Let $X_1$ and $X_2$ be two independent random variables, each with pdf $f(x) = e^{-x}$, $0<x<\infty$ consider $Y1=X1-X2$ and $Y2=X1 + X2$. Find the joint pdf of $Y_1$ and $Y_2$.
I made the transform and afterwards used the Jacobian and got $1/2$ however the answer to the question is $1/2 e^{-y_2}$. My question is where is the $e^{-y_2}$ coming from? I understand I had to change $e^{-x}$ in terms of $y$. I just don't understand where $-y_2$ is coming from as the transform...not sure if I'm making much sense thanks in advance for the help!
Because $Y_1=X_1-X_2$ and $Y_2=X_1+X_2$, then $Y_1+Y_2=2X_1$ and $Y_2-Y_1=2X_2$
Thusly, and since $X_1,X_2$ are independent, we have:
$$\begin{align}f_{Y_1,Y_2}(y_1,y_2) &= \left\lVert\frac{\partial(\frac{y_1+y_2}2,\frac{y_2-y_1}2)}{\partial(y_1,y_2)}\right\rVert \cdot f_{X_1,X_2}(\frac{y_1+y_2}2,\frac{y_2-y_1}2)\\[1ex] &= \tfrac 12\cdot f_{X_1}(\frac{y_1+y_2}2)\cdot f_{X_2}(\frac{y_2-y_1}2) \\[1ex]\end{align}$$
Having correctly obtained the Jacobian's absolute determinant you should have reached this point . The rest is just substitution and algebra.