# Show that the sample Mean is not complete

Suppose that $$X_1, ..., X_n$$ is a sample from a $$\mathcal{N}(-\frac12 \sigma^2, \sigma^2)$$ density. Show that the statistic $$\bar{X}$$ = $$n^{-1} \sum_{i=1}^n X_i$$ is NOT complete.

I am struggling to find a function of $$\bar{X}$$.

I was thinking let $$g(Q) = \bar{X} + \frac n2{S^2}$$ $$\Rightarrow E(g(Q)) = -\frac 12 \sigma^2 + \frac{n}{2n} \sigma^2 = 0$$,

but $$S^2$$ is not a function of only $$\bar{X}$$, since it is also a function of $$X_i$$.

Any hints/suggestions on what other function I can use?

• Where does this exercise come from? Nov 5 at 0:37
• This is a past qualifying exam question from 2009. Nov 5 at 0:51
• This is very closely related to stats.stackexchange.com/questions/631935. One might generally consider the same question for any positive-codimensional submanifold of a multiparameter exponential family.
– whuber
Nov 22 at 15:39

$$n\bar{X} \sim \mathcal{N}(-n\sigma^2/2, n\sigma^2)$$, so its mgf is
$$M_{n\bar{X}}(t) = \mathbb{E}[e^{tn\bar{X}}] = \exp\left[-nt\sigma^2/2 + n\sigma^2 t^2/2\right].$$ At $$1$$ we have $$\mathbb{E}[\exp(n\bar{X})] = 1$$. So $$g(\bar{X}) := \exp(n\bar{X}) -1$$ has expectation $$0$$ for all $$\sigma^2 > 0$$ yet it isn't $$0$$ wp1.
• Just to check your work, at t = 1, $E[exp(n \bar{X})] = e^0 = 1$, and so it should be $g(\bar{X}) = exp(n \bar{X}) - 1$. Which has expectation 0 for all $\sigma^2 > 0$ and isn't 0 wp1? Nov 22 at 3:08