What the MGF Records—and Where It Exists
A probability generating function is designed for nonnegative integer-valued variables. For a general real-valued random variable, the moment generating function (MGF) instead weights outcomes exponentially. Its domain is part of its definition: an expression that is finite for some values of the argument need not be finite for all real values.
Let \(X:\Omega\to\mathbb{R}\) be a random variable on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). Since \(e^{tX}\) is nonnegative, its expectation is defined as an extended expectation and may equal \(+\infty\).
In particular, being finite at zero is automatic and says little about the tails of the distribution. The useful condition is finiteness on an open interval containing zero: there must be some \(a,b>0\) such that \(M_X(t)<\infty\) for every \(t\in(-a,b)\). The interval need not be symmetric; the positive and negative sides can have different ranges.
The Domain Is an Interval
Finiteness at two arguments forces finiteness at every argument between them. This fact makes the domain easier to describe, but it does not decide whether a finite endpoint belongs to the domain.
Proof. For \(\theta=0\) or \(\theta=1\), the assertion is just the assumed finiteness at \(s\) or \(t\). Suppose \(0<\theta<1\), and set \(r=(1-\theta)s+\theta t\). Pointwise, \(e^{rX}=(e^{sX})^{1-\theta}(e^{tX})^\theta\). Apply Hölder’s inequality with exponents \(1/(1-\theta)\) and \(1/\theta\) to the two nonnegative factors. It gives \(\mathbb{E}[e^{rX}]\leq(\mathbb{E}[e^{sX}])^{1-\theta}(\mathbb{E}[e^{tX}])^\theta\). Both expectations on the right are finite by assumption, so \(M_X(r)<\infty\). Thus every point between any two points of \(D_X\) lies in \(D_X\), which is the interval property. Since \(M_X(0)=1\), the interval contains zero. \(\square\)
The inequality also says that the MGF is log-convex wherever it is finite: taking logarithms gives a convexity inequality for \(\log M_X(t)\). The theorem does not assert that every endpoint is finite. That must be checked from the distribution itself.
Worked Examples: Finding the Domain
Worked Example: A Two-Point Distribution
Let \(X\) equal \(1\) and \(-1\), each with probability \(1/2\). Compute its MGF and domain.
By summing over the two possible values, for every real \(t\), \(M_X(t)=\tfrac12 e^t+\tfrac12 e^{-t}\). Both terms are finite for every real \(t\). Therefore \(D_X=\mathbb{R}\). In the usual notation, the displayed expression is \(\cosh(t)\).
Check: At \(t=0\), the expression gives \(\tfrac12 e^0+\tfrac12 e^0=\tfrac12+\tfrac12=1\), as required. The domain is all of \(\mathbb{R}\), not merely an interval near zero.
Worked Example: An Exponential Distribution and an Excluded Endpoint
Let \(X\) have density \(f_X(x)=2e^{-2x}\) for \(x\geq0\), and density \(0\) for \(x<0\). Find the exact domain of its MGF.
For any real \(t\), the defining integral is \(M_X(t)=\int_0^\infty e^{tx}2e^{-2x}\,dx=\int_0^\infty 2e^{-(2-t)x}\,dx\). If \(t<2\), then \(2-t>0\), so integration gives \(M_X(t)=\left[-\frac{2}{2-t}e^{-(2-t)x}\right]_{0}^{\infty}=\frac{2}{2-t}\). If \(t=2\), the integrand is constantly \(2\), and its integral over \([0,\infty)\) is infinite. If \(t>2\), the integrand \(2e^{(t-2)x}\) grows with \(x\), so its integral is also infinite. Hence \(D_X=(-\infty,2)\).
Check: Substituting \(t=0\) into the finite formula gives \(2/(2-0)=1\). The endpoint \(2\) is excluded because its defining integral is \(\int_0^\infty 2\,dx=\infty\).
Worked Example: A Pareto Tail with No Positive MGF Argument
Let \(X\) have density \(f_X(x)=2x^{-3}\) for \(x\geq1\), and density \(0\) for \(x<1\). Determine its MGF domain. This example has a finite first moment, but that does not ensure a positive MGF argument is finite.
For \(t\leq0\), \(e^{tx}\leq1\) on \(x\geq1\). Thus \(M_X(t)=\int_1^\infty 2e^{tx}x^{-3}\,dx\leq\int_1^\infty2x^{-3}\,dx=1\). The integral is therefore finite, including at \(t=0\), where its value is exactly \(1\).
