Bounds, Continuity, and Conjugate Symmetry

Why These Four Properties Matter

A characteristic function is built from complex exponentials, but its values cannot vary arbitrarily. Its probability-law origin forces a fixed value at the origin, a uniform bound, a relation between values at opposite arguments, and continuity. These facts are basic consistency checks: a proposed characteristic function must satisfy all four.

Each property follows from the same simple observation: the complex exponential \(e^{itx}\) has modulus one for every pair of real numbers \(t\) and \(x\). The first three properties are immediate consequences of this observation and the definition of expectation. Continuity requires one additional step, using dominated convergence.

Definition and Conventions

Let \(X\) be a real-valued random variable on a probability space, and let \(\mu\) denote its probability law on \(\mathbb R\). The characteristic function of \(X\), or equivalently of \(\mu\), is the function \(\varphi_X:\mathbb R\to\mathbb C\) defined by the expectation of the complex exponential.

The expectation exists for every real \(t\), even if \(X\) has no finite moments. Indeed, \(|e^{itX}|=1\) everywhere, so the integrand is bounded and therefore integrable with respect to a probability measure. This boundedness is also what will give the universal modulus bound.

The Four Fundamental Properties

Proof. At \(t=0\), the integrand is \(e^{i0X}=1\), so its expectation is the expectation of the constant one. Since the underlying measure is a probability measure, this expectation equals one:

For any real \(t\), the triangle inequality for the expectation of an integrable complex random variable gives \(|\mathbb E[e^{itX}]|\leq\mathbb E[|e^{itX}|]\). Since the absolute value of the integrand is one, the right-hand side is \(\mathbb E[1]=1\). Thus \(|\varphi_X(t)|\leq1\).

For conjugate symmetry, use \(\overline{e^{itX}}=e^{-itX}\). Complex conjugation commutes with expectation for an integrable complex random variable; this follows by writing the variable as its real part plus \(i\) times its imaginary part and taking expectations of those two real variables. Therefore

It remains to prove continuity. Fix \(t\in\mathbb R\), and let \((t_n)\) be any sequence of real numbers converging to \(t\). For every outcome at which \(X\) is real-valued, continuity of the exponential gives \(e^{it_nX}\to e^{itX}\). Also, \(|e^{it_nX}|=1\) for every \(n\), and the constant function one is integrable on a probability space. The dominated convergence theorem therefore gives \(\mathbb E[e^{it_nX}]\to\mathbb E[e^{itX}]\). In other words, \(\varphi_X(t_n)\to\varphi_X(t)\). The sequential criterion for continuity on \(\mathbb R\) now shows that \(\varphi_X\) is continuous at \(t\). Since \(t\) was arbitrary, it is continuous on the real line. \(\square\)

Worked Examples

What Continuity Does—and Does Not—Require

The continuity proof has a useful feature: it imposes no moment assumption on \(X\). The random variable need not have a finite mean or variance. The proof uses only pointwise convergence of the exponentials and their common bound by one. A finite expectation of \(|X|\) is not needed.

The argument also explains why continuity is a property of the whole function, not just a formula checked at the origin. To prove continuity at a particular \(t\), one takes any sequence \(t_n\to t\), checks convergence of the integrands, and applies dominated convergence. Nothing in that reasoning depends on \(t\) being zero.

A common pitfall is to infer continuity merely because a displayed expression looks familiar or smooth. Characteristic functions are defined by expectations, and their continuity is guaranteed by the theorem above. When manipulating a formula, special care is needed at points where an algebraic expression has a zero denominator: the integral definition still gives a value there, and continuity must be checked against that value. The uniform example illustrates precisely this issue at the origin.

How to Use the Properties

These properties serve both as tools and as quick diagnostics. When a candidate formula is proposed for a characteristic function, check the normalization at zero and the unit bound first. Then check the relation between opposite arguments. Finally, continuity rules out jumps, including at the origin. Failure of any one check proves that the candidate cannot be a characteristic function of a real-valued random variable.

The checks are necessary conditions, not a claim that every continuous function satisfying these three algebraic conditions is a characteristic function. Characterizing which functions arise from probability laws requires additional results. Here the point is narrower: every law necessarily produces a function with these four properties, and the proofs depend directly on the expectation definition.