Consider the ordered sample $X_{(1)} < X_{(2)} < X_{(3)}$ from a distribution with PDF $f_X(x) = 2x, 0<x<1$. Show that $Y_1 = \frac{X_{(1)}}{X_{(2)}}$, $Y_2 = \frac{X_{(2)}}{X_{(3)}}$ and $Y_3 = X_{(3)}$ are mutually independent.
So far I have found that $f_{Y_1, Y_2, Y_3}(y_1, y_2, y_3) = 48y_1y_2^3y_3^5$. I know I need to show that that $f_{Y_1, Y_2, Y_3}(y_1, y_2, y_3) = f_{Y_1}(y_1) f_{Y_2}(y_2) f_{Y_3}(y_3)$ for mutual independence.
My first approach was to use the joint distribution of the order statistics $$f_{X_{(1)}, X_{(2)}}(x_1, x_2) = 3!(1 - x_2^2)2x_12x_2 = 24x_1(x_2 - x_2^3)$$
and then perform a bivariate transformation with $Y_1 = \frac{X_{(1)}}{X_{(2)}}$ and $W_2 = X_{(1)}$. However, when I do this and then find the marginal $f_{Y_1}(y_1)$ of $f_{{Y_1}, {W_1}}(y_1, w_1)$ I end up with a ridiculous marginal PDF. Could someone point me in the right direction? Is my approach for finding $f_{Y_1}(y_1)$ incorrect?
You can get the marginal distributions directly from the joint distribution by integrating out the other variables, e.g.
$$ f_{Y_1}(y_1)=\int_0^1\mathrm dy_2\int_0^1\mathrm dy_3\,f_{Y_1,Y_2,Y_3}(y_1,y_2,y_3)=2y_1\;, $$
and likewise $f_{Y_2}(y_2)=4y_2^3$ and $f_{Y_3}(y_3)=6y_3$. In fact, you don't even have to perform these integrations, since it's already clear from the fact that $f_{Y_1,Y_2,Y_3}$ factors into functions of $y_1$, $y_2$ and $y_3$ that it's the product of the marginal distributions; the only thing you find out by performing the integrations is how to split up the normalization factor.