Show that $Y_1=\frac{X_{(1)}}{X_{(2)}},Y_2=\frac{X_{(2)}}{X_{(3)}},\dots, Y_{n-1}=\frac{X_{(n-1)}}{X_{(n)}}$, and $Y_{(n)}=X_{(n)}$ are independent I am into order statistics lately, and I have a problem here.
Let $X_1,X_2,..,X_n$ be a random sample from $f(x)=1 , 0<x<1$. Show that $Y_1=\frac{X_{(1)}}{X_{(2)}},Y_2=\frac{X_{(2)}}{X_{(3)}},\dots, Y_{n-1}=\frac{X_{(n-1)}}{X_{(n)}}$, and $Y_n=X_{(n)}$ are independent.
I know all the procedures to do the problem.The only thing is that the Jacobian of transformation is coming pretty bad and is quite tedious to calculate.
Do you have a simpler method of calculating the Jacobian?
 A: [I think my previous idea still works. So I modified based on the deleted post here.]
Note that the joint density of $X_{(j)}$'s reads
$$
f_{X_{(1)},\cdots,X_{(n)}}(x_1,\cdots,x_n)=n!\mathbb{1}_{\left\{0\le x_1\le\cdots\le x_n\le 1\right\}}.
$$
Define $Z_j=\log X_{(j)}$ for $j=1,\cdots,n$, and the joint density of $Z_j$'s goes that
$$
f_{Z_1,\cdots,Z_n}(z_1,\cdots,z_n)=n!e^{z_1+\cdots+z_n}\mathbb{1}_{\left\{z_1\le\cdots\le z_n\le 0\right\}}.
$$
Define $W_j=\log Y_j$ for $j=1,\cdots,n$, and the relation between $Y_j$'s and $X_{(j)}$'s implies that
\begin{align}
W_1&=Z_1-Z_2,\\
W_2&=Z_2-Z_3,\\
&\cdots\\
W_{n-1}&=Z_{n-1}-Z_n,\\
W_n&=Z_n,
\end{align}
whose inverse reads
\begin{align}
Z_1&=W_1+\cdots+W_n,\\
Z_2&=W_2+\cdots+W_n,\\
&\cdots\\
Z_{n-1}&=W_{n-1}+W_n,\\
Z_n&=W_n.
\end{align}
Therefore, the joint density of $W_j$'s could be figured out as
\begin{align}
f_{W_1,\cdots,W_n}(w_1,\cdots,w_n)&=n!e^{w_1+2w_2+\cdots+nw_n}\mathbb{1}_{\left\{w_1+\cdots+w_n\le w_2+\cdots+w_n\le\cdots\le w_n\le 0\right\}}\\
&=n!e^{w_1+2w_2+\cdots+nw_n}\mathbb{1}_{\left\{w_1\le 0,w_2\le 0,\cdots,w_n\le 0\right\}}.
\end{align}
Finally, the joint distribution of $Y_j$'s could be recovered as
\begin{align}
f_{Y_1,\cdots,Y_n}(y_1,\cdots,y_n)&=n!y_1y_2^2\cdots y_n^n\mathbb{1}_{\left\{0\le y_1\le 1,0\le y_2\le 1,\cdots,0\le y_n\le 1\right\}}\\
&=n!\prod_{j=1}^ny_j^j\mathbb{1}_{\left\{0\le y_j\le 1\right\}}.
\end{align}
Based on this last result, we may conclude that $Y_j$'s are independent.
