When the median beats the mean for the center

Question

Sensor errors follow a Laplace (double-exponential) distribution centered at $\mu$ with scale $b$ , so the density is $f(x) = \frac{1}{2b}e^{-|x-\mu|/b}$ . You estimate the center $\mu$ either with the sample mean $\bar X$ or the sample median $\tilde X$ .

Which is more efficient here, and what is the asymptotic relative efficiency of the sample mean versus the median?

Accepted Answer

The sample mean. A Laplace variable has variance \operatorname{Var}(X) = 2b^2, so \operatorname{Var}(\bar X) = \frac{2b^2}{n}. The sample median. Use \operatorname{Var}(	ilde X) \approx \frac{1}{4n f(m)^2}, where the median equals the center m = \mu. The density there is f(\mu) = \frac{1}{2b}e^{0} = \frac{1}{2b}, \qquad f(\mu)^2 = \frac{1}{4b^2}, so \operatorname{Var}(	ilde X) \approx \frac{1}{4n\cdot 1/(4b^2)} = \frac{b^2}{n}. Compare. The median is the more efficient estimator here (it is in fact the maximum likelihood estimator for Laplace data). The efficiency of the mean relative to th…

When the median beats the mean for the center

Your answer

Discussion

Learn the concepts

Related questions