Proving $E_{\theta}[T(X)] = \frac{\psi'(\theta)}{\eta'(\theta)}$

Question

I'm trying to understand how to prove the following theorem:

Let $\{P_{\theta}, \theta \in \Theta\}$ be a family of distributions in the one parameter exponential family with density (pmf)

$\hspace{15mm} f(x|\theta) = e^{\eta(\theta)T(x) - \psi(\theta)}h(x)$.

Then, at any $\theta$ at which $\eta'(\theta) \neq $0,

$\hspace{15mm}E_{\theta}[T(X)] = \frac{\psi'(\theta)}{\eta'(\theta)}$

and

$\hspace{15mm}\operatorname{Var}_{\theta}[T(X)] = \frac{\psi''(\theta)}{[\eta'(\theta)]^2}-\frac{\psi'(\theta)\eta''(\theta)}{[\eta'(\theta)]^3}$.

I know this is incorrect, but I believe this involves taking the natural log and manipulating the individual parts of the equation:

$$\ln (\exp\{\eta(\theta)T(x) - \psi(\theta)\} \rightarrow T(x) = \frac{\psi(\theta)}{\eta(\theta)}.$$

However, I'm not quite sure why the derivatives are there.

Answer 1

$\ln[f(x|\theta)] = \eta(\theta)T(X) - \psi(\theta) + \ln h(X)$

$E_\theta[T(X)]= \frac{d}{d(\theta)}\ln[f(x|\theta)] = \eta'(\theta)T(X) - \psi'(\theta)$ = 0

$\eta'(\theta)T(X) = \psi'(\theta)$

$T(X) = \frac{\psi'(\theta)}{\eta'(\theta)}$