Let $(X_1,X_2,X_3)$ be a random vector with joint density $$\begin{cases} 8x_1x_2x_3, & x_1,x_2,x_3 \in (0,1) \\ 0, & x_1,x_2,x_3 \in (0,1)^c \end{cases}$$
Let $Y_1 =X_1$, $Y_2=X_1X_2$ and $Y_3=X_1X_2X_3$ and $Y=(Y_1,Y_2,Y_3)$. I have managed to find the density of $Y$, which is $f_Y(y)=\frac{8y_3}{y_1y_2}$, but now i must prove that this really is a density function and i'm having some troubles with the boundaries of the triple integral. Any hint? Thanks!
The conditions were
$$0<X_1<1\\ 0<X_2<1\\ 0<X_3<1$$
If we solve for $X_1,X_2,X_3$ in terms of $Y_1,Y_2,Y_3$ we get
$$X_1=Y_1\\ X_2=Y_2/X_1=Y_2/Y_1\\ X_3=Y_3/(X_1X_2)=Y_3/Y_1(Y_1/Y_2)=Y_3/Y_2$$
Thus the limits are
$$0<Y_1<1\\ 0<Y_2/Y_1<1\\ 0<Y_3/Y_2<1$$
which can be written as
$$0<Y_1<1\\ 0<Y_2<Y_1\\ 0<Y_3<Y_2$$
Thus you could integrate as
$$\int_0^1\int_0^{y_1}\int_0^{y_2} f(y_1,y_2,y_3)dy_3dy_2dy_1$$