Existence of complete sufficient statistics

Question

Suppose $X_1,\ldots,X_n$ are iid r.v.'s, each with pdf $f_{\theta}(x)=\frac{1}{\theta}I\{\theta<x<2\theta\}$. I find the minimal sufficient statistics $(X_{(1)},X_{(n)})$. I am trying to prove it is complete. Can someone give me hint? Also are there any complete sufficient statistics in this model?

Answer 1

We have $$\operatorname{E} (X_{(1)}) = \theta + \dfrac \theta {n+1} = \dfrac{n+2}{n+1} \theta$$ and $$\operatorname{E}(X_{(n)}) = 2\theta - \dfrac{\theta}{n+1} = \dfrac {2n+1} {n+1} \theta,$$ so $$ \operatorname{E} \left( \frac{n+1}{n+2} X_{(1)} - \frac{n+1}{2n+1} X_{(n)} \right) = 0 $$ regardless of the value of $\theta>0$.

Therefore, the statistic $(X_{(1)}, X_{(n)})$ is not complete.

Answer 2

Find $E(X_{(1)})$ and $E(X_{(n)})$. Play with them to make it $0$ i.e. find $a,b$ such that $E[aX_{(1)}+bX_{(n)}]=0$. Call $g(T)=aX_{(1)}+bX_{(n)}$. Then we have $E[g(T)]=0$ but does that mean $g(T)=0$ a.e.? No right?