Is this a sufficient statistic for uniform distribution? I'm having trouble proving that given a uniform distribution $X_i \sim U(0,\theta)$ with $\theta$ unknown, the statistics $2\bar X$ and $\bar X$ are not sufficient; any ideas?
Thanks for your help.
 A: How to spot an insufficient statistic.
Suppose that $u({\bf y})$ is a function of the data vector ${\bf y}=(y_1,\dots, y_n).$
The factorization theorem says that if $u$ is sufficient for $\theta$, then
for any two parameters $\theta$ and $\theta^*$, and any two 
data vectors ${\bf y}$ and ${\bf y^*}$ with $u({\bf y})=u({\bf y^*})$, we have
$L(\theta,{\bf y})\,L(\theta^*,{\bf y^*})= L(\theta,{\bf y^*})\,L(\theta^*,{\bf y}).$
In your problem, provided  $n>1$,  it is easy to find parameters and data vectors so that this equation fails. 
A: Here's a hint: $ \operatorname{E}(2\bar X) = \theta,$ so $2\bar X$ is an unbiased estimator of $\theta$, but if, for example, $(X_1,X_2,X_3) = (1,2,12)$ then the estimate of $\theta$ is actually smaller than the largest of the three observations.  That shows there is more information about $\theta$ in the sample than there is in $\bar X$.
Here's somewhat more than a hint: The conditional distribution of $X_1,X_2,X_3$ given that $\bar X = 1$ and $\theta = 2$ is supported in some subset of $[0,2]^3$, but the conditional distribution of $X_1,X_2,X_3$ given that $\bar X = 1$ and $\theta=20$ has the point $(0,0,3)$ in its support.  (Note that $(0+0+3)/3 = \bar X = 1$.)  That shows that $\bar X$ is not sufficient without helping you find any one-dimensional sufficient statistic.  In fact, one exists.
