# Finding a UMVU estimator for the mean

I am struggling with the following problem.

We are given an i.i.d sample of size $$n,$$ with the form $$X_{i}=\mu+n_{i}$$, where $$\mu$$ is a deterministic unknown constant, and $$n_{i}$$ is a noise with a known distribution and mean $$0.$$

The purpose is to find a noise distribution, for which the UMVU estimator for the mean $$\hat{\mu}$$ dominates the sample average estimator: (for finite $$n$$)

$$\forall \mu:\operatorname{Var}\left(\hat{\mu}\right) \leq \operatorname{Var} \left(\frac{1}{n}\sum_{i=1}^n X_i\right).$$

Does anybody know a noise distribution for which it holds?

• You don't have $n$ i.i.d. samples; rather you have an i.i.d. sample of size $n. \qquad$ – Michael Hardy Nov 16 '18 at 19:37
• I might think about $\displaystyle \frac 1 2 e^{-|x|} \, dx.$ My suspicion is that the sample median is the UMVU, but I'm not sure. $\qquad$ – Michael Hardy Nov 16 '18 at 19:48
• @MichaelHardy Thank you! Do you know whether the expected value of the sample median is guaranteed to be zero in this case? – Or Raveh Nov 17 '18 at 14:33
• Symmetry shows that the expected value of the median of an i.i.d. sample from $\dfrac 1 2 e^{-|x|} \, dx$ is $0. \qquad$ – Michael Hardy Nov 18 '18 at 21:56

Suppose $$n_i$$ is uniformly distributed on the interval $$[-1,1].$$ Then \begin{align} & \operatorname{var}\left( \frac{n_1+\cdots+n_n} n \right) = \frac 1 {3n} \\[10pt] \text{and} \quad & \operatorname{var} \left( \frac{\max\{n_1,\ldots,n_n\} + \min\{n_1,\ldots+n_n\}} 2 \right) = \frac{4n}{(n+1)(n+2)^2} \end{align} and the former exceeds the latter when $$n\ge 6.$$