Dependence and Independence in Elliptical Models

Why Zero Covariance Is Not Enough

For a bivariate normal vector, zero covariance between the coordinates guarantees independence. It is tempting to extend this rule to every distribution with elliptical contours. That extension is false: elliptical symmetry can make covariance vanish while the coordinates remain dependent. The special role of the normal distribution can be identified precisely by examining the characteristic function.

Throughout, consider a bivariate elliptical vector \(X=(X_1,X_2)^\mathsf{T}\) with finite, positive coordinate variances. Use the characteristic-function representation \( \phi_X(t)=\exp(i t^\mathsf{T}\mu)\psi(t^\mathsf{T}\Omega t) \), where \(t\in\mathbb{R}^2\), \(\mu\) is the location vector, \(\Omega\) is a symmetric positive-definite scatter matrix, and \(\psi(0)=1\) is the characteristic generator. The scatter matrix determines the shape in this representation; when the covariance exists, it is a positive scalar multiple of \(\Omega\).

Definition: Two random variables are uncorrelated if their covariance exists and \(\operatorname{Cov}(X_1,X_2)=0\). They are independent if \(P(X_1\in A,X_2\in B)=P(X_1\in A)P(X_2\in B)\) for every pair of Borel sets \(A,B\). Independence implies zero covariance when the relevant second moments are finite; the converse does not hold in general.

The Independence Criterion for Elliptical Coordinates

With finite second moments, the covariance matrix of an elliptical vector is proportional to its scatter matrix. Thus, for positive coordinate variances, zero covariance is equivalent to zero off-diagonal scatter: \(\Omega_{12}=0\). In that case the characteristic function becomes \( \phi_X(s,t)=\exp(i(s\mu_1+t\mu_2))\psi(\Omega_{11}s^2+\Omega_{22}t^2) \). The coordinates are independent exactly when this joint characteristic function factors into the product of the two marginal characteristic functions.

Theorem: Suppose \(X\) is a bivariate elliptical vector with finite, positive coordinate variances and diagonal scatter matrix \(\Omega=\operatorname{diag}(\Omega_{11},\Omega_{22})\). Then \(X_1\) and \(X_2\) are independent if and only if \( \psi(u+v)=\psi(u)\psi(v) \) for every \(u,v\geq 0\). If they are independent, \(X\) is bivariate normal. Conversely, a bivariate normal vector with diagonal covariance matrix has independent coordinates.

Proof. The marginal characteristic functions, obtained by setting the other argument to zero in the joint characteristic function, are \( \phi_{X_1}(s)=\exp(is\mu_1)\psi(\Omega_{11}s^2) \) and \( \phi_{X_2}(t)=\exp(it\mu_2)\psi(\Omega_{22}t^2) \). The coordinates are independent if and only if the joint characteristic function equals \(\phi_{X_1}(s)\phi_{X_2}(t)\) for all real \(s,t\). Cancelling the location factor gives \( \psi(\Omega_{11}s^2+\Omega_{22}t^2)=\psi(\Omega_{11}s^2)\psi(\Omega_{22}t^2) \). Because both diagonal entries are positive, \(\Omega_{11}s^2\) ranges over all \(u\geq0\) and \(\Omega_{22}t^2\) ranges over all \(v\geq0\). This proves the stated equivalence.

To identify the generators satisfying the equation, first note that \(\psi\) is continuous at zero and \(\psi(0)=1\). The equation implies \(\psi(u)=\psi(u/2)^2\), so \(\psi(u)\geq0\). It cannot vanish at any \(u>0\): if \(\psi(u)=0\), repeated use of the equation gives \(\psi(u/2^k)=0\) for every positive integer \(k\), contradicting continuity at zero. Therefore \(\psi(u)>0\). Define \(g(u)=\log\psi(u)\). Then \(g\) is continuous and satisfies \(g(u+v)=g(u)+g(v)\) for nonnegative \(u,v\). The continuous additive functions on the nonnegative reals are linear, so \(g(u)=au\) for some real \(a\). Since a characteristic function has absolute value at most one, \(\psi(u)=\exp(au)\) cannot have \(a>0\); hence \(a\leq0\). Write \(a=-c\), with \(c\geq0\). It follows that \( \phi_X(t)=\exp(i t^\mathsf{T}\mu-c\,t^\mathsf{T}\Omega t) \), the characteristic function of a normal vector with covariance \(2c\Omega\). Positive coordinate variances rule out \(c=0\). Conversely, if \(X\) is normal with diagonal covariance, its characteristic function factors into the two marginal characteristic functions, which proves independence. \(\square\)

Worked Examples

Worked Example: Diagonal-Covariance Normal Coordinates

Let \(X\sim N(\mu,\Sigma)\), where \(\mu=(2,-1)^\mathsf{T}\) and \(\Sigma=\operatorname{diag}(4,9)\). Its characteristic function at \((s,t)\) is \( \phi_X(s,t)=\exp(i(2s-t)-\tfrac12(4s^2+9t^2)) \). Separating the terms gives

\phi_X(s,t) = \exp(2is-2s^2)\, \exp(-it-\tfrac{9}{2}t^2).

The first factor is the characteristic function of \(X_1\sim N(2,4)\), and the second is that of \(X_2\sim N(-1,9)\). The joint characteristic function is their product, so the coordinates are independent. Their covariance is zero, but the conclusion follows here because the vector is normal—not merely because its covariance vanishes.

