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 nontrivial probability distribution for which min($X_1$,...,$X_n$) is an unbiased estimator? (Nontrivial meaning the variables can have more than one potential value).

$\begingroup$ 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

$\begingroup$ @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 HancheOlsen Nov 13 '09 at 1:33

$\begingroup$ 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

$\begingroup$ 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
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.
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).

$\begingroup$ Jonathan: I find your second paragraph somewhat vague. For the family of normal distributions, it is demonstrable that the minimumvariance unbiased estimator of the population mean is the sample mean. (The proof actually relies on the onetoone nature of the twosided Laplace transform.) $\endgroup$ – Michael Hardy Jun 2 '10 at 3:22

$\begingroup$ Hi Michael. That was a clear typo that I corrected with an edit $\endgroup$ – Jonathan Kariv Jun 29 '10 at 16:04

$\begingroup$ Um corrected with an edit in response to your pointing our my mistake $\endgroup$ – Jonathan Kariv Jun 29 '10 at 16:04