Mean or median for an exponential service time?

Question

Service times at a counter are i.i.d. Exponential with unknown mean $\theta$ . Since an exponential has median $m = \theta\ln 2$ , you could estimate $\theta$ either by the sample mean $\bar X$ or by rescaling the sample median, $\hat\theta_{\text{med}} = \tilde X / \ln 2$ .

Which estimator is more efficient, and by how much? Report the asymptotic relative efficiency of the median-based estimator versus the mean.

Accepted Answer

The sample mean. For an exponential, \operatorname{Var}(X) = 	heta^2, so \operatorname{Var}(\bar X) = \frac{	heta^2}{n}. The sample median. The general large-sample result for a sample median 	ilde X is \operatorname{Var}(	ilde X) \approx \frac{1}{4 n f(m)^2}, where f is the density and m the population median. For the exponential with rate \lambda = 1/	heta, the density at its own median is f(m) = \lambda e^{-\lambda m} = \frac{1}{	heta}e^{-\ln 2} = \frac{1}{2	heta}, so f(m)^2 = 1/(4	heta^2) and \operatorname{Var}(	ilde X) \approx \frac{1}{4n\cdot 1/(4	heta^2)} = \frac{	heta^2}{n}…

Mean or median for an exponential service time?

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