# Proving that a statistic is not sufficient (uniform case).

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

• 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! Jul 1 '20 at 15:07

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.

• Could you tell me more about that? Feb 23 '16 at 19:46
• 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.
– user93238
Feb 23 '16 at 19:53

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 $$T(a_1)=.3\neq T(a_2)=.25$$