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:
$$ f_{X, Y}(x, y) =f_X(x) f_Y(y) = \begin{cases} 2e^{-x-2y} \ & \mbox{ if } x \geq 0, y \geq 0; \\ 0 \ & \mbox{ elsewhere}. \end{cases} $$
However I don't know how to use this to calculate $f_{Z, W}$
Best Answer
The map $g:(x,y) \mapsto (x-2y,x)$ is a differentiable and invertible function between $(0,\infty)\times (0,\infty)$ and $R=\{(z,w) | z< w \text{ and } w>0\}$, 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 $$ \frac{dg^{-1}}{d(z,w)}(z,w) = \begin{pmatrix}0 & 1 \\ -\frac12 & \frac12 \end{pmatrix},$$ 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 $(z,w) \in \{(z,w) | z< w \text{ and } w>0\}$. Just to verify, that this is in fact a valid density we compute $$ \int_0^\infty \int_{-\infty}^w e^{z-2w} dzdw = 1$$