When the median beats the mean: heavy tails

Question

You have n i.i.d. draws from a Laplace (double-exponential) distribution centered at \mu, with density f(x) = \frac{1}{2b}e^{-x-\mu/b}. You want to estimate the center \mu using either the sample mean \bar X or the sample median 	ilde X. Which is more efficient here, and by exactly how much?

Accepted Answer

Sample mean. The Laplace variance is \operatorname{Var}(X) = 2b^2, so \operatorname{Var}(\bar X) = \frac{2b^2}{n}. Sample median. The general large-sample variance of a median is \frac{1}{4nf(m)^2}, evaluated at the population median m = \mu. The Laplace density peaks there at f(\mu) = \frac{1}{2b}, so f(\mu)^2 = \frac{1}{4b^2} and \operatorname{Var}(	ilde X) \approx \frac{1}{4n\cdot \frac{1}{4b^2}} = \frac{b^2}{n}. Relative efficiency. The median's variance (b^2/n) is half the mean's (2b^2/n): 	ext{ARE of the mean vs the median} = \frac{\operatorname{Var}(	ilde X)}{\operatorname{Var}(\bar…

When the median beats the mean: heavy tails

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