# Is the min function ever an unbiased estimator for the mean?

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).

• What do you consider trivial? For instance, any distribution that's nonzero only on a set of measure zero will have this property. – TerronaBell Nov 13 '09 at 0:07
• @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). – Harald Hanche-Olsen Nov 13 '09 at 1:33
• 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. – Claudiu Nov 13 '09 at 2:02
• 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." – 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.