Joint density of two exponential random variables For independent random variables X ∼ Exp(1) and Y ∼ Exp(2), find the density of (Z, W) = (X-2Y, X).

My approach:

Since for any exponential distribution with parameter $\lambda$ the function is
$f(x) = \lambda e^{-\lambda x}$

$f_X(x) =  e^{-x}$

$f_Y(y) = 2e^{-2y}$

Therefore the joint density function is: 
$$ f_{X, Y}(x, y) =f_X(x) f_Y(y) = \begin{cases} 2e^{-x-2y} \ &amp; \mbox{ if } x \geq 0, y \geq 0;  \\ 0 \ &amp; \mbox{ elsewhere}. \end{cases} $$ 

However I don't know how to use this to calculate $f_{Z, W}$
The map $g:(x,y) \mapsto (x-2y,x)$ is a differentiable and invertible function between $(0,\infty)\times (0,\infty)$ and $R=\{(z,w) | z&lt; w \text{ and } w&gt;0\}$, so first of all we get that the support for $(Z,W)=(X-2Y,X)$ must be $R$. 

The transformation theorem for probability densities states that:

$$f_{Z,W}(z,w) = f_{X,Y}(g^{-1}(z,w)) |det(\frac{dg^{-1}}{d(z,w)}(z,w))|,$$
where $\frac{dg^{-1}}{d(z,w)}(z,w)$ is the jacobian of $g^{-1}$.

(see https://en.wikipedia.org/wiki/Probability_density_function#Vector_to_vector) 

We first compute $g^{-1}(z,w)= (w,\frac{w-z}{2})$ and the jacobian
$$ \frac{dg^{-1}}{d(z,w)}(z,w) = \begin{pmatrix}0 &amp; 1 \\ -\frac12 &amp; \frac12 \end{pmatrix},$$
which has determinant $\frac12$ for all $z,w$. We now plugin, and get
$$ f_{Z,W}(z,w) = \frac12 f_{X,Y} ((w,\frac{w-z}{2})) = e^{-w}e^{-2\frac{w-z}{2}}=e^{z-2w}.$$
for all $(z,w) \in \{(z,w) | z&lt; w \text{ and } w&gt;0\}$. Just to verify, that this is in fact a valid density we compute 
$$ \int_0^\infty \int_{-\infty}^w e^{z-2w} dzdw = 1$$
If I'm not wrong, you've formed the joint density function under the assumption that the random variables $X$ and $Y$ are independent.

In the formula for $f_{X,Y}(x,y)$, let us put $X-2Y$ in place of $X$, $X$ in place of $Y$, and let us also put $x-2y$ in place of $x$ and $x$ in place of $y$ to obtain
$$ f_{Z, W}(z, w) = 2e^{- (x-2y) - x } = 2 e^{2y-2x} $$
if $x-2y \geq 0$ and $x\geq 0$, that is, if $x \geq \max(2y, 0)$, and $0$ otherwise.

However, I'm not completely sure if my solution is correct.