Transforming Order Statistics Assume random variables $X_1, ... , X_n$ and $Y_1, ..., Y_n$ are independent and $U(0,a)$-distributed. Show that $Z_n= n\log\frac{\max(Y_{(n)},X_{(n)})}{\min(Y_{(n)},X_{(n)})}$ has an $\text{Exp}(1)$ distribution.
I've started this problem by setting $\{X_1,...,X_n,Y_1,...Y_n\} = \{Z_1,...,Z_n\}$ Then the $\max(Y_n,X_n)= Z_{(2n)}$ would be distributed as $(\frac{z}{a})^{2n}$ and $\min(Y_n,X_n)= Z_{(1)}$ would be distributed as $1 - (1 - \frac{z}{a})^{2n}$
The densities can be found easily as $f_{Z_{1}}(z) = (2n)(1-\frac{z}{a})^{2n-1}\frac{1}{a}$ and $f_{Z_{(2n)}}(z) = (2n)(\frac{z}{a})^{2n-1} \frac{1}{a}$
This is where I'm having a hard time knowing where to go next now that these are calculated. I'm thinking it has to do something with a transformation, but I'm unsure...
 A: I will sketch the solution, here using a computer algebra system to do the nitty gritties ...
Solution
If $X_1, ... , X_n$ is a sample of size $n$ on parent  $X \sim \text{Uniform}(0,a)$, then the pdf of the sample maximum is:  $$f_{n}(x) = \frac{n}{a^n} x^{n-1}$$ and similarly for $Y$. 
Approach 1: Find the joint pdf of $(X_{(n)},Y_{(n)})$
Since $X$ and $Y$ are independent, the joint pdf of the 2 sample maximums $(X_{(n)},Y_{(n)})$ is simply the product of the 2 pdf's, say $f_{(n)}(x,y)$:

Given  $Z_n= n\log\frac{\max(Y_{(n)},X_{(n)})}{\min(Y_{(n)},X_{(n)})}$. Then, the cdf of $Z_n$ is $P(Z_n < z)$ is:

where I am using the Prob function from the mathStatica package for Mathematica to automate. Differentiating the cdf wrt $z$ yields the pdf of $Z_n$ as standard Exponential.

Approach 2: Order statistics
We can use order statistics to 'by-pass' the mechanics of having to deal with the Max and Min functions.
Once again:  If $X_1, ... , X_n$ is a sample of size $n$ on parent  $X \sim \text{Uniform}(0,a)$, then the pdf of the sample maximum $W = X_{(n)}$ is, say, $f_n(w)$:  

The sample maximums $X_{(n)}$ and $Y_{(n)}$ are just two independent drawings from this distribution of $W$; i.e. the $1^{st}$ and $2^{nd}$ order statistics of $W$ (in a sample of size 2) are just what we are looking for:


*

*$W_{(1)} = \min(Y_{(n)},X_{(n)})$ 

*$W_{(2)} = \max(Y_{(n)},X_{(n)})$ 
The joint pdf of $(W_{(1)}, W_{(2)})$, in a sample of size 2, say $g(.,.)$, is:

Given  $Z_n= n\log\frac{\max(Y_{(n)},X_{(n)})}{\min(Y_{(n)},X_{(n)})}$. Then, the cdf of $Z_n$ is $P(Z_n < z)$ is:

The advantage of this approach is that the probability calculation no longer involves the max/min functions, which may make the derivation (especially by hand) somewhat easier to express.
Other
As per my comment above, it appears you have misinterpreted the question ... 
We are asked to find:
$$Z_n= n\log\frac{\max(Y_{(n)},X_{(n)})}{\min(Y_{(n)},X_{(n)})}$$
where the denominator is min(xMax, yMax), ... not the minimum of all the $X$'s and $Y$'s. 
A: This problem can be solved from the definitions alone: the only advanced calculation is the integral of a monomial.

Preliminary observations
Let's work with the variables $X_i/a$ and $Y_i/a$ throughout: this does not change $Z_n$ but it makes $(X_1, \ldots, Y_n)$ iid with Uniform$(0,1)$ distributions, eliminating all distracting appearances of $a$ in the calculations.  Thus we may assume $a=1$ without any loss of generality.
Note that the independence of the $Y_i$ and their uniform distribution imply that for any number $y$ for which $0\le y \le 1$,
$$\Pr(y \ge Y_{(n)}) = \Pr(y \ge Y_1 , \ldots, y \ge Y_n) = \Pr(y \ge Y_1)\cdots \Pr(y \ge Y_n) = y^n,$$
with an identical result holding for $X_{(n)}$. For future reference, this allows us to compute
$$\mathbb{E}(2X_{(n)}^n) = \int_0^1 2x^n\mathrm{d}(x^n) = \int_0^1 2nx^{2n-1}\mathrm{d}x = 1.$$

Solution
Let $t$ be a positive real number.  To find the distribution of $Z_n$, substitute its definition and simplify the resulting inequality:
$$\eqalign{
\Pr(Z_n \gt t) &= \Pr(Z_n / n \gt t/n) = \Pr\left(\exp(Z_n/n) \gt e^{t/n}\right) \\
&=\Pr\left(\frac{\max(X_{(n)}, Y_{(n)})}{\min(X_{(n)}, Y_{(n)})} \gt e^{t/n}\right) \\
&= \Pr\left(e^{-t/n}{\max(X_{(n)}, Y_{(n)})} \gt {\min(X_{(n)}, Y_{(n)})}\right).
}$$
This event breaks into two equiprobable cases, depending on whether $X_{(n)}$ or $Y_{(n)}$ is the smaller of the two (and their intersection, with zero probability, can be ignored). Thus we need only compute the chance of one of these cases (say where $Y_{(n)}$ is the smaller) and double it.  Since $t\ge 0$, $0 \le e^{-t/n}X_{(n)} \le 1$, allowing us (upon letting $e^{-t/n}X_{(n)}$ to play the role of $y$) to apply the computations in the preliminary section:
$$\Pr(Z_n \gt t) =2\Pr\left(e^{-t/n}X_{(n)} \gt {Y_{(n)}}\right)
=2 \mathbb{E}\left[\left(e^{-t/n}{X_{(n)}}\right)^n\right] = e^{-t} \mathbb{E}\left[2{X_{(n)}^n}\right] = e^{-t}.
$$
That's what it means for $Z_n$ to have an Exp$(1)$ distribution.
