2
$\begingroup$

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?

$\endgroup$
4
  • $\begingroup$ You don't have $n$ i.i.d. samples; rather you have an i.i.d. sample of size $n. \qquad$ $\endgroup$ 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. $\qquad$ $\endgroup$ 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
    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 $0. \qquad$ $\endgroup$ Nov 18, 2018 at 21:56

1 Answer 1

1
$\begingroup$

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 $n\ge 6.$

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.