Why Use a Characteristic Function?
A moment generating function may fail to exist away from zero: the expectation of an exponential can be infinite. A characteristic function avoids that obstacle by using a complex exponential whose modulus is always one. It is therefore defined at every real argument for every real-valued random variable. Its elementary properties—boundedness, continuity, its value at zero, and conjugate symmetry—provide basic checks for calculations and a foundation for later uses of transforms.
Let \(X:\Omega\to\mathbb{R}\) be a real-valued random variable on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). Here \(i\) denotes the imaginary unit, with \(i^2=-1\), and \(t\) is a real number. For a complex-valued random variable with integrable real and imaginary parts, expectation is defined by taking the expectations of those parts separately.
The defining expression is an average of points on the unit circle in the complex plane. The average need not itself have modulus one, but it cannot have modulus greater than one. The next result collects the basic properties and proves each directly from the definition.
Proof. First, for any integrable complex-valued random variable \(Z\), \(|\mathbb{E}Z|\le\mathbb{E}|Z|\). To verify this inequality, put \(w=\mathbb{E}Z\). If \(w=0\), the inequality follows from \(\mathbb{E}|Z|\ge0\). If \(w\ne0\), let \(u=\overline{w}/|w|\), so \(|u|=1\) and \(uw=|w|\). Consequently, \(|w|=\operatorname{Re}(\mathbb{E}[uZ])=\mathbb{E}[\operatorname{Re}(uZ)]\le\mathbb{E}|uZ|=\mathbb{E}|Z|\). Apply this with \(Z=e^{itX}\). Its modulus is one, so \(|\varphi_X(t)|\le\mathbb{E}1=1\).
For continuity, fix \(t\in\mathbb{R}\) and let \(s\) approach \(t\). For real \(a,b\), the inequality \(|e^{ia}-e^{ib}|\le|a-b|\) follows, for example, by integrating the derivative \(ie^{ix}\), whose modulus is one, between \(a\) and \(b\). Also, the difference between two points of modulus one has modulus at most two. Given \(\varepsilon>0\), choose \(K\ge1\) so that \(\mathbb{P}(|X|>K)<\varepsilon/4\). Such a \(K\) exists because the events \(\{|X|>m\}\), for positive integers \(m\), decrease to a probability-zero event when \(X\) is real-valued, and probability is continuous for decreasing events. Splitting according to \(|X|\le K\) and \(|X|>K\) gives
The second term is less than \(\varepsilon/2\). If \(|t-s|<\varepsilon/(2K)\), the first term is less than \(\varepsilon/2\), so \(|\varphi_X(t)-\varphi_X(s)|<\varepsilon\). This proves continuity at \(t\); since \(t\) was arbitrary, it proves continuity everywhere.
Finally, \(e^{i0X}=1\), so \(\varphi_X(0)=\mathbb{E}1=1\). For real \(tX\), \(e^{-itX}=\overline{e^{itX}}\). Expectation commutes with complex conjugation, as follows by writing a complex random variable as its real part plus \(i\) times its imaginary part. Therefore \(\varphi_X(-t)=\mathbb{E}[\overline{e^{itX}}]=\overline{\mathbb{E}[e^{itX}]}=\overline{\varphi_X(t)}\). This proves all four claims. \(\square\)
The continuity proof is useful beyond this particular transform: it does not assume that \(X\) has a density, finite moments, or bounded support. The tail probability controls the part where \(X\) may be large, while on a bounded range of \(X\), changing \(t\) slightly changes the exponential by a uniformly small amount.
Worked Calculations
Worked Example: A Constant Random Variable
Suppose \(X=c\) with probability one, where \(c\) is any fixed real number. Compute its characteristic function and check the basic properties.
Because \(X=c\) almost surely, the random quantity \(e^{itX}\) equals \(e^{itc}\) almost surely. Thus
Check: At \(t=0\), this gives \(e^{i0c}=1\). Its modulus is \(|e^{itc}|=1\), so it satisfies the bound. Also, \(\varphi_X(-t)=e^{-itc}=\overline{e^{itc}}=\overline{\varphi_X(t)}\). The formula is continuous in \(t\), as is the complex exponential.
Worked Example: An Unequally Weighted Two-Point Variable
Let \(X\) take the value \(-1\) with probability \(\tfrac13\) and the value \(2\) with probability \(\tfrac23\). These probabilities are nonnegative and sum to \(\tfrac13+\tfrac23=1\). Find \(\varphi_X(t)\), then evaluate it at \(t=\pi\).
