Question

Given the pdf $$f(x ; \theta)=\frac{1}{\pi\left[1+(x-\theta)^{2}\right]}, \quad-\infty

Short answer

The Rao-Cramér lower bound is \(2 / n\). Asymptotically as \(n\) goes to infinity, the distribution of \(\sqrt{n}(\widehat{\theta}-\theta)\) becomes standard normal \(N(0,1)\).

Step-by-step solution

Calculation of Fisher Information

From the given p.d.f., we first compute the first derivative of the log-likelihood \(\log f(x ; \theta)\) w.r.t. \(\theta\). Then calculate the Fisher Information by taking expected value of its square.

Find the Rao-Cramér Lower Bound

Use the formula for Rao-Cramér Lower Bound which is the reciprocal of the Fisher Information. We compute this by substituting the calculated Fisher Information into this formula.

Find the MLE

To compute the MLE, we first write down the likelihood function, which is the product of the density functions. Then take a log and differentiate w.r.t. \(\theta\). Equate this to zero and solve for \(\theta\). The result is \(\widehat{\theta}\).

Determine the Asymptotic Distribution

We can use standard results regarding the asymptotic properties of the MLE, which state that as \(n\) approaches infinity, \(\sqrt{n}(\widehat{\theta}-\theta)\) converges in distribution to a normal distribution with mean 0 and variance equal to the reciprocal of the Fisher Information.