Worked Example: Uniform Distribution on a Disk

Let \((X_1,X_2)\) be uniform on the unit disk \(x^2+y^2\leq1\). Its density is \(1/\pi\) on the disk and zero outside. Reflection symmetry gives \(E[X_1]=E[X_2]=E[X_1X_2]=0\), so the coordinates have zero covariance. To test independence, compare the mixed second moment of their squares with the product of the separate second moments.

Write \(X_1=r\cos\theta\), \(X_2=r\sin\theta\). In polar coordinates the density times area element is \(r\,dr\,d\theta/\pi\), for \(0\leq r\leq1\) and \(0\leq\theta<2\pi\). Thus

E[X_1^2] =\frac{1}{\pi}\int_0^1 r^3\,dr\int_0^{2\pi}\cos^2\theta\,d\theta =\frac14, \qquad E[X_2^2]=\frac14.

Likewise,

E[X_1^2X_2^2] =\frac{1}{\pi}\int_0^1 r^5\,dr\int_0^{2\pi}\cos^2\theta\sin^2\theta\,d\theta =\frac{1}{\pi}\cdot\frac16\cdot\frac{\pi}{4} =\frac{1}{24}.

If \(X_1\) and \(X_2\) were independent, then \(E[X_1^2X_2^2]=E[X_1^2]E[X_2^2]=1/16\), since these moments are finite. But \(1/24\ne1/16\); indeed, \(\operatorname{Cov}(X_1^2,X_2^2)=1/24-1/16=-1/48\). The coordinates are dependent despite having zero covariance. Their possible values are linked by the disk constraint: large absolute values of one coordinate leave less room for the other.

Worked Example: Uncorrelated but Dependent Student Coordinates

Consider the bivariate Student \(t\) distribution with five degrees of freedom, location zero, and scatter \(I_2\), the \(2\times2\) identity matrix. Its density is

f(x,y)=\frac{1}{2\pi} \left(1+\frac{x^2+y^2}{5}\right)^{-7/2}.

The coefficient normalizes the density. In polar coordinates, its integral is

\int_0^{2\pi}\int_0^\infty \frac{1}{2\pi}\left(1+\frac{r^2}{5}\right)^{-7/2}r\,dr\,d\theta =\int_0^\infty\left(1+\frac{r^2}{5}\right)^{-7/2}r\,dr =\frac52\int_1^\infty u^{-7/2}\,du=1,

where \(u=1+r^2/5\), so \(r\,dr=(5/2)\,du\). The law is rotationally symmetric and has covariance \((5/3)I_2\), so the coordinates have finite variances and zero covariance. (For example, the radial integral gives \(E[X_1^2+X_2^2]=10/3\); symmetry splits this equally between the coordinates.)

To show dependence directly, suppose the coordinates were independent. Their positive continuous joint density would then factor into marginal densities, implying \(f(x,y)f(0,0)=f(x,0)f(0,y)\). For \(x=y=1\), after cancelling the common coefficient, the left-hand side has power factor \((7/5)^{-7/2}\), while the right-hand side has \((6/5)^{-7}\). These are unequal: equality would require \((7/5)^{1/2}=6/5\), but squaring gives \(7/5\ne36/25\). Therefore the coordinates are dependent. The example has circular, and hence elliptical, contours and uncorrelated coordinates; neither property guarantees independence.

What the Geometry Does—and Does Not—Tell Us

An elliptical contour describes how the joint density or other distributional level sets are shaped. When the scatter matrix is diagonal, its principal directions align with the coordinate axes, and finite covariance makes the coordinates uncorrelated. Neither fact describes how probability within those contours is allocated. In particular, a shared radial constraint or a non-Gaussian elliptical generator can link the coordinates even when the axes are aligned.

The theorem gives a precise boundary: among nondegenerate bivariate elliptical distributions with finite coordinate variances and diagonal scatter, independent coordinates require a normal law. This does not mean every nonnormal elliptical model must have dependent coordinates in every possible representation; the stated conclusion concerns independent coordinates under the theorem's conditions. It does mean that zero covariance alone cannot be used as a general independence test outside the normal setting.

Common Mistake

Incorrect reasoning: “The contours are circular and the covariance matrix is diagonal, so the coordinates are independent.” This confuses symmetry and lack of linear association with factorization of the joint distribution.
Correction: Check whether the joint characteristic function factors into the actual marginal characteristic functions, or verify independence by another valid argument. For elliptical coordinates with diagonal scatter, the generator must satisfy \(\psi(u+v)=\psi(u)\psi(v)\); under the theorem's conditions this forces the normal case.
Another common error: A joint characteristic function has been written as two expressions multiplied together, so it must establish independence. Each expression must depend on only its own coordinate's argument and equal that coordinate's marginal characteristic function.

Key takeaway: For finite-variance elliptical coordinates, zero covariance identifies diagonal scatter, not independence. In the nondegenerate bivariate case, independence with diagonal scatter forces a normal distribution; uniform-disk and Student examples show how dependence can remain when covariance is zero.

Check Your Understanding

Answer each question using the distinction between covariance, scatter, and characteristic-function factorization.

For a bivariate elliptical vector with diagonal scatter, write the functional equation on \(\psi\) that is equivalent to independence.
Why does the equation \(\psi(u+v)=\psi(u)\psi(v)\), together with continuity and the characteristic-function bound, lead to an exponential generator?
In the uniform-disk example, which calculation disproves independence even though the coordinate covariance is zero?
What is the correctly factored characteristic function for a bivariate normal vector with mean \((\mu_1,\mu_2)^\mathsf{T}\) and diagonal covariance \(\operatorname{diag}(\sigma_1^2,\sigma_2^2)\)?
Why do circular contours and zero covariance fail to establish independence for the bivariate Student example?