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Working through a HW problem, and a hint is that for a decision rule

$$T(X) = \frac{X_{(1)} + X_{(n)}}{2}$$

Then

$$T - \bar{X} $$

is ancillary.

Intuitively this makes complete sense, but I am failing to see how to show this. I thought about going to the pdf of an ordered statistic:

$$g_i(x) = \frac{n!}{(i-1)!(n-i)!}[F(x)]^{i-1}[1-F(x)]^{n-i}f(x)$$ which for i = {1,n} reduces down to $$g_1(x) = n[1-F(x)]^{n-i}f(x), \:\:\:\: g_n(x) = n[F(x)]^{n-i}f(x)$$ But I would probably need the convolution from there to get the pdf of T. I feel like I'm way off and there is something simpler here to show that subtracting off $\bar{X}$ makes T independent of theta.

By the way $X \sim N(\theta, \sigma^2)$, $\sigma$ known

In response to comments:

Then $$T - \theta = \frac{1}{2} \left[ (X_{(1)} - \theta) + (X_{(n)} - \theta) \right]$$ and since $X_{(1)}$ is a location family $$X_{(1)} = Z_{(1)} + \theta$$ where $Z_{(1)}$ is the first ordered statistic of a standard normal vector, which implies $$X_{(1)} - \theta = Z_{(1)} \sim \left[ 1 - \int\limits_{-\infty}^z (2\pi)^2\exp(\frac{-z^2}{2})dz \right]^{n-1} (2\pi)^{-1/2}\exp(\frac{-z^2}{2})$$ which does not depend on theta. Similaryly $X_{(n)} - \theta$ also does not depend on $\theta \Longrightarrow T - \theta$ is ancillary for $\theta$.

One big problem though, I am subtracting $\theta$, not $\bar{X}$. Unless were talking about a first order ancillary statistic (where $E\bar{X} = \theta$), I am once again confused.

Xi'an
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FAS
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    @Andre "HW" = "homework." (I added [tag:self-study] to clarify this.) FAS: it is important that $\sigma$ be known. That reduces your problem to showing that the distribution of $T-\bar{X}$ does not depend on $\theta$. Although this could be done (painfully) using the method you describe, a simple demonstration is possible based on observing that your family of distributions is a *location family.* – whuber Nov 13 '13 at 17:07
  • Thanks! As usual this highlights something that I thought I understood (location family) where I obviously do not. If possible, please see updated/edited question. Thanks. – FAS Nov 14 '13 at 02:06
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    Re the edit: instead of subtracting $\theta$, subtract $\bar{X}$ from $T$. It might help to observe that $T-\bar{X} = (T-\theta) - (\bar{X}-\theta).$ – whuber Nov 14 '13 at 07:42

1 Answers1

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To repeat the comments, considering $$T(X) - \bar X = \frac{X_{(1)} + X_{(n)}}{2}-\frac{1}{n}\sum_{i=1}^n X_i$$ is identical to considering $$T(X) - \bar X =\{T(X) -\theta\}- \{\bar X-\theta\} = \frac{X_{(1)}-\theta + X_{(n)}-\theta}{2}-\frac{1}{n}\sum_{i=1}^n \{X_i-\theta\}$$ with the lhs being independent of $\theta$ and the rhs not requiring the knowledge of $\theta$. The reason is that both quantities are equivariant under translation.

Xi'an
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