Characteristic Functions of Elliptical Vectors

Why a Characteristic Function Reveals Elliptical Shape

A characteristic function records how a distribution responds to every linear direction. For an elliptical vector, that directional information has a special structure: the location contributes a phase factor, while the remaining dependence on direction is summarized by a quadratic form. Consequently, one scalar function describes how the distribution behaves across all directions.

This structure follows from the affine representation of an elliptical vector as a location shift and linear transformation of a spherical vector. Spherical symmetry makes the characteristic function depend only on the length of its argument; the linear transformation turns that length into a quadratic form. No moment assumptions are needed.

Definition: For a random vector \(X\) in \(\mathbb{R}^p\), its characteristic function is \(\phi_X(t)=E[\exp(i t'X)]\), for \(t\in\mathbb{R}^p\). A random vector \(Z\) is spherical if \(QZ\) has the same distribution as \(Z\) for every \(p\times p\) orthogonal matrix \(Q\), meaning \(Q'Q=I\). An elliptical vector with location \(\mu\) and positive definite scatter matrix \(\Sigma\) has a representation \(X=\mu+AZ\), where \(Z\) is spherical and \(AA'=\Sigma\).

The matrix \(A\) need not be a particular square root: different choices can give the same \(\Sigma\). The characteristic-function formula below depends on \(A\) only through \(AA'\). The scatter matrix describes the shape and scale in the quadratic form; its name does not, by itself, assert that it is a covariance matrix.

The Generator and the Elliptical Formula

For a spherical vector, rotations can move any argument of a given length to any other argument of that length. Its characteristic function therefore depends only on the squared length. It is convenient to record that dependence using a function on the nonnegative real numbers.

Definition: If \(Z\) is spherical, its characteristic-function generator is the function \(\psi:[0,\infty)\to\mathbb{R}\) defined by \(\psi(u)=\phi_Z(\sqrt{u}\,e_1)\), where \(e_1=(1,0,\ldots,0)'\). Then \(\phi_Z(s)=\psi(s's)\) for every \(s\in\mathbb{R}^p\). In particular, \(\psi(0)=1\).

The generator is real-valued because a spherical distribution is unchanged by reflection \(Z\mapsto -Z\). Thus \(Z\) and \(-Z\) have the same distribution, and \(\phi_Z(s)=E[\cos(s'Z)]\): the sine term has expectation zero. The generator need not be nonnegative at every argument. As with any characteristic function, it satisfies \(|\psi(u)|\leq 1\); it is also continuous. Not every arbitrary function on \([0,\infty)\) is a valid generator.

Theorem: If \(X=\mu+AZ\), where \(Z\) is spherical and \(AA'=\Sigma\), then \[ \phi_X(t)=\exp(i t'\mu)\,\psi(t'\Sigma t),\qquad t\in\mathbb{R}^p, \] where \(\psi\) is the characteristic-function generator of \(Z\).

Proof. For every \(t\in\mathbb{R}^p\), the definition of a characteristic function and the affine representation give \[ \phi_X(t)=E[\exp(i t'(\mu+AZ))] =\exp(i t'\mu)\,E[\exp(i (A't)'Z)] =\exp(i t'\mu)\,\phi_Z(A't). \] Since \(Z\) is spherical, \(\phi_Z(s)=\psi(s's)\). Substituting \(s=A't\) yields \[ \phi_X(t)=\exp(i t'\mu)\,\psi((A't)'(A't)) =\exp(i t'\mu)\,\psi(t'AA't) =\exp(i t'\mu)\,\psi(t'\Sigma t). \] This proves the formula. \(\square\)

The same reasoning also gives a characterization when \(\Sigma\) is positive definite. If an actual random vector \(X\) has characteristic function \(\exp(i t'\mu)\psi(t'\Sigma t)\), set \(Y=\Sigma^{-1/2}(X-\mu)\). Substitution shows \(\phi_Y(s)=\psi(s's)\), which is unchanged when \(s\) is replaced by \(Q's\) for any orthogonal \(Q\). Hence \(QY\) and \(Y\) have the same characteristic function and, by uniqueness of characteristic functions, the same distribution. Thus \(Y\) is spherical and \(X\) is an affine image of a spherical vector. The formula is a characterization for a valid characteristic function; a proposed \(\psi\) cannot be assumed valid merely because it is defined on nonnegative numbers.

Worked Examples

Worked Example: A Gaussian Elliptical Vector

Let \(Z\sim N(0,I)\) in two dimensions, let \(\mu=(1,-2)'\), and let \(A=\operatorname{diag}(2,3)\). Find the characteristic function of \(X=\mu+AZ\).

The standard bivariate normal characteristic function is \(\phi_Z(s)=\exp(-s's/2)\), so its generator is \(\psi(u)=\exp(-u/2)\). The scatter matrix is \(\Sigma=AA'=\operatorname{diag}(4,9)\). For \(t=(t_1,t_2)'\), \[ t'\mu=t_1-2t_2,\qquad t'\Sigma t=4t_1^2+9t_2^2. \] Substitution in the elliptical formula gives \[ \phi_X(t)=\exp(i(t_1-2t_2))\exp\!\left(-\frac{4t_1^2+9t_2^2}{2}\right). \] The location appears in the phase, and the direction-dependent quadratic form appears as the argument of the generator.

Worked Example: A Spherical Vector with Fixed Radius

In \(\mathbb{R}^2\), let \(\Theta\) be uniform on \([0,2\pi)\) and set \(Z=2(\cos\Theta,\sin\Theta)'\). Derive its characteristic function and generator.

