Continuity and Uniform Continuity of Characteristic Functions

Why Continuity Matters

A characteristic function records a distribution through expectations of complex exponentials. Its values can vary with the argument, and continuity tells us that small changes in that argument cannot cause sudden jumps. For statistical theory, it is important that this regularity holds for every probability distribution, not only distributions with a density or finite moments.

In fact, characteristic functions have a stronger property: they are uniformly continuous on the whole real line. The key is to compare two arguments by their difference. An elementary bound for complex exponentials then gives control that does not depend on where those arguments are located.

Definition: If \(X\) is a real-valued random variable, its characteristic function is \(\phi_X(t)=E[e^{itX}]\), for \(t\in\mathbb{R}\). The expectation exists for every \(t\), since \(|e^{itX}|=1\) and thus \(E[|e^{itX}|]=1\).

An Elementary Bound for Complex Exponentials

For any real \(u\), the two complex numbers \(e^{iu}\) and \(1\) lie on the unit circle, so their distance is at most \(2\). A second bound follows by integrating the derivative of the exponential along the real interval from \(0\) to \(u\):

|e^{iu}-1|\leq \min(2,|u|).

Indeed, the bound by \(2\) follows from \(|e^{iu}-1|\leq |e^{iu}|+1=2\). For the other bound, the fundamental theorem of calculus gives

e^{iu}-1=\int_0^u i e^{is}\,ds, \qquad |e^{iu}-1| \leq \left|\int_0^u i e^{is}\,ds\right| \leq \int_{\min(0,u)}^{\max(0,u)} |i e^{is}|\,ds =|u|.

Combining these bounds and substituting \(u=hX\) gives the pointwise inequality

|e^{ihX}-1|\leq \min(2,|hX|).

The bound by \(2\) is valid regardless of the size of \(X\); the bound by \(|hX|\) is especially useful when \(X\) is not too large.

Continuity at Every Argument

Theorem: For every real-valued random variable \(X\), the characteristic function \(\phi_X\) is continuous at every \(t\in\mathbb{R}\).

Proof. Fix \(t\in\mathbb{R}\), and let \((t_n)\) be any real sequence with \(t_n\to t\). For every outcome at which \(X\) is finite, \(e^{it_nX}\to e^{itX}\). A real-valued random variable is finite almost surely, so this convergence holds almost surely. Also, \(|e^{it_nX}|=1\) for every \(n\), and the constant bound \(1\) is integrable on a probability space. The bounded convergence theorem, applied to the real and imaginary parts (or equivalently to the bounded complex-valued functions), therefore yields

\phi_X(t_n)=E[e^{it_nX}] \longrightarrow E[e^{itX}]=\phi_X(t).

Since this holds for every sequence \(t_n\to t\), the sequential criterion for continuity on \(\mathbb{R}\) proves continuity at \(t\). The choice of \(t\) was arbitrary, so \(\phi_X\) is continuous everywhere. \(\square\)

This argument needs no density, no special form for the distribution, and no moment assumption. The integrable bound is simply \(1\), which comes from the modulus of a complex exponential.

Worked Example: A Bernoulli Characteristic Function

Let \(X\) have a Bernoulli distribution with \(P(X=1)=p\) and \(P(X=0)=1-p\), where \(0\leq p\leq1\). Its characteristic function is

\phi_X(t) =E[e^{itX}] =(1-p)e^{it\cdot 0}+p e^{it\cdot 1} =1-p+p e^{it}.

For an increment \(h\), subtract the values at \(t+h\) and \(t\):

|\phi_X(t+h)-\phi_X(t)| =p|e^{i(t+h)}-e^{it}| =p|e^{it}||e^{ih}-1| \leq p\min(2,|h|).

The right-hand side tends to zero as \(h\to0\), proving continuity. More strongly, it does not depend on \(t\), so it directly gives uniform continuity. When \(p=0\), the characteristic function is the constant \(1\); when \(p=1\), the same bound remains valid and is sharp as a bound of this form.

Uniform Continuity on the Whole Line

Theorem: For every real-valued random variable \(X\), the characteristic function \(\phi_X\) is uniformly continuous on \(\mathbb{R}\). In particular, for every \(t,h\in\mathbb{R}\), \[ |\phi_X(t+h)-\phi_X(t)| \leq E[\min(2,|hX|)]. \]

Proof. Linearity of expectation and the triangle inequality for the modulus of an expectation give

\begin{aligned} |\phi_X(t+h)-\phi_X(t)| &=\left|E\left[e^{i(t+h)X}-e^{itX}\right]\right|\\ &\leq E\left[\left|e^{itX}(e^{ihX}-1)\right|\right]\\ &=E[|e^{ihX}-1|]\\ &\leq E[\min(2,|hX|)]. \end{aligned}

Here \(|e^{itX}|=1\), which is why the bound has no dependence on \(t\). To see that its right-hand side tends to zero as \(h\to0\), take any sequence \(h_n\to0\). Almost surely, \(h_nX\to0\), and \(0\leq\min(2,|h_nX|)\leq2\). The constant \(2\) is integrable on the probability space, so the bounded convergence theorem implies

E[\min(2,|h_nX|)]\longrightarrow 0.

