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Suppose $X_1,X_2,\ldots,X_n$ is a random sample drawn from a distribution with pdf $$f_{\theta}(x)=\small\frac{\ln\theta}{\theta-1}\theta^x\,\mathbf1_{0<x<1}\quad,\,\theta>1$$

Does there exist uniformly minimum variance unbiased estimator (UMVUE) of $\theta$? If not, can it be derived which functions of $\theta$ admit a UMVUE ?

The pdf $f_{\theta}$ is a member of the one-parameter exponential family, from which it follows that a complete sufficient statistic for $\theta$ based on the sample is $T=\sum\limits_{k=1}^n X_k$. So of course $T$ is the UMVUE of its expectation which can be found easily.

But I could not find any trivial unbiased estimator of $\theta$ for example, which leads me to ask which functions of $\theta$ are unbiasedly estimable in the first place. If $g(\theta)$ is unbiasedly estimated by some $h(X_1,\ldots,X_n)$, then certainly its UMVUE is given by $E(h\mid T)$.

If I can identify the sampling distribution of $T$ my job would be easier, but I am not certain about going down that road without coming up with any particular transformation that results in a common distribution.

I also wish to know if this distribution has a name, and more importantly, if it is related to a standard distribution (the probability integral transform is not much helpful here).

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  • $\begingroup$ To calibrate how we should approach this, is this a textbook problem or something we should expect is easy to answer in principle? $\endgroup$
    – guy
    Jan 27, 2019 at 15:40
  • $\begingroup$ @guy Not a textbook problem but someone asked this on the math site claiming that they were asked to find UMVUE of $\theta$ in an exam. I could not find an unbiased estimator of $\theta$, so I am curious to know if there exists one. $\endgroup$ Jan 27, 2019 at 15:44
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    $\begingroup$ This distribution is a truncated exponential $\mathcal{E}(\ln\theta)$ when restricted to the unit interval. $\endgroup$
    – Xi'an
    Feb 19, 2019 at 9:06
  • $\begingroup$ I do not think there is an unbiased estimator because the estimator need diverge as $\theta$ grows to infinity, while the distribution does not degenerate in the limit, since it is the Uniform $(0,1)$ distribution. $\endgroup$
    – Xi'an
    Feb 19, 2019 at 9:33

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