Let $X=(X_1,...,X_n)$ be i.i.d. $U(0,\theta)$. How to show that $$\frac{2}{n}\sum_{i=1}^{n}X_i$$ is not a sufficient statistic?
I have already proven that $\max_{i=1,...,n}X_i$ is a sufficient statistic. Also the PDF of the distribution function of the density can be described as $$P_{\theta}(X_n<x)=\left(\frac{x}{\theta_0}\right)^n\cdot I\left(\max_{i=1,...,n}x_i<\theta\right).$$ Can I just use the factorisation theorem and state that $$P_{\theta}(X_n<x)=\left(\frac{x}{\theta_0}\right)^n\cdot I\left(\max_{i=1,...,n}x_i<\theta\right)$$ cannot be written in terms of $\frac{2}{n}\sum_{i=1}^{n}X_i$?
A rigourous way to do this is to first show that $\max X_i$ is a minimal sufficient statistic by a corollary of the factorisation theorem, and then it follows immediately that $\frac{2}{n} \sum_{i=1}^n X_i$ is not sufficient, as the minimal sufficient statistic is not a function of it.