Sample mean or sample median, which is more efficient?

Question

You have n i.i.d. draws from N(\mu, \sigma^2) and want to estimate the center \mu. Two candidates: the sample mean \bar X and the sample median 	ilde X. Which is more efficient, and by exactly how much? When would you switch to the median?

Accepted Answer

Sample mean. \operatorname{Var}(\bar X) = \sigma^2/n. For the normal, \bar X is the MLE and attains the Cramér–Rao bound, it is fully efficient. Sample median. The general large-sample distribution of a sample median is \sqrt{n}\,(	ilde X - m) \ \xrightarrow{d}\ N\!\left(0,\ \frac{1}{4\,f(m)^2}ight). For N(\mu,\sigma^2) the median equals the mean, m = \mu, and the density there is f(\mu) = \frac{1}{\sigma\sqrt{2\pi}} \quad\Rightarrow\quad f(\mu)^2 = \frac{1}{2\pi\sigma^2}. Hence \operatorname{Var}(	ilde X) \approx \frac{1}{n}\cdot\frac{1}{4 f(\mu)^2} = \frac{1}{n}\cdot\frac{2\pi\sigma^2}{…

Sample mean or sample median, which is more efficient?

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