# Showing that a statistic is ancillary for a parameter

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.

• @Andre "HW" = "homework." (I added 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, 2013 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, 2013 at 2:06
• 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, 2013 at 7:42

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.