Why the Domain Matters
The moment-generating function (MGF) is useful not only because it encodes moments, but also because its logarithm provides a natural object for studying cumulants. That logarithm is meaningful as a finite real-valued function only where the MGF itself is finite. In particular, knowing the MGF’s value at zero is not enough to justify taking derivatives there: derivatives in the usual two-sided sense require the function to be defined around zero.
The key issue is therefore local. An MGF can be finite at zero—and even have derivatives from one side there—without being finite on any open interval containing zero. We will define the cumulant-generating function on its finite domain, prove that this domain is an interval, and show why finiteness on a neighborhood of zero gives a well-behaved function for derivative-based work.
Definition and Finite Domain
The finite domain records exactly where the ordinary real-valued log MGF exists. Some conventions extend \(K_X\) by setting it equal to \(+\infty\) outside \(D_X\); here, statements about \(K_X\) as a real-valued function refer to \(D_X\). The set \(D_X\) need not be all of \(\mathbb{R}\), nor must it contain an open interval around zero.
Proof. Let \(s,t\in D_X\), and let \(0<\lambda<1\). Then $$ e^{(\lambda s+(1-\lambda)t)X} =(e^{sX})^\lambda(e^{tX})^{1-\lambda}. $$ Hölder’s inequality, with conjugate exponents \(1/\lambda\) and \(1/(1-\lambda)\), gives $$ M_X(\lambda s+(1-\lambda)t) \leq M_X(s)^\lambda M_X(t)^{1-\lambda}<\infty. $$ Thus every strict convex combination of \(s\) and \(t\) is in \(D_X\); the endpoints \(s,t\) are in \(D_X\) by assumption. Therefore \(D_X\) is an interval. Separately, \(M_X(0)=E[1]=1\), so \(0\in D_X\). The argument does not imply that a finite endpoint must itself be in the domain; that depends on the distribution. \(\square\)
For derivative-based cumulants, the relevant condition is stronger than \(0\in D_X\). We say the MGF is finite in a neighborhood of zero if there is some \(\varepsilon>0\) such that \(M_X(t)<\infty\) for every \(t\in(-\varepsilon,\varepsilon)\). Equivalently, zero is an interior point of \(D_X\). The interval property means that the domain can extend far in one direction while failing this two-sided condition.
Why an Open Neighborhood Supports Derivatives
Finiteness near zero controls both tails of \(X\). If the MGF is finite at \(b\) and \(-b\) for some \(b>0\), then $$ E[e^{b|X|}]\leq E[e^{bX}]+E[e^{-bX}]=M_X(b)+M_X(-b)<\infty. $$ This exponential bound dominates every fixed power of \(|X|\). It is the reason that a finite interval around zero permits derivatives and a power-series expansion, rather than merely supplying the value \(M_X(0)=1\).
Proof. Choose \(b\) with \(0<b<\varepsilon\). Both \(M_X(b)\) and \(M_X(-b)\) are finite, so the inequality above shows that \(E[e^{b|X|}]<\infty\). For any \(|t|<b\), the exponential series satisfies $$ \sum_{k=0}^{\infty}\frac{|tX|^k}{k!}=e^{|t||X|}\leq e^{b|X|}. $$ The right-hand side is integrable. Tonelli’s theorem therefore permits summing the nonnegative absolute terms under expectation, and absolute integrability permits exchanging expectation with the signed series. Hence $$ M_X(t)=E\left[\sum_{k=0}^{\infty}\frac{(tX)^k}{k!}\right] =\sum_{k=0}^{\infty}\frac{E[X^k]}{k!}t^k,\qquad |t|<b. $$ This also shows that every absolute moment is finite: for each \(k\), \(|X|^k\leq k!b^{-k}e^{b|X|}\). Thus \(M_X\) is analytic around zero. Since \(M_X(0)=1\), continuity gives a smaller interval around zero on which \(M_X(t)>0\). The logarithm is analytic on the positive real numbers, so its composition with \(M_X\) is real analytic on that smaller interval. \(\square\)
This result explains the local-domain requirement. When the MGF is finite around zero, the log MGF is an ordinary smooth function there, and its derivatives at zero can be used to define cumulants. The next tutorial develops those derivative formulas. Here, the essential point is that an open interval gives a genuine function to differentiate on both sides of zero, along with the integrability needed for its local expansion.
