# How to prove: Moment Generating Function Uniqueness Theorem

Many results are based on the fact of the Moment Generating Function (MGF) Uniqueness Theorem, that says:

If $X$ and $Y$ are two random variables and equality holds for their MGF's: $m_X(t) = m_Y(t)$ then $X$ and $Y$ have the same probability distribution: $F_X(x) = F_Y(y)$.

The proof of this theorem is never shown in textbooks, and I cannot seem to find it online or in any book I have access to.

Can someone show me the proof or tell me where to look it up?

• Got something from an answer below?
– Did
Apr 11, 2014 at 10:27
• Indeed, the formulas are very handy, but I wish there were more "meat" on them (details). Apr 11, 2014 at 12:27
• Which part(s) are you unable to complete after 30 seconds of thinking?
– Did
Apr 11, 2014 at 13:43
• Both, I don't see why the equalities are true - there need some intermediary calculations. Apr 12, 2014 at 13:20
• Aug 12, 2017 at 4:03

Let us first clarify the assumption. Denote the moment generating function of $$X$$ by $$M_X(t)=Ee^{tX}$$.

Uniqueness Theorem. If there exists $$\delta>0$$ such that $$M_X(t) = M_Y(t) < \infty$$ for all $$t \in (-\delta,\delta)$$, then $$F_X(t) = F_Y(t)$$ for all $$t \in \mathbb{R}$$.

To prove that the moment generating function determines the distribution, there are at least two approaches:

• To show that finiteness of $$M_X$$ on $$(-\delta,\delta)$$ implies that the moments $$X$$ do not increase too fast, so that $$F_X$$ is determined by $$(EX^k)_{k\in\mathbb{N}}$$, which are in turn determined by $$M_X$$. This proof can be found in Section 30 of Billingsley, P. Probability and Measure.

• To show that $$M_X$$ is analytic and can be extended to $$(-\delta,\delta)\times i\mathbb{R} \subseteq \mathbb{C}$$, so that $$M_X(z)=Ee^{zX}$$, so in particular $$M_X(it)=\varphi_X(t)$$ for all $$t\in\mathbb{R}$$, and then use the fact that $$\varphi_X$$ determines $$F_X$$. For this approach, see Curtiss, J. H. Ann. Math. Statistics 13:430-433 and references therein or Roja's answer.

At undergraduate level, it is interesting to work with the moment generating function and state the above theorem without proving it. One possible proof requires familiarity with holomorphic functions and the Identity Theorem from complex analysis, which restricts the set of students to which it can be taught.

In fact, the proof is so advanced that, at such a point it usually makes more sense to accept working with complex numbers, forget about moment generating function and work with the charachteristic function $$\varphi_X(t)=Ee^{itX}$$ instead. Almost every graduate textbook takes this path and proves that the characteristic function determines the distribution as a corollary of the inversion formula.

This proof of the inversion formula is bit long, but it only requires Fubini Theorem to switch an expectation with an integral and Dominated Convergence Theorem to switch an integral with a limit. A direct proof of uniqueness without inversion formula is shorter and simpler, and it only requires Weierstrass Theorem to approximate a continuous function by a trigonometric polynomial.

Side remark. If you only admit random variables whose support are contained in $$\mathbb{Z}_+$$, then the probability generating function $$G_X(z)=Ez^X$$ determines $$p_X$$ (and thus $$F_X$$). This elementary result is proved in most undergraduate textbooks and is mentioned in Did's answer. If you only admit random variables whose support are contained in $$\mathbb{Z}$$, then it is simpler to show that $$\varphi_X$$ determines $$p_X$$, as also mentioned in Did's answer, and the proof uses Fubini.

• This is a super helpful answer. thanks. Apr 4, 2018 at 22:19

$$(\forall n\geqslant0)\qquad \left.\frac{\mathrm d^n}{\mathrm ds^n}\mathbb E[s^X]\right|_{s=0}=n!\cdot\mathbb P[X=n]$$ $$(\forall x\in\mathbb R)\qquad \int_0^{2\pi}\mathbb E[\mathrm e^{\mathrm itX}]\,\mathrm e^{-\mathrm itx}\,\mathrm dt=2\pi\cdot\mathbb P[X=x]$$

• I guess Did is giving a formula for recovering a discrete probability distribution from its MGF. Aug 4, 2013 at 4:31
• Then $E[s^X]=ps^j+(1-p)s^k$. Alternatively, in the general case, $E[s^X]=M_X(\log s)$.
– Did
Aug 5, 2013 at 10:21
• Does your formula works for negative values of X ? What is X is -16 ? Aug 6, 2013 at 19:22
• Let me add that it is a tad unsettling to see that, when it is suggested that you perform yourself a trivial computation which would allow you to understand the general case, you do not even acknowledge the suggestion but fall back on demands already answered a long time ago. If you think maths can be learned as a spectator, you are wrong (but hey, this is your call).
– Did
Apr 12, 2014 at 18:41
• "If you think maths can be learned as a spectator, you are wrong " well put Nov 1, 2015 at 11:54

The following is a proof with two steps. It requires familiarity with Cauchy-Riemann equations and the identity theorem from complex analysis. Dominated convergence theorem is also applied. Thus, it fits for any advanced undergraduate or graduate student who is familiar with these topics.

