}r\,dr =\pi\int_0^\infty e^{-(q-2)^2}\,dq, $$ which is positive and finite because the integrand is bounded on finite intervals and decays like a Gaussian in $q$ as $q\to\infty$.

At density height $a=c_g e^{-1}$, the equation $g(q)=e^{-1}$ becomes $\exp(-(q-2)^2)=e^{-1}$, or $(q-2)^2=1$. Its nonnegative solutions are $q=1$ and $q=3$. Since $S=I$, these radial levels are circles centered at the origin with radii $1$ and $\sqrt{3}$. Thus one density height corresponds to two contours, not one. The density is at least $c_g e^{-1}$ when $1\leq q\leq3$, which is the annulus between those circles.

Common Pitfall: Confusing the Scatter Matrix with the Contour

A common mistake is to read the entries of $S$ as direct axis lengths, or to assume that every constant-density set is one ellipsoid. Neither conclusion follows from the definition. The axes come from the eigenvectors and eigenvalues, and their lengths also depend on the radial level $r$. Moreover, a nonmonotone generator can assign the same density to multiple values of $q$, as the annulus example shows.

Another tempting but incorrect shortcut is to call $q(x)$ ordinary squared Euclidean distance from $\mu$. That is true only when $S=I$. For a general scatter matrix, $q$ weights directions differently; in eigenvector coordinates it is the sum of squared coordinates divided by their corresponding eigenvalues. Using the Euclidean distance alone would erase both the stretching and rotation of the contours.

Key takeaway: The density formula separates radial behavior from geometry. The generator determines which quadratic-form levels share a density value; the positive-definite scatter matrix determines the orientation and relative lengths of the ellipsoidal axes at each level.

Check Your Understanding

Use the quadratic form and the generator separately when answering these questions.

For a contour $q(x)=r^2$, how do the eigenvectors and eigenvalues of $S$ determine its orientation and semiaxis lengths?
If all eigenvalues of $S$ are equal, what shape do its radial contours have?
Why can a nonmonotone generator produce more than one contour at the same density height?
For $S=\operatorname{diag}(9,4)$ and $q(x)=1$, what are the semiaxis lengths and their directions?
What property of an invertible matrix $B$ ensures that $BSB^{\mathsf T}$ remains positive definite?

Elliptical Densities and Ellipsoidal Contours

What You'll Learn