most recent 30 from http://mathoverflow.net 2013-05-25T11:53:41Z http://mathoverflow.net/feeds/question/5282 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/5282/is-the-min-function-ever-an-unbiased-estimator-for-the-mean Claudiu 2009-11-12T23:37:29Z 2010-06-29T16:03:22Z

<p>Given n iid variables X1 to Xn 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(X1,...,Xn) is an unbiased estimator? (Non-trivial meaning the variables can have more than one potential value).</p>

http://mathoverflow.net/questions/5282/is-the-min-function-ever-an-unbiased-estimator-for-the-mean/5311#5311 Harald Hanche-Olsen 2009-11-13T04:09:05Z 2009-11-13T04:09:05Z

<p>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.</p>

http://mathoverflow.net/questions/5282/is-the-min-function-ever-an-unbiased-estimator-for-the-mean/5314#5314 Jonathan Kariv 2009-11-13T06:14:47Z 2010-06-29T16:03:22Z

<p>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 <em>mean</em> (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.</p> <p>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).</p>