Cramer Rao lower bound in Cauchy distribution I need to calculate the Cramer Rao lower bound of variance for the parameter $\theta$ of the distribution $$f(x)=\frac{1}{\pi(1+(x-\theta)^2)}$$


How do I proceed I have calculated $$4 E\frac{(X-\theta)^2}{1+X^2+\theta^2-2X\theta}$$

Can somebody help
Suppose $X\sim C(\theta,1)$, a Cauchy distribution with location $\theta$ and scale unity.

For $\theta\in\mathbb R$, the pdf of $X$ is 

$$f_{\theta}(x)=\frac{1}{\pi(1+(x-\theta)^2)}\qquad,\,x\in\mathbb R$$

Clearly, $$\frac{\partial}{\partial\theta}\ln f_{\theta}(x)=-\frac{2(x-\theta)}{1+(x-\theta)^2}$$

Therefore,

$$E_{\theta}\left[\frac{\partial}{\partial\theta}\ln f_{\theta}(X)\right]^2=4E_{\theta}\left[\frac{X-\theta}{1+(X-\theta)^2}\right]^2\tag{1}$$

Now for every $\theta$,

\begin{align}
E_{\theta}\left[\frac{X-\theta}{1+(X-\theta)^2}\right]^2
&amp;=\frac{1}{\pi}\int_{\mathbb R}\left[\frac{x-\theta}{1+(x-\theta)^2}\right]^2\frac{1}{1+(x-\theta)^2}\,\mathrm{d}x
\\&amp;=\frac{1}{\pi}\int_{\mathbb R}\frac{(x-\theta)^2}{(1+(x-\theta)^2)^3}\,\mathrm{d}x\\&amp;=\frac{2}{\pi}\int_0^\infty\frac{t^2}{(1+t^2)^3}\,\mathrm{d}t
\\&amp;=\frac{1}{\pi}\int_0^\infty\frac{\sqrt u}{(1+u)^3}\,\mathrm{d}u
\\&amp;=\frac{1}{\pi}B\left(\frac{3}{2},\frac{3}{2}\right)
\\&amp;=\frac{1}{8}
\end{align}

So from $(1)$, we have the Fisher information

$$I(\theta)=E_{\theta}\left[\frac{\partial}{\partial\theta}\ln f_{\theta}(X)\right]^2=\frac{1}{2}\quad,\forall\,\theta$$

And Cramer-Rao lower bound for $\theta$ is $$\text{CRLB}(\theta)=\frac1{I(\theta)}=2 \quad,\forall\,\theta$$

In case $X_1,X_2,\ldots,X_n$ are i.i.d with pdf $f_{\theta}$, Fisher information in $\mathbf X=(X_1,\ldots,X_n)$ is

$$ I_{\mathbf X}(\theta)=n I(\theta)=\frac n2 \quad,\forall\,\theta$$

The Cramer-Rao bound for $\theta$ is then $\frac2n$ for every $\theta$.
First of all you should notice that there's no suficient estimator for the center of the bell $\theta$. Let's see this. 
The likelihood for the Cauchy distribution is
$$L(x;\theta) = \prod _i^n \frac{1}{\pi\left [ 1+(x_i-\theta)^2 \right ]}, $$
and its logarithm is
$$\ln L(x;\theta) = -n \ln \pi -\sum_i^n\ln\left [ 1+(x_i-\theta)^2 \right ].$$
The estimator will maximize the likelihood and if there's a suficient estimator it's derivate can be factoriced i.e.:
$$\frac{\partial L(x;\theta)}{\partial \theta} = A(\theta)\left[t(x)-h(\theta)-b(\theta)\right], $$
where $A(\theta)$ is a function exclusively from the parameter, $t(x)$ is function exclusively of your data, $h(\theta$) is what you want to estimate and $b(\theta)$ is a possible bias.

If you derivate you should notice that the Cauchy distribution can not be factorized, but Cramer-Rao lets you find a bound for the variance, that is 
$$ Var(t) \geq \frac{\left(\frac{\partial}{\partial \theta}(h+b)\right)^2}{E\left[(\frac{\partial}{\partial \theta}\ln L)^2\right]},$$
where the equality hold only if $\frac{\partial \ln L}{\partial \theta}$ can be factorized.

The most you can do is calculate the bound but it has no analytic closed solution for $\theta$