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Let $X$ be an observation from a distribution with probability mass function:$f(x;\theta) = \left(\frac{\theta}{2}\right)^{|x|}(1-\theta)^{1-|x|}I_{\{-1,0,1\}}(x), \, \theta \in (0,1).$ Use Rao-Blackwell theorem to find the Minimum-Variance Unbiased Estimator (MVUE).

To begin with, the PMF can also be written as $$ f(x;\theta) = I_{\{-1,0,1\}}(x) (1-\theta)\exp \left\{|x| \ln\left(\frac{\theta}{2(1-\theta)}\right)\right\} $$ and therefore $f$ belongs to the Exponential Family. According to Pitman-Koopman, there exists a sufficient function for the parameter $\theta$.

Now, from R-B: $W = E[R = |x| \big| T]$ (where $R = |x|$ is an Unbiased estimator of $\theta$ and $T$ is a sufficient and complete function for $\theta$).

The problem I face is that I don't get how we construct a $T$ which is sufficient and complete for $\theta$ and how does one use it to find an explicit formula for $W$. Can someone provide me with these two steps for finding the MVUE?

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    $\begingroup$ You need the self-study tag. We can't give you the complete answer but we can give you guidance or hints if you add the tag. $\endgroup$ – Michael R. Chernick Apr 28 '19 at 20:26
  • $\begingroup$ By Factorisation theorem, $|X|$ is sufficient (and complete since this is exponential family). $\endgroup$ – StubbornAtom Apr 28 '19 at 21:41
  • $\begingroup$ @MichaelChernick Okey, I just added it. $\endgroup$ – Andrew Apr 29 '19 at 7:08
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Hint: Write out the sampling distribution for the statistic $|X|$. It is an extremely simple and well-known distributional form. Once you have found this, finding the MVUE should not present much problem.

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