Characteristic Function of a Gaussian Vector

From Gaussian Projections to a Vector Transform

A multivariate Gaussian vector can be studied through its one-dimensional projections: for each fixed vector \(t\), the quantity \(t^\mathsf{T}X\) is normal. This fact gives a direct route to the vector’s characteristic function. Rather than integrating a multivariate density—which may not even exist in a singular case—we evaluate one scalar normal characteristic function at a time.

The result is compact, but its two terms have distinct roles. The mean vector contributes a linear expression in \(t\), while the covariance matrix contributes a quadratic expression. Deriving the formula makes clear why every covariance, including an off-diagonal covariance, matters.

Use the linear-combination definition of multivariate normality: \(X\) is Gaussian if every scalar projection \(a^\mathsf{T}X\), for \(a\in\mathbb{R}^p\), has a univariate normal distribution, with a constant allowed as a degenerate normal variable. In the parameterization used here, the projection in direction \(a\) has mean \(a^\mathsf{T}\mu\) and variance \(a^\mathsf{T}\Sigma a\), where \(\mu\in\mathbb{R}^p\) and \(\Sigma\) is the covariance matrix. The latter quantity is nonnegative for every \(a\).

The Scalar Normal Characteristic Function

The vector calculation depends on the characteristic function of a univariate normal variable. We derive it first, including the case of zero variance. This also makes explicit the analytic step used in the calculation.

Proof. First let \(Z\sim N(0,1)\), with density \(g(x)=(2\pi)^{-1/2}e^{-x^2/2}\), and write \(\psi(s)=E[e^{isZ}]\). Since \(|x|g(x)\) is integrable, differentiation under the integral is justified by dominated convergence: the absolute value of the derivative of \(e^{isx}g(x)\) with respect to \(s\) is \(|x|g(x)\), independently of \(s\). Thus

Because \(g'(x)=-xg(x)\), integration by parts gives

The boundary term vanishes because \(g(x)\to0\) as \(x\to\pm\infty\), while \(|e^{isx}|=1\). It follows that \(\psi'(s)=-s\psi(s)\), with \(\psi(0)=1\). Solving this differential equation gives \(\psi(s)=e^{-s^2/2}\). If \(v>0\), write \(Y=m+\sqrt{v}\,Z\); then

If \(v=0\), \(Y=m\) with probability one, so \(E[e^{iuY}]=e^{ium}\), which is the same formula with \(v=0\). \(\square\)

Deriving the Gaussian Vector Formula

Proof. Fix an arbitrary \(t\in\mathbb{R}^p\). By the linear-combination definition, \(Y=t^\mathsf{T}X\) is a scalar normal variable. Its mean is \(t^\mathsf{T}\mu\) and its variance is \(t^\mathsf{T}\Sigma t\). The scalar result applies even if this variance is zero. By the definition of the vector characteristic function,

Since \(t\) was arbitrary, this holds for every \(t\in\mathbb{R}^p\). \(\square\)

The formula also follows directly from the parameters of the projection. In the scalar normal characteristic function, set \(u=1\), \(m=t^\mathsf{T}\mu\), and \(v=t^\mathsf{T}\Sigma t\). The frequency \(t\) is the direction of the projection as well as the argument of the vector characteristic function; it is not a random vector.

How the Mean and Covariance Enter

The mean term \(i\,t^\mathsf{T}\mu\) is linear in \(t\). It determines the oscillatory, or phase, part of the characteristic function. The covariance term is real and nonpositive in the exponent, since \(t^\mathsf{T}\Sigma t\geq0\). Consequently,

Thus the covariance controls the magnitude of the characteristic function, while the mean affects its phase. The matrix expression is not merely a sum of marginal variances: it includes the covariances between components.

Common Pitfall and Why the Formula Matters

A frequent error is to omit the off-diagonal entries of \(\Sigma\) when evaluating \(t^\mathsf{T}\Sigma t\). For a symmetric matrix, expansion gives

The cross-terms can increase or decrease the variance of a projection, depending on the signs of the coefficients and covariances. In the correlated-pair example, dropping the cross-term would give \(5\), not the correct value \(3\).

• Wrong reasoning: “The quadratic term is the sum of squared coefficients times component variances, so covariance entries do not enter.”
• Correction: That sum accounts only for the diagonal terms. The variance of \(t^\mathsf{T}X\) is \(t^\mathsf{T}\Sigma t\), which includes the cross-terms.
• Another check: At \(t=0\), the formula must give \(\varphi_X(0)=1\). At any \(t\), its magnitude must be at most \(1\), as the exponent’s real part is nonpositive.

The projection-based derivation is useful whenever a calculation asks for the characteristic function in a particular direction. It also works for Gaussian vectors whose distributions are concentrated on a lower-dimensional set, because it never requires a multivariate density.