Let the joint distribution of $X$ and $Y$ be given by:
$f(x,y) = e^{-x}$ if $0 < y \leq x < \infty$
Define $Z = X+Y$ and $W = X-Y$
- Find the joint pdf of $Z$ and $W$
- Calculate $f_{ZW} (0.1,0.5)$
Attempt
I think I should calculate the pdf of $Z=X+Y$ and then the pdf of $W=X-Y$. I can then multiply the two pdf's if they are independent, i think they are independent but I am not sure.
With $X$ and $Y$ two continuous r.v.'s with joint pdf $f_{XY}(x,y)$ and you want to transform to two new variables
$$z=g(x,y) \:\: \text{and} \:\: w=h(x,y) \:\: \:\: \:\:\:\:\:\: (1)$$
then if the transformations is one-to-one and has inverse transformations $x=q(z,w)$ and $y=r(z,w)$ then the joint pdf of $Z$ and $W$ is
$$f_{ZW}(z,w) = f_{XY}(x,y)|J(x,y)|^{-1} \:\:\:\:\:\:\:\:\:(2)$$
where $x=q(z,w)$ and $y=r(z,w)$ and
$$J(x,y)= \begin{vmatrix} \frac{\partial{g}}{\partial{x}} & \frac{\partial{g}}{\partial{y}} \\ \frac{\partial{h}}{\partial{x}} & \frac{\partial{h}}{\partial{y}} \\ \end{vmatrix} = \begin{vmatrix} \frac{\partial{z}}{\partial{x}} & \frac{\partial{z}}{\partial{y}} \\ \frac{\partial{w}}{\partial{x}} & \frac{\partial{w}}{\partial{y}} \\ \end{vmatrix} $$
which is the Jacobian of the transformation $(1)$. If we define
$$\overline{J}(x,y)= \begin{vmatrix} \frac{\partial{q}}{\partial{z}} & \frac{\partial{q}}{\partial{w}} \\ \frac{\partial{r}}{\partial{z}} & \frac{\partial{r}}{\partial{w}} \\ \end{vmatrix} = \begin{vmatrix} \frac{\partial{x}}{\partial{z}} & \frac{\partial{x}}{\partial{w}} \\ \frac{\partial{y}}{\partial{z}} & \frac{\partial{y}}{\partial{w}} \\ \end{vmatrix} $$
then $(2)$ can be expressed as
$$f_{ZW}(z,w) = f_{XY}[q(z,w),r(z,w)]|\overline{J}(x,y)| $$
And it should be straight forward to finish from here.