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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 $\min(X_1,\ldots,X_n)$ 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. $\endgroup$ Nov 13, 2009 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). $\endgroup$ Nov 13, 2009 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, 2009 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." $\endgroup$ Nov 13, 2009 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.

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Suppose $X_1,\ldots,X_n \sim \text{i.i.d.} \operatorname{Uniform}(\theta,0),$ where $\theta$ could be any negative number. Then $\min\{X_1,\ldots,X_n\}\cdot (n+1)/(2n)$ is a better unbiased estimator of the population mean $\theta/2$ than is the sample mean $(X_1+\cdots+X_n)/n,$ in the sense that it has a (much) smaller variance. In fact, it can be shown to have a smaller variance than all other unbiased estimators of the population mean.

In fact, it is better in another sense as well: There is positive probability that at least one of the observations is less than twice the sample mean. In such an event, it would be a certainty that the population mean $\theta/2$ would be less than the sample mean, so use of the sample mean as an estimator of the population mean would be clearly wrong.

Not exactly what you asked but it's a case in which using the sample minimum is a better way to estimate the population mean than is using any other statistic.

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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 functions 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 unbiased estimate of the mean. (I think $[(n+1)/2]\min(X_i)$ works for a $\mathrm{Uniform}(0,\theta)$ 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.) $\endgroup$ Jun 2, 2010 at 3:22
  • $\begingroup$ Hi Michael. That was a clear typo that I corrected with an edit $\endgroup$ Jun 29, 2010 at 16:04
  • $\begingroup$ Um corrected with an edit in response to your pointing our my mistake $\endgroup$ Jun 29, 2010 at 16:04
  • $\begingroup$ In this case you're not only throwing away relevant data if you use the sample minimum, but you're also throwing away relevant data if you use the sample mean. For example, there are samples in which some of the observations are more than twice the sample mean, whereas it is impossible for any observation to be more than twice the population mean. If such an observation were in the sample, then using the sample mean as an estimate of the population mean would be seriously mistaken. $\endgroup$ Jul 18, 2023 at 13:13

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