What does it mean for a moment generating function to exist in a neighborhood of 0?

Question

It is my understanding that a moment generating function (MGF) uniquely determines a distribution with bounded or unbounded support only if the "MGF exists in a neighborhood of zero." But what does it mean exactly for an MGF to exist in a neighborhood of zero?

Answer 1

Adding to Bey's answer, there's a reason you might care about this. The idea is that the MGF is a Laplace transform, and in this case it requires that your (continuous) probability density $f(x)$ decreases at least exponentially fast for large $x$, i.e. $e^{tx}f(x)\rightarrow 0$ for $x\rightarrow\infty$. This can be somewhat weakened but the main idea survives.

Anyways, it's usually the case that if $t$ is too large, this becomes false. So for example if $f(x)=2e^{-2x}$, then the MGF exists (i.e. is finite), for $t\in[0,2)$. As long as $f(t)$ is a density, everything is fine for $t<0$ but it turns out $t>0$ contains a wealth more of information. In general, saying that the MGF exists in a neighborhood of $0$ means that there is some $\epsilon>0$ such that your MGF is finite for all $t\in[0,\epsilon)$. Once your MGF exists, by abstract nonsense it corresponds to a unique distribution (your $f(x)$) and you can exploit all of it's nice properties, for example use it to bound probabilities. In similar vein to characteristic functions (i.e. fourier transforms), the regularity of your MGF near $t=0$ is intimately connected to the rate of decay of your density $f(x)$ as $x\rightarrow\infty$, an example of which you can see in the last link.

Perhaps more familiar to you, derivatives of the MGF, evaluated at $t=0$ give you back the moments of your distribution, so perhaps you can believe why you really only need to know what your $MGF$ looks likes near $t=0$ to extract almost everything about your random variable.

Answer 2

The MGF of a random variable $X$ is given by:

$$\mathrm{MGF}_X = \mathbb{E}[e^{tX}]$$

Where it is a function of $t$.

As you can see, the MGF at $t=0$ is always $1$, even if $X$ is, say, a Cauchy random variable, which has no moments. So, basically, when mathematicians say "a neighborhood", they mean:

$$\exists \epsilon>0: \forall t \in (-\epsilon,\epsilon),\;\;\mathbb{E}[e^{tX}]< \infty$$

Note, it just needs to be some positive constant.

Answer 3

I thought I'd chime in with an example that illustrates when we might have to worry about this.

Suppose $X \sim Exp(\lambda)$ so that the probability density function is: $$f(x\vert \lambda) = \lambda e^{-\lambda x}$$

Then the moment generating function is: $$M_X(t) = E(e^{Xt}) = \int_0^\infty e^{xt} \lambda e^{-\lambda x} dx = \lambda \int_0^\infty e^{x(t - \lambda)}dx$$

Notice that if $t - \lambda \geq 0$, then this integral diverges. $$M_X(t) = \lambda\int_0^\infty e^{x(t - \lambda)}dx = \begin{cases} \infty, & t \geq \lambda \\ \frac{\lambda}{\lambda - t}, & t < \lambda \end{cases}$$

In otherwords, $M_X(t)$ is finite only when $t < \lambda$. But since $\lambda$ is strictly positive, there exists a neighborhood $N_\lambda(0)$ for which the MGF exists and we can use it in the usual way.