I'm working in a self study fashion preparing for a course I'm going to take this semester in generalised linear models. The question is, given that the Y random variable belongs to the exponential family, show that: $$ E\left(\frac{\partial L}{\partial \theta}\right) = 0 $$
$$ E\left(\frac{\partial^2 L}{\partial \theta^2}\right) = -E\left(\left(\frac{\partial L}{\partial \theta}\right)^2\right) $$
I'm a bit rusty in this type of exercise, but this is what I've managed so far.
For the first part, it is easy to differentiate $L(\theta)$, where $L$ is the log likelihood. The exact parametrization of the exponential family I'm using is (treating $\phi$ as known) the following:
$$ f(y; \theta, \phi) = \exp[\phi(y\theta - b(\theta)) + c(y;\phi)] $$
And $Y$ is the random variable distributed by $f$.
I can arrive at $\frac{\partial L}{\partial \theta} = \phi y - \phi b'(\theta)$ (the functions $b$ and $c$ are differentiable). However, in order to conclude that $E(\frac{\partial L}{\partial \theta}) = 0$ I need to assume that $b'(\theta) = E(Y) = \mu$ so that I can use the properties of expectation an eliminate it altogether. And it feels like I'm cheating, since I don't have this assumption in the first place.
Calculating $E(Y) = \int_{\mathbb{R}}yf(y)dy$ just doesn't work out nicely.
The second part also culminates in me having to calculate $E(b''(\theta))$ in the same fashion.
In McCullagh and Nelder's book [1], they say the the relations $E(\frac{\partial L}{\partial \theta}) = 0$ and $E(\frac{\partial^2 L}{\partial \theta^2}) = -E((\frac{\partial L}{\partial \theta})^2)$ are well known (p. 28) and use it to establish $E(Y)$, so the result I'm trying to prove apparently precedes the $E(Y)$ calculation.
1: Generalized Linear Models, 2nd edition P. McCullagh and. J.A. Nelder (1989)