Independence and Order Statistics I have a problem at hand, which I am  not being able to proceed. Can someone help me begin?
$Y_1<Y_2<Y_3$ :An order statistic of size 3 from distribution having pdf
$$ f(x)=2x\ \ \ 0<x<1$$ Also, define $$U_1={Y_1\over Y_2} \ \ \text{and }\ \ \ U_2={Y_2\over Y_3}$$
The task is to compute the joint pdf of $U_1\  \&\ U_2$.
My work:
I have found out the marginal of $U_1\ \&\ U_2$. 
$$P(U_1\le u_1)=\int_0^1\int_0^{u_1y_2}f_{Y_1,Y_2}(y_1,y_2)dy_1dy_2$$
$$P(U_2\le u_2)=\int_0^1\int_0^{u_2y_3}f_{Y_2,Y_3}(y_2,y_3)dy_2dy_3$$
What step am I to do next?
 A: Here is an exact symbolic solution which traces out the steps required ... here using automated tools to do the nitty gritties
Let $(X_1, X_2, X_3)$ denote a sample of size 3 from parent pdf $f(x)$:

Then, the joint pdf of the ordered sample $(X_{(1)}, X_{(2)}, X_{(3)})$ is say $g(x_1,x_2,x_3)$:

where I am using the OrderStat function form the mathStatica package for Mathematica.
The joint cdf of $(U_1, U_2)$ is $P\big(\frac{X_{(1)}}{X_{(2)}}<u_1, \,\frac{X_{(2)}}{X_{(3)}}<u_2\big)$:

The joint pdf of $(U_1, U_2)$ is derived by simply differentiating the cdf wrt $u_1$ and $u_2$:

Finally, as a quick Monte Carlo check, here is a comparison of:


*

*the exact theoretical solution derived (the joint pdf - the orange surface)

*plotted against an empirical Monte Carlo simulated joint pdf (3D histogram):

A: Here is a guide to solving this problem (and others like it).  I use simulated values to illustrate, so let's begin by simulating a large number of independent realizations from the distribution with density $f$.  (All the code in this answer is written in R.)
n <- 4e4 # Number of trials in the simulation
x <- matrix(pmax(runif(n*3), runif(n*3)), nrow=3)

# Plot the data
par(mfrow=c(1,3))
for (i in 1:3) {
  hist(x[i, ], freq=FALSE, main=paste("i =", i))
  curve(f(x), add=TRUE, col="Red", lwd=2)
}


The histograms show $40,000$ independent realizations of the first, second, and third elements of the datasets.  The red curves graph $f$.  That they coincide with the histograms confirms the simulation is working as intended.
You need to work out the joint density of $(Y_1, Y_2, Y_3)$. Since you're studying order statistics, this should be routine--but the code gives some clues, because it plots their distributions for reference.
y <- apply(x, 2, sort)

# Plot the order statistics.
f <- function(x) 2*x
ff <- function(x) x^2
for (i in 1:3) {
  hist(y[i, ], freq=FALSE, main=paste("i =", i))
  k <- factorial(3) / (factorial(3-i)*factorial(1)*factorial(i-1))
  curve(k * (1-ff(x))^(3-i) * f(x) * ff(x)^(i-1), add=TRUE, col="Red", lwd=2)
}


The same data have been reordered within each of the $40,000$ datasets.  On the left is the histogram of their minima $Y_1$, on the right their maxima $Y_3$, and in the middle their medians $Y_2$.
Next, compute the joint distribution of $(U_1, U_2)$ directly.  By definition this is
$$F(u_1, u_2) = \Pr(U_1 \le u_1, U_2 \le u_2) = \Pr(Y_1 \le u_1 Y_2, Y_2 \le u_2 Y_3).$$
Since you have computed the joint density of $(Y_1, Y_2, Y_3)$, this is a routine matter of doing the (triple) integral expressed by the right-hand probability.  The region of integration must be $$0 \le Y_1 \le u_1 Y_2,\ 0 \le Y_2 \le u_2 Y_3,\ 0 \le Y_3 \le 1.$$
The simulation can give us an inkling of how $(U_1, U_2)$ are distributed: here is a scatterplot of the realized values of $(U_1, U_2)$.  Your theoretical answer should describe this density.
par(mfrow=c(1,1))
u <- cbind(y[1, ]/y[2, ], y[2, ]/y[3, ])
plot(u, pch=16, cex=1/2, col="#00000008", asp=1)


As a check, we may look at the marginal distributions and compare them to the theoretical solutions.  The marginal densities, shown as red curves, are obtained as $\partial F(u_1, 1)/\partial u_1$ and  $\partial F(1, u_2)/\partial u_2$.
par(mfrow=c(1,2))
hist(u[, 1], freq=FALSE); curve(2*x, add=TRUE, col="Red", lwd=2)
hist(u[, 2], freq=FALSE); curve(4*x^3, add=TRUE, col="Red", lwd=2)
par(mfrow=c(1,1))


It is curious that $U_1$ has the same distribution as the original $X_i$.
