Cramer Rao lower bound in Cauchy distribution

Question

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

Answer 1

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 $\theta$