Consequently, for any \(\varepsilon>0\), there is a \(\delta>0\) such that \(|h|<\delta\) implies \(E[\min(2,|hX|)]<\varepsilon\). The bound above then shows that whenever \(s,t\in\mathbb{R}\) satisfy \(|s-t|<\delta\), \(|\phi_X(s)-\phi_X(t)|<\varepsilon\): set \(h=s-t\) and use the bound with base argument \(t\). The same \(\delta\) works for every base argument. This is uniform continuity. \(\square\)

There is also a direct epsilon argument that makes the role of the tail of \(X\) visible. Since \(X\) is finite almost surely, \(P(|X|>M)\to0\) as \(M\to\infty\). For any \(M>0\), split the expectation according to whether \(|X|\leq M\):

E[\min(2,|hX|)] \leq |h|M+2P(|X|>M).

Given \(\varepsilon>0\), first choose \(M>0\) so that \(2P(|X|>M)<\varepsilon/2\). Then choose \(\delta=\varepsilon/(2M)\). If \(|h|<\delta\), the displayed bound is less than \(\varepsilon\). This again gives a single increment threshold that works for every \(t\), and it does not require a finite first moment.

Worked Example: Rademacher Distribution

Suppose \(P(X=1)=P(X=-1)=1/2\). Then

\phi_X(t) =\tfrac12 e^{it}+\tfrac12 e^{-it} =\cos t.

For any \(t,h\in\mathbb{R}\), the general increment inequality gives

|\cos(t+h)-\cos t| =|\phi_X(t+h)-\phi_X(t)| \leq E[\min(2,|hX|)] =\min(2,|h|).

The last equality holds because \(|X|=1\) with probability one. Thus \(|h|<\varepsilon\) guarantees that the change in \(\cos t\) is less than \(\varepsilon\), uniformly over all \(t\). For \(|h|\geq2\), the bound by \(2\) remains valid, though it is not needed to establish continuity.

Worked Example: Uniform Continuity Without a Finite Mean

Define a positive-integer-valued random variable by \(P(X=k)=1/[k(k+1)]\) for \(k=1,2,\ldots\). These probabilities sum to one, since

\sum_{k=1}^{\infty}\frac{1}{k(k+1)} =\sum_{k=1}^{\infty}\left(\frac1k-\frac1{k+1}\right) =1.

Its first absolute moment is infinite:

E[|X|] =\sum_{k=1}^{\infty}k\frac{1}{k(k+1)} =\sum_{k=1}^{\infty}\frac{1}{k+1} =\infty.

Nevertheless, its characteristic function is uniformly continuous. For an integer \(M\geq1\), the tail probability telescopes exactly:

P(X>M) =\sum_{k=M+1}^{\infty}\left(\frac1k-\frac1{k+1}\right) =\frac{1}{M+1}.

Splitting the increment bound at \(X=M\), where \(X\) is integer-valued, gives

|\phi_X(t+h)-\phi_X(t)| \leq E[\min(2,|hX|)] \leq |h|M+2P(X>M) =|h|M+\frac{2}{M+1}.

Given \(\varepsilon>0\), choose an integer \(M\geq1\) large enough that \(2/(M+1)<\varepsilon/2\), and then take \(|h|<\varepsilon/(2M)\). The final bound is less than \(\varepsilon\), independently of \(t\). This verifies uniform continuity despite the infinite first absolute moment.

What the Result Does—and Does Not—Require

A finite first absolute moment gives a useful stronger estimate. If \(E[|X|]<\infty\), then the exponential bound \(|e^{iu}-1|\leq |u|\) yields

|\phi_X(t+h)-\phi_X(t)| \leq E[|hX|] =|h|E[|X|].

Thus \(\phi_X\) is Lipschitz continuous in that case: its change is bounded by a fixed constant times the size of the increment. But this estimate uses the moment condition. Uniform continuity itself does not. The preceding heavy-tailed example has no finite first absolute moment and is still uniformly continuous.

A common mistake is to use \(|e^{ihX}-1|\leq |hX|\), take expectations, and conclude that uniform continuity requires \(E[|X|]<\infty\). The inequality is correct, but that particular expectation bound may be infinite and is not needed. The always-valid bound \(\min(2,|hX|)\) handles large values of \(|X|\) by \(2\), while bounded convergence—or the explicit tail split—controls their probability. The distinction is between a convenient Lipschitz bound and the more general uniform continuity result.

Continuity is useful throughout the study of characteristic functions because it is a general property available before any distribution-specific calculation. It also prevents a mistaken expectation that regularity must come from smoothness of a density or existence of moments. The guarantee proved here is continuity, not differentiability; no derivative claim follows merely from the arguments above.

Check Your Understanding

Use the exponential bounds and the proofs above to answer these questions.

Why is \(E[e^{itX}]\) defined for every real \(t\), even if \(X\) has no finite moments?
In the proof of pointwise continuity, what integrable bound permits the use of bounded convergence?
Explain why the increment bound for \(|\phi_X(t+h)-\phi_X(t)|\) is independent of \(t\).
For \(P(X=0)=1-p\) and \(P(X=1)=p\), derive the bound \( |\phi_X(t+h)-\phi_X(t)|\leq p\min(2,|h|)\).
Why does a finite first absolute moment give a Lipschitz bound, while it is not required for uniform continuity?