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Let $X_1 , \ldots , X_n$ be iid with pdf $f(x \mid \theta) = \theta x^{-2}$, $0 < \theta \leq x < \infty$. Find a low dimensional sufficient statistic for $\theta$ and the MLE of $\theta$

So for the sufficient stat i just used the Lehmann and Scheffe theorem and found that $X_{(1)}$ being the minimum of ordered $X$'s is a minimal SUfficient stat so that makes it a sufficient stat too ya? how else would someone have gone about this one?

How do i find the MLE with this indicator function applied to their products? should i just ignore the fact that $\theta < x$ and find my $L(\theta$|x) and go about the usual method to maximize?

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The theorem you presumably used for finding the sufficient statistic was the factorization theorem and not Lehmann-Scheffe. To find the MLE we only have to look at the likelihood function:

$$ L(\theta) = \theta^n \prod_{i=1}^{n} x_i^{-2} I_{[\theta, \infty)} (x_{(1)}) $$

where $I_{[\theta, \infty)}(s) = 1$ if $s \in [\theta, \infty)$ otherwise $I_{[\theta, \infty)}(s) = 0$. Can you solve it from here?

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  • $\begingroup$ so from there you would just assume $\theta$ hat = w/e it equals assuming $\mathbb{I}$=1 and i do the whole derivative/log ordeal depending which is easier. Or if $\mathbb{I}$=0 there is nothing to say>? $\endgroup$ – qqq2 Mar 19 '16 at 16:47
  • $\begingroup$ There's no calculus involved. Just think about how this function behaves as $\theta$ changes. $\endgroup$ – dsaxton Mar 19 '16 at 23:06

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