Q:

Fisher Information for general one parameter exponential family (inconsistent with Poisson distribution)

For one of my hw questions, I was asked to derive Fisher Information for one parameter exponential family. 
Here's my approach:
$$L(\theta) = f(x\mid\theta) = e^{\theta T(x) - \eta(\theta)}h(x)$$
$$\log L = \theta T(x) - \eta(\theta)$$
$$\frac{\delta \log L}{\delta \theta} = T(x) - \eta '(\theta)$$
$$\frac{\delta^{2}\log L}{\delta^{2}\theta} = -\eta ''(\theta)$$
Then $$I(\theta) = -E(-\eta ''(\theta)) = E(\eta ''(\theta))$$
Assuming this is correct, I tried this on Poisson distribution where $f(x\mid\theta) = \frac{e^{-\theta}\theta^{x}}{x!}$ and we can rewrite this as $\frac{e^{-\theta}e^{\log(\theta)x}}{x!}$, then $T(x) = x, \eta(\theta) = \theta$, and $\log(\theta) = \theta$.
If I use $E(\eta ''(\theta))$ to calculate Fisher information for poisson, then the fisher information should be 0.  But if I directly calculate $I(\theta)$ I get $1/\theta$. Can someone explain to me what I did wrong to get inconsistent answers?  

A:

Assuming all regularity conditions, one-parameter exponential family is usually expressed in the form 
$$f(x\mid\theta)=e^{\eta(\theta) T(x)-A(\theta)}h(x)$$
This gives $$\frac{\partial^2}{\partial\theta^2}\ln f(x\mid\theta)=\eta''(\theta)T(x)-A''(\theta)$$
Fisher information is $$I(\theta)=E_{\theta}\left[\frac{\partial}{\partial\theta}\ln f(X\mid\theta)\right]^2=-E_{\theta}\left[\frac{\partial^2}{\partial\theta^2}\ln f(X\mid\theta)\right]$$
For $X\sim \mathsf{Poisson}(\theta)$, you have $\eta(\theta)=\ln\theta$, $A(\theta)=\theta$ and $T(x)=x$.
This is consistent with the direct calculation $I(\theta)=\frac{1}{\theta}$.

