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Given random sample $\left\{ { X }_{ 1 },{ X }_{ 2 },...,{ X }_{ n } \right\} $ from $ U(0,\theta)$.

Let ${Y}_{i}$ be the order statistics.

Then the sufficient statistic for $\theta$ is ${ Y }_{ n }$.

My question is how to prove it by Factorization Theorem.

I can only factor the joint p.d.f of the random sample to this form,

${ \left( \frac { 1 }{ \theta } \right) }^{ n }{ I }_{ \left[ 0,{ Y }_{ n } \right] }({ y }_{ 1 }){ I }_{ \left[ { Y }_{ 1 },\theta \right] }\left( { y }_{ n } \right) $ where $I$ is the indicator function.

I cannot figure out why this is independent of ${Y}_{1}$.

Need some help. Thanks in advance.

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Check the Fisher–Neyman factorization theorem. –  беркай Dec 6 '13 at 22:23
    
Yes...I have mentioned factorization theorem in my question...and that's where I got stuck. More elaborations, please? –  Aiden Dec 7 '13 at 2:24

1 Answer 1

Just figured out!

Factorize it in another way!

${ \left( \frac { 1 }{ \theta } \right) }^{ n }{ I }_{ \left[ { Y }_{ 1 },\theta \right] }\left( { y }_{ n } \right)\prod _{ i=1 }^{ n }{ {I}_{\left[0,\infty\right)} }({x}_{i}) $

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