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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

$$L(\theta)=\pi^{-n}\prod_i^n \frac{1}{1+(x_i-\theta)^2}$$ $$\Rightarrow l(\theta)=-n\operatorname{log}\pi-\sum_i^n\operatorname{log}(1+(x_i-\theta)^2)$$ $$\Rightarrow l'(\theta)= 2\sum_i^n \frac{x_i-\theta}{1+(x_i-\theta)^2}$$ $$\Rightarrow -l''(\theta)=2\sum_i^n \frac{1-(x_i-\theta)^2}{(1+(x_i-\theta)^2)^2}$$

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    Backslash (\ ), not slash ( `/` ), for $\LaTeX$ commands.2012-02-06
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    @Arturo Magidin: sorry about that.. Wrote it on my phone as my Internet is down :(2012-02-06
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    Hint: $E_\theta(u(X_i-\theta))$ is independent on $\theta$. Validity check: $I(\theta)=n/\pi$.2012-02-06
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    @Didier Piau: thank you! But what do we do about the denominator concerning linearity of the expectation?2012-02-06
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    ??? Can you compute $I(\theta)$ when $n=1$?2012-02-06
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    @didier piau: No, that is basically what I'm unsure about2012-02-06
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    Why not? This is $E(u(X))$ and you know the function $u$ and the distribution of $X$. What is stopping you?2012-02-06
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    @Didier Piau: It won't integrate and I can't see errors in my working, is this the right function?2012-02-06
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    Can you show your workings which lead to this?2012-02-06
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    If it's not integrable, that's not surprising. If it is, then the sum of the $n$ terms is $n$ times the sum of any one of them, since they're obviously all equal.2012-02-06
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    @DidierPiau: Here are my working, sorry for the late reply, my internet is now up again.2012-02-07
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    Right. Now you must compute $E((1-(X-\theta)^2)/(1+(X-\theta)^2)^2)$ and you know that the distribution of $X$ has density $f(x)=1/(\pi(1+(x-\theta)^2))$. Thus you are after the expectation of a known function of a random variable with known density. Nothing to stop you here...2012-02-07
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    $E \big[ \frac{1-(x_i-\theta)^2}{(1+(x_i-\theta)^2)^2} \big]=\int_{-\infty}^\infty \frac{1-(x_i-\theta)^2}{\pi(1+(x_i-\theta)^2)^3} dx=\frac{1}{4}$ Hmm, ok, thanks!2012-02-08

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