The estimator whose efficiency is exactly zero

Question

Measurements are i.i.d. from a Cauchy distribution centered at $\mu$ with scale $\gamma$ , whose density is $f(x) = \frac{1}{\pi\gamma}\cdot\frac{1}{1 + ((x-\mu)/\gamma)^2}$ . You consider estimating the center $\mu$ with the sample mean $\bar X$ .

What is the asymptotic efficiency of the sample mean here, and what should you use instead?

Accepted Answer

The mean does not even converge. The Cauchy distribution has no finite mean or variance (the tails are too heavy for the defining integrals to converge). A striking fact: the average of n i.i.d. Cauchy variables is itself Cauchy with the same scale, for every n: \bar X \sim 	ext{Cauchy}(\mu, \gamma) \quad	ext{for all } n. Averaging never sharpens the distribution. So \bar X does not converge in probability to \mu, it is not even consistent, and its variance is infinite. Its efficiency is 0. The median works fine. The sample median is consistent and asymptotically normal. Using \operatorname…

The estimator whose efficiency is exactly zero

Your answer

Discussion

Learn the concepts

Related questions