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How can I show that the geometric mean $( \prod_{i=1}^{n} x_i )^{1/n}$ of a random sample of size $n$ from a distribution with pdf $f(x;\theta)=\theta x^{\theta-1},0<x<1,$, zero elsewhere, and $\theta>0$ is a sufficient statistic for $\theta$?

By the Factorization Theorem we can see that:

$$\prod_{i=1}^{n} f(x_i;\theta)=\theta^n \left[ \prod_{i=1}^n x_i \right]^{\theta-1} $$ and therefore $\prod x_i$ is a sufficient statistic for $\theta$. Now, I know that sufficient statistics are not unique and thus one does not exclude the other.

This distribution is an exponential family, which means its density can be rewritten as:

$$ \exp \{n\log\theta +(\theta-1) \sum \log x_i \}$$

but afterwards, in order to arrive at the geometric mean, there seems to be some subtle manipulation that I am missing.

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    $\begingroup$ Hint: Based on your last formula, it looks like the only way in which the data enter into the likelihood is via the expression $\sum\log x_i$. How is that related to the geometric mean? $\endgroup$ – whuber Jan 28 '14 at 16:00
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    $\begingroup$ Almost: you are moving between exponentials and logs too quickly and have mixed them up, but you have the right idea. $\endgroup$ – whuber Jan 28 '14 at 16:26
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    $\begingroup$ Exchange the position of the two power-coefficients to obtain the geometric mean -you could have done that without using the "exponential class" route. $\endgroup$ – Alecos Papadopoulos Jan 28 '14 at 19:26
  • $\begingroup$ What does this have to do with the beta distribution? $\endgroup$ – Neil G Jun 10 '17 at 12:35
  • $\begingroup$ @NeilG Maybe because $f(x;\theta)=\theta x^{\theta-1},0<x<1$ is the beta density? I don't know, I am just spitballing here. $\endgroup$ – JohnK Jun 10 '17 at 12:47

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