Why Scatter and Covariance Are Different
For an elliptical distribution, the scatter matrix describes the shape and orientation of its ellipsoidal contours. Covariance also describes variation and orientation, but its scale depends on the distribution’s radial behavior. The two matrices are therefore related, but they are not automatically equal.
The connection is especially simple in a stochastic radial representation: a random radius controls distance from the center, while an independent random direction controls orientation. When the radial second moment is finite, the covariance is a scalar multiple of the scatter matrix. The scalar is determined by the average squared radius.
The scatter matrix is positive definite under the nonsingularity assumption on \(A\). Its role in the representation is distinct from that of \(R\): \(A\) transforms directions, while \(R\) sets their distance from the center. In particular, the representation does not require \(E[R^2]=p\), the condition that would make scatter equal covariance.
The Second Moment of a Uniform Direction
The calculation depends on the second-moment matrix of \(U\). Its value follows from symmetry and the fact that \(U\) always has unit length.
To see why, the uniform distribution is unchanged by changing the sign of any one coordinate. Consequently each component of \(E[U]\) is zero, and each off-diagonal entry \(E[U_iU_j]\), for \(i\neq j\), is zero. The distribution is also unchanged by permuting coordinates, so all diagonal entries \(E[U_i^2]\) are equal. Since \(U'U=1\),
There are \(p\) equal terms in this sum, so each is \(1/p\). Thus the diagonal entries are \(1/p\), proving the result. This argument also shows why a single coordinate has second moment \(1/p\), not one: the total squared length is spread equally across the \(p\) coordinates.
Deriving Covariance from the Radius
A finite radial second moment guarantees that the mean exists as well. Indeed, the Cauchy–Schwarz inequality gives \(E[R]\leq \sqrt{E[R^2]}<\infty\). Independence of \(R\) and \(U\), together with \(E[U]=0\), then gives \(E[X]=\mu\).
Proof. Since \(E[R^2]<\infty\), \(R\) has a finite mean. Independence and \(E[U]=0\) imply
We can therefore calculate the covariance by taking the expectation of \((X-\mu)(X-\mu)'\). Independence allows the scalar factor \(R^2\) to be separated from \(UU'\), and the second-moment result for \(U\) gives
All entries in this calculation are integrable: \(E[R^2]\) is finite, and the coordinates of \(U\) are bounded in absolute value by one. Hence the covariance is well-defined, and the factorization by independence is valid. \(\square\)
The theorem identifies the scalar linking scatter and covariance as \(c=E[R^2]/p\). A finite second moment is essential: without it, the displayed covariance need not exist. When it does exist, it is positive because \(A\) is nonsingular and \(R\geq 0\) with \(E[R^2]>0\), unless \(R=0\) almost surely.
Worked Examples
Worked Example: A Gaussian Vector Has Covariance Equal to Scatter
Let \(p=2\), let $$ \Sigma=\begin{pmatrix}4&1\\1&2\end{pmatrix}, $$ and suppose \(X=\mu+RAU\), where \(AA'=\Sigma\), \(U\) is uniform on the unit circle, and \(R=\sqrt{G_1^2+G_2^2}\) for independent \(G_1,G_2\sim N(0,1)\), independent of \(U\). For this radial construction, \(E[R^2]=E[G_1^2]+E[G_2^2]=2\). Thus $$ \operatorname{Cov}(X)=\frac{2}{2}\Sigma =\begin{pmatrix}4&1\\1&2\end{pmatrix}. $$
This is the usual Gaussian radial representation: the squared radius is the sum of two squared standard normal variables. The covariance equals scatter here because the expected squared radius is the dimension, \(p=2\). The matrix is positive definite since its leading diagonal entry is \(4>0\) and its determinant is \(4(2)-1^2=7>0\).
Worked Example: A Two-Point Radius Changes the Scale
Let \(p=3\), \(\mu=0\), and $$ \Sigma=\begin{pmatrix}1&0&0\\0&2&0\\0&0&3\end{pmatrix}. $$ Take \(R=1\) or \(R=3\), each with probability \(1/2\), independently of \(U\). The radial second moment is $$ E[R^2]=\frac12(1^2)+\frac12(3^2)=\frac12+\frac92=5. $$
The covariance is therefore $$ \operatorname{Cov}(X)=\frac{5}{3}\Sigma =\begin{pmatrix}5/3&0&0\\0&10/3&0\\0&0&5\end{pmatrix}. $$ For instance, the variance of the first coordinate is \(5/3\), not the scatter entry \(1\). The variance of the third coordinate is \(5\), not \(3\). The direction distribution determines the relative shape, while the larger average squared radius multiplies every direction’s second moment by the same factor.
