Let $X_1, X_2, X_3, X_4$ be independent and $\operatorname{Exp}( \lambda )$. We have,
$$Y_1 := X_1+X_2+ X_3+ X_4$$
$$Y_2:= \frac{X_1 }{X_1+X_2}$$
$$Y_3:= \frac{X_1 + X_2 }{X_1+X_2+ X_3}$$
$$Y_4:=\frac{X_1 + X_2 + X_3}{X_1+X_2+ X_3+ X_4}$$
We are now supposed to show that these, $Y_i$ are independent and determine their joint distribution (of $(Y_1, Y_2, Y_3, Y_4)$).
My initial approach is to use the fact that,
$$Y_i = g_i(X_1,\dots,X_n)$$
$$X_i = h_i(Y_1,\dots,Y_n)$$
$$f_\mathbf{Y}(y_1,\dots,y_n) = f_\mathbf{X}(h_1(y_1,\dots,y_n),\dots,h_n(y_1,\dots,y_n)) \cdot |J|$$
$J$ being the Jacobian determinant. I start of with re-writing each $X_i$ as a function of $(Y_1,\dots, Y_1)$ and get, after some manipulations that,
$$X_1 = Y_1 Y_2 Y_3 Y_4$$
$$X_2 = Y_1 Y_3 Y_4 (1-Y_2)$$
$$X_3 = Y_1 Y_4(1-Y_3)$$
$$X_4 = Y_1(1-Y_4)$$
Computing all partial derivatives,$\frac{\partial x_j}{\partial y_i}$ , yields me the following matrix,
\begin{bmatrix} y_2y_3y_4 & y_1y_3y_4 & y_1y_2y_4 & y_1y_2y_3 \\ y_3y_4(1-y_2) & -y_1y_3y_4 & y_1y_4(1-y_2) & y_1y_3(1-y_2) \\ y_4(1-y_3) & 0 & -y_1y_4 & y_1(1-y_3) \\ 1-y_4 & 0 & 0 & -y_1 \end{bmatrix}
My linear algebra is a little rusty, but I believe the doing the following is valid. The aim is to perform row/column operations to obtain a triangular determinant. First I swap the columns: 1. Swap column #1 and #4. 2. Swap column #2 and #1, 3. Swap column #2 and #3. This yields
\begin{bmatrix} y_1y_3y_4 & y_1y_2y_4 & y_1y_2y_3 & y_2y_3y_4 \\ -y_1y_3y_4 & y_1y_4(1-y_2) & y_1y_3(1-y_2)& y_3y_4(1-y_2) \\ 0 & -y_1y_4 & y_1(1-y_3) & y_4(1-y_3) \\ 0 & 0 & -y_1 & 1-y_4 \end{bmatrix}
I then: 1. Add row #1 to row #2. 2. Add row #2 to row #3. 3. Add row #3 to row #4. I then have,
\begin{bmatrix}
y_1y_3y_4 & y_1y_2y_4 & y_1y_2y_3 & y_2y_3y_4\\
0 & y_1y_4 & y_1y_3 & y_3y_4 \\
0 & 0 & y_1 & y_4 \\
0 & 0 & 0 & 1
\end{bmatrix}
If I've done this correctly then,
$$
|J| = y_1^3 y_4^2 y_3
$$
We have that the joint p.d.f of $(X_1,X_2,X_3,X_4)$ is,
$$
f_\mathbf{X} = \lambda^4 e^{-\lambda(x_1+x_2+x_3+x_4)}
$$
We then get, after some manipulation $(x_1 + x_2 + x_3 + x_4 = y_1)$, that,
$$
f_\mathbf{Y} = \lambda^4 e^{-\lambda y_1} y_1^3 y_4^2 y_3 $$
The question is now, have I done this correctly (i.e. computed the joint distribution of $(Y_1, Y_2, Y_3, Y_4)$? If so, how do I show independence for the $Y_i$'s? Do I have to compute all the marginal distributions or is there a simpler approach?
Any hints on how to proceed would be greatly appreciated!
Thanks.