Characteristic Functions of Random Vectors

From a Probability Law to a Fourier Transform

A random vector’s distribution can be studied through a function defined on the same Euclidean space as its frequency argument. This function, the characteristic function, exists whether the law is discrete, continuous, mixed, or concentrated on a lower-dimensional set. Its definition therefore belongs to the probability law itself, not to a particular choice of density.

Let \(X\) be an \(\mathbb{R}^d\)-valued random vector, let \(F\) denote its probability law, and let \(t=(t_1,\ldots,t_d)^T\in\mathbb{R}^d\). Write \(t^T X=\sum_{j=1}^d t_jX_j\) for the usual inner product. The complex exponential \(e^{it^T X}\) has modulus one, which will ensure that its expectation is always well-defined.

Definition: The characteristic function of \(X\), or equivalently of its law \(F\), is the function \(\phi_X:\mathbb{R}^d\to\mathbb{C}\) defined by

\phi_X(t)=E\!\left[e^{it^T X}\right] =\int_{\mathbb{R}^d} e^{it^T x}\,dF(x).

If \(X\) has a density \(f\) with respect to \(d\)-dimensional Lebesgue measure, the same definition can be written as \(\phi_X(t)=\int_{\mathbb{R}^d}e^{it^T x}f(x)\,dx\). For a discrete vector with point masses \(p_m\) at \(x_m\), it becomes \(\phi_X(t)=\sum_m p_m e^{it^T x_m}\).

The notation \(dF(x)\) means integration with respect to the probability law; it does not require that a density exist. In the density formula, \(t\) is a fixed argument of the function, while \(X\) is random. The characteristic function is generally complex-valued even though the underlying probability law is real-valued.

Worked Calculations

Worked Example: A Discrete Vector with Three Possible Values

Suppose \(X\in\mathbb{R}^2\) has probabilities \(1/2\), \(1/3\), and \(1/6\) at \((1,0)^T\), \((0,2)^T\), and \((0,0)^T\), respectively. These probabilities are nonnegative and sum to \(1/2+1/3+1/6=1\), so they define a probability law.

For \(t=(t_1,t_2)^T\), the inner products at the three possible values are \(t_1\), \(2t_2\), and \(0\). Taking the expectation as a probability-weighted sum gives

\phi_X(t_1,t_2) =\frac12e^{it_1}+\frac13e^{2it_2}+\frac16.

At the origin, this is \(\phi_X(0,0)=1/2+1/3+1/6=1\). For example, at \(t=(\pi,0)^T\), it is \(-1/2+1/3+1/6=0\). The value need not be positive or real, even though all three probabilities are positive.

Worked Example: A Uniform Vector on a Rectangle

Let \(X=(X_1,X_2)^T\) have the uniform density \(f(x_1,x_2)=1/(ab)\) on \(0\leq x_1\leq a,\ 0\leq x_2\leq b\), where \(a,b>0\), and density zero elsewhere. The rectangle has area \(ab\), so this density integrates to \(1\). Direct integration gives

\begin{aligned} \phi_X(t_1,t_2) &=\frac{1}{ab}\int_0^a\int_0^b e^{i(t_1x_1+t_2x_2)}\,dx_2\,dx_1\\ &=\left(\frac1a\int_0^a e^{it_1x_1}\,dx_1\right) \left(\frac1b\int_0^b e^{it_2x_2}\,dx_2\right). \end{aligned}

For \(t_1\ne0\), the first factor is \((e^{iat_1}-1)/(iat_1)\); at \(t_1=0\), its integral is \(1\). The second factor is \((e^{ibt_2}-1)/(ibt_2)\) for \(t_2\ne0\), and \(1\) at \(t_2=0\). Thus the formula, with these values at zero, is

\phi_X(t_1,t_2)= \begin{cases} \dfrac{e^{iat_1}-1}{iat_1},&t_1\ne0,\\[6pt] 1,&t_1=0 \end{cases} \begin{cases} \dfrac{e^{ibt_2}-1}{ibt_2},&t_2\ne0,\\[6pt] 1,&t_2=0. \end{cases}

In particular, \(\phi_X(0,0)=1\). The separate factors here follow from evaluating this particular rectangular integral; the definition itself does not require a product density.

Worked Example: A Standard Bivariate Normal Vector

Let \(X\) have density \(f(x_1,x_2)=(2\pi)^{-1}\exp\{-(x_1^2+x_2^2)/2\}\) on \(\mathbb{R}^2\). The exponent separates into a term in \(x_1\) and a term in \(x_2\), so the defining integral separates into two one-dimensional Gaussian integrals. To evaluate them, define

I(u)=\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}e^{iux}e^{-x^2/2}\,dx.

Differentiation under the integral is valid because the absolute value of the differentiated integrand is bounded by \(|x|e^{-x^2/2}/\sqrt{2\pi}\), an integrable function. Integration by parts, with the Gaussian factor vanishing at both infinite endpoints, yields

I'(u)=\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}ix e^{iux}e^{-x^2/2}\,dx =-uI(u).

Indeed, \(xe^{-x^2/2}=-(e^{-x^2/2})'\), so integration by parts turns the integral of \(ix e^{iux}e^{-x^2/2}\) into \(-u\int e^{iux}e^{-x^2/2}\,dx\). Since \(I(0)=1\), solving this differential equation gives \(I(u)=e^{-u^2/2}\). Applying it to each coordinate,

\phi_X(t_1,t_2)=I(t_1)I(t_2) =e^{-(t_1^2+t_2^2)/2}.

