Suppose that $X_1,...,X_n$ are i.i.d. with probability density function
$f_{\theta}(x)= \, \frac{1}{2\sqrt{\theta x}} \quad 0\lt x \lt \theta$
Show that $X_{(n)}$ is a sufficient for $\theta$ .
I am going with the original definition of sufficient statistic.
Where it says that T(x) is sufficient iff $L(\theta,x_1,x_2,...x_n| T(X)= t)$ should be independent of $ \theta$.
first finding the pdf of T(X) :-
$$f(X_{(n)} ) = \frac{n!}{(n-1)!}\, F(x)^{n-1}\, f(x) $$
=$$ f(X_{(n)} ) =n {(\frac{\sqrt x}{\sqrt \theta})}^{n-1} \frac{1}{2\sqrt{\theta x}}$$
$$\therefore T(X=t) = f(X_{(n)} =t )=n {(\frac{\sqrt t}{\sqrt \theta})}^{n-1} \frac{1}{2\sqrt{\theta t}} \Rightarrow n {(\frac{\sqrt t}{\sqrt \theta})}^{n} \frac{1}{2\sqrt{\theta t}}$$
Here is the part I am not so sure about $$L(\theta,x_1,x_2,...x_n|T(x)=t)=L(\theta,x_1,x_2,...x_n|f(X_{(n)} =t ))$$ $$L(\theta,x_1,x_2,...x_n|f(X_{(n)} =t )) \Rightarrow \frac{P(X_1=x_1,X_2=x_2,...X_n=x_n \cap f(X_{(n)} =t ))}{f(X_{(n)} =t )} $$ $$\Rightarrow\frac {P(X_1\le t,X_2 \le t,...X_n \le t )}{f(X_{(n)} =t )}$$
After substituting the values I got this :- $L(\theta,x_1,x_2,...x_n|f(X_{(n)} =t ))= \frac{1}{2t}$ which is clearly independent of $\theta$.
Is my method correct or is there any other way to approach this problem
Thanks.