The Uniqueness Theorem for MGFs

Why MGF Uniqueness Matters

A moment generating function summarizes a probability distribution through the expectations \(E[e^{tX}]\). It can be easier to compare or manipulate these functions than to compare probability mass functions or densities directly. The key question is whether an MGF contains enough information to determine the full distribution, rather than only selected features such as the mean and variance.

The answer is yes when the MGFs of two random variables are both finite on a common open interval around zero and agree throughout that interval. The interval condition matters: an MGF is not guaranteed to exist for every real \(t\), and matching a few values or derivatives is not the same as matching the function on an interval. This result is aligned with the SOA VEE Mathematical Statistics syllabus.

Definition: The MGF of a random variable \(X\) is \(M_X(t)=E[e^{tX}]\) at every real value \(t\) for which this expectation is finite. A common open interval around zero means there is some \(a>0\) such that both MGFs are finite for every \(t\in(-a,a)\).

The Uniqueness Theorem

Theorem: Let \(X\) and \(Y\) be random variables. Suppose there is an \(a>0\) such that \(M_X(t)\) and \(M_Y(t)\) are finite for every \(t\in(-a,a)\). If \(M_X(t)=M_Y(t)\) for every \(t\in(-a,a)\), then \(X\) and \(Y\) have the same probability distribution. Equivalently, their cumulative distribution functions agree at every real number.

“Same probability distribution” does not mean that \(X\) and \(Y\) must be the same random variable on a particular probability space, or that their outcomes are equal in every trial. It means that they assign the same probabilities to all events of the form \(\{X\leq x\}\) and \(\{Y\leq x\}\). For discrete variables, this gives the same probability at every possible value; for continuous variables, it gives the same distribution function, whether or not a density exists.

The interval must be open and contain zero. Since every MGF has \(M_X(0)=1\), agreement at zero alone says nothing about whether the distributions match. Agreement at a finite number of other points is not enough either. The theorem uses agreement at every point in an interval where both functions are finite.

Why Equality on an Interval Determines the Distribution

A short justification uses two standard facts: an MGF that is finite near zero extends to a complex analytic function in a vertical strip, and characteristic functions uniquely determine probability distributions. The connection between these facts explains why the theorem works.

Extend each MGF into a strip.
For a complex number \(z\), define \(F_X(z)=E[e^{zX}]\) when the expectation exists. If \(|\operatorname{Re}(z)|<a\), then \(|e^{zX}|=e^{\operatorname{Re}(z)X}\), whose expectation is finite by the assumed MGF condition. Thus \(F_X\) is defined in the strip \(|\operatorname{Re}(z)|<a\); it is analytic there. The same applies to \(Y\).

Use equality on the real interval.
The analytic functions \(F_X\) and \(F_Y\) agree on the real interval \((-a,a)\), which has accumulation points inside the strip. By the identity theorem for analytic functions, they agree throughout the connected strip.

Reach the characteristic functions.
For real \(u\), set \(z=iu\). Then \(F_X(iu)=E[e^{iuX}]\), the characteristic function of \(X\), and likewise for \(Y\). Since the two analytic functions agree in the strip, their characteristic functions agree for every real \(u\).

Conclude equality of distributions.
The uniqueness theorem for characteristic functions states that two probability distributions with the same characteristic function at every real argument are identical. Therefore \(X\) and \(Y\) have the same distribution.

The analyticity step is justified by the exponential moments available on both sides of zero. In any smaller strip, derivatives of \(E[e^{zX}]\) can be taken under the expectation: the exponential terms dominate the factors involving powers of \(|X|\). This permits the usual local power-series representation. The crucial point is that equality on the real interval identifies the analytic extensions, including their values on the imaginary axis.

The theorem is stronger than the claim that MGFs give moments. If the MGF exists near zero, its derivatives at zero give moments, but the uniqueness argument does not stop at comparing a few derivatives. It uses the whole function on an interval, then connects that function to the characteristic function, which determines the full distribution.

Worked Example: Two Descriptions of a Discrete Distribution

Let \(X\) have probability \(1/3\) at \(-1\) and probability \(2/3\) at \(2\). Independently of how \(X\) is represented, define \(I\sim\operatorname{Bernoulli}(2/3)\) and \(Y=3I-1\). Use the MGF method to verify that \(X\) and \(Y\) have the same distribution. Both variables are discrete with the stated probabilities; there is no interest rate or mortality assumption in this probability model.

Method. Compute each MGF from the possible values, establish equality on a common open interval, and apply the uniqueness theorem. This method avoids assuming that differently written random variables must have different distributions.

For \(X\), the two possible contributions to the expectation are

M_X(t)=\frac{1}{3}e^{-t}+\frac{2}{3}e^{2t}.

For \(Y\), if \(I=0\), then \(Y=-1\), and if \(I=1\), then \(Y=2\). Therefore

M_Y(t)=E[e^{t(3I-1)}] =\frac{1}{3}e^{-t}+\frac{2}{3}e^{2t}.

Both MGFs are finite for every real \(t\), because each variable has only two finite possible values. In particular, they are finite on, for example, \((-1,1)\), and the calculations show that they agree at every \(t\) in this interval. The uniqueness theorem gives the same distribution for \(X\) and \(Y\). As a direct check, \(P(Y=-1)=P(I=0)=1/3\) and \(P(Y=2)=P(I=1)=2/3\), exactly matching \(X\).

Worked Example: A Normal Variable and a Scaled Representation

Suppose \(X\sim N(3,4)\), where the second parameter is the variance. Let \(Z\sim N(0,1)\), and define \(Y=3+2Z\). Verify by MGFs that \(X\) and \(Y\) have the same distribution. Both descriptions are normal; \(X\) has mean \(3\) and variance \(4\), while \(Z\) has mean \(0\) and variance \(1\). No interest or mortality assumptions are involved.