Now take \(t>0\). Since \(e^{tx}\geq (tx)^2/2\) for \(x\geq1\), the defining integral satisfies \(M_X(t)\geq\int_1^\infty 2x^{-3}(tx)^2/2\,dx=t^2\int_1^\infty x^{-1}\,dx=\infty\). Therefore \(D_X=(-\infty,0]\): it includes zero but contains no open interval around zero.
Check: At \(t=0\), the density integrates to \(\int_1^\infty2x^{-3}\,dx=[-x^{-2}]_1^\infty=1\). For every \(t>0\), the lower bound is a positive constant times the divergent integral of \(1/x\), so the MGF cannot be finite there.
Why Finiteness Around Zero Matters
Finiteness on an interval around zero supplies exponential control in both directions. The positive side controls large positive values of \(X\); the negative side controls large negative values. This differs from finiteness only for \(t>0\), which need not control the left tail.
Proof. For every real \(x\), \(e^{a|x|}\leq e^{ax}+e^{-ax}\): if \(x\geq0\), the first term on the right equals the left, and if \(x<0\), the second does. Taking expectations gives \(\mathbb{E}[e^{a|X|}]\leq M_X(a)+M_X(-a)<\infty\). For \(k=0\), \(|X|^k=1\) and its expectation is \(1\). For \(k\geq1\), the function \(y^k e^{-ay}\) on \(y\geq0\) is bounded: its derivative is \(y^{k-1}e^{-ay}(k-ay)\), so its maximum occurs at \(y=k/a\), with value \((k/(ae))^k\). Therefore \(|X|^k\leq(k/(ae))^k e^{a|X|}\). Taking expectations proves the claim. \(\square\)
There is also a direct tail interpretation. If \(t>0\), then on \(\{X\geq x\}\), for \(x>0\), \(e^{tX}\geq e^{tx}\). Thus, when \(M_X(t)<\infty\), \(\mathbb{P}(X\geq x)\leq e^{-tx}M_X(t)\). If \(t<0\), then on \(\{X\leq-x\}\), \(e^{tX}\geq e^{-tx}\), giving \(\mathbb{P}(X\leq-x)\leq e^{tx}M_X(t)\). The bounds use the same nonnegative-variable inequality in each case: if \(Y\geq c>0\) on an event \(A\), then \(\mathbb{E}[Y]\geq c\mathbb{P}(A)\). A finite MGF on both sides of zero can therefore bound both tails exponentially.
The Pareto example shows why the interval condition is substantive. Its MGF is finite at zero and at every negative argument, but it is infinite at every positive argument. It cannot provide the two-sided exponential control above. By contrast, the two-point example is finite everywhere, and the exponential example is finite on an interval around zero even though its positive endpoint is excluded.
A Common Domain Error
Pitfall: Applying an integrated formula without retaining the condition that made the integral converge. In the exponential example, the calculation \(M_X(t)=2/(2-t)\) was obtained only for \(t<2\). Substituting \(t=3\) into that expression gives \(2/(2-3)=-2\), which cannot be an expectation of a nonnegative random variable. The error is not that the expectation is negative; the integral formula was used outside its valid range. Directly, at \(t=3\) the integrand is \(2e^x\), and \(\int_0^\infty2e^x\,dx=\infty\). Always retain the convergence condition from the integral when reporting an MGF formula.
When finding a domain, a reliable order is to write the expectation or integral first, identify where it converges, and then check boundary points separately. The expression at zero is always \(1\), but that single value does not settle whether any neighborhood of zero belongs to the domain.
Check Your Understanding
For each question, distinguish finiteness at a point from finiteness on an open interval around zero.
- If \(X\) takes values \(0\) and \(2\), each with probability \(1/2\), find \(M_X(t)\) and its domain.
- For the exponential density \(2e^{-2x}\) on \(x\geq0\), is \(M_X(2)\) finite? Verify using the defining integral.
- Suppose \(M_X(s)\) and \(M_X(t)\) are finite, where \(s<t\). What does the interval property imply for arguments strictly between \(s\) and \(t\)?
- Why does finiteness at zero alone give no information about whether \(M_X(t)\) is finite for any \(t>0\)?
- If both \(M_X(a)\) and \(M_X(-a)\) are finite for some \(a>0\), what can be concluded about \(\mathbb{E}[e^{a|X|}]\)?