Characteristic Functions of Gaussian Vectors

From Gaussian Projections to a Vector Formula

A multivariate Gaussian vector can be studied through its one-dimensional projections. Given a vector of coefficients \(u\), the projection \(u^{\mathsf T}X\) is a scalar random variable. Its mean and variance follow directly from the mean vector and covariance matrix of \(X\). Since a Gaussian projection is normally distributed, its scalar characteristic function gives the characteristic function of the whole vector at \(u\).

Let \(X=(X_1,\ldots,X_d)^{\mathsf T}\) be a random vector on a probability space \((\Omega,\mathcal{F},\mathbb{P})\), with state space \(\mathbb{R}^d\). Assume its coordinates have finite second moments. Write its mean vector as \(m=\mathbb{E}[X]\) and its covariance matrix as \(\Sigma=\operatorname{Cov}(X)\), whose \((j,k)\) entry is \(\operatorname{Cov}(X_j,X_k)\). For \(u\in\mathbb{R}^d\), the characteristic function of \(X\) at \(u\) is \(\phi_X(u)=\mathbb{E}[\exp(iu^{\mathsf T}X)]\), where \(i^2=-1\).

For any fixed \(u\), linearity of expectation and the definition of covariance give \(\mathbb{E}[u^{\mathsf T}X]=u^{\mathsf T}m\) and \(\operatorname{Var}(u^{\mathsf T}X)=u^{\mathsf T}\Sigma u\). The latter quantity is nonnegative because it is a variance. We first derive the scalar characteristic function needed to use these facts.

Proof. First take a standard normal random variable \(Z\), with density \(g(x)=(2\pi)^{-1/2}\exp(-x^2/2)\), and let \(F(t)=\int_{\mathbb{R}}\exp(itx)g(x)\,dx\). Differentiation under the integral is valid because the absolute value of the differentiated integrand is \(|x|g(x)\), which is integrable. Since \(g'(x)=-xg(x)\), integration by parts, with vanishing boundary terms as \(x\to\pm\infty\), yields \(\int_{\mathbb{R}}x\exp(itx)g(x)\,dx=-\int_{\mathbb{R}}\exp(itx)g'(x)\,dx=itF(t)\). Consequently \(F'(t)=i\int_{\mathbb{R}}x\exp(itx)g(x)\,dx=-tF(t)\). Together with \(F(0)=1\), this differential equation gives \(F(t)=\exp(-t^2/2)\). If \(v>0\), write \(Y\) in distribution as \(a+\sqrt{v}Z\). Then its characteristic function is \(\exp(ita)F(t\sqrt{v})=\exp(ita-\tfrac12vt^2)\). If \(v=0\), \(Y=a\) almost surely, so its characteristic function is \(\exp(ita)\), which is the same formula with \(v=0\). \(\square\)

The Characteristic Function of a Gaussian Vector

Proof. Fix \(u\in\mathbb{R}^d\). By the definition of a Gaussian vector, \(u^{\mathsf T}X\) is normal, with mean \(u^{\mathsf T}m\) and variance \(u^{\mathsf T}\Sigma u\). By the definition of the vector characteristic function, \(\phi_X(u)=\mathbb{E}[\exp(iu^{\mathsf T}X)]\), which is the scalar characteristic function of \(u^{\mathsf T}X\) evaluated at \(t=1\). Applying the lemma with \(a=u^{\mathsf T}m\) and \(v=u^{\mathsf T}\Sigma u\) gives the stated formula. This argument also covers \(u^{\mathsf T}\Sigma u=0\), by the lemma’s zero-variance case. \(\square\)

The formula makes explicit how the mean contributes a linear term in \(u\) and the covariance contributes a quadratic term. It does not require \(\Sigma\) to be invertible. In particular, one must not assume a density exists: a singular covariance matrix can describe a vector concentrated on a lower-dimensional subset of \(\mathbb{R}^d\).

Proof. By the definition of a Gaussian vector, \(c^{\mathsf T}X\) is normal. Linearity of expectation gives its mean \(c^{\mathsf T}m\), and the definition of covariance gives its variance \(c^{\mathsf T}\Sigma c\). A random variable with variance zero equals its mean almost surely: indeed, \(\mathbb{E}[(L-\mathbb{E}L)^2]=0\), and a nonnegative random variable with expectation zero is zero almost surely. This proves both the stated distribution parameters and the zero-variance conclusion. \(\square\)

Equivalently, substitute \(u=tc\) into the vector formula. The characteristic function of \(L\) is then \(\mathbb{E}[\exp(itL)]=\exp(itc^{\mathsf T}m-\tfrac12t^2c^{\mathsf T}\Sigma c)\), which is the scalar normal characteristic function with exactly those parameters.

Common Pitfall: Dropping Covariance Terms

When expanding \(c^{\mathsf T}\Sigma c\), each off-diagonal covariance contributes twice, once from each symmetric position in the matrix. In the first worked example, a mistaken calculation might use only \(4(2^2)+1((-1)^2)=17\) and omit the cross term, concluding that \(2X_1-X_2\) has variance \(17\). The correction is to expand the full quadratic form: \(4(2^2)+1((-1)^2)+2(2)(-1)(1)=16+1-4=13\). Substituting into \(\phi_X(2r,-r)\) also gives the coefficient \(13\) of \(r^2\), so the missing cross term is detectable by the characteristic-function check.

The projection method is useful whenever the question concerns a sum, contrast, or other linear combination of Gaussian coordinates. It avoids finding a joint density and remains valid in degenerate cases. The formula also provides a direct check: evaluate the vector characteristic function along the line \(u=tc\), and compare the result with the scalar normal parameters computed from the mean and covariance.