The range is an ancillary statistic for location families I am having trouble understanding why the range is considered an ancillary statistic for location family pdfs.... I will try to set up the proof for this and then point out where I am having trouble understanding.
1) Note that a statistic $S(X)$ whose distribution does not depend on parameter $\theta$ is called ancillary. (This is opposed to a sufficient statistic, that contains all information about theta within a sample). 
Now suppose that you have any number of location-family distributions. I will omit examples of exact pdfs, but just note that a location family of distributions is any distribution with a common SHAPE pdf, that's location is determined by the parameter $\theta\$
For each pdf, let $X_{(1)} < ... < X_{(n)}$ be the ordered sample. And define $Y_{I} = X_{(n)} - X_{(1)}, I=1,...,n-1$
Verify the set $(Y_{I}.....Y{n-1})$ is ancillary for $\theta$. 
We know that our pdfs are all members of the same location family. 
For location families we have that $X$ $=$ in distribution to $Z + \theta, z \sim f_X(x| \theta=0)$
The proof of this, usings the MGF technique, follows:
$M_X(t)$ $=$ $\mathbb{E}(e^{tx})$ $=$ $\int_-a^a \! f(x-\theta)e^{tx} \, \mathrm{d}x$ =$= \int_-a^a \! f(z)e^{t(z+\theta)} \, \mathrm{d}x$ $=$ $\mathbb{E}(e^{t(z+\theta)})$ $=$ $M_{Z+\theta}(t)$ where $a=\infty$
If $Y_{I} = X_{(n)} - X_{(1)}$, then the above condition $X=\theta+Z$ implies $Y_{I} = Z_{(n)}-Z_{(1)}$, and we know from this that z does not depend on $\theta$.  
MY QUESTION IS: even though the distribution of Z itself is given as $ z \sim f_X(x| \theta=0)$ ie, z itself doesn't depend of theta, doesn't the location condition $Z= X - \theta$ impose some kind of dependence? Even though we proved that the MGFs are equivalent, the last MGF also contains the parameter $\theta$ . Or is $\theta$ itself, even when taken into the MGF, and taken into the location condition $X=\theta + Z$, somehow not factored in any way into X and Z's distribution?
Hope this makes sense...! 
 A: The key idea is exposed in a more general result.  Let $s:\mathbb{R}^n\to\mathbb{R}$ be any translation-invariant function; that is,
$$s(x_1, x_2, \ldots, x_n) = s(x_1 + \theta, x_2 + \theta, \ldots, x_n+\theta)$$ for all real numbers $\theta$.  (There are plenty of such functions. For instance, any function of $(x_2-x_1, x_3-x_1, \ldots, x_n-x_1)$ will be translation-invariant.)
Suppose $X_1, X_2, \ldots, X_n$ is an iid sample from a distribution $F_\theta$ in some location-invariant distribution family whose location is parameterized by $\theta$.  This means that for any numbers $\theta$ and $\phi$ and all real numbers $x$,
$$F_\theta(x) = F_\phi(x - \theta + \phi).$$
Supposing $s$ is measurable, consider the distribution function of the random variable $s(X_1, X_2, \ldots, X_n)$.  This is defined to be the function $F_s$ for which
$$F_{s,\theta}(x) = {\Pr}_{F_\theta}(s(X_1, X_2, \ldots, X_n) \le x)$$
for real numbers $x$.  Suppose $\phi$ is a real number.  Notice that
$$\eqalign{
F_{s,\theta}(x) &= {\Pr}_{F_\theta}(s(X_1, X_2, \ldots, X_n) \le x) \\
& = {\Pr}_{F_\phi}(s(X_1-\theta+\phi, X_2-\theta+\phi, \ldots, X_n-\theta+\phi) \le x) \\
&=  {\Pr}_{F_\phi}(s(X_1, X_2, \ldots, X_n) \le x) \\
&= F_{s,\phi}(x).
}$$
We owe the equality on the second line to the location invariance of the family $\{F_\theta|\theta\in\mathbb{R}\}$ and the next inequality to the translation invariance of $s$.  Consequently, $F_{s,\theta}$ does not depend on $\theta$.  The interplay between the location parameter and the translation invariance of $s$ is clear.
Applying this result to the range $s(x_1, x_2, \ldots, x_n) = \max(x_i) - \min(x_i)$, which obviously is translation-invariant (and measurable), shows its distribution does not depend on $\theta$: it is an ancillary statistic.
