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.