Method. Evaluate the MGF for each description and compare the resulting functions on a common interval. The calculation avoids confusing the standard deviation \(2\) with the variance \(4\).

For a normal variable \(X\sim N(\mu,\sigma^2)\), the MGF is \(M_X(t)=e^{\mu t+\sigma^2t^2/2}\), finite for all real \(t\). Substituting \(\mu=3\) and \(\sigma^2=4\) gives

M_X(t)=e^{3t+4t^2/2}=e^{3t+2t^2}.

For \(Y=3+2Z\), directly use the definition and the MGF of \(Z\):

\begin{aligned} M_Y(t) &=E[e^{t(3+2Z)}]\\ &=e^{3t}E[e^{(2t)Z}]\\ &=e^{3t}M_Z(2t)\\ &=e^{3t}e^{(2t)^2/2}\\ &=e^{3t+2t^2}. \end{aligned}

Both MGFs are finite for every real \(t\), and the expressions agree for every real \(t\), hence on any open interval around zero. The theorem confirms that \(X\) and \(Y\) have the same distribution. As a check on the parameters, \(E[Y]=3+2E[Z]=3\), and \(\operatorname{Var}(Y)=2^2\operatorname{Var}(Z)=4\), consistent with \(N(3,4)\).

What Matching Moments Does—and Does Not—Show

A frequent exam trap is to replace “the MGFs agree on an interval” with “the means and variances agree.” The latter does not imply equality of distributions. In fact, even agreement of several moments is not the theorem’s condition. The next example shows two MGFs that exist everywhere and have matching first two moments, but are not equal.

Worked Example: Same Mean and Variance, Different Distributions

Let \(A\) take the values \(-1\) and \(1\), each with probability \(1/2\). Let \(B\) take the values \(-\sqrt{2}\), \(0\), and \(\sqrt{2}\), with probabilities \(1/4\), \(1/2\), and \(1/4\), respectively. Compare their means, variances, and MGFs. Both variables are discrete with the probabilities just specified; no interest or mortality assumptions are involved.

Method. Calculate the first two moments and then compare the full MGF expressions. This method avoids the mistake of applying the uniqueness theorem to matching moments rather than matching MGFs.

By symmetry, \(E[A]=0\) and \(E[B]=0\). Their second moments are

E[A^2]=\frac{1}{2}(1)+\frac{1}{2}(1)=1, \qquad E[B^2]=\frac{1}{4}(2)+\frac{1}{2}(0)+\frac{1}{4}(2)=1.

Thus both variances are \(1\). But their MGF expressions are

M_A(t)=\frac{1}{2}e^{-t}+\frac{1}{2}e^t=\cosh(t)

and

M_B(t)=\frac{1}{4}e^{-\sqrt{2}t}+\frac{1}{2}+\frac{1}{4}e^{\sqrt{2}t} =\frac{1}{2}+\frac{1}{2}\cosh(\sqrt{2}t).

Both MGFs are finite for every real \(t\). To see that they do not agree near zero, expand through the fourth-order term:

M_A(t)=1+\frac{t^2}{2}+\frac{t^4}{24}+\cdots, \qquad M_B(t)=1+\frac{t^2}{2}+\frac{t^4}{12}+\cdots.

The coefficients of \(t^4\) differ, so the MGFs are not equal on any open interval around zero. Correspondingly, \(E[A^4]=1\), whereas \(E[B^4]=2\). The distributions are different, despite their matching means and variances. The theorem is not contradicted: its equality condition is not satisfied.

Common Exam Traps

Checking only at zero. Every MGF that exists at zero has value \(1\). That single equality does not identify a distribution.
Checking only a finite set of values. Equality at several selected arguments does not establish equality throughout an open interval.
Comparing just the mean and variance. These are only two features of a distribution. Different distributions can share both, as the example shows.
Ignoring the common domain. The theorem requires both MGFs to be finite on a common open interval around zero. If an MGF does not exist near zero, this theorem cannot be applied as stated.
Misreading the conclusion. The theorem establishes identical distributions, not that the variables are equal outcome by outcome.

Key takeaway: To invoke MGF uniqueness, verify two conditions: both MGFs are finite on a common open interval containing zero, and their values agree at every point of that interval. Then the probability distributions are identical. Equality of only a few moments or MGF values is insufficient.

Check Your Understanding

For each problem, decide whether the uniqueness theorem applies and state the conclusion justified by the given information.

MGFs \(M_X\) and \(M_Y\) are finite on \((-0.4,0.4)\) and equal at every point of that interval. What can be concluded? (Answer: \(X\) and \(Y\) have the same probability distribution.)
Two variables have equal means, equal variances, and MGFs finite near zero, but no information about whether their MGFs agree elsewhere. Does the theorem establish equal distributions? (Answer: No; equality of the first two moments does not establish equality of the MGFs on an interval.)
Both MGFs are finite on \((-2,1)\) and agree at every point of \((-0.2,0.2)\). Does the theorem apply, despite the larger domains being different in width from a symmetric interval? (Answer: Yes; \((-0.2,0.2)\) is a common open interval around zero, so the distributions are the same.)
\(M_X\) is finite on \((-1,1)\), \(M_Y\) is finite on \((-2,2)\), and the functions agree throughout \((-0.5,0.5)\). What follows? (Answer: The variables have the same probability distribution, because the stated interval is common, open, and contains zero.)
Two MGFs are finite near zero and have the same values at \(t=-1,0,1\), but equality on any interval is not given. Is the uniqueness theorem enough to conclude that the distributions match? (Answer: No; agreement at three points does not establish equality throughout an open interval.)

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