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?


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.


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

  • 2
    $\begingroup$ And if you had $U(a,b)$ then you would want to take both $\endgroup$ – Henry Jun 13 '16 at 7:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.