Moment-Generating Functions and Their Domains

Why the MGF Domain Matters

The moment-generating function (MGF) packages information about a distribution into expectations of exponentials. But the expression \(E[e^{tX}]\) is not necessarily finite for every real \(t\). For an unbounded random variable, one tail may make the expectation infinite for positive \(t\), the other tail may do so for negative \(t\), or both may restrict the range.

The domain records exactly which values of \(t\) give a finite expectation. Its behavior around zero matters: a value at zero is automatic, whereas finiteness on an interval around zero is a substantially stronger condition. We will identify these cases using tail behavior, without assuming that the MGF can be differentiated.

Definition and Basic Domain Facts

Definition: For a real-valued random variable \(X\), its moment-generating function is the extended expectation \(M_X(t)=E[e^{tX}]\), for real \(t\). Since \(e^{tX}\) is nonnegative, this expectation is defined in \([0,\infty]\), and may be infinite. The finiteness domain is \(D_X=\{t\in\mathbb{R}:M_X(t)<\infty\}\).

At \(t=0\), \(e^{tX}=1\) for every outcome, so \(M_X(0)=1\) for every real-valued \(X\). In particular, \(0\in D_X\) without any moment or tail assumption. In the usual sense, an MGF exists in a neighborhood of zero if there is some \(\epsilon>0\) for which it is finite for every \(t\in(-\epsilon,\epsilon)\). Thus, merely having a finite value at zero does not establish neighborhood existence.

The exponential weights explain why the two tails have different roles. For \(t>0\), \(e^{tX}\) grows as \(X\) tends to \(+\infty\) but is bounded by \(1\) when \(X\leq 0\). For \(t<0\), it grows as \(X\) tends to \(-\infty\) and is bounded by \(1\) when \(X\geq 0\). More precisely, for any real \(t\),

M_X(t) = E[e^{tX}\mathbf{1}_{\{X\geq 0\}}] + E[e^{tX}\mathbf{1}_{\{X<0\}}].

This split is valid even when one term is infinite: it is the additivity of the integral of a nonnegative function over two disjoint events. For a random variable with density \(f\), the same criterion can be read as finiteness of \(\int_0^\infty e^{tx}f(x)\,dx\) and \(\int_{-\infty}^0 e^{tx}f(x)\,dx\) together. For a discrete distribution, the corresponding integrals are sums over the two parts of the support.

Theorem: The finiteness domain \(D_X\) is an interval containing zero. The interval need not be open or closed at either endpoint. In particular, if \(M_X(a)\) and \(M_X(b)\) are finite for \(a<b\), then \(M_X(t)\) is finite for every \(t\in[a,b]\).

Proof. The endpoints \(a\) and \(b\) are finite by hypothesis. For \(t\in(a,b)\), write \(t=\lambda a+(1-\lambda)b\), where \(\lambda=(b-t)/(b-a)\) lies strictly between zero and one. Then

e^{tX}=(e^{aX})^\lambda(e^{bX})^{1-\lambda}.

Hölder’s inequality, with conjugate exponents \(1/\lambda\) and \(1/(1-\lambda)\), gives

E[e^{tX}] \leq \big(E[e^{aX}]\big)^\lambda \big(E[e^{bX}]\big)^{1-\lambda} <\infty.

The endpoints themselves are finite by assumption. Thus every point between any two members of \(D_X\) also belongs to \(D_X\), which is exactly the interval property. Also \(0\in D_X\), as shown above. \(\square\)

This result does not decide whether a finite boundary point belongs to the domain. That depends on the distribution’s tail at the boundary rate: the exponential factor there may still leave an integrable tail, or it may not.

Neighborhood Existence and One-Sided Finiteness

A neighborhood of zero requires control of both tails. This can be expressed in a useful equivalent form: \(M_X(t)\) is finite throughout some open interval containing zero if and only if there is an \(\epsilon>0\) such that \(E[e^{\epsilon|X|}]<\infty\).

To verify the equivalence, suppose first that \(E[e^{\epsilon|X|}]<\infty\). For every \(t\) with \(|t|<\epsilon\), \(e^{tX}\leq e^{|t||X|}\leq e^{\epsilon|X|}\), so \(M_X(t)<\infty\). Conversely, suppose there is an \(\epsilon>0\) such that \(M_X(\epsilon)\) and \(M_X(-\epsilon)\) are finite. Pointwise,

e^{\epsilon|X|} = e^{\epsilon X}\mathbf{1}_{\{X\geq 0\}} + e^{-\epsilon X}\mathbf{1}_{\{X<0\}} \leq e^{\epsilon X}+e^{-\epsilon X}.

Taking expectations shows \(E[e^{\epsilon|X|}]<\infty\). Finiteness at both endpoints also implies finiteness throughout \([-\epsilon,\epsilon]\) by the interval theorem.

By contrast, finiteness only for \(t\) in a right-hand interval \([0,\epsilon)\) controls the positive tail but places no corresponding exponential-integrability condition on the negative tail. Finiteness on a left-hand interval \((-\epsilon,0]\) controls the negative tail instead. A domain may also extend indefinitely on one side.

Worked Example: A Bounded Distribution

Let \(X\) have the uniform distribution on \([-1,2]\), with density \(f(x)=1/3\) on that interval. For \(t\neq 0\), direct integration gives

M_X(t) =\frac{1}{3}\int_{-1}^{2}e^{tx}\,dx =\frac{e^{2t}-e^{-t}}{3t}.

