Characteristic Functions That Exist Without Moment Generating Functions

Two Transforms, Different Existence Requirements

A characteristic function and a moment generating function are both expectations built from a random variable, but their integrands behave very differently in the tails. The characteristic-function integrand has constant modulus, while a moment-generating-function integrand can grow exponentially in one tail. This distinction is decisive for heavy-tailed distributions: a characteristic function is always available, even when no moment generating function exists on any neighborhood of zero.

Definition: For a real-valued random variable \(X\), its characteristic function is \(\varphi_X(t)=E[e^{itX}]\), for real \(t\). Its moment generating function is \(M_X(t)=E[e^{tX}]\) at those real \(t\) for which this expectation is finite. In the usual use of an MGF to encode moments, it must be finite on an open interval containing zero. Every MGF has \(M_X(0)=1\), but finiteness at zero alone does not meet this condition.

The expectation defining \(\varphi_X(t)\) is a complex expectation, meaning that the real and imaginary parts are integrated separately. Both are integrable because \(|\cos(tX)|\leq 1\) and \(|\sin(tX)|\leq 1\). By contrast, \(e^{tX}\) is nonnegative but can become arbitrarily large as \(X\) moves into one tail. Its expectation can therefore be infinite.

Why Every Characteristic Function Exists

The existence claim needs no moment assumption: \(X\) need only be a real-valued random variable. In particular, \(E[X]\) need not exist.

Theorem: For every real-valued random variable \(X\) and every \(t\in\mathbb{R}\), the characteristic function \(\varphi_X(t)\) exists and satisfies \(|\varphi_X(t)|\leq 1\).

Proof. Since \(|e^{itX}|=1\) for every outcome, \(e^{itX}\) is integrable as a complex-valued random variable: its real and imaginary parts have absolute value at most one. Thus its expectation exists. The triangle inequality for expectations gives \(|\varphi_X(t)|=|E[e^{itX}]|\leq E[|e^{itX}|]=1\). This holds for every real \(t\), with no assumption about moments of \(X\). \(\square\)

The bound is about the transform, not about the size of \(X\). Oscillation of \(e^{itX}\) cannot make its modulus exceed one. Exponential weighting has no such protection: for positive \(t\), large positive values of \(X\) are amplified, and for negative \(t\), large negative values are amplified.

Worked Examples: Heavy Tails and Transform Domains

Worked Example: The Standard Cauchy Law

Let \(X\) have the standard Cauchy density \(f(x)=1/(\pi(1+x^2))\), for \(x\in\mathbb{R}\). Its characteristic function, obtained by direct integration, is \(\varphi_X(t)=e^{-|t|}\). It is finite for every real \(t\), as the theorem also guarantees.

Now examine the MGF. At zero, \(M_X(0)=\int_{-\infty}^{\infty}f(x)\,dx=1\). If \(t>0\), its positive-tail contribution is \(\frac{1}{\pi}\int_1^\infty e^{tx}/(1+x^2)\,dx\). For \(x\geq 1\), \(1+x^2\leq 2x^2\), so this integral is at least \(\frac{1}{2\pi}\int_1^\infty e^{tx}/x^2\,dx\). The latter diverges: on each interval \([n,n+1]\), its integral is at least \(e^{tn}/(n+1)^2\), and these lower bounds do not tend to zero. Hence \(M_X(t)=\infty\) for \(t>0\).

If \(t<0\), write \(y=-x\) in the negative tail. There \(e^{tx}=e^{|t|y}\) when \(x=-y\), so the same argument shows divergence. Thus the Cauchy MGF is finite only at \(t=0\). In particular, it is not finite on any open interval around zero.

The Cauchy law also lacks an ordinary mean: \(E[|X|]=\frac{2}{\pi}\int_0^\infty x/(1+x^2)\,dx=\infty\). Symmetry can make a Cauchy principal value for the signed integral look like zero, but that principal value is not the expectation \(E[X]\).

Worked Example: A Pareto Law with a One-Sided MGF Domain

Let \(X\) have the Pareto density \(f(x)=2x^{-3}\) for \(x\geq1\), and zero otherwise. This is a probability density because \(\int_1^\infty 2x^{-3}\,dx=[-x^{-2}]_1^\infty=1\). Its characteristic function exists for every real \(t\), since \(|E[e^{itX}]|\leq E[|e^{itX}|]=1\); the defining integral is also absolutely bounded by \(\int_1^\infty2x^{-3}\,dx=1\).

For \(t\leq0\), \(0<e^{tX}\leq1\), so \(M_X(t)\leq1\) and is finite. For \(t>0\), \(M_X(t)=\int_1^\infty 2e^{tx}x^{-3}\,dx\). On \([n,n+1]\), the integrand is at least \(2e^{tn}/(n+1)^3\). These lower bounds grow without bound as \(n\) increases, so the integral diverges. Its finite MGF domain is therefore \((-\infty,0]\), which does not contain an open interval around zero.

This example differs from the Cauchy case: some nonzero MGF arguments are allowed, but there is still no MGF neighborhood of zero. A finite MGF on only one side is not enough for the usual two-sided local moment-generating role.

