Probability, Variable change, diffeomorphism

Question

Let $(X,Y)$ be a random couple, such that : $X$ and $Y$ are independent. and $X$ has an uniform distribution on $[0,1]$, and $Y$ has an exponential distribution on $[0,+\infty[$ with parameter $(\lambda >0)$.

If we define two new random variables $U=X+Y$ and $V=X-Y$

And we wanted to define the desity of $(U,V)$.

My idea :

Since $X$ and $Y$ are independent, then $f_{(X,Y)}(x,y)=f_X(x).f_Y(y)=\lambda \exp{(-\lambda y)}.1_{[0,+\infty[}(y).1_{[0,1]}(x)$ and the diffeomorphism that we'll be using for variable change is :

$\phi(x,y)=(x+y,x-y)$, and it's inverse $\phi^{-1}(u,v)=(\frac{u+v}{2},\frac{u-v}{2})$

But I could not define the diffeomorphism's domain $D$ and $\phi (D)$.

Answer 1

$$0\leq X \leq 1 \quad 0\leq Y$$

so

$$U=X+Y\geq 0 \quad \quad -\infty < V=X-Y\leq 1 \quad \quad (1)$$

also $|J|=\frac{1}{2}$

$$f_{(U,V)}(u,v)=|J|f_{(X,Y)}(\frac{u+v}{2},\frac{u-v}{2})=\frac{1}{2}f_X(\frac{u+v}{2}).f_Y(\frac{u-v}{2})$$

$$=1_{[0,1]}(\frac{u+v}{2}) \times \lambda e^{(-\lambda \frac{u-v}{2})}\times 1_{[0,\infty)}(\frac{u-v}{2})$$

so

$$0\leq \frac{u+v}{2} \leq 1 \quad 0\leq \frac{u-v}{2} \quad \quad (2)$$

so by (1) and (2)

so

for $0 < u< 1 \quad \quad$ $-u<v<u$

for $ 1<u \quad \quad$ $-u<v<2-u$