I am doing an exercise in the book "Applied Stochastic Analysis" by E-Li-Vanden-Eijnden, and I meet this problem (on Page 26):
(Exercise 1.19) Prove that if the moment generating function $M_X(t)$ can be defined on an open set $U$, then $M_X(t)\in C^{\infty}(U)$.
Here is my approach:
Use Taylor's expansion at $t=t_0$, $$\frac{1}{t}\mathbb{E}[e^{(t_0+t)X}-e^{t_0X}]=\mathbb{E}[e^{t_0X}X]+o(t)\mathbb{E}[e^{t_0X}]$$ Let $t\to 0$, $$\frac{d}{dt}M_X(0)=\mathbb{E}[e^{t_0X}X]$$ Repeat differentiation $n$ times, $$\frac{d^n}{dt^n}M_X(0)=\mathbb{E}[e^{t_0X}X^n]$$ Since $n$ is arbitrary, as $n\to\infty$, we prove the claim.
Can anyone help me check whether my approach is valid? I'm not quite confident, since the exercise allows only on open sets. Alternative approaches are also fine. Thank you.
Edit: OKay, I think I figure this out. Choose $U=(-\epsilon,\epsilon)$. For $|t|<\epsilon$, $$\frac{|t|^n|X|^n}{n!}\leq e^{t|X|}\leq e^{tX}+e^{-tX}<\infty$$ and since $$\left|\sum_{k=0}^{n}\frac{(tX)^k}{k!}\right|\leq\sum_{k=0}^{n}\frac{|tX|^k}{k!}\leq e^{|tX|}\leq e^{tX}+e^{-tX}$$ By dominated convergence theorem, $$M_X(t)=\mathbb{E}\left[\sum_{k=0}^{\infty}\frac{(tX)^k}{k!}\right]=\sum_{k=0}^{\infty}\frac{t^k}{k!}\mathbb{E}[X^k]$$ $M_X(t)$ coincides with the power series on $U=(-\epsilon,\epsilon)$, and hence it is infinitely differentiable on $U$, and $M_X^{(k)}(0)=\mathbb{E}[X^k]$.
Answer for the case $U=(-r,r)$: First note that $Ee^{t|X|} \leq E(e^{tX}+e^{-tX}) <\infty$ for $0< t <r$. Let us show that $M_X$ is differentiable. Using similar argumnets and induction we can show that it is infinitely differentiable. Let $|t|<r$ and choose $u$ with $|t|<u<r$. Consider $M_X'(t)=\lim_{h \to 0} E{e^{tX}}\frac {e^{hX-1}} h$. By MVT we get $ |{e^{tX}}\frac {e^{hX-1}} h| \leq |X| e^{u|X|}$ provided $|h| <u-|t|$. If we show that $E|X|e^{u|X|} <\infty $ it follows by DCT that $M_X$ is differentiable are $t$ and $M_X'(t)=EXe^{tX}$. Now pick $v$ such that $u <v<r$ and note that there exists a finite constant $C$ such that $|x|e^{u|x|} \leq Ce^{v|x|}$ for all real numbers $x$. It follows that $E|X|e^{u|X|} <\infty $ so we are done.