Discussion
Suppose \(\{X_{1}, X_{2}, \ldots, X_{n}\}\) is a random sample from Cauchy \(C(\theta, 1)\) distribution. A test procedure for testing \(H_{0} : \theta = 0\) against the alternative \(H_{1} : \theta \neq 0\) is not consistent if the test statistic is based on :
Answer: B
A test based on the sample mean is not consistent.
For a Cauchy sample the sample mean \(\overline{X}_{n}\) has exactly the same \(C(\theta, 1)\) distribution as a single observation, whatever \(n\) is: the characteristic function of the mean is \((e^{i\theta t/n - |t|/n})^{n} = e^{i\theta t - |t|}\). So its distribution never concentrates, and a test based on it cannot have power tending to 1 against a fixed \(\theta \neq 0\).
Sample quantiles behave well for the Cauchy. The sample median converges in probability to \(\theta\), and the first and third quartiles converge to \(\theta - 1\) and \(\theta + 1\), so tests built on any of them are consistent. That is why the median is the standard location estimator for the Cauchy.
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