Suppose $X_1,\ldots,X_n$ are iid with $f(x,\theta)=\frac{1}{\pi[1+(x-\theta)^2]}$ With $x$ real.
I'm trying to find the Fisher information, $I(\theta)$. And a method for finding a 95% confidence interval, can anyone help with this? It would be greatly appreciated!
So far I have found that $I(\theta)=2\sum_i^n E \left[ \frac{1-(x_i-\theta)^2}{(1+(x_i-\theta)^2)^2} \right]$
EDIT:
Here are my workings
$L(\theta)=\pi^{-n}\prod_i^n \frac{1}{1+(x_i-\theta)^2}$ $\Rightarrow l(\theta)=-n\operatorname{log}\pi-\sum_i^n\operatorname{log}(1+(x_i-\theta)^2)$ \Rightarrow l'(\theta)= 2\sum_i^n \frac{x_i-\theta}{1+(x_i-\theta)^2} \Rightarrow -l''(\theta)=2\sum_i^n \frac{1-(x_i-\theta)^2}{(1+(x_i-\theta)^2)^2}