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

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


$$T - \bar{X} $$

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.

  • 3
    $\begingroup$ @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. $\endgroup$
    – whuber
    Nov 13, 2013 at 17:07
  • $\begingroup$ 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. $\endgroup$
    – FAS
    Nov 14, 2013 at 2:06
  • 2
    $\begingroup$ Re the edit: instead of subtracting $\theta$, subtract $\bar{X}$ from $T$. It might help to observe that $T-\bar{X} = (T-\theta) - (\bar{X}-\theta).$ $\endgroup$
    – whuber
    Nov 14, 2013 at 7:42

1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.