Given $n$ i.i.d. variables $X_1$ to $X_n$ with an unknown probability distribution, the sample average is an unbiased estimator for the mean of the distribution. Is there some non-trivial probability distribution for which min($X_1$,...,$X_n$) is an unbiased estimator? (Non-trivial meaning the variables can have more than one potential value).
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$\begingroup$ What do you consider trivial? For instance, any distribution that's nonzero only on a set of measure zero will have this property. $\endgroup$ – TerronaBell Nov 13 '09 at 0:07
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$\begingroup$ @fuzzytron: Your comment makes little sense to me. If you want to use the language of distributions, clearly the intended meaning is a distribution (i.e., a probability measure on $\mathbb{R}$) whose support is not a point. Anyway, the question seems too easy to me, it looks almost like homework (hint: compare the two suggested estimators). $\endgroup$ – Harald Hanche-Olsen Nov 13 '09 at 1:33
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$\begingroup$ The homework was to find out whether the min is biased for an exponential distribution. It was, of course. This is me being curious if there is a distribution where it's not biased. $\endgroup$ – Claudiu Nov 13 '09 at 2:02
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$\begingroup$ Sorry - was far too sloppy there. Consider a distribution on the unit interval that is one everywhere except for on a set of measure zero. This (probability) distribution satisfies the criteria above, except that you may consider it "trivial." $\endgroup$ – TerronaBell Nov 13 '09 at 15:32
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No. The minimum as always smaller than or equal to the arithmetic mean, and is strictly smaller with positive probability (i.e., when not all the $X_i$ have the same value). Hence its expected value is strictly smaller than that of the mean.
answered Nov 13 '09 at 4:09
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Not unless n=1 (sorry couldn't resist). Not sure why you're asking this but there do exist f(n,min(X_i)) that work for given distributions. (That is funtions of n and min(X_i) that work). So given only the mean (edit meant min here) and a parametric form of a distribution you can get an unbiassed estimate of the mean. (I think [(n+1)/2]*min(X_i) works for a Uniform(0,theta) for example.
Of course these are going to be much worse (higher variance) estimators than the arithmetic mean because you've thrown away information (the other data).
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$\begingroup$ Jonathan: I find your second paragraph somewhat vague. For the family of normal distributions, it is demonstrable that the minimum-variance unbiased estimator of the population mean is the sample mean. (The proof actually relies on the one-to-one nature of the two-sided Laplace transform.) $\endgroup$ – Michael Hardy Jun 2 '10 at 3:22
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$\begingroup$ Hi Michael. That was a clear typo that I corrected with an edit $\endgroup$ – Jonathan Kariv Jun 29 '10 at 16:04
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$\begingroup$ Um corrected with an edit in response to your pointing our my mistake $\endgroup$ – Jonathan Kariv Jun 29 '10 at 16:04
