TITLE: Derivative of moment generating function
QUESTION [10 upvotes]: If the moment generating function of $X$ exists, i.e., $$M_X(t)=E[e^{tX}],$$ then the derivative with respect to $t$ is usually taken as $$\frac{dM_X(t)}{dt}=E[Xe^{tX}].$$ Usually, if we want to change the order of derivative and calculus, there are some conditions need to verified. Why the derivative goes inside for the moment generating function?

REPLY [2 votes]: This is a general result for power series. For the power series 
$$
g(x) = \sum_{n=0}^{\infty} a_n (x-b)^n
$$
with radius of convergence $R>0$, then for any $x \in (b-R, b+R)$, the derivative of $g$ exists and is given by
$$
g'(x) = \sum_{n=1}^{\infty} n a_n (x-b)^{n-1}.
$$
You can use this fact along with the result that for any real valued random variable $X$ with mgf $m_X(t)$, $|t|< \delta$. Then $\mathbb{E}X^n$ exists and is finite for all $n \in \mathbb{N}$, and 
$$
m_X(t) = \sum_{n=0}^{\infty} \frac{t^n \mathbb{E} X^n}{n!}, \quad |t| < \delta. 
$$
This is Theorem 4.8 in the text Elements of Distribution Theory by Severini