Let $X\sim N(\mu ,\sigma^2)$ and g : $\Bbb R$ $\to$ $\Bbb R$ be a differentiable function and $E[|g'(X)|] < \infty$.
a) Show that $E[g'(X)] = \sigma^{-2}E[g(X)(X-\mu)]$
b) Make use of the above identity to derive the n-th moment of $X$, for $n\ge3$ (recursive form is enough).
My attempt:
a) $$g'(X)=\frac{-(X-\mu)}{\sigma^2}g(X)$$
and
$$E[g(X)(X-\mu)]=\int_{-\infty}^{\infty}{(x-\mu)}g(x)f_X(x)dx$$ $$= \int_{-\infty}^{\infty}-\sigma^2(\frac{-(x-\mu)}{\sigma^2}g(x))f_X(x)dx$$
then I put the $g'(X)$ in and get $E[g(X)(X-\mu)]=-\sigma^2E[g'(X)]$ which is wrong because I got a negative sign.
b) $n^{th}$ moment of $X$ is $E[X^n] = M_X^{(n)}(0)$ and $M_X=E[e^{tX}]$, but I can't think of a way to define the MGF.
You can use integration by part : \begin{align*} \mathbb E[g'(X)] &= \int_{-\infty}^{\infty} g'(x) \frac{1}{\sqrt{2\pi \sigma^2}} \exp\left( -\frac{(x-\mu)^2}{2\sigma^2} \right)\\ &= -\left[ g(x)\frac{1}{\sqrt{2\pi \sigma^2}} \exp\left( -\frac{(x-\mu)^2}{2\sigma^2} \right) \right]_{-\infty}^\infty + \int_{-\infty}^\infty g(x) \frac{x-\mu}{\sigma^2\sqrt{2\pi \sigma^2}} \exp\left( -\frac{(x-\mu)^2}{2\sigma^2} \right)\\ &= 0+\frac{1}{\sigma^2}\mathbb E[g(X) (X-\mu)] \end{align*} The first term goes to zero, otherwise $g'$ is not absolutely integrable.
For the n'th moment you can use the function $g(x)=(x-\mu)^{n-1}$ with derivative $(n-1) (x-\mu)^{n-2}$ to obtain $\mathbb E[(x-\mu)^n]=\sigma^2(n-1) \mathbb E[(x-\mu)^{n-2}]$