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Given $X_1,\ldots, X_7$ are i.i.d r.vs which follow $N(\mu, \sigma^2)$. Find the UMVUE of $\frac{\mu}{\sigma^2}$.

My thought: First, we realize that $E(S^2) = \sigma^2$ where $S^2 = \frac{\sum_{i=1}^{7} (X_i - \overline{X})^2}{6}$ is a sample variance. Thus, $E(\frac{1}{S^2}) = \frac{A}{\sigma^2}$ (I could not see how to compute this constant $A$). Since the sample mean and sample variance of $X_1,\ldots, X_7$ are independent, we have: $E(\frac{\overline{X}}{S^2}) = E(\overline{X})E(\frac{1}{S^2}) = \frac{\mu}{A}$, so $\frac{\overline{X}}{AS^2}$ is the unbiased estimator of $\frac{\mu}{\sigma^2}$.

Now, this unbiased estimator could be expressed as a complete sufficient statistics $(\sum_{i=1}^{7} X_i, \sum_{i=1}^{7} X_i^2)$ because $S^2 = \frac{\sum_{i=1}^{7} X_i^2 - \frac{(\sum_{i=1}^{7} X_i)^2}{7}}{6}$. This implies $\fbox{$\frac{\overline{X}}{AS^2}$}$ is a UMVUE of $\frac{\mu}{\sigma^2}$

My question: Could anyone please help me determine the constant $A$ above to complete this proof? I feel so shameful not to be able to compute $E(\frac{1}{S^2})$.

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    $\begingroup$ You should probably add the self-study tag $\endgroup$ – kjetil b halvorsen Apr 8 '17 at 17:29
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    $\begingroup$ I will provide the solutions in class tonight. I do not like that my students are using this site to get unauthorized help on homework problems. Why not ask me for a hint? C. Sutton $\endgroup$ – C Sutton Apr 12 '17 at 19:11
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    $\begingroup$ @CSutton: I've added the self study tag to this question - according to our site policy that means it should receive hints rather than a completely worked-out answer. I'd encourage people using CV for help with homework to be open about it with their teachers. Knowing where students get stuck is very useful feedback. $\endgroup$ – Scortchi - Reinstate Monica Apr 12 '17 at 19:56
  • $\begingroup$ Cross posted at math.stackexchange.com/questions/2212726/…. $\endgroup$ – StubbornAtom Oct 27 '18 at 5:39

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