Distribution of ratio of minimum and maximum of uniform I have a problem in which $X_1, X_2, .., X_n$ follows $U(0, \theta)$. We have $X_{(n)}$ as the maximum and $X_{(1)}$ as the minimum. I am required to compute the correlation between $X_{(n)}$ and the ratio $\frac{X_{(1)}}{X_{(n)}}$.
To do that, I will  first need the distributions of $X_{(n)}$ and $X_{(1)}$. I know I can find the distribution using the formula:
$F(x_{n}) = n[F(x)]^{n}$ and $F(x_{1}) = n[1-F(x)]^{n}$. Even if I have these two, How will I obtain the distribution of $\frac{X_{(1)}}{X_{(n)}}$. One way could have  been to multiply the densities of max and min and then obtain the joint and then obtain the densities of the ratio by doing jacobian transformation but i am not sure we are allowed to do that here because these two densities are ordered statistics (hence dependent) and we can't just multiply them to get the joint $X_{(n)}$ and $X_{(1)}$. Then, How can we compute the distribution of the ratio?
 A: You need the joint distribution of $X_{(1)}$ and $X_{(n)}$.
see here for the derivation in the case of Unif(0,1).
The joint density is
$$f_{X_{1},X_{n}}(u,v)=n! \frac{(v-u)^{n-2}}{(n-2)!} \theta^{-n}$$
You can find the expected value of any function of the two random variables by integrating the function of $u$ and $v$ over the two-dimensional integral with limits $0<u<v<\theta$.
For example,
$$E\left[\frac{X_{1}}{X_{n}} \right]=\int_{0}^\theta \int_{u}^\theta \frac{u}{v} f_{X_{1},X_{n}}(u,v)dv du=\frac{1}n$$
The correlation is the covariance divided by the square root of the product of the two variances.
$$\text{Cov}\left(X_{n},\frac{X_{1}}{X_{n}} \right)=
E\left[X_{1} \right]-E\left[X_{n} \right] E\left[\frac{X_{1}}{X_{n}} \right]$$
A: To answer the question about correlation, we can note that $\frac{X_{(1)}}{X_{(n)}}=\frac{X_{(1)}/\theta}{X_{(n)}/\theta}=\frac{Y_{(1)}}{Y_{(n)}}$ where $Y_1,\ldots,Y_n$ are i.i.d uniform on $(0,1)$. As such the distribution of the ratio is free of $\theta$, hence it is an ancillary statistic. Moreover $X_{(n)}$ is a complete sufficient statistic for the $U(0,\theta)$ family of distributions with $\theta\in \mathbb R^+$. Hence by Basu's theorem, $\frac{X_{(1)}}{X_{(n)}}$ and $X_{(n)}$ must be independent.
We can also use the independence to find the expectation of the ratio:
$$E\left[X_{(1)}\right]=E\left[\frac{X_{(1)}}{X_{(n)}}\cdot X_{(n)}\right]=E\left[\frac{X_{(1)}}{X_{(n)}}\right]E\left[X_{(n)}\right]\,,$$
so that $$E\left[\frac{X_{(1)}}{X_{(n)}}\right]=\frac{E\left[X_{(1)}\right]}{E\left[X_{(n)}\right]}=\frac{E\left[Y_{(1)}\right]}{E\left[Y_{(n)}\right]}$$
For the exact distribution of $\frac{X_{(1)}}{X_{(n)}}=\frac{Y_{(1)}}{Y_{(n)}}$, one can just use the joint density of $(Y_{(1)},Y_{(n)})$:
$$f_{Y_{(1)},Y_{(n)}}(x,y)=n(n-1)(y-x)^{n-2}\mathbf 1_{0<x<y<1}$$
