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I'm trying self-study some inference and now I'm trying to understand how to solve some problems on this topic but I found this basic problem that I'm not being able to solve.

Problem:

Let $X_{1},...,X_{n}$ i.i.d $f(x | \theta)$, where $f(x | \theta) = \theta x ^{\theta -1 }$.

Is $\sum_{i}^{n} X_{i}$ a sufficent statistic for $\theta$? How can I find a complete sufficient statistic?

Im really confused actually. I know these problems can be solved with the factorization theorem, so we have that $f(x_{1},...,x_{n} | \theta) = \prod \theta x_{i} ^{\theta -1} = \theta ^{n} \prod x_{i}^{\theta} \prod x_{i}^{-1}$, then $h(x) = \prod x_{i} ^{-1}$ and $g[\theta,T(x_{i})] = \theta^{n}\prod x_{i}^{\theta}$. So in this case $T(x_{1},...,x_{n}) = \prod x_{i}$ is a sufficient statistic right?

But I don't know hot to check if $\sum X_{i}$ is in fact a suff. statistic and I don't know how to prove that the statistic I found is complete.

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  • $\begingroup$ Are you asking about $\sum x_i$ or $\prod x_i$ ? Is $X_i$ restricted to $[0,1]$? $\endgroup$
    – Henry
    Commented Jul 21, 2021 at 9:12
  • $\begingroup$ Im refering to $\sum x_{i}$ and yes, X_{i} is restricted to $[0,1]$. Thanks for your comment :) $\endgroup$
    – user1trill
    Commented Jul 21, 2021 at 9:36
  • $\begingroup$ Then you seem to have shown $\sum x_i$ is not a sufficient statistic but that $\prod x_i$ is - or some 1-1 function of it like $\sum \log(x_i)$ $\endgroup$
    – Henry
    Commented Jul 21, 2021 at 9:39
  • $\begingroup$ To check whether or not $T(X)$ is a sufficient statistic you may consider the conditional distribution of $X$ given $T(X)$. $\endgroup$
    – Xi'an
    Commented Jul 21, 2021 at 11:34
  • $\begingroup$ Or you may consider whether or not the likelihood ratio$$\ell(x)/\ell(x^\prime)$$is constant in $\theta$ when$$T(x)=T(x^\prime)$$ $\endgroup$
    – Xi'an
    Commented Jul 21, 2021 at 11:36

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