1
$\begingroup$

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}

  • 0
    $E \big[ \frac{1-(x_i-\theta)^2}{(1+(x_i-\theta)^2)^2} \big]=\int_{-\infty}^\infty \frac{1-(x_i-\theta)^2}{\pi(1+(x_i-\theta)^2)^3} dx=\frac{1}{4}$ Hmm, ok, thanks!2012-02-08

0 Answers 0