Order statistic and independence "If $Y_1, Y_2, ..., Y_n$ are independent, uniformly distributed random variables on the interval $[0, \theta]$, show that $U=Y_{(1)}/Y_{(n)}$ and $Y_{(n)}$ are independent."
I have already found $$f_{Y_{(1)}}(y) = \frac{n}{\theta}(1-\frac{1}{\theta}y)^{n-1}$$ and $$f_{Y_{(n)}}(y) = \frac{n}{\theta^n}y^{n-1}$$
I was thinking of showing independence by showing $f(u, y_{(n)}) = f_U(u)f_{Y_{(n)}}(y)$, but I'm not sure if this even makes sense?
Thanks for any help.
 A: Approach 1: I would start with the joint distribution of $Y_{(1)}$ and $Y_{(n)}$.  (Since you know how to find the individual order statistic distributions you can use a similar argument to get the joint distribution.)   This is $$f_{Y_{(1)},Y_{(n)}}(y_1, y_n) = \frac{n(n-1)}{\theta^n}(y_n-y_1)^{n-2}, \:\:\:\: 0 < y_1 < y_n < \theta.$$
Then do a bivariate transformation to obtain $f_{U,Y_{(n)}}(u,y_n)$.  The Jacobian of the transformation is just $Y_{(n)}$, and so you get
$$f_{U,Y_{(n)}}(u,y_n) = \frac{n(n-1)}{\theta^n}(y_n - uy_n)^{n-2} y_n, \:\:\: 0 < u < 1, \: 0 < y_n < \theta.$$
Since $f_{U,Y_{(n)}}(u,y_n)$ factors into a function of $u$ and a function of $y_n$, $U$ and $Y_{(n)}$ must be independent.
You can fill in the details, but this is the basic argument for this approach.

Approach 2: Because I can't stop myself, let me also give the argument described by did. :)
Obtain the conditional distribution $f_{Y_{(1)}|Y_{(n)}}(y_1|y_n)$ by dividing $f_{Y_{(1)},Y_{(n)}}(y_1,y_n)$ by the marginal distribution for $f_{Y_{(n)}}$.  This yields
$$f_{Y_{(1)}|Y_{(n)}}(y_1|y_n) = \frac{(n-1)(y_n-y_1)^{n-2}}{y_n^{n-1}}, \:\:\: 0 < y_1 < y_n.$$
Then calculate $P(U < u | Y_{(n)})$ from $f_{Y_{(1)}|Y_{(n)}}(y_1|y_n)$.  This is
$$\begin{align}
P(U < u | Y_{(n)}) &= P(Y_{(1)} < u Y_{(n)} | Y_{(n)} = y_n) \\
&= \int_0^{u y_n} \frac{(n-1)(y_n-y_1)^{n-2}}{y_n^{n-1}} \, dy_1 \\ 
&= \frac{-1}{y_n^{n-1}} \left[(y_n - y_1)^{n-1} \right]_0^{u y_n} \\
&= \frac{-1}{y_n^{n-1}} \left[(y_n - uy_n)^{n-1} - y_n^{n-1} \right] \\
&= \frac{-1}{y_n^{n-1}} y_n^{n-1} \left[(1 - u)^{n-1} - 1\right]\\
&= 1 - (1-u^{n-1}).
\end{align}$$
Since $P(U < u | Y_{(n)})$ does not depend on $Y_{(n)}$, $U$ and $Y_{(n)}$ are independent.
