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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?

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  • $\begingroup$ You don't have $n$ i.i.d. samples; rather you have an i.i.d. sample of size $n. \qquad$ $\endgroup$
    – Michael Hardy
    Commented Nov 16, 2018 at 19:37
  • $\begingroup$ 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. $\qquad$ $\endgroup$
    – Michael Hardy
    Commented Nov 16, 2018 at 19:48
  • $\begingroup$ @MichaelHardy Thank you! Do you know whether the expected value of the sample median is guaranteed to be zero in this case? $\endgroup$
    – Tanakak
    Commented Nov 17, 2018 at 14:33
  • $\begingroup$ Symmetry shows that the expected value of the median of an i.i.d. sample from $\dfrac 1 2 e^{-|x|} \, dx$ is $0. \qquad$ $\endgroup$
    – Michael Hardy
    Commented Nov 18, 2018 at 21:56

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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 $n\ge 6.$

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