Worked Example: The Standard Normal as a Contrast

Let \(X\sim N(0,1)\), with density \(f(x)=(2\pi)^{-1/2}e^{-x^2/2}\). For any real \(t\), complete the square: \(-x^2/2+tx=-(x-t)^2/2+t^2/2\). Therefore \(M_X(t)=e^{t^2/2}\int_{-\infty}^{\infty}(2\pi)^{-1/2}e^{-(x-t)^2/2}\,dx\). The integral is one, since its integrand is a normal density shifted by \(t\); hence \(M_X(t)=e^{t^2/2}\), finite for every real \(t\).

The normal distribution illustrates that an MGF can exist everywhere, but this is a property of that distribution, not a requirement for defining a characteristic function. For the Cauchy and Pareto laws above, the characteristic function exists at every real argument even though the MGF fails to exist around zero.

Moments, Derivatives, and Exponential Integrability

A characteristic function can exist without moments, but when absolute moments do exist they control its derivatives. The precise condition below is absolute integrability of the relevant power of \(X\).

Theorem: If \(E[|X|^k]<\infty\) for an integer \(k\geq1\), then \(\varphi_X\) has derivatives through order \(k\), and \(\varphi_X^{(j)}(t)=E[(iX)^j e^{itX}]\) for \(0\leq j\leq k\) and every real \(t\). In particular, \(\varphi_X^{(j)}(0)=i^jE[X^j]\).

Proof. For \(k=1\), the difference quotient of \(e^{itX}\), for nonzero \(h\), is \(e^{itX}(e^{ihX}-1)/h\). The elementary inequality \(|e^{iu}-1|\leq |u|\) bounds its modulus by \(|X|\). As \(h\to0\), it converges pointwise to \(iXe^{itX}\). Since \(E[|X|]<\infty\), the dominated convergence theorem permits taking the limit inside the expectation, giving \(\varphi_X'(t)=E[iXe^{itX}]\).

For higher orders, suppose the derivative formula holds through order \(j-1\), where \(j\leq k\). The difference quotient for \(E[(iX)^{j-1}e^{itX}]\) is bounded in absolute value by \(|X|^j\), using the same inequality, and converges pointwise to \(E[(iX)^j e^{itX}]\)'s integrand. The assumed finite \(j\)th absolute moment supplies an integrable dominating random variable. Dominated convergence proves the formula at order \(j\). Induction gives all orders through \(k\); substituting \(t=0\) gives the moment identity. \(\square\)

The converse warning is important: existence of \(\varphi_X(t)\) for all \(t\) does not guarantee even a first moment, as the Cauchy example shows. Its characteristic function \(e^{-|t|}\) is not differentiable at zero, in keeping with the absence of a finite absolute first moment.

There is also a useful implication in the other direction, from MGF existence to moments. If \(M_X(a)\) and \(M_X(-a)\) are finite for some \(a>0\), then \(e^{a|x|}\leq e^{ax}+e^{-ax}\) for every real \(x\), and hence \(E[e^{a|X|}]\leq M_X(a)+M_X(-a)<\infty\). For every integer \(k\geq1\), the function \(x^k e^{-ax}\) is bounded on \([0,\infty)\); call an upper bound \(C\). Then \(|X|^k\leq C e^{a|X|}\), so \(E[|X|^k]\leq C E[e^{a|X|}]<\infty\). Thus an MGF finite at both \(a\) and \(-a\) guarantees all absolute moments. The Cauchy law cannot satisfy this condition.

Common Mistakes and Why the Distinction Matters

Assuming every transform needs moments. A characteristic function is an expectation of a bounded complex-valued quantity; it needs no finite mean or variance.
Treating \(M_X(0)=1\) as proof of an MGF near zero. Every real-valued \(X\) has this value at zero, including the Cauchy law. Neighborhood finiteness must be checked separately.
Using symmetry to assign a nonexistent mean. A symmetric improper integral may have principal value zero while \(E[|X|]=\infty\). The ordinary expectation is then undefined.
Assuming an MGF finite on one side is enough for all moment arguments. The Pareto example has a finite MGF for \(t\leq0\), but not on an open interval containing zero.

The distinction is useful in distribution theory because characteristic functions remain available for the full range of probability laws, including heavy-tailed ones. MGFs can provide convenient moment calculations when they exist near zero, but their failure does not obstruct characteristic-function arguments or make a distribution inaccessible to transform methods.

Key takeaway: The characteristic function always exists because its integrand has modulus one. An MGF can diverge because exponential weighting magnifies a tail; even its existence at zero says nothing about whether it is finite nearby.

Check Your Understanding

Use the definitions and examples above to answer the following.

Which bound guarantees that \(E[e^{itX}]\) exists for every real-valued \(X\)?
For the standard Cauchy law, why does a negative \(t\) make the negative tail of \(e^{tX}\) problematic?
What is the finite MGF domain for the Pareto density \(2x^{-3}\), \(x\geq1\), and why does it not provide a neighborhood of zero?
Why does a symmetric principal-value integral of \(x f(x)\) not establish that \(E[X]=0\) for the Cauchy law?
If \(E[|X|^2]<\infty\), what are \(\varphi_X'(0)\) and \(\varphi_X''(0)\) in terms of moments?