Characteristic Functions of Elliptical Laws

From a Radial Representation to a Characteristic Function

An elliptical distribution can be described by the geometry of its scatter matrix, but its characteristic function reveals that geometry directly in the direction \(t\). The resulting formula separates location from radial behavior: location contributes a phase factor, while scatter enters through a quadratic form. This makes the formula useful for studying symmetry, moments, and whether different parameter descriptions represent the same law.

Use the stochastic radial representation established for elliptical vectors. Let \(X\) be \(p\)-dimensional, let \(\mu\in\mathbb{R}^p\), and let \(\Sigma\) be a symmetric positive semidefinite \(p\times p\) scatter matrix. Choose a matrix \(A\) satisfying \(AA^{\mathsf T}=\Sigma\). Suppose \(X=\mu+RAU\), where \(R\geq 0\) is independent of \(U\), and \(U\) is uniform on the unit sphere in \(\mathbb{R}^p\). Singular \(\Sigma\) is allowed.

Definition: The characteristic function of \(X\) is \(\phi_X(t)=E[\exp(i t^{\mathsf T}X)]\), for \(t\in\mathbb{R}^p\). For the radial representation above, define its scalar characteristic generator by \(\psi(s)=E[\exp(i\sqrt{s}\,R U_1)]\), \(s\geq 0\), where \(U_1\) is the first coordinate of \(U\). Then the elliptical characteristic-function form is \(\phi_X(t)=\exp(i t^{\mathsf T}\mu)\psi(t^{\mathsf T}\Sigma t)\).

To derive the formula, condition on \(R\) and use the characteristic function definition: \(E[\exp(i t^{\mathsf T}X)]=\exp(i t^{\mathsf T}\mu)E[\exp(iR t^{\mathsf T}AU)]\). The vector \(A^{\mathsf T}t\) has squared length \(t^{\mathsf T}AA^{\mathsf T}t=t^{\mathsf T}\Sigma t\). Rotational invariance of the uniform distribution on the sphere means that the projection of \(U\) onto any fixed direction has the same distribution as \(U_1\). Thus, for \(A^{\mathsf T}t\ne 0\), the expectation in the last expression is \(\psi(t^{\mathsf T}\Sigma t)\). When \(A^{\mathsf T}t=0\), that expectation is \(1=\psi(0)\), so the same formula holds.

Since \(U\) and \(-U\) have the same distribution, the imaginary part of \(E[\exp(i\sqrt{s}RU_1)]\) is zero. Consequently \(\psi\) is real-valued, \(\psi(0)=1\), and \(\phi_X(t)\) is the location phase multiplied by a real radial function. The argument depends on rotational invariance, not on the existence of a density.

Worked Example: A Uniform Distribution on a Circle

Let \(p=2\), \(\mu=(1,-1)^{\mathsf T}\), \(\Sigma=I_2\), and \(R=2\) with probability one. Then \(X=\mu+2U\) is uniform on the circle of radius \(2\) centered at \(\mu\). Write \(U=(\cos\Theta,\sin\Theta)^{\mathsf T}\), where \(\Theta\) is uniform on \([0,2\pi)\).

For \(t=(t_1,t_2)^{\mathsf T}\), the quadratic form is \(t^{\mathsf T}\Sigma t=t_1^2+t_2^2\). Rotating the direction of \(t\) inside the circle gives \(\psi(s)=\frac{1}{2\pi}\int_0^{2\pi}\exp(i\,2\sqrt{s}\cos\theta)\,d\theta\). The imaginary part integrates to zero, so equivalently \(\psi(s)=\frac{1}{2\pi}\int_0^{2\pi}\cos(2\sqrt{s}\cos\theta)\,d\theta\). Therefore \(\phi_X(t)=\exp(i(t_1-t_2))\psi(t_1^2+t_2^2)\).

The radial function depends only on the length of \(t\), not its direction. Also, \(E[R^2]=4\), so the covariance calculation derived below gives \(\operatorname{Cov}(X)=\frac{4}{2}I_2=2I_2\). The circle has fixed distance \(2\) from its center, but its covariance is not \(4I_2\): variation is shared across the two coordinates according to the uniform direction.

