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We are given that $ X_i \sim U( 0 , \theta )$ where $\theta$ is unknown. We need to find a sufficient statistic.

First we write the conditional distribution :

$$ f( x_1 ,\ldots ,x_n \mid \theta ) = \frac 1 {(\theta)^n}, \text{ where } ( x_i \leq \theta , i=1,2,\ldots,n).$$

So , we write the conditional distribution as follows :

$$ f( x_1 ,\ldots ,x_n \mid \theta ) = {(\theta)^{-n}} \quad ( x_i \leq \theta , i=1,2,\ldots,n) \tag 1$$

where $(1)$ is taken as an indicator function which gives the value zero if $ x_i \notin [0,\theta]$

So if $x_i \in [0,\theta]$ then $ \max(x_i) \in U(0,\theta)$, so,

$$ f( x_1 ,\ldots ,x_n \mid \theta ) = {(\theta)^{-n}} \quad ( \max(x_i) \leq \theta)\tag 1$$

And hence the sufficient statistic $\max(x_i)%$ is taken.

Now my question is, in a very similar manner $\min(x_i)$ could also be taken, would that be wrong?

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No, we cannot consider the $\min(X_i)$ because only the $\max(X_i)$ bounds all the variables-since we have an increeasing function-thus containing more information about the distribution.

Note

If we had the $U(\theta, \infty)$ then we should have taken $\min(X_i)$ for the same reasons outlined above.

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    $\begingroup$ And if you had $U(a,b)$ then you would want to take both $\endgroup$ – Henry Jun 13 '16 at 7:47

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