Discussion
Suppose \(\{X_{1}, X_{2}, \ldots, X_{n}\}\) is a random sample from normal \(N(\mu, \sigma^{2})\) distribution. Suppose \(M_{n}\) is the sample median. Which of the following is true ?
Answer: B
\(\sqrt{n}(M_{n} - \mu) \xrightarrow{d} N\left(0, \frac{\pi\sigma^{2}}{2}\right)\).
For a sample median from a density \(f\) with median \(m\), \(\sqrt{n}(M_{n} - m) \to N\left(0, \frac{1}{4f(m)^{2}}\right)\). For \(N(\mu, \sigma^{2})\), \(m = \mu\) and \(f(\mu) = \frac{1}{\sigma\sqrt{2\pi}}\), so
\[ \begin{aligned} \frac{1}{4f(\mu)^{2}} &= \frac{2\pi\sigma^{2}}{4} \\ &= \frac{\pi\sigma^{2}}{2} \end{aligned} \]
The asymptotic variance \(\pi\sigma^{2}/2 \approx 1.57\sigma^{2}\) is larger than the \(\sigma^{2}\) of the sample mean, which is why the median has asymptotic relative efficiency \(2/\pi \approx 0.64\) at the normal. \(2\sigma^{2}\) is not the right constant, \(\pi/(2\sigma^{2})\) inverts \(\sigma^{2}\), and \(\sigma^{2}/(2\pi)\) inverts the \(\pi\).
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