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Let $X_1,\ldots,X_n$ be a random sample from $N(\mu,\sigma^2)$ with both parameters unknown. How can I show that $(\bar{X}, S^2)$ is independent of $(X_{(n)}-\bar{X})/S$?

Work:

I am quite confident that I should use Basu's Theorem here. I can see that $(\bar{X}, S^2)$ is a complete sufficient statistic, but I am having trouble showing it. I assume that I will need to write $N(\mu,\sigma^2)$ as an exponential family to show this statistic is complete sufficient (for use in Basu's Theorem).

$$f(x\mid \mu,\sigma)=\frac{1}{\sqrt{2\pi\sigma^2}}\exp\left(\frac{-1}{2\sigma^2}(x-\mu)^2\right)=\frac{1}{\sqrt{2\pi\sigma^2}}\exp\left(\frac{-\mu^2}{2\sigma^2}\right)\exp\left(\frac{-x^2}{2\sigma^2} + \frac{x\mu}{\sigma^2}\right)$$

With this work, $T(\mathbf{X})=(\sum x_i, \sum x_i^2)$ is a complete statistic for $\mu, \sigma$, which is not $T(\mathbf{X})=(\bar{X}, S^2)$. I see that it closely resembles what I need, so how can I continue this work?

Additionally, we would need to show that $Y=(X_{(n)}-\bar{X})/S$ is ancillary, meaning that $Y$'s distribution is free of both $\mu$ and $\sigma$. However, I am not too sure on how to show this. Is this a case of applying a location-scale transformation?

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    $\begingroup$ One approach: what happens to the distribution of $(X_{(n)}-\bar X)/S$ when you change $\mu$ to $\mu + \lambda$? When you change $\sigma^2$ to $\lambda^2\sigma^2$? $\endgroup$
    – whuber
    Feb 6, 2020 at 5:15
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    $\begingroup$ How about you write $Y$ as $Y=\frac{\frac{X_{(n)}-\mu}{\sigma}-\frac{\bar X-\mu}{\sigma}}{S/\sigma}$? $\endgroup$ Feb 6, 2020 at 7:11
  • $\begingroup$ @StubbornAtom so you're approach is suggesting a scale transformation, whereas whuber's is suggesting a location-scale, correct? Is it typical to use location-scale transformations when showing a statistic is ancillary? It seems like the "direct" method of finding the joint pdf of these three variables and doing several transformations is not as efficient. So, I will attempt your scale transformation and get back to you. Thanks! $\endgroup$
    – Ron Snow
    Feb 6, 2020 at 17:20
  • $\begingroup$ @whuber if you make those suggested changes, then we have a location-scale transformation. From there, the typical line of attack is setting $Y_i = \lambda x_i + \lambda$, right? $\endgroup$
    – Ron Snow
    Feb 6, 2020 at 17:22

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To expand on my comment, you can write $Y$ as $$Y=\frac{\frac{X_{(n)}-\mu}{\sigma}-\frac{\overline X-\mu}{\sigma}}{S/\sigma}$$

From the distributions of $\overline X$ and $S$ it should be clear that those of $\frac{\overline X-\mu}{\sigma}$ and $S/\sigma$ are free of $(\mu,\sigma^2)$. As for the order statistic, notice that $\frac{X_{(n)}-\mu}{\sigma}$ has the same distribution as $Z_{(n)}$ where $Z_1,\ldots,Z_n$ are i.i.d standard normal. Hence its distribution is also free of $(\mu,\sigma^2)$. You can show this from the distribution function if you want.

This shows $Y$ is ancillary for $(\mu,\sigma^2)$. Note that @whuber's suggestion is along the same lines.

And the pdf of $X_1,X_2,\ldots,X_n$ is indeed a member of a full-rank (regular) exponential family, which shows that $\left(\sum\limits_{i=1}^n X_i,\sum\limits_{i=1}^n X_i^2\right)$ and equivalently its one-to-one transform $(\overline X,S^2)$ is a complete sufficient statistic for $(\mu,\sigma^2)$.

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  • $\begingroup$ Thank you for providing this detail. I understand the ancillary statistics for sure. I have two remaining questions for you. How can I show that $S^2$ is a one-to-one transformation of $\sum ^n_{i=1}X_i^2$? Also, how did you identify how to rewrite $Y$? $\endgroup$
    – Ron Snow
    Feb 7, 2020 at 1:40
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    $\begingroup$ $S^2=\frac1n\sum X_i^2-\overline X^2$ (does not matter if divisor is $n-1$) is a function of $\sum X_i^2$ and $\sum X_i$, making the pair $T_2=(\overline X,S^2)$ a function of $T_1=(\sum X_i,\sum X_i^2)$. The transformation $T_1\mapsto T_2$ is clearly one-to-one because there is no loss of information going from one pair to the other. You can also use the Factorization theorem to directly get $T_2$ instead of $T_1$. As for rewriting $Y$, it is just a location-scale transformation as you mentioned. $\endgroup$ Feb 9, 2020 at 15:50
  • $\begingroup$ Would it also be possible to let $X_i=\sigma Z_i + \mu$, so $(X_{(n)}-\bar{X})/S=\frac{\sqrt{n-1}(Z_{(n)}-\bar{Z})}{\sqrt{(n-1)S^2/\sigma^2}}$, whose distribution doesn't really on $(\mu,\sigma^2)$? $\endgroup$
    – Ron Snow
    Feb 12, 2020 at 2:54

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