Step 1:

Assume that $$X$$ is a random variable such that there exists $$\delta\in(0,\infty]$$ for which

$$M_X(s)=Ee^{sX}<\infty\ , \ \forall s\in(-\delta,\delta)\,.$$

Now, let's denote $$\Omega\equiv\left\{z\in\mathbb{C};\text{Re}(z)\in(-\delta,\delta)\right\}$$ and define

$$\phi_X(z)\equiv Ee^{zX}\ ,\ \forall z\in\Omega\,.$$

Let $$z=s+it\in\Omega$$ and observe that

$$\phi_X(z)=E\ \overset{u(s,t;X)}{\overbrace{e^{sX}\cos (tX)}}+iE\ \overset{v(s,t;X)}{\overbrace{e^{sX}\sin (tX)}}\,.$$

It is possible to apply dominated convergence theorem in order to show that $$Eu(\cdot;X)$$ and $$Ev(\cdot;X)$$ are differentiable on $$(-\delta,\delta)\times\mathbb{R}$$. To see this, notice that for every $$(s,t)\in(-\delta,\delta)\times\mathbb{R}$$

$$|\frac{\partial u(s,t;X)}{\partial t}|\ ,\ |\frac{\partial u(s,t;X)}{\partial s}|\ ,\ |\frac{\partial v(s,t;X)}{\partial t}|\ ,\ |\frac{\partial u(s,t;X)}{\partial s}|$$ are all dominated by the random variable $$Y=|X|e^{sX}$$. In addition notice that for every $$\epsilon>0$$

$$|X|=\epsilon^{-1}|\epsilon X|\leq\epsilon^{-1}\sum_{k=0}^\infty\frac{|\epsilon X|^k}{k!}$$

$$=\epsilon^{-1}e^{\epsilon |X|}\leq\epsilon^{-1}\left(e^{-\epsilon X}+e^{\epsilon X}\right)$$ and hence

$$|Y|\leq \frac{e^{sX}}{\epsilon}\left(e^{-\epsilon X}+e^{\epsilon X}\right)=\epsilon^{-1}\left(e^{(s-\epsilon) X}+e^{(s+\epsilon)X}\right)\,.$$

Therefore, since $$M_X(\cdot)$$ is finite on $$(-\delta,\delta)$$, then by taking small enough $$\epsilon>0$$ deduce that

$$E|Y|\leq\epsilon^{-1}\left[M_X(s-\epsilon)+M_X(s+\epsilon)\right]<\infty\,.$$

Thus dominated convergence theorem can be carried out to show that $$Eu(\cdot;X)$$ and $$Ev(\cdot;X)$$ have partial derivatives with respect to both of their coordinates. Moreover, these partial derivatives are obtained by differentiating inside the expectation. In addition, dominated convergence theorem can be applied once more (with the same dominating random variable) in order to show that these partial derivatives are continuous on $$(-\delta,\delta)\times\mathbb{R}$$, i.e., $$Eu(\cdot;X)$$ and $$Ev(\cdot;X)$$ are differentiable on $$(-\delta,\delta)\times\mathbb{R}$$. Now, it is straightforward to verify that the Cauchy-Riemann equations are satisfied on $$(-\delta,\delta)\times\mathbb{R}$$ and hence $$\phi_X(\cdot)$$ is holomorphic on $$\Omega$$.

Step 2: Assume that $$X$$ and $$Y$$ are two random variables such that there exists $$\delta\in(0,\infty)$$ for which $$M_X(s)=M_Y(s)<\infty$$ for every $$s\in(-\delta,\delta)$$. Then, define

$$H(z)\equiv\phi_X(z)-\phi_Y(z)\ , \ \forall z\in\Omega$$

and by the result of Step 1, $$H(\cdot)$$ is holomorphic on $$\Omega$$ which is an open connected subset of $$\mathbb{C}$$. Now, denote a sequence $$z_n=\frac{\delta}{2n}\in\Omega,\forall n\in\mathbb{N}$$ and note that $$z_n\rightarrow 0\in\Omega$$ as $$n\to\infty$$. In addition, $$H(z_n)=0,\forall n\in\mathbb{N}$$ and hence using the identity theorem deduce that $$H(z)=0,\forall z\in\Omega$$. Then, as a special case, for every $$t\in\mathbb{R}$$ set $$z=it$$ and get

$$Ee^{itX}=Ee^{itY}$$

which means that $$F_X(u)=F_Y(u),\forall u\in\mathbb{R}$$ due to the uniqueness property of characteristic function (There are plenty of sources for the proof of this property and as mentioned by others its proof requires no advanced staff but dominated convergence and Fubini's theorems).

• Why the sequence $z_n=\delta/2n$. Can we take another sequence thats not zero as $n$ goes to infinity? Oct 6, 2020 at 1:04
• You could choose any sequence $(z_n)_{n=1}^\infty$ for which (1) $z_n\in\Omega,\forall n\in\mathbb{N}$, (2) $H(z_n)=0,\forall n\in\mathbb{N}$ and (3) $\exists\lim_{n\to\infty}z_n\in\Omega$. For more accurate details you may see the identity theorem for holomorphic functions, e.g., en.wikipedia.org/wiki/Identity_theorem.
– Roja
Oct 6, 2020 at 15:49

In the case where $X$ has density function $\phi(x)$, $$M_X(it) = E(e^{itX}) = \int_{-\infty}^\infty e^{itx}\phi(x)dx,$$ which is the Fourier transform of $\phi(x)$. Therefore $\phi(x)$ can be recovered from its MGF using the Fourier inversion formula.

The function $M_X(it)$ is called the characteristic function of $X$. See Chapter 6 of Kai Lai Chung's book A Course in Probability Theory for more details.

• Thanks alot. So in order to understand the proof I must look into Fourier transformation theory ? Aug 3, 2013 at 12:19
• I guess Did's answer above works nicely for random variables on a finite probability space. Aug 5, 2013 at 10:40
• @AmritanshuPrasad Actually, for every nonnegative integer valued random variable.
– Did
Aug 6, 2013 at 19:27
• @Did Actually, you prove uniqueness restricted to that space. You don't rule out the possibility that other random variables happen to have the same MGF. Jul 28, 2017 at 3:45