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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 $\frac{2}{n}\sum_{i=1}^{n}X_i$?

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  • $\begingroup$ What is $\theta_0$ in $P_{\theta}(X_n<x)=\left(\frac{x}{\theta_0}\right)^n\cdot I\left(\max_{i=1,...,n}x_i<\theta\right).$ I think the PDF calculation is not valid! $\endgroup$
    – Masoud
    Jul 1 '20 at 15:07
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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.

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  • $\begingroup$ 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. $\endgroup$
    – user93238
    Feb 23 '16 at 19:53
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If you want to show a statistic is not a sufficient statistic , you can compare it with minimal sufficient statistic. Use the fact that a minimal sufficient statistic is a function of any sufficient statistic.

Define $T=\max (X_1,\cdots ,X_n)$ and $U=\frac{2}{n}\sum_{i=1}^{n} X_i$. We want to prove $U$ is not a sufficient statistics.

Since $T$ is minimal sufficient statistic, so it is a function of any sufficient statistic. It is enough to show that $T$ is not a function of $U$. So conclude $U$ is not a sufficient statistic, that is, the approach is:

  1. Find a minimal sufficient $T$

  2. Show that the minimal sufficient is not a function of $U$

  3. Compare with the fact that a minimal sufficient statistic is a function of any sufficient statistic. So conclude $U$ is not a sufficient statistic.

Obviously $T=\max (X_1,\cdots ,X_n)$ is not a function of $U=\frac{2}{n}\sum_{i=1}^{n} X_i$ but we prove it. $T$ is a function of $U$ if $U(a_1)=U(a_2) \Longrightarrow T(a_1)=T(a_2)$.

So it is enough to find two points that $U(a_1)=U(a_2)$ but $T(a_1)\neq T(a_2)$ , and hence $T$ is not a function of $U$ and hence $U$ is not a sufficient statistic.

Lets $a_1=(.2,.2,\cdots,.2,.1,.3)$ and $a_2=(.2,.2,\cdots,.2,.15,.25)$ so $U(a_1)=U(a_2)$ but $T(a_1)=.3\neq T(a_2)=.25$

See sufficient-statistic-for-poisson , can-this-statistic-be-shown-not-to-be-sufficient-for-theta and sufficient-statistics-when-mean-is-known

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