The angle is unchanged in distribution by any rotation, so \(Z\) is spherical. Write \(t=(t_1,t_2)'\) and \(r=\sqrt{t_1^2+t_2^2}\). When \(r>0\), choose an angle \(\alpha\) such that \(t=r(\cos\alpha,\sin\alpha)'\). Then \(t'Z=2r\cos(\Theta-\alpha)\). Since shifting a uniform angle by \(\alpha\) preserves its distribution, \[ \phi_Z(t)=\frac{1}{2\pi}\int_0^{2\pi}\exp(i\,2r\cos\theta)\,d\theta =\frac{1}{2\pi}\int_0^{2\pi}\cos(2r\cos\theta)\,d\theta. \] The imaginary part is zero: replacing \(\theta\) by \(\theta+\pi\) changes the sign of the sine term and leaves the integral unchanged. This expression also holds at \(r=0\), where it equals 1. As \(r^2=t't\), the generator is \[ \psi(u)=\frac{1}{2\pi}\int_0^{2\pi}\cos(2\sqrt{u}\cos\theta)\,d\theta,\qquad u\geq 0. \] This example shows that a generator need not have the Gaussian exponential form. The integral gives the characteristic function directly, without requiring a density.

Worked Example: An Affine Image of a Radial Scale Mixture

Let \(G\sim N(0,I)\) in two dimensions, and independently let \(S=1\) or \(S=2\), each with probability \(1/2\). Set \(Z=SG\), then \(X=(1,-1)'+AZ\), where \(A=\operatorname{diag}(2,1)\). Find \(\psi\) and \(\phi_X\).

Conditional on \(S=s\), \(Z=sG\), so \(\phi_{Z\mid S=s}(t)=\exp(-s^2 t't/2)\). Averaging over the two possible values of \(S\) gives \[ \phi_Z(t)=\tfrac12\exp(-t't/2)+\tfrac12\exp(-4t't/2) =\tfrac12\exp(-t't/2)+\tfrac12\exp(-2t't). \] Thus \(\psi(u)=\tfrac12\exp(-u/2)+\tfrac12\exp(-2u)\). The vector \(Z\) is spherical because each conditional Gaussian distribution is spherical; mixing those distributions preserves rotational invariance. Also, \(\Sigma=AA'=\operatorname{diag}(4,1)\). For \(t=(t_1,t_2)'\), the elliptical formula gives \[ \phi_X(t)=\exp(i(t_1-t_2)) \left\{\tfrac12\exp\!\left(-\frac{4t_1^2+t_2^2}{2}\right) +\tfrac12\exp\!\left(-2(4t_1^2+t_2^2)\right)\right\}. \] The same quadratic form appears inside both terms because the affine transformation acts on the argument of the spherical characteristic function.

Recognizing Spherical and Gaussian Cases

A centered elliptical vector with \(\Sigma=I\) has characteristic function \(\psi(t't)\), which depends only on the length of \(t\), as expected for a spherical vector. More generally, \(\Sigma=cI\) for \(c>0\) also gives a spherical distribution about \(\mu\): its characteristic function depends on \(t\) only through \(t't\), apart from the location phase. For a spherical vector centered at the origin, take \(\mu=0\).

For the Gaussian case, \(\psi(u)=\exp(-u/2)\), and the formula becomes \[ \phi_X(t)=\exp(i t'\mu-t'\Sigma t/2). \] This is the characteristic function of \(N(\mu,\Sigma)\). Other generators give other elliptical laws, so the quadratic-form structure alone does not imply normality.

A characteristic-function generator is not the same object as a density generator. A density generator describes a density as a function of a squared distance from the center. A characteristic-function generator describes the Fourier transform as a function of the quadratic form in frequency space. They are related through the distribution, but their functional forms are not interchangeable.

Common Pitfalls and Interpretation

Using the wrong argument for the generator. The input is \(t'\Sigma t\), not \(t'\Sigma t\) with a separate square root inserted. For a spherical vector before transformation, the corresponding expression is \(\psi(t't)\).
Assuming every radial-looking function is a generator. A proposed function must come from a valid characteristic function in the dimension under consideration. The formula’s shape does not establish validity on its own.
Confusing density and characteristic-function generators. One describes a density in the observation space; the other describes a characteristic function in the frequency argument.
Treating \(\Sigma\) as uniquely specified by the formula. For any \(c>0\), the same quadratic-form dependence can be written using \(c\Sigma\) and \(\psi_c(u)=\psi(u/c)\). The generator and scatter matrix therefore depend on a scale convention unless an additional normalization is imposed.
Assuming the formula requires moments. Characteristic functions exist for every probability distribution. No expectation or covariance assumption is needed to derive the elliptical form.

Key takeaway: For an elliptical vector, \(\phi_X(t)=\exp(i t'\mu)\psi(t'\Sigma t)\). The location creates the phase, the scatter matrix turns direction into a quadratic form, and the generator carries the remaining radial distributional shape.

Check Your Understanding

Use the definitions and derivation above to answer each question.

Why does the characteristic function of a spherical vector depend only on the length of its argument?
If \(AA'=\Sigma\), show where \(t'\Sigma t\) enters when computing the characteristic function of \(X=\mu+AZ\).
For \(Z\sim N(0,I)\), what is its characteristic-function generator?
What does \(\Sigma=cI\), with \(c>0\), imply about the shape of the characteristic function apart from its location phase?
Why does the elliptical characteristic-function formula not establish that \(\Sigma\) is a covariance matrix?