# Uniform convergence of Empirical Moment Generating Function

In the article, "The Empirical Moment Generating Function" by Csörgö, the author defines the empirical moment generating function for a sample of $n$ variables $X_1,X_2, \dots, X_n$ as: $$$$M_n (t) = \frac{1}{n}\sum\limits_{i = 1}^{n} e^{t X_i} = \int\limits_{-\infty}^{\infty} e^{tx} dF_n (x),$$$$ where $F_n (x)$ denotes the empirical distribution function. He then states, without proof, the following proposition:

Let $M(t)$ be the moment generating function of $X$ and assume that $M$ is defined for all $t$ in a non-degenerate interval $J$ then: $$\sup_{t \hspace{1mm} \in \hspace{1mm} J} | M_n (t) - M(t) | \to 0, \quad \textrm{as } \quad n \to \infty.$$

He establishes that the proof can be done by noticing that: $$M_n(t) - M(t) = \int\limits_{-\infty}^{\infty} e^{tx} d\big(F_n (x) - F(x)\big),$$ and then dividing the integral "into two parts $\int\limits_{|x| > A} + \int\limits_{|x| \leq A}$ and making use of the Glivenko–Cantelli theorem and another classical result, Dini's theorem" ($A$ is never specified in the article).

I am not interested in other proofs of this (such as the one in p. 459 here that uses convexity) but in understanding how Csörgö did it. Could someone please provide a more detailed guideline as how Csörgö's proof would go?

Thanks!

• This is a really intriguing question. Csörgö has given some indication of the proof, but not quite enough detail to understand what he was actually thinking of (perhaps, to Csörgö, it seemed too trivial to write down?). If $X$ is bounded then I can see how to use Glivenko-Cantelli to prove the result. Likewise, if the integrand is bounded (as for characteristic functions) then life is easy. Dec 1, 2015 at 20:55
• For unbounded $X$, I can see how Dini's theorem might be used to show that $E[e^{Xt}1_{|X|>A}]$ is small, uniformly in $t$, for suitably large $A$. But not sure how to handle $\int\limits_{|x| > A}$ for the empirical mgf. Dec 1, 2015 at 20:56
• I'm not sure if this helps; but If $A$ is greater than the maximum of the sample (i.e. $A > max \left \{ \left| X_i \right| : i = 1, \dots, n \right \}$ ) then $F_n(x) = 0$ if $x < A$ and $F_n(x) = 1$ if $x > A$. In addition, for all $\epsilon$ there exists an $A$ such that for all $x > A$, $F(x)$ is pretty close to $1$: $\left| F(x) - 1 \right| <\epsilon$ and for all $x < A$, $F(x)$ is almost $0$: $\left| F(x) - 0 \right| < \epsilon$. The difference $\left| F(x) - F_n(x) \right|$ is therefore $< 2\epsilon$ for $\left| x \right| > A$. Dec 1, 2015 at 22:42
• I can't see how this helps, but maybe I've missed something. If we ask that $A > max \left \{ \left| X_i \right| : i = 1, \dots, n \right \}$ then $A$ will potentially increase without bound as $n\to\infty$, which would (I think) prevent useful application of Glivenko Cantelli theorem to the $\int\limits_{|x| <A}$. Maybe you could ask this question on mathoverflow if you don't get a better response on here? Dec 2, 2015 at 22:21
• Thanks, I have now made the question in mathoverflow. Dec 3, 2015 at 1:50