Let $X=(X_1,...,X_n)$ be i.i.d. $N(0,\sigma^2)$. 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 used 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 $\frac{2}{n}\sum_{i=1}^{n}X_i$?
I think it's easier if you just find the minimal sufficient statistic $T(X_1,\dots,X_n)$ for a $N(0,\sigma^2)$, which is
$$T(X_1,\dots,X_n)=\sum_{i=1}^nX_i^2$$
and then you show that this statistic is not a function of $\frac{2}{n}\sum_{i=1}^nX_i$. To prove that, you just need to take two different sample points which give the same image under one of the statistics, but a different image under the other one. For example, the points $(1,1,0,\dots,0)$ and $(\sqrt2,0,\dots,0)$ do the trick.