How to calculate the central moment giving function of a distribution

Question

Is there a function which gives the central moments instead of just moments of a distribution and if so how to calculate this function for a distribution e.g. the normal distribution.

Answer 1

For the moments, you may use the characteristic function: $$\varphi_X(t) = \mathbb{E}[e^{itX}] \tag{1}$$ that gives $$ \varphi_X^{(n)}(t) = \mathbb{E}[(iX)^n e^{itX}],\qquad \mathbb{E}[X^n] = i^{-n} \varphi_X^{(n)}(0).\tag{2} $$ The characteristic function of a normal $X=N(\mu,\sigma^2)$ variable is $e^{it\mu-\frac{1}{2}\sigma^2 t^2}$, hence in such a case:

$$ \mathbb{E}[(X-\mu)^n] = i^{-n}\frac{d^n}{dt^n}\left. e^{-\frac{1}{2}\sigma^2 t^2}\right|_{t=0}\tag{3} $$ and since: $$ e^{-\frac{\sigma^2}{2}t^2} = \sum_{n\geq 0}\frac{(-1)^n \sigma^{2n}}{2^n n!}t^{2n}\tag{4} $$ it follows that: $$ \mathbb{E}[(X-\mu)^{2m}]=\frac{\sigma^{2m}(2m)!}{2^m m!}.\tag{5}$$