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.

  • $\begingroup$ Could you tell me more about that? $\endgroup$ – marco11 Feb 23 '16 at 19:46
  • $\begingroup$ The minimal sufficiency criterion states that a statistic $T = t(X)$ is minimal sufficient if and only if $f(x|\theta)/f(y|\theta)$ is independent of $\theta$ $\iff t(x) = t(y)$, where $f$ is the likelihood function. The wikipedia page en.wikipedia.org/wiki/Sufficient_statistic#Minimal_sufficiency) explains the concept pretty well - feel free to post here if you have more questions. $\endgroup$ – Mark Feb 23 '16 at 19:53

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