# Proving that a minimal sufficient statistic is not complete

## Problem

Let $$X_1, \ldots, X_n$$ be a sample from normal distribution $$N(\theta, \theta^2)$$, where $$\theta > 0$$. I am to find minimal sufficient statistic and prove that it is not complete.

## Finding the minimal sufficient statistic

First of all I did some transformations which led me to the following expression $$f(x|\theta) = \gamma(\theta) \exp \bigg \{ -\bigg(\frac{1}{2\theta^2} \sum_{i=1}^nx_i^2 - \sum_{i=1}^nx_i + \frac{n}{2} \bigg) \bigg \} \tag{1}.$$ Using $$(1)$$ I can write that my sufficient statistic would be $$T(X) = \bigg(\sum_{i=1}^nx_i^2, \sum_{i=1}^nx_i \bigg) \tag{2}.$$ To show that $$(2)$$ is minimal I calculate the following fraction $$\frac{f(x|\theta)}{f(y|\theta)} \tag{3}.$$ It's easy to notice that $$(3)$$ is independent from $$\theta$$ if and only if $$x_i = y_i$$. Thus T(X) is minimal.

## Doubts

Are my calculation and reasoning correct? How can I prove that $$(2)$$ is not complete?

• First part looks good. There is a general theorem for k-dimensional exponential family that if your natural parameter space has k+1 vectors s.d. $\eta_i - \eta_0$, $i=1,\dots,k$, are linearly independent then your $T(X)$ is minimally sufficient. That's easy to verify here because your natural parameter space is the curve $(-1/2\theta^2, 1/\theta)$. – James Yang Nov 27 '18 at 17:02