TITLE: Complete Statistic: Uniform distribution
QUESTION [15 upvotes]: Take a random sample $X_1, X_2,\ldots X_n$ from the distribution
$f(x;\theta)=1/\theta$ for $0\le x\le 
\theta$.
I need to show that $Y=\max(X_1,X_2,...,X_n)$ is complete.
Now, I know I should multiply the sample distribution of $Y$ and multiply it with a function of $Y$, then integrate over the range of $\theta$ and equate them to zero. But how do I get the sampling distribution of $Y$?

REPLY [2 votes]: The p.d.f of $Y=X_{(n)}$ is $$\delta_{\theta}(y)=\frac{n}{\theta^n}y^{n-1},y\in[0,\theta]$$
For any function $f$ satisfying $E_{\theta}\left[f(y)\right]=0,\forall\theta\in\mathbb{R}^+$, we have $$\int_0^\theta f(y)y^{n-1}dy=0,\forall\theta\in\mathbb{R}^+$$
Since $[0,\theta]$ generate the Borel $\sigma$-algebra of $\mathbb{R}^+$, we have $$\int_A f(y)y^{n-1}dy=0, \text{ for all Borel set }A\subset\mathbb{R}^+$$
Therefore $$f(y)y^{n-1}= 0, a.e.\iff f(y)\equiv0, a.e.$$
This means $Y=X_{(n)}$ is complete.