(In)sufficient statistics I have the following problem. I am given a random sample $X_1, X_2, \dots X_n$, which follows a distribution with density function $f_\theta (x) = \frac{ e^\frac{x}{\theta}}{\theta (e-1)},  \hspace{0.1cm} 0\leq x \leq \theta$. They asked me two questions:
Which is a sufficient statistic for $\theta$? Is the maximum likelihood estimator a sufficient statistic for $\theta$?

I have find the maximun likelihood estimator which is the maximum, $X_{(n)}$. I think it's not sufficient because I wasn't able to prove it by the factorisation theorem. Which tools can I use to prove that a given statistic is not sufficient? Is there a natural way to find a sufficient statistic?
Thanks in advance
 A: Factorisation theorem give nesessary and sufficient conditions for sufficiency, so impossibility to write likelihood function in the product form implies that statistics is not sufficient. To prove it, one can take two different values of $\theta$ and consider the ratio
$$
\frac{f_{\theta_1}(X_1,\ldots,X_n)}{f_{\theta_2}(X_1,\ldots,X_n)}
$$
which should be a function of $\theta_1,\theta_2$ which depends on samples only through sufficient statistics. 
$$
f_\theta (X_1,\ldots,X_n) = \frac{ e^\frac{\sum_{i=1}^n X_i}{\theta}}{\theta^n (e-1)^n}\mathbf 1_{\{0\leq X_{(1)}\}}\mathbf 1_{\{X_{(n)} \leq \theta\}}
$$
On the event $\{0\leq X_{(1)}\leq X_{(n)}\leq \min(\theta_1,\theta_2)\}$, the ratio of likelihood functions equals to
$$
\frac{f_{\theta_1}(X_1,\ldots,X_n)}{f_{\theta_2}(X_1,\ldots,X_n)} =\frac{\theta_2^n}{\theta_1^n} e^{\sum_{i=1}^n X_i\left(\frac{1}{\theta_1}-\frac{1}{\theta_2}\right)}.
$$
If this ratio depends on samples only through statistics $X_{(n)}$, then for two different samples $x_1,\ldots, x_n$ and $y_1,\ldots, y_n$ with the same value of $x_{(n)}=y_{(n)}$, this ratio should be the same. But this is not the case: for samples with different sums but equal maximal values this ratio differ.
Sure, the statistics $(\bar X, X_{(n)})$ is minimal sufficient here.