Symmetry and Covariance

The radial form gives central symmetry without requiring a density. Because \(U\) and \(-U\) are identically distributed and \(R\) is independent of \(U\), \(RAU\) and \(-RAU\) have the same distribution. Thus \(X-\mu\) and \(-(X-\mu)\) have the same distribution: the law is centrally symmetric about \(\mu\).

A second useful consequence is a precise relationship between scatter and covariance. It requires a finite second radial moment; without that condition, a scatter matrix need not be a covariance matrix.

Theorem: If \(E[R^2]<\infty\), then \(E[X]=\mu\) and \(\operatorname{Cov}(X)=\frac{E[R^2]}{p}\Sigma\). Moreover, the right derivative of the scalar characteristic generator at zero exists and satisfies \(\psi'(0+)=-\frac{E[R^2]}{2p}\), so equivalently \(\operatorname{Cov}(X)=-2\psi'(0+)\Sigma\).

Proof. Rotational invariance gives \(E[U]=0\). It also gives \(E[UU^{\mathsf T}]=I_p/p\): off-diagonal entries vanish by reflection symmetry, all diagonal entries are equal by coordinate symmetry, and their sum is \(E[U^{\mathsf T}U]=1\). Independence then implies \(E[RAU]=0\) and \(E[(RAU)(RAU)^{\mathsf T}]=E[R^2]A(I_p/p)A^{\mathsf T}\). The finite second moment ensures these expectations exist. Hence \(E[X]=\mu\) and the covariance formula follows from \(AA^{\mathsf T}=\Sigma\).

For the derivative, symmetry makes \(\psi(s)=E[\cos(\sqrt{s}RU_1)]\). For \(s>0\), \(\frac{\psi(s)-1}{s}=E[\frac{\cos(\sqrt{s}RU_1)-1}{s}]\). The integrand converges pointwise to \(-R^2U_1^2/2\) as \(s\) decreases to zero. The inequality \(0\leq 1-\cos x\leq x^2/2\) bounds its absolute value by \(R^2U_1^2/2\), which is integrable because \(E[R^2]<\infty\) and \(|U_1|\leq1\). Dominated convergence therefore gives \(\psi'(0+)=-E[R^2U_1^2]/2=-E[R^2]/(2p)\), using independence and \(E[U_1^2]=1/p\). Substitution proves the equivalent covariance expression. \(\square\)

Worked Example: Reading Covariance from a Radial Moment

Let \(p=3\), \(\Sigma=\operatorname{diag}(1,2,3)\), and suppose the independent radius satisfies \(P(R=1)=P(R=3)=1/2\). Then \(E[R^2]=(1+9)/2=5\). The theorem gives \(\operatorname{Cov}(X)=\frac{5}{3}\operatorname{diag}(1,2,3)=\operatorname{diag}(5/3,10/3,5)\).

The generator derivative is \(\psi'(0+)=-5/(2\cdot3)=-5/6\). Checking the alternate formula gives \(-2\psi'(0+)\Sigma=(5/3)\operatorname{diag}(1,2,3)\), the same covariance. The check also makes clear why \(\Sigma\) itself is not the covariance here.

Worked Example: The Gaussian Characteristic Generator

Let \(Z\sim N(0,I_p)\). Its distribution is rotationally invariant, so it can be written \(Z=RU\), with \(R=\lVert Z\rVert\) independent of the uniform direction \(U\). Since \(E[R^2]=E[Z^{\mathsf T}Z]=p\), the representation has unit covariance.

For any fixed \(t\), the standard normal characteristic function gives \(E[\exp(i t^{\mathsf T}Z)]=\exp(-t^{\mathsf T}t/2)\). Comparing with the radial form when \(\Sigma=I_p\) shows that \(\psi(s)=\exp(-s/2)\). For \(X=\mu+AZ\), where \(AA^{\mathsf T}=\Sigma\), this yields \(\phi_X(t)=\exp(i t^{\mathsf T}\mu)\exp(-t^{\mathsf T}\Sigma t/2)\), and the covariance theorem gives \(\operatorname{Cov}(X)=\Sigma\).

Equality of Elliptical Distributions

Characteristic functions determine probability distributions uniquely. For two radial representations, this gives an exact equality test that does not require either distribution to have a density.

