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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?

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  • $\begingroup$ Where does this exercise come from? $\endgroup$
    – Zhanxiong
    Nov 5 at 0:37
  • $\begingroup$ This is a past qualifying exam question from 2009. $\endgroup$
    – Stats_Rock
    Nov 5 at 0:51
  • $\begingroup$ 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. $\endgroup$
    – whuber ♦
    Nov 22 at 15:39

1 Answer 1

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$n\bar{X} \sim \mathcal{N}(-n\sigma^2/2, n\sigma^2)$, 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.

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    $\begingroup$ Awesome! I didn't think to use MGF. Very elegant solution! $\endgroup$
    – Stats_Rock
    Nov 5 at 3:38
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    $\begingroup$ 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? $\endgroup$
    – Stats_Rock
    Nov 22 at 3:08

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