Existence and Domain of an MGF

Why the Domain of an MGF Matters

The formula \(M_X(t)=E[e^{tX}]\) is an expectation, and an expectation need not be finite for every real value of \(t\). The set of arguments for which it is finite is the domain of the moment generating function (MGF). Finding this domain is part of specifying an MGF: a formula that is valid only for \(t<2\), for example, must not be used at \(t=2\) or beyond.

A random variable always has a finite MGF value at \(t=0\), because \(e^{0X}=1\). That fact alone does not mean its MGF exists in the standard sense used in probability theory. In this context, the MGF is said to exist when it is finite throughout some open interval containing zero. This two-sided neighborhood is important in results that use the MGF to study moments and characterize distributions.

Definition: The finite domain of \(M_X\) is the set of real numbers \(t\) for which \(E[e^{tX}]<\infty\). The MGF is said to exist if there is an \(\varepsilon>0\) such that \(E[e^{tX}]<\infty\) for every \(t\) satisfying \(|t|<\varepsilon\). Since \(M_X(0)=1\), zero is always in the finite domain, but the domain may contain no open interval around zero.

For a continuous random variable with density \(f_X\), the expectation is evaluated as \(\int_{-\infty}^{\infty} e^{tx}f_X(x)\,dx\); for a discrete random variable, it is a sum over possible values. In either case, the integrand or summands are nonnegative. The question is whether their total is finite, not merely whether the expression can be written down.

A Normal Distribution Has a Domain of All Real Numbers

A normal variable provides an important contrast with heavy-tailed models: its MGF is finite for every real argument. The following calculation shows both the formula and why the integral converges.

Worked Example: Finding the Domain for a Normal Variable

Suppose a hypothetical measurement \(X\), in units, has distribution \(N(3,4)\): its mean is \(3\) units and its variance is \(4\) square units. Find the domain of its MGF and evaluate \(M_X(0.5)\).

Method. Write the density integral and complete the square in its exponent. This also checks convergence for every real \(t\).

The standard deviation is \(\sqrt{4}=2\), so the density is

f_X(x)=\frac{1}{2\sqrt{2\pi}}\exp\left(-\frac{(x-3)^2}{8}\right).

For any real \(t\), combine the exponents and complete the square:

tx-\frac{(x-3)^2}{8} =-\frac{(x-(3+4t))^2}{8}+3t+2t^2.

Therefore,

M_X(t) =\int_{-\infty}^{\infty}e^{tx}f_X(x)\,dx =e^{3t+2t^2} \int_{-\infty}^{\infty} \frac{1}{2\sqrt{2\pi}} \exp\left(-\frac{(x-(3+4t))^2}{8}\right)\,dx =e^{3t+2t^2}.

The integral is \(1\): its integrand is a normal density with mean \(3+4t\) and variance \(4\). This remains a valid density for every real \(t\), so the MGF is finite for every real \(t\). Its domain is \((-\infty,\infty)\), which includes an open interval around zero.

At \(t=0.5\), the exponent is \(3(0.5)+2(0.5)^2=1.5+0.5=2\). Thus \(M_X(0.5)=e^2\), approximately \(7.3891\) to four decimal places. Check: at zero, the formula gives \(M_X(0)=e^0=1\), as required.

Common mistake avoided: Treating the variance \(4\) as the standard deviation would give an incorrect density and exponent. Here the standard deviation is \(2\); the \(N(\mu,\sigma^2)\) notation places the variance in its second argument.

Finiteness on an Interval and Its Endpoints

The finite domain has a useful structural property: if the MGF is finite at two arguments, it is finite at every argument between them. This follows from Hölder's inequality, which bounds an expectation of a product using the expectations of its factors.

Result: The finite domain of an MGF is an interval, possibly with one or both endpoints included. It contains zero, but it need not contain any open interval around zero.

To verify the interval property, suppose \(M_X(s)\) and \(M_X(t)\) are finite, and let \(0<\lambda<1\). The argument between \(s\) and \(t\) is \(r=\lambda s+(1-\lambda)t\). Since \(e^{rX}=(e^{sX})^\lambda(e^{tX})^{1-\lambda}\), Hölder's inequality gives

M_X(r) =E\left[(e^{sX})^\lambda(e^{tX})^{1-\lambda}\right] \leq \{M_X(s)\}^{\lambda}\{M_X(t)\}^{1-\lambda}<\infty.

Thus all arguments strictly between two finite-domain points are also in the domain. The endpoints require separate examination: an integral may converge at a boundary, or it may diverge there. The next example illustrates a boundary that is excluded.

Worked Example: An Exponential MGF and Its Excluded Boundary

Let \(X\) be a hypothetical waiting time in years with an exponential distribution of rate \(2\) per year. Its density is \(f_X(x)=2e^{-2x}\) for \(x\geq 0\). Determine its MGF domain, including the boundary, and evaluate \(M_X(1)\).

Method. Substitute the density into the expectation and test convergence of the resulting improper integral. Here \(t\) has units of inverse years, so it can be compared with the rate \(2\) per year.

M_X(t)=\int_0^\infty e^{tx}2e^{-2x}\,dx =2\int_0^\infty e^{-(2-t)x}\,dx.

