# Complete sufficient statistic

I've recently started studying statistical inference. I've been working through various problems and this one has me completely stumped.

Let $X_1,\dots,X_n$ be a random sample from a discrete distribution which assigns with probability $\frac{1}{3}$ the values $\theta-1,\space\theta,\space\text{or}\space\theta+1$, where $\theta$ is an integer. Show that there does not exist a complete sufficient statistic.

Any ideas?

• What do you have so far? – gung - Reinstate Monica Jun 18 '13 at 3:09
• I can write the likelihood as: $(\frac{1}{3})^n$ times the product of the indicator functions that each observation is equal to either $\theta-1,\space\theta,\space\text{or}\space\theta+1$. From this it looks like the sufficient statistic is the order statistics. I've been thinking about this for days, it's like nothing I've seen before. – Tony Jun 18 '13 at 3:21
• What do you know about completeness? – Glen_b Jun 18 '13 at 4:59
• A statistic, $T$, is complete if it satisfies the condition that, for some function $g(T)$, if $E[g(T)]=0$, then $g(T)=0$ $a.e.$ – Tony Jun 18 '13 at 5:06
• So you need to find a counter-example ... what clearly ancillary statistic can you find from the sample minimum & maximum? – Scortchi - Reinstate Monica Jun 18 '13 at 12:56

(1) Show that for a sample size $n$, $T=\left(X_{(1)}, X_{(n)}\right)$, where $X_{(1)}$ is the sample minimum & $X_{(n)}$ the sample maximum, is minimal sufficient.
(2) Find the sampling distribution of the range $R=X_{(n)}-X_{(1)}$ & hence its expectation $\newcommand{\E}{\operatorname{E}}\E R$. It will be a function of $n$ only, not of $\theta$ (which is the important thing, & which you can perhaps show without specifying it exactly).
(3) Then simply let $g(T)=R-\E R$. It's not a function of $\theta$, & its expectation is zero; yet it's not certainly equal to zero: therefore $T$ is not complete. As $T$ is minimal sufficent, it follows from Bahadur's theorem that no sufficient statistic is complete.
• Thinking about it, it would've been simpler just to say that $T$, being minimal sufficient, is some function $f(\cdot)$ of any sufficient statistic $S$, & therefore $g(T)=g(f(S))$ also goes to show the incompleteness of $S$. – Scortchi - Reinstate Monica Jul 24 '18 at 20:44