Heavy Tails With Finite Moments but No MGF

Moments and MGFs Are Different Tail Questions

A moment of order \(k\) weights values of a random variable by a power such as \(X^k\). An MGF weights them by an exponential, \(e^{tX}\). Because exponential growth can outpace polynomial growth, finite moments do not by themselves guarantee that the MGF is finite near zero. The distinction is especially clear when comparing a standard Cauchy random variable, which has no finite positive integer moments, with a lognormal random variable, which has finite moments of every positive integer order.

The relevant question is not just whether a distribution has a long tail, but how its tail compares with the function being integrated. We will check that comparison directly for both distributions.

Definition: For a random variable \(X\) and \(k\geq 0\), the absolute moment of order \(k\) is \(E[|X|^k]\), when this expectation is finite. For an integer \(k\geq 1\), the ordinary raw moment \(E[X^k]\) is well-defined and finite if \(E[|X|^k]<\infty\). The moment-generating function is \(M_X(t)=E[e^{tX}]\) at every real \(t\) for which this expectation is finite. An MGF exists in a neighborhood of zero if there is some \(\epsilon>0\) such that \(M_X(t)<\infty\) for every \(t\) with \(|t|<\epsilon\).

Since \(e^{0X}=1\), every probability distribution has \(M_X(0)=1\). That value alone does not establish that the MGF exists in a neighborhood of zero. In fact, each distribution below has a finite MGF at zero and fails to have one on any open interval containing zero.

The Standard Cauchy: Even the First Absolute Moment Fails

A standard Cauchy random variable has density \(f(x)=1/(\pi(1+x^2))\) for \(x\in\mathbb{R}\). Its density has tails proportional to \(1/x^2\). Multiplying that tail by a positive power \(x^p\) gives a function proportional to \(x^{p-2}\), whose integral over the positive half-line converges only when \(p<1\).

Worked Example: A Finite Fractional Moment of a Cauchy Variable

Let \(X\) be standard Cauchy. Check whether \(E[|X|^{1/2}]\) is finite. Symmetry of the density gives

E[|X|^{1/2}] =\frac{2}{\pi}\int_0^\infty \frac{x^{1/2}}{1+x^2}\,dx.

On \(0<x\leq 1\), \(1+x^2\geq 1\), so the integrand inside the integral is at most \(x^{1/2}\). Its integral over this range is \(2/3\), which is finite. On \(x\geq 1\), \(1+x^2\geq x^2\), so the integrand is at most \(x^{-3/2}\). Its integral from 1 to infinity is 2. Thus the full integral is bounded above by \(2/3+2\), and \(E[|X|^{1/2}]<\infty\).

This does not mean the Cauchy has a finite mean. The half-power moment places a lighter weight on large values than the first absolute moment does.

Proposition: If \(X\) is standard Cauchy, then \(E[|X|^p]<\infty\) for \(0\leq p<1\), and \(E[|X|^p]=\infty\) for \(p\geq 1\).

Proof. By symmetry, for \(p\geq 0\),

E[|X|^p]=\frac{2}{\pi}\int_0^\infty \frac{x^p}{1+x^2}\,dx.

Near zero, \(1+x^2\geq 1\), so the integrand is at most \(x^p\), which is integrable on \((0,1]\). For \(x\geq1\), \(x^2\leq1+x^2\leq2x^2\); therefore

\frac{1}{2}x^{p-2} \leq \frac{x^p}{1+x^2} \leq x^{p-2}.

The integral of \(x^{p-2}\) from 1 to infinity is finite exactly when \(p-2<-1\), or \(p<1\). The lower bound shows divergence when \(p\geq1\), including the logarithmic divergence at \(p=1\). This proves the claim. \(\square\)

In particular, \(E[|X|]=\infty\), so the Cauchy mean is not defined as an ordinary expectation. Symmetry may suggest cancellation between the positive and negative tails, but expectations require integrability; cancellation in a symmetric improper integral does not supply it.

Worked Example: The Cauchy MGF at a Nonzero Argument

For standard Cauchy \(X\), consider \(t>0\). The part of \(E[e^{tX}]\) contributed by \(x\geq1\) is

\frac{1}{\pi}\int_1^\infty \frac{e^{tx}}{1+x^2}\,dx \geq \frac{1}{2\pi}\int_1^\infty \frac{e^{tx}}{x^2}\,dx.

Since \(tx-2\ln x\longrightarrow\infty\), there is an \(R\geq1\) such that \(e^{tx}/x^2\geq1\) for all \(x\geq R\). The integral on the right is therefore at least \((1/(2\pi))\int_R^\infty 1\,dx=\infty\). Hence \(M_X(t)=\infty\) for every \(t>0\).

For \(t<0\), set \(u=-x\) on the negative half-line. There, \(e^{tx}=e^{|t|u}\) for \(u=-x>0\), and the same tail comparison proves divergence. Thus \(M_X(t)=\infty\) for every \(t\ne0\), even though \(M_X(0)=1\).

The Lognormal: All Positive Integer Moments Are Finite

Let \(Z\sim N(\mu,\sigma^2)\), where \(\sigma>0\), and define \(X=e^Z\). Then \(X\) is lognormal and is strictly positive. Its moment calculation uses the Gaussian density and a completed square; this also shows that every nonnegative real-order moment is finite.