If \(t<2\), then \(2-t>0\), and the exponential decays as \(x\) increases. Evaluating the integral gives

M_X(t)=2\left[\frac{-e^{-(2-t)x}}{2-t}\right]_0^\infty =\frac{2}{2-t}.

At \(t=2\), the integrand in the original expression is \(2\), so \(2\int_0^\infty 1\,dx=\infty\). For \(t>2\), the integrand grows exponentially, so the integral also diverges. The domain is therefore \((-\infty,2)\), with the endpoint \(2\) excluded. This domain contains an open interval around zero.

At \(t=1\), \(M_X(1)=2/(2-1)=2\). Check: substituting \(t=0\) into the formula gives \(2/2=1\).

Common mistake avoided: Using the formula \(2/(2-t)\) at \(t=2\) is invalid; it has a zero denominator because the defining expectation diverges there. A formula obtained by integration on one range must be paired with its convergence condition.

Heavy Tails Can Prevent MGF Existence

For a nonnegative loss with a heavy, polynomially decreasing tail, negative arguments can still give finite expectations: \(e^{tX}\leq 1\) when \(t<0\). Positive arguments are different. The factor \(e^{tX}\) grows exponentially, and can overwhelm a density that decreases only as a power of \(X\). The result can be a finite domain that ends at zero, leaving no open interval around zero.

Worked Example: A Pareto Loss with No MGF Near Zero

Suppose a hypothetical loss \(X\), measured in thousands of dollars, has Pareto survival function \(P(X>x)=(1+x/2)^{-3}\) for \(x\geq 0\). The scale is \(2\) thousand dollars and the shape is \(3\). Determine its MGF domain and decide whether its MGF exists.

Method. Differentiate the survival function to obtain the density, then compare the exponential factor in the MGF integral with the density's polynomial tail. The argument \(t\) is measured in inverse thousands of dollars.

Differentiating the survival function gives

f_X(x) =-\frac{d}{dx}(1+x/2)^{-3} =\frac{3}{2}(1+x/2)^{-4} =\frac{24}{(x+2)^4},\qquad x\geq 0.

For \(t=0\), the expectation is \(M_X(0)=1\). For \(t<0\), \(0<e^{tX}\leq 1\), so

0<M_X(t)=\int_0^\infty e^{tx}f_X(x)\,dx \leq\int_0^\infty f_X(x)\,dx=1.

Thus the MGF is finite for every negative \(t\). For \(t>0\), its integrand is \(24e^{tx}/(x+2)^4\). Exponential growth eventually exceeds polynomial growth, so this integrand eventually exceeds \(1\); its integral over an infinite interval diverges. Therefore the domain is \((-\infty,0]\).

This domain includes zero but does not include any open interval around zero: every such interval contains positive values of \(t\), where the expectation is infinite. The MGF does not exist in the standard neighborhood-of-zero sense, even though \(M_X(0)=1\) and it is finite for all negative \(t\).

Common mistake avoided: Concluding that the MGF exists because it is finite at zero, or because it is finite on one side of zero, misses the requirement for an open interval containing zero. Check: the domain includes \(0\), as \(M_X(0)=1\), but includes no positive value.

What the Open Interval Requirement Does

A finite value at an isolated point provides very little information about how \(E[e^{tX}]\) behaves as \(t\) changes. Finiteness on an open interval around zero supplies control on both sides of zero. That local control underlies standard uses of MGFs to obtain moments and to identify distributions. Those uses require appropriate results; one should not assume that differentiating an expectation is valid just because the expectation is finite at zero.

The domain also makes assumptions visible. For a normal model, every real argument is allowed. For an exponential model, the domain has a finite upper boundary. For the Pareto loss above, only nonpositive arguments are allowed. A useful exam habit is to state the domain alongside any MGF formula, then test the boundary directly in the defining expectation rather than relying on the formula outside its range.

Key takeaway: Always distinguish \(M_X(0)=1\) from existence of an MGF near zero. Determine convergence on both sides of zero, check any proposed endpoint in the original expectation, and state the full finite domain.

Check Your Understanding

For each model, find the finite domain and check whether the MGF exists in an open interval around zero.

A bounded random variable \(X\) takes values only in \([-4,6]\). What is its MGF domain? (Answer: All real \(t\); it exists near zero.)
A waiting time \(X\), in months, has an exponential distribution with rate \(3\) per month. Give the domain and determine whether the boundary \(t=3\) is included. (Answer: \((-\infty,3)\), with \(t=3\) excluded; the MGF exists near zero.)
A variable has distribution \(N(-2,9)\). Find the domain and calculate \(M_X(1)\), rounding to four decimal places. (Answer: All real \(t\); \(M_X(1)=e^{2.5}\approx 12.1825\), so the MGF exists near zero.)
A nonnegative loss has survival function \(P(X>x)=(1+x/5)^{-2}\) for \(x\geq0\). Determine the MGF domain by considering negative and positive \(t\). (Answer: \((-\infty,0]\); the MGF does not exist near zero.)
A variable has density \(f_X(x)=\tfrac12e^{-x}\) for \(x\geq0\), and \(f_X(x)=e^{2x}\) for \(x<0\). Find the MGF domain, checking both tails and both boundary points. (Answer: \((-2,1)\); both endpoints are excluded, and the MGF exists near zero.)

What You'll Learn