Validity Conditions for Transform-Based Identification

When Does a Transform Identify a Distribution?

A transform can turn a distributional question into an algebraic one. For independent random variables, for example, products of transforms often describe sums. But an algebraic identity identifies a law only when the transforms are being used on valid domains and the transform itself uniquely determines that law. A formula evaluated outside its domain, or an equality checked at just one argument, does not establish equality of distributions.

This tutorial compares three transforms used in distribution theory: the probability-generating function for nonnegative integer-valued variables, the moment-generating function where it is finite, and the characteristic function, which exists for every real-valued random variable. The central distinction is that characteristic functions always exist and uniquely identify a law, whereas a moment-generating function needs a finiteness condition near zero for its standard uniqueness result.

The PGF is finite on \([0,1]\), and its power-series representation is \(G_N(s)=\sum_{k=0}^{\infty}P(N=k)s^k\). An MGF may be finite only on part of the real line; its domain is not automatically all real \(t\). By contrast, \(|e^{itX}|=1\), so the expectation defining the CF exists for every real \(t\). The existence of an expectation and the uniqueness of a distribution from its values are separate questions.

Three Uniqueness Conditions

The interval conditions in the first two statements are important: a transform’s value at one argument, even when that value is well-defined, is not generally enough. The MGF condition also explicitly requires finiteness near zero. The CF statement needs no such finiteness assumption.

Why the Uniqueness Conditions Hold

For the PGF, the power series \(\sum_{k=0}^{\infty}P(N=k)z^k\) converges absolutely for every complex \(z\) with \(|z|<1\), because \(\sum_k P(N=k)|z|^k\leq 1\). It therefore defines an analytic function on the open unit disk. If two PGFs agree on an open real interval inside that disk, the identity theorem for analytic functions implies they agree throughout the disk. Their power-series coefficients are then equal, so \(P(N_1=k)=P(N_2=k)\) for each \(k\geq0\). This proves the first statement.

For the MGF statement, let \(J\) be the intersection of open intervals containing zero on which the two MGFs are finite. This intersection is itself an open interval containing zero. On the complex strip \(S=\{z:\operatorname{Re}(z)\in J\}\), define \(M_X(z)=E[e^{zX}]\). This expectation is finite in absolute value because \(|e^{zX}|=e^{\operatorname{Re}(z)X}\). It is analytic on the strip: on any compact subset, choose real endpoints slightly farther out within \(J\); the exponential moments at those endpoints dominate the integrands and their derivatives. More explicitly, for each derivative order \(m\), the factor \(|X|^m e^{\operatorname{Re}(z)X}\) is bounded by a constant times the sum of exponentials at those farther endpoints, which have finite expectations. Differentiation under the expectation is therefore justified locally.

If the two MGFs agree on an open real interval containing zero, their analytic extensions agree on that interval. The identity theorem makes them equal throughout the connected strip \(S\), in particular on its imaginary axis. Thus \(\varphi_{X_1}(t)=M_{X_1}(it)=M_{X_2}(it)=\varphi_{X_2}(t)\) for every real \(t\). The CF uniqueness result proved below then gives equality in distribution. This establishes the MGF statement, subject to that result.

Here is a proof of CF uniqueness. For \(a>0\), let \(Z\sim N(0,1)\) be independent of \(X\). The random variable \(X+\sqrt a Z\) has density

The Fourier integral for the Gaussian density gives

Interchanging expectation and integration here is valid: the absolute value of the integrand before taking expectation is bounded by \(e^{-at^2/2}\), whose integral over \(\mathbb R\) is finite. Consequently, if two CFs agree everywhere, the densities of their Gaussian-smoothed variables agree for every \(a>0\).

As \(a\) decreases to zero, \(\sqrt a Z\) converges to zero in probability. To see what this implies for distribution functions, for any \(\epsilon>0\),

At a continuity point \(x\) of the distribution function of \(X\), first let \(a\) decrease to zero and then let \(\epsilon\) decrease to zero. These bounds show that \(P(X+\sqrt a Z\leq x)\) tends to \(P(X\leq x)\). If two CFs agree, their smoothed distribution functions agree for every \(a\), so the original distribution functions agree at their common continuity points. Each distribution function has at most countably many discontinuities; common continuity points are dense. Right-continuity then gives equality everywhere. This proves CF uniqueness. \(\square\)

Worked Applications

Common Mistakes and Why the Conditions Matter

• Using a transform outside its domain. An algebraic expression derived from an expectation remains valid only where that expectation is finite. Verify the domain before substituting an argument.
• Equating one value with equality of laws. For every random variable, \(M_X(0)=1\) and \(\varphi_X(0)=1\). Agreement at zero says nothing beyond that universal identity. The uniqueness conditions require agreement on an interval for PGFs or MGFs, or on the entire real line for CFs.
• Assuming a few moments identify a distribution. Derivatives of an MGF at zero, when they exist, give moments; matching a finite number of moments does not invoke any of the uniqueness results above. Do not replace a transform-uniqueness argument with a comparison of only a few derivatives.
• Multiplying transforms without independence. The identity \(M_{X+Y}(t)=M_X(t)M_Y(t)\), when the expectations are finite, follows from independence. Without independence, the expectation of \(e^{tX}e^{tY}\) need not factor. The same caution applies to characteristic functions.

The practical sequence is to identify which transform applies, check its domain, verify the required equality over the appropriate set of arguments, and then invoke the matching uniqueness result. This makes transform calculations a valid route from an algebraic expression to a distributional conclusion rather than just a plausible pattern match.