Theorem: If \(X=e^Z\) with \(Z\sim N(\mu,\sigma^2)\) and \(\sigma>0\), then for every real \(r\geq0\), \(E[X^r]=\exp(r\mu+r^2\sigma^2/2)\). Its MGF is finite for \(t\leq0\) and infinite for \(t>0\). In particular, it does not exist on any neighborhood of zero.

Proof of the moment formula. Since \(X^r=e^{rZ}\), integrate against the normal density and complete the square:

\begin{aligned} E[X^r] &=\frac{1}{\sigma\sqrt{2\pi}} \int_{-\infty}^{\infty} \exp\!\left(rz-\frac{(z-\mu)^2}{2\sigma^2}\right)\,dz\\ &=\exp\!\left(r\mu+\frac{r^2\sigma^2}{2}\right) \frac{1}{\sigma\sqrt{2\pi}} \int_{-\infty}^{\infty} \exp\!\left(-\frac{(z-\mu-r\sigma^2)^2}{2\sigma^2}\right)\,dz\\ &=\exp\!\left(r\mu+\frac{r^2\sigma^2}{2}\right). \end{aligned}

The last integral, including its prefactor, is 1: it is a normal density integrated over the real line. Thus the formula is finite for every \(r\geq0\), including every positive integer.

Proof of the MGF claim. When \(t\leq0\), \(0<e^{tX}\leq1\) because \(X>0\); therefore \(0<M_X(t)\leq1\). To examine \(t>0\), the change of variable \(x=e^z\) gives the lognormal density \(f(x)=\exp(-(\ln x-\mu)^2/(2\sigma^2))/(x\sigma\sqrt{2\pi})\) for \(x>0\). Consequently,

M_X(t)=\frac{1}{\sigma\sqrt{2\pi}} \int_0^\infty \exp\!\left(tx-\frac{(\ln x-\mu)^2}{2\sigma^2}\right)\frac{dx}{x}.

The logarithm of the integrand, including the constant prefactor, is \(tx-(\ln x-\mu)^2/(2\sigma^2)-\ln x-\ln(\sigma\sqrt{2\pi})\). As \(x\to\infty\), this expression tends to \(+\infty\): the positive linear term \(tx\) grows faster than \((\ln x-\mu)^2\) and \(\ln x\). The integrand is therefore at least 1 for all sufficiently large \(x\), so its integral over that infinite tail diverges. Hence \(M_X(t)=\infty\) for every \(t>0\). No open interval around zero can consist entirely of finite MGF values. \(\square\)

Worked Example: Moments and MGF of a Lognormal Variable

Let \(Z\sim N(0,1)\) and \(X=e^Z\), so \(X\) is lognormal with \(\mu=0\) and \(\sigma^2=1\). Substituting \(r=1,2,3,4\) into the moment formula gives

E[X]=e^{1/2},\qquad E[X^2]=e^2,\qquad E[X^3]=e^{9/2},\qquad E[X^4]=e^8.

Each is finite. The variance follows from \(\operatorname{Var}(X)=E[X^2]-(E[X])^2\):

\operatorname{Var}(X)=e^2-e\approx4.67077.

Nevertheless, for \(t>0\), \(M_X(t)=\infty\), while for \(t\leq0\), \(M_X(t)\leq1\). The example has finite moments of every positive integer order but no MGF on a neighborhood of zero.

What the Comparison Teaches

These distributions fail the neighborhood condition for different reasons. The Cauchy tail is so heavy that absolute moments of order at least 1 diverge, and either sign of a nonzero exponential tilt magnifies one of its tails. The lognormal tail is light enough for every power moment, but its positive tail is still too heavy for any positive exponential tilt. Its MGF is finite on the nonpositive half-line, but that half-line does not contain an open interval around zero.

A common mistake is to argue that the lognormal has an MGF near zero because all its ordinary moments are finite. The conclusion does not follow: the moment condition integrates powers of \(X\), whereas the MGF for positive \(t\) integrates \(e^{tX}\). Exponential growth eventually outruns polynomial growth. Conversely, an MGF being finite at a single point—especially at \(t=0\)—does not mean it exists in a neighborhood of that point.

For the Cauchy: \(E[|X|^p]\) is finite for \(0\leq p<1\), but \(E[|X|]\) and all higher absolute moments diverge; \(M_X(t)\) is finite only at \(t=0\).
For the lognormal: all nonnegative real-order moments are finite; \(M_X(t)\) is finite for \(t\leq0\) and infinite for \(t>0\).

The practical lesson is to check the particular tail-weighted integral required by the quantity of interest. Knowing that some moments exist, or even that every integer moment exists, is not a substitute for checking the MGF’s domain.

Check Your Understanding

Use the tail comparisons and formulas from this tutorial to answer the following questions.

For a standard Cauchy variable, is \(E[|X|^{3/4}]\) finite? Which tail comparison decides?
Why does symmetry of the standard Cauchy density not make \(E[X]\) a well-defined expectation?
For \(X=e^Z\) with \(Z\sim N(\mu,\sigma^2)\), write \(E[X^5]\).
For a lognormal variable, explain why \(M_X(-2)\) is finite but \(M_X(2)\) is infinite.
What is the difference between an MGF being finite at zero and existing in a neighborhood of zero?

Heavy Tails With Finite Moments but No MGF

What You'll Learn