5
$\begingroup$

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?

$\endgroup$
3
  • $\begingroup$ Where does this exercise come from? $\endgroup$
    – Zhanxiong
    Nov 5, 2023 at 0:37
  • $\begingroup$ This is a past qualifying exam question from 2009. $\endgroup$
    – Stats_Rock
    Nov 5, 2023 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, 2023 at 15:39

1 Answer 1

12
$\begingroup$

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

$\endgroup$
2
  • 2
    $\begingroup$ Awesome! I didn't think to use MGF. Very elegant solution! $\endgroup$
    – Stats_Rock
    Nov 5, 2023 at 3:38
  • 1
    $\begingroup$ 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, 2023 at 3:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.