This calculation gives \(\phi_X(0,0)=1\), and the function approaches \(1\) as \((t_1,t_2)\) approaches the origin.

Existence and Basic Properties

The definition is meaningful for every probability law, without moment or density assumptions. Several further properties follow directly from the bounded complex exponential.

Theorem: For every \(\mathbb{R}^d\)-valued random vector \(X\) and every \(t\in\mathbb{R}^d\), the characteristic function exists and satisfies \(\phi_X(0)=1\), \(|\phi_X(t)|\leq1\), and \(\phi_X(-t)=\overline{\phi_X(t)}\).

Proof. For each outcome, \(|e^{it^T X}|=1\). Hence \(e^{it^TX}\) is bounded and integrable under the probability law, proving existence. At \(t=0\), its value is identically \(1\), so \(\phi_X(0)=1\). The triangle inequality for expectations gives

|\phi_X(t)|=\left|E[e^{it^TX}]\right| \leq E[|e^{it^TX}|]=1.

Finally, complex conjugation commutes with expectation for an integrable complex random variable, and \(\overline{e^{it^TX}}=e^{-it^TX}\). Therefore

\overline{\phi_X(t)} =\overline{E[e^{it^TX}]} =E[e^{-it^TX}] =\phi_X(-t).

This proves all three claims. \(\square\)

A further structural property is positive definiteness. It is a constraint on values at several arguments at once, stronger than the pointwise bound \(|\phi_X(t)|\leq1\).

Theorem: For any positive integer \(m\), any \(t_1,\ldots,t_m\in\mathbb{R}^d\), and any \(c_1,\ldots,c_m\in\mathbb{C}\),

\sum_{j=1}^m\sum_{k=1}^m \overline{c_j}c_k\,\phi_X(t_j-t_k)\geq0.

The displayed quantity is real, and the inequality means it is a nonnegative real number.

Proof. Substitute the expectation definition into the finite sum. A finite sum can be interchanged with expectation, giving

\begin{aligned} \sum_{j,k}\overline{c_j}c_k\phi_X(t_j-t_k) &=E\!\left[\sum_{j,k}\overline{c_j}c_k e^{i(t_j-t_k)^TX}\right]\\ &=E\!\left[\left|\sum_{k=1}^m c_k e^{-it_k^TX}\right|^2\right]\geq0. \end{aligned}

The random variable inside the final expectation is a nonnegative real number and is bounded above by \((\sum_k|c_k|)^2\), so its expectation exists. This proves the claim. \(\square\)

Behavior at the Origin

Every characteristic function equals \(1\) at the origin and is continuous there. In fact, it is continuous everywhere. For fixed \(t,h\in\mathbb{R}^d\),

|\phi_X(t+h)-\phi_X(t)| \leq E\!\left[|e^{ih^TX}-1|\right].

As \(h\to0\), the integrand tends pointwise to zero. Its absolute value is at most \(2\), which is integrable under a probability law. The dominated convergence theorem therefore shows that the expectation tends to zero. This proves continuity at every \(t\), and in particular \(\phi_X(h)\to\phi_X(0)=1\) as \(h\to0\). No moment condition is needed.

When \(E[\|X\|]<\infty\), where \(\|\cdot\|\) is the Euclidean norm, there is also a useful quantitative bound. The elementary inequality \(|e^{iu}-1|\leq |u|\) for real \(u\), together with \(|h^T X|\leq\|h\|\|X\|\), gives

|\phi_X(h)-1| \leq E[|e^{ih^TX}-1|] \leq E[|h^TX|] \leq \|h\|E[\|X\|].

The first-moment condition is needed for this bound to be finite, but not for continuity. This distinction matters: the characteristic function always exists, even when moments do not.

Why the Definition Matters—and a Common Pitfall

A characteristic function expresses a probability law through its Fourier transform while remaining defined for every law. If a density exists, a Fourier integral using that density is convenient; for a discrete law, a weighted sum is convenient. Both are instances of the same expectation with respect to \(F\).

Common pitfall: requiring a density or moments. One might reason that a Fourier transform can be written only by integrating a density, or that an expectation must fail when the random vector has no finite mean. The correction is to use the probability-law definition: \(|e^{it^TX}|=1\), so this expectation is finite for every probability distribution. A density is needed only for the density-specific integral formula, and a finite moment is needed only for additional bounds such as the first-moment bound above.

A second useful check is the origin: any proposed characteristic function must take the value \(1\) at \(t=0\), because a probability law has total mass \(1\). The bound, conjugate symmetry, positive-definiteness condition, and continuity provide further checks on calculations.

Check Your Understanding

Answer these questions using the definition and results above.

Why does the characteristic function exist even if a random vector has no density and no finite first moment?
A discrete vector has masses \(p_r\) at points \(x_r\). Write its characteristic function as a sum and state its value at the origin.
What does the conjugate-symmetry identity say about \(\phi_X(-t)\) and \(\phi_X(t)\)?
In the proof of continuity, which bound permits the use of the dominated convergence theorem?
Which additional condition makes the bound \(|\phi_X(h)-1|\leq\|h\|E[\|X\|]\) finite?