Theorem: Suppose \(X_1\) and \(X_2\) have elliptical characteristic functions \(\phi_j(t)=\exp(i t^{\mathsf T}\mu_j)\psi_j(t^{\mathsf T}\Sigma_jt)\), for \(j=1,2\). Their distributions are equal if and only if \(\exp(i t^{\mathsf T}\mu_1)\psi_1(t^{\mathsf T}\Sigma_1t)=\exp(i t^{\mathsf T}\mu_2)\psi_2(t^{\mathsf T}\Sigma_2t)\) for every \(t\in\mathbb{R}^p\).

Proof. If the distributions are equal, expectations of the bounded measurable function \(x\mapsto\exp(i t^{\mathsf T}x)\) are equal for every \(t\), so their characteristic functions agree. Conversely, if the characteristic functions agree for every \(t\), the uniqueness theorem for characteristic functions implies that the probability distributions agree. \(\square\)

The theorem compares complete characteristic functions; it does not say that equal distributions must have equal scatter matrices when the generators are allowed to differ. For example, a normal law with location \(\mu\) and scatter \(\Sigma\) has generator \(\psi(s)=\exp(-s/2)\). The same characteristic function is obtained using scatter \(c\Sigma\) and generator \(\psi_c(s)=\exp(-s/(2c))\), for any \(c>0\): \(\exp(i t^{\mathsf T}\mu)\psi_c(t^{\mathsf T}(c\Sigma)t)=\exp(i t^{\mathsf T}\mu)\exp(-t^{\mathsf T}\Sigma t/2)\). Thus scatter scale and generator scale can trade off without changing the law.

Worked Example: Equality with Different Scatter Scales

Take \(p=2\), \(\mu=0\), and \(\Sigma=I_2\). The characteristic function with generator \(\psi(s)=\exp(-s/2)\) is \(\phi(t)=\exp(-t^{\mathsf T}t/2)\). Now describe a second elliptical law with scatter \(4I_2\) and generator \(\psi_4(s)=\exp(-s/8)\). For every \(t\), \(\exp(i t^{\mathsf T}0)\psi_4(t^{\mathsf T}(4I_2)t)=\exp(-(4t^{\mathsf T}t)/8)=\exp(-t^{\mathsf T}t/2)\).

The characteristic functions, and therefore the distributions, are identical, although the scatter matrices differ by a factor of four. In both descriptions the covariance is \(I_2\): in the second representation the generator's radial scale changes to compensate for the larger scatter. A scatter matrix becomes identifiable only after an appropriate scale convention is fixed, for example by fixing the radial normalization; when the second moment exists, the covariance itself is fixed by the distribution.

Common Pitfall and Practical Use

A frequent mistake is to read \(\Sigma\) in the radial characteristic function as the covariance matrix automatically. That is correct for a normalization such as the Gaussian example, but not in general. The covariance formula has the factor \(E[R^2]/p\), and it applies only when the second radial moment is finite. If that moment does not exist, the characteristic function still has its radial form, but the covariance may be undefined.

The radial form is useful because it separates three questions that should not be conflated: the center \(\mu\), the shape and scale encoded jointly by \(\Sigma\) and \(\psi\), and the moment properties implied by the behavior of \(\psi\) near zero. In particular, symmetry follows from the radial representation even when covariance does not exist, while equality of laws is decided by agreement of the full characteristic functions.

Key takeaway: An elliptical radial representation yields \(\phi_X(t)=\exp(i t^{\mathsf T}\mu)\psi(t^{\mathsf T}\Sigma t)\). It gives central symmetry directly and gives \(\operatorname{Cov}(X)=-2\psi'(0+)\Sigma\) when the second radial moment is finite. Equality concerns the full characteristic function; scatter alone is not generally identifiable without a scale normalization.

Check Your Understanding

Answer these questions using the radial representation and characteristic-function results.

Why does rotational invariance reduce \(E[\exp(iR t^{\mathsf T}AU)]\) to a function of \(t^{\mathsf T}\Sigma t\) alone?
Which symmetry of \(U\) implies that the scalar generator is real-valued?
If \(p=4\) and \(E[R^2]=12\), what is \(\operatorname{Cov}(X)\) in terms of \(\Sigma\), assuming the required representation conditions?
Can two elliptical descriptions with different scatter matrices define the same distribution? Explain how the generator can compensate.
Does the characteristic-function formula alone guarantee that \(X\) has finite covariance? State the condition used for the covariance result.