Find the Fisher information $I(\theta)$ of the gamma distribution with $\alpha=4$ and $\beta=\theta>0$ Let $X$ have a gamma distribution with $\alpha = 4$ and $\beta = \theta > 0$. Find the Fisher information $I(\theta)$. I have found the second derivative of the log of the likelihood function and then to find the information, I did:$E((-4/o^2)+(x/o^2))^2$. The answer is $\frac{4}{o^2}$ but I don't know how to get here.  
 A: I'm doing this to work through this myself as much as help you.  Lets give it a go.
PDF of a Gamma = $\frac{X^{\alpha-1}}{\Gamma(\alpha)\theta^{\alpha}}e^{\frac{X}{\theta}}$.  Log likelihood is then:
\begin{align}
L(\theta) &= (\alpha - 1) \Sigma \log X_i - n \log(\Gamma (\alpha)) - n\alpha \log(\theta) - \frac{1}{\theta} \Sigma X_i  \\[5pt]
\frac{\partial}{\partial \theta} &= -\frac{n\alpha}{\theta} + \frac{\Sigma X_i}{\theta^2}  \\[5pt]
\frac{\partial^2}{\partial \theta^2} &= \frac{n\alpha}{\theta^2} - \frac{2\Sigma X_i}{\theta^3}
\end{align} 
What is the expectation of a gamma dist? (looks like $\alpha \theta$)
\begin{align}
-E \frac{\partial^2}{\partial \theta^2} &= -\frac{n\alpha}{\theta^2} + \frac{2\alpha n}{\theta^2} \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad  \\[5pt]
 &=  \frac{n\alpha}{\theta^2}  \\[5pt]
\alpha &= 4 \text{ so;}  \\[5pt]
 &= \frac{4n}{\theta^2}
\end{align}
so if $n = 1$ (i.e., a single observation from a gamma distribution, like this problem seems to be asking), then in fact the answer is: 
$$ = \frac{4}{\theta^2}\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad$$
Feel free to correct/critique my errors.
