Using your Head is Permitted

November 2015 riddle

Note: This month, because the question is well-known, no Internet searching whatsoever, please.

Let X1,...,Xn be independent random variables, all distributed uniformly in [a, b] for some unknown a and b.

This month's question: find (with proof) the minimum variance unbiased estimator of the expectation of the n random variables. (Or prove it does not exist.)

Some explanations regarding the terminology:

  1. A function from the values of random variables X1,...,Xn that are drawn from a distribution characterised by (among others) some parameter, θ, is called an unbiased estimator of θ if for any setting of the distribution parameters its expectation over all X1,...,Xn equals θ. In our case, we are looking for a function, f, such that the expectation of f(X1,...,Xn) equals (a+b)/2.
  2. An unbiased estimator of θ is called a minimum variance unbiased estimator if for every setting of the distribution parameters the value of the estimator (which, in itself is a random variable) has the lowest variance, as compared to any other unbiased estimator. Note that if there is no single estimator that achieves this for the entire range of all possible parameters, then the minimum variance unbiased estimator does not exist. There are many cases where such estimators do not exist.
For simplicity, you may restrict yourself to considering only estimators that are continuous functions of X1,...,Xn.

List of solvers:

Radu-Alexandru Todor (8 November 11:23)
Jan Fricke (22 November 01:47)
Bart De Vylder (24 November 19:40)

Elegant and original solutions can be submitted to the puzzlemaster at Names of solvers will be posted on this page. Notify if you don't want your name to be mentioned.

The solution will be published at the end of the month.


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