I am working on problem in my book, and I am having trouble determining the domain on which the transformed density function is defined. The random vector ((X, Y, Z)) has the density function given by:
$$ f(x, y, z) = (y - z) e^{-x}, \quad 0 \leqslant x, \ 0 \leqslant y \leqslant x, \ 0 \leqslant z \leqslant y. $$
I am asked to find the distribution of the variable (W = X - Y + Z) after a transformation ((u, v, w) = (x, y, -u + v + w_i)). The Jacobian of this transformation is found to be (1), and the density function of the transformed vector ((U, V, W)) is given by:
$$ g(u, v, w) = f(u, v, -u + v + w) = (u - v) e^{-w}. $$ However, I am having difficulty determining the domain on which (g) is defined. The inequalities are as follows (this is solution, but I can't figure how to do it for other excersises):
$$ \begin{align*} &0 \leqslant x \quad 0 \leqslant u \quad 0 \leqslant w \\ &0 \leqslant y \leqslant x \Longleftrightarrow \quad 0 \leqslant v \leqslant u \quad \omega \leqslant u \\ &0 \leqslant z \leqslant y \quad u - v \leqslant w \leqslant u \quad u - w \leqslant v \leqslant u \\ \end{align*} $$
I have calculated the marginal distribution with respect to (w) as:
$$ g_w(w) = \int_{-\infty}^{\infty} du \int_{-\infty}^{\infty} g(u, v, w) dv = \int_w^{\infty} du \int_{u-w}^u (u - w) e^{-w} dv = \int_w^{\infty} w(u - w) e^{-u} du = w e^{-w}, \quad w \geqslant 0. $$
Any guidance on how determin the valid range for (w) or any corrections to my approach would be greatly appreciated. Thank you!