Worked Examples of MGF Domains
Worked Example: A Normal Distribution Has Full Domain
Let \(X\sim N(\mu,\sigma^2)\), with \(\sigma^2>0\). Completing the square in the density integral gives $$ \begin{aligned} M_X(t) &=E[e^{tX}]\\ &=\frac{1}{\sqrt{2\pi\sigma^2}}\int_{-\infty}^{\infty} \exp\left(tx-\frac{(x-\mu)^2}{2\sigma^2}\right)\,dx\\ &=\exp\left(\mu t+\frac{\sigma^2t^2}{2}\right) \frac{1}{\sqrt{2\pi\sigma^2}}\int_{-\infty}^{\infty} \exp\left(-\frac{(x-\mu-\sigma^2t)^2}{2\sigma^2}\right)\,dx\\ &=\exp\left(\mu t+\frac{\sigma^2t^2}{2}\right). \end{aligned} $$ The last integral, including its normalizing factor, is \(1\), since it is a normal density integrated over the real line. The displayed expression is finite for every real \(t\), so \(D_X=\mathbb{R}\) and $$ K_X(t)=\mu t+\frac{\sigma^2t^2}{2},\qquad t\in\mathbb{R}. $$ In particular, zero is an interior point of the domain.
Worked Example: An Exponential Distribution Has a Finite Endpoint
Let \(X\) have an exponential distribution with rate \(\lambda>0\), so its density is \(\lambda e^{-\lambda x}\) for \(x\geq0\). Direct integration gives $$ M_X(t)=\int_0^\infty e^{tx}\lambda e^{-\lambda x}\,dx =\lambda\int_0^\infty e^{-(\lambda-t)x}\,dx. $$ If \(t<\lambda\), the exponent has a negative coefficient of \(x\), and the integral equals \(\lambda/(\lambda-t)\). If \(t=\lambda\), the integrand is the constant \(\lambda\), and the integral diverges. If \(t>\lambda\), the integrand grows with \(x\), so it also diverges. Consequently, $$ D_X=(-\infty,\lambda),\qquad K_X(t)=\log\left(\frac{\lambda}{\lambda-t}\right),\quad t<\lambda. $$ Although the finite endpoint \(\lambda\) is excluded, the domain contains an open interval around zero because \(\lambda>0\). Thus the CGF is well-defined near zero.
Worked Example: A Lognormal Distribution Has No Open MGF Neighborhood
Let \(Y\sim N(\mu,\sigma^2)\) with \(\sigma^2>0\), and set \(X=e^Y\). For \(t\leq0\), \(0<e^{tX}\leq1\), so \(M_X(t)\leq1\); the expectation is finite. For \(t>0\), writing the expectation as a normal-density integral gives $$ M_X(t)=\frac{1}{\sqrt{2\pi\sigma^2}}\int_{-\infty}^{\infty} \exp\left(te^y-\frac{(y-\mu)^2}{2\sigma^2}\right)\,dy. $$ As \(y\) tends to infinity, \(te^y\) grows faster than the quadratic term \((y-\mu)^2/(2\sigma^2)\). The integrand therefore tends to infinity, so its integral over a sufficiently far tail diverges. Hence $$ D_X=(-\infty,0]. $$ The MGF is finite at zero and on the entire negative half-line, but not for any positive \(t\). It has no open neighborhood of finiteness around zero. Nevertheless, the lognormal has finite moments of every positive integer order: \(E[X^k]=E[e^{kY}]=\exp(k\mu+k^2\sigma^2/2)\). This example separates having all moments from having an MGF finite near zero. Its MGF does not provide a two-sided real function around zero whose ordinary derivatives there define the usual derivative-based cumulants.
Common Mistake: Confusing a Value at Zero With a Neighborhood
A tempting but incorrect argument is: “Every MGF satisfies \(M_X(0)=1\), so its logarithm must be differentiable at zero.” The first statement is true, but it only establishes that zero belongs to the finite domain. It says nothing about whether the domain extends to positive and negative values near zero. For the lognormal example, \(M_X(0)=1\), yet \(M_X(t)=\infty\) for every \(t>0\). The ordinary two-sided derivative of the finite log MGF at zero is therefore not available.
The correction is to check local finiteness, not just the value at zero. A finite open interval around zero is a sufficient condition for the analytic, derivative-based framework: it controls the tails, gives finite moments of every order, and makes the log MGF analytic near zero. Having moments alone, even moments of every integer order, does not guarantee that the MGF has such a neighborhood.
Check Your Understanding
Use the definition and domain arguments from this tutorial.
- Why is \(K_X(0)=0\) whenever \(X\) is a real-valued random variable?
- Suppose \(M_X(s)\) and \(M_X(t)\) are finite. What does Hölder’s inequality imply about \(M_X\) between \(s\) and \(t\)?
- For an exponential random variable with rate \(\lambda\), is the endpoint \(t=\lambda\) included in the MGF’s finite domain? Explain using the integral.
- Why does finiteness of \(M_X(t)\) on an open interval around zero imply that \(E[e^{b|X|}]<\infty\) for some \(b>0\)?
- How can a distribution have moments of every positive integer order but lack an MGF finite on an open interval around zero?