Worked Example: A Finite Mean Does Not Guarantee Covariance
Let \(p=2\), \(\mu=0\), and \(\Sigma=I_2\). Take \(U\) uniform on the unit circle and independent of a Pareto radius \(R\) with density \(f(r)=\tfrac32 r^{-5/2}\) for \(r\geq1\), and \(f(r)=0\) otherwise. This is a probability density because $$ \int_1^\infty \frac32 r^{-5/2}\,dr =\frac32\left[\frac{-2}{3}r^{-3/2}\right]_1^\infty=1. $$
Its mean radius is finite: $$ E[R]=\int_1^\infty \frac32 r^{-3/2}\,dr =\frac32\left[-2r^{-1/2}\right]_1^\infty=3. $$ Thus \(E[X]=0\). But the second moment diverges: $$ E[R^2]=\int_1^\infty \frac32 r^{-1/2}\,dr=\infty. $$ Since \(E[U_1^2]=1/2\), independence gives \(E[X_1^2]=E[R^2]E[U_1^2]=\infty\), and likewise for \(X_2\). The covariance matrix does not exist as a finite matrix, despite the finite mean and the perfectly well-defined scatter matrix \(I_2\).
When Can Scatter Be Rescaled to Equal Covariance?
If \(0<E[R^2]<\infty\), the theorem gives a precise rescaling. Defining \(c=E[R^2]/p\), the covariance is \(c\Sigma\). Thus multiplying the scatter matrix by \(c\) gives the covariance. Equivalently, scatter itself equals covariance when the radius is normalized so that \(E[R^2]=p\).
This scale depends on the radial convention. For any \(a>0\), replacing \(\Sigma\) by \(\widetilde{\Sigma}=a\Sigma\) and \(R\) by \(\widetilde{R}=R/\sqrt{a}\) leaves the represented vector unchanged: $$ \widetilde{R}\,\widetilde{A}U =\frac{R}{\sqrt{a}}(\sqrt{a}A)U =RAU, \qquad \widetilde{A}\widetilde{A}'=a\Sigma. $$ The new radial second moment is \(E[\widetilde{R}^2]=E[R^2]/a\); consequently $$ \frac{E[\widetilde{R}^2]}{p}\widetilde{\Sigma} =\frac{E[R^2]/a}{p}(a\Sigma) =\frac{E[R^2]}{p}\Sigma. $$ Covariance is unchanged, although the representation’s scatter and radius have been rescaled. A normalization convention for the radius is therefore needed if scatter is to have a fixed scale.
Common Pitfall and Practical Interpretation
A common mistake is to read the scatter matrix as the covariance matrix simply because both describe the elliptical shape. That reasoning overlooks the radius. For example, in the two-point-radius example, the scatter entry in the first coordinate is \(1\), but that coordinate’s variance is \(5/3\). The correction is to calculate \(E[R^2]/p\) before identifying the matrices.
A second mistake is to infer that a finite mean ensures a finite covariance. The Pareto-radius example has \(E[R]<\infty\) but \(E[R^2]=\infty\). The mean calculation uses a first moment; covariance requires a second moment. When the second radial moment is infinite, one cannot assign a finite covariance merely by multiplying scatter by a large or undefined scalar.
This distinction matters when interpreting elliptical models: scatter provides the shape matrix in a chosen representation, while covariance additionally reflects radial dispersion. It also matters when comparing parametrizations, since two representations can describe the same distribution with differently scaled scatter matrices. Always check the radius convention and moment condition before interpreting a scatter parameter as a covariance.
Check Your Understanding
Use the radial representation and the second-moment condition to answer these questions.
- For \(p=4\) and \(E[R^2]=12\), what scalar relates covariance to scatter?
- Why does \(E[UU']=I_p/p\) rather than \(I_p\)?
- If \(E[R^2]=p\), what is the relationship between covariance and scatter?
- Can a finite mean guarantee a finite covariance for an elliptical vector? Explain.
- What happens to the radius and scatter if the scatter matrix is multiplied by a positive constant while the represented vector is kept unchanged?