At \(t=0\), the defining expectation is \(M_X(0)=1\); the formula with denominator \(t\) is not used at zero. For every fixed real \(t\), the integrand is continuous and bounded on the finite interval \([-1,2]\), so the integral is finite. Therefore \(D_X=\mathbb{R}\): the MGF is finite at zero, on both sides of zero, and on every neighborhood of zero.

The conclusion follows from bounded support, not from the particular formula. If \(|X|\leq c\) almost surely, then \(e^{tX}\leq e^{|t|c}\) for every real \(t\), so \(M_X(t)\leq e^{|t|c}<\infty\).

Worked Example: An Exponential Right-Tail Boundary

Let \(X\) have an exponential distribution with rate \(\lambda>0\), so \(f(x)=\lambda e^{-\lambda x}\) for \(x\geq0\). For \(t<\lambda\),

M_X(t) =\int_0^\infty e^{tx}\lambda e^{-\lambda x}\,dx =\lambda\int_0^\infty e^{-(\lambda-t)x}\,dx =\frac{\lambda}{\lambda-t}.

The integral is finite because \(\lambda-t>0\); evaluating the integral over \([0,R]\) and letting \(R\to\infty\) gives \((1-e^{-(\lambda-t)R})/(\lambda-t)\), which tends to \(1/(\lambda-t)\). At \(t=\lambda\), the integrand in the last integral is the constant \(\lambda\), so the integral diverges. For \(t>\lambda\), the integrand grows exponentially and the integral also diverges. Thus \(D_X=(-\infty,\lambda)\).

This MGF is finite at zero and throughout a neighborhood of zero, since \(\lambda>0\). It is finite for all negative \(t\) as well. Its positive boundary is excluded because the exponential decay of the density is exactly canceled at \(t=\lambda\).

Worked Example: A Pareto Distribution Finite Only on One Side

Let \(X\) have density \(f(x)=\alpha x^{-\alpha-1}\) for \(x\geq1\), where \(\alpha>0\). This is a probability density because

\int_1^\infty \alpha x^{-\alpha-1}\,dx =\left[-x^{-\alpha}\right]_1^\infty=1.

For \(t\leq0\), \(0<e^{tx}\leq1\) on the support, and hence \(0<M_X(t)\leq\int_1^\infty f(x)\,dx=1\). At \(t=0\), the value is exactly one. For \(t>0\), choose \(R\geq1\) large enough that \(tx\geq(\alpha+1)\log x\) for all \(x\geq R\). Such an \(R\) exists because \(x/\log x\to\infty\). On that range, \(e^{tx}\geq x^{\alpha+1}\), so

\int_R^\infty e^{tx}\alpha x^{-\alpha-1}\,dx \geq \int_R^\infty \alpha\,dx =\infty.

Consequently, \(D_X=(-\infty,0]\). The MGF is finite at zero and on a left-hand interval, but on no right-hand interval and no open neighborhood of zero. The polynomially decaying right tail cannot offset any positive exponential factor.

Reading the Domain From the Tails

For a distribution with density \(f\), determining the domain means checking exponential integrability in the two tail integrals. Positive \(t\) places the factor \(e^{tx}\) on the right tail; negative \(t\) places the growing factor on the left tail. A lighter tail may permit a wider range of \(t\), but the boundary itself requires a separate check.

Start at zero.
Record \(M_X(0)=1\). This establishes only that zero belongs to the domain.

Check each tail for the sign of \(t\).
For positive \(t\), test the right-tail integral or sum; for negative \(t\), test the left-tail integral or sum.

Check candidate endpoints directly.
At a boundary value, evaluate the corresponding exponential-weighted tail; do not infer endpoint inclusion from finiteness at interior points.

Classify the result around zero.
Decide whether the domain includes an open interval around zero, extends only to one side, or contains zero alone.

A domain can be unbounded in one direction and have a finite boundary in the other, as for the exponential distribution. It can also include a boundary point, as the Pareto example includes zero, or exclude one, as the exponential example excludes \(\lambda\). The interval theorem permits all these possibilities.

Common Mistake: Treating \(M_X(0)=1\) as Neighborhood Existence

A tempting but incorrect argument is: “The MGF is finite at zero, so it exists near zero.” The premise is true for every real-valued random variable, but it says nothing about exponential integrability away from zero. The Pareto example above has \(M_X(0)=1\) and is finite for negative \(t\), yet it is infinite for every positive \(t\). Therefore it has no open interval around zero on which the MGF is finite.

The correction is to test both tails. Neighborhood existence requires exponential integrability in both directions for some positive rate; one-sided finiteness is not enough. Also, when a candidate boundary is reached, its inclusion must be checked from the defining expectation rather than assumed from the fact that the domain is an interval.

Key takeaway: \(M_X(0)=1\) always, but the finiteness domain away from zero is determined by the tails. Finiteness on an open neighborhood of zero requires exponential control of both tails; one-sided finiteness requires control in only the corresponding direction.

Check Your Understanding

For each question, focus on the finiteness domain rather than on differentiating the MGF.

Why is \(M_X(0)=1\) for every real-valued random variable, including one whose MGF is infinite for all \(t\neq0\)?
If \(M_X(a)\) and \(M_X(b)\) are finite for \(a<b\), what does the interval theorem imply for values strictly between \(a\) and \(b\)?
For an exponential distribution with rate \(\lambda\), why is the boundary \(t=\lambda\) excluded from the MGF domain?
For the Pareto density in the worked example, why does \(t<0\) give a finite expectation while every \(t>0\) gives an infinite one?
State an equivalent exponential-integrability condition for an MGF to be finite on some open neighborhood of zero.