Sum the exponential over the two possible values, weighting each by its probability:
At \(t=\pi\), \(e^{-i\pi}=-1\) and \(e^{2i\pi}=1\). Substitution gives
Check: The result is real, so \(\varphi_X(-\pi)=\overline{\varphi_X(\pi)}=\tfrac13\). Direct substitution into the original formula gives \(\varphi_X(-\pi)=\tfrac13 e^{i\pi}+\tfrac23 e^{-2i\pi}=-\tfrac13+\tfrac23=\tfrac13\), as required. Its modulus is \(\tfrac13\le1\). At zero, the formula gives \(\tfrac13+\tfrac23=1\).
Worked Example: A Variable with a Density
Let \(X\) have density \(f(x)=2x\) for \(0\le x\le1\), and density zero elsewhere. This is a probability density because \(f(x)\ge0\) on its support and \(\int_0^1 2x\,dx=[x^2]_0^1=1\). Compute its characteristic function for \(t\ne0\), and evaluate it at \(t=\pi\).
For a variable with density \(f\), expectation is integration against \(f\). Thus, for \(t\ne0\),
In the integration by parts, the boundary term at \(x=1\) is \(e^{it}/(it)\), the boundary term at \(x=0\) is zero, and \(\int_0^1 e^{itx}\,dx=(e^{it}-1)/(it)\). These substitutions yield the displayed expression.
At \(t=\pi\), use \(e^{i\pi}=-1\) and \((i\pi)^2=-\pi^2\):
Check: Evaluate the defining integral at \(t=\pi\) by separating real and imaginary parts. Integration gives \(\int_0^1 2x\cos(\pi x)\,dx=-4/\pi^2\) and \(\int_0^1 2x\sin(\pi x)\,dx=2/\pi\). Hence the integral is \(-4/\pi^2+2i/\pi\), matching the formula. At \(t=0\), use the definition rather than the expression with denominators: \(\varphi_X(0)=\int_0^1 2x\,dx=1\).
Interpreting the Basic Properties
The bound \(|\varphi_X(t)|\le1\) is a general constraint, not a claim that the function is real or nonnegative. The two-point example has a complex-valued expression for general \(t\), and the density example has a nonzero imaginary part at \(t=\pi\). Conjugate symmetry describes how the values at opposite arguments are related: knowing \(\varphi_X(t)\) determines \(\varphi_X(-t)\), but it does not generally make either value real.
The value at zero is a useful normalization check. Any proposed characteristic function of a probability distribution must equal one there. Continuity is another necessary check, including at zero; an algebraic expression obtained for \(t\ne0\) may have a removable singularity at zero, so its value there should be evaluated from the defining expectation or established by a limit. The density example illustrates why the definition remains the reliable starting point at a special argument.
Common mistake: moving expectation into an exponential. In general, \(\mathbb{E}[e^{itX}]\) is not \(e^{it\mathbb{E}X}\). For the two-point variable, \(\mathbb{E}X=(-1)(\tfrac13)+2(\tfrac23)=-\tfrac13+\tfrac43=1\). At \(t=\pi\), the incorrect expression would give \(e^{i\pi\mathbb{E}X}=e^{i\pi}=-1\). The correct computation is \(\varphi_X(\pi)=\tfrac13(-1)+\tfrac23(1)=\tfrac13\). The mistake replaces the average of the exponentials by the exponential of the average, which is not a valid rule for expectation.
These basic properties are practical diagnostics. When calculating a transform, check that its value at zero is one, its modulus never exceeds one, and its values at \(t\) and \(-t\) are conjugates. Such checks can expose sign errors or missing probability weights. They are necessary properties; by themselves, however, they are not a proof that an arbitrary function is the characteristic function of a probability distribution.
Check Your Understanding
Use the definition and the proved properties to answer each question. When a formula is proposed, check it by substituting into the defining expectation.
- If \(X=c\) almost surely, what are \(\varphi_X(0)\) and \(|\varphi_X(t)|\) for real \(t\)?
- A variable takes values \(0\) and \(1\) with probabilities \(\tfrac14\) and \(\tfrac34\). Write its characteristic function and evaluate it at \(t=\pi\).
- Explain why the expectation defining \(\varphi_X(t)\) exists even if \(X\) has no finite moment generating function at nonzero arguments.
- For the density \(f(x)=2x\) on \([0,1]\), why should the formula derived for \(t\ne0\) not be substituted at \(t=0\) without checking the original definition?
- State the relation between \(\varphi_X(-t)\) and \(\varphi_X(t)\), and explain why it does not imply that \(\varphi_X(t)\) is always real.