Estimating the top of a uniform: max or twice the mean?

Question

You have n i.i.d. draws from U(0, 	heta) and want to estimate the unknown upper endpoint 	heta. Two candidates: the method-of-moments estimator \hat	heta{MM} = 2\bar X (since E[X] = 	heta/2), and the (bias-corrected) maximum \hat	heta{\max} = \frac{n+1}{n}\maxi Xi. Which is more efficient, and by how much? Notice anything unusual about the convergence rate.

Accepted Answer

Twice the mean. For U(0,	heta), \operatorname{Var}(X) = 	heta^2/12, so \operatorname{Var}(\hat	heta{MM}) = 4\,\operatorname{Var}(\bar X) = 4\cdot\frac{	heta^2/12}{n} = \frac{	heta^2}{3n}. This shrinks like 1/n, the usual rate. The corrected maximum. The raw maximum satisfies E[\max] = \frac{n}{n+1}	heta (biased low), and multiplying by \frac{n+1}{n} removes the bias. Its variance is \operatorname{Var}(\hat	heta{\max}) = \Big(\frac{n+1}{n}\Big)^2 \cdot \frac{n}{(n+1)^2(n+2)}	heta^2 = \frac{	heta^2}{n(n+2)} \approx \frac{	heta^2}{n^2}. This shrinks like 1/n^2, dramatically faster. Com…

Estimating the top of a uniform: max or twice the mean?

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