Let $X_1$, $X_2$ be independent random variables with the density $f(x)=e^{-x} \mathbb{1}_{(x>0)}(x)$. I would like to ask if I have evaluated correctly the joint distibution of $(Y_1,Y_2)=(X_1-X_2,X_1+X_2).$
My solution
I used the formula $$ f_{Y_1,Y_2}(y_1,y_2)=f_{X_1,X_2}(x_1,x_2)\cdot \left|J(x_1,x_2)\right|^{-1}. $$
We have $X_1\sim \mathcal{Exp}(1)$, $X_2\sim \mathcal{Exp}(1)$, $Y_1=X_1-X_2$, $Y_2=X_1+X_2$,
$J(x_1,x_2)=\begin{vmatrix} \dfrac{\partial Y_1}{\partial X_1} & \dfrac{\partial Y_1}{\partial X_2} \\ \dfrac{\partial Y_2}{\partial X_1} & \dfrac{\partial Y_2}{\partial X_2} \end{vmatrix}=\begin{vmatrix} \dfrac{\partial (X_1-X_2)}{\partial X_1} & \dfrac{\partial (X_1-X_2)}{\partial X_2} \\ \dfrac{\partial (X_1+X_2)}{\partial X_1} & \dfrac{\partial (X_1+X_2)}{\partial X_2} \end{vmatrix}=\begin{vmatrix} 1 & -1 \\ 1 & 1 \end{vmatrix}=1\cdot 1-1\cdot(-1)=1+1=2$.
Because of the fact that $X_1=\dfrac{Y_1+Y_2}{2}$, $X_2=\dfrac{Y_2-Y_1}{2}$ we have
$f_{Y_1,Y_2}(y_1,y_2)=f_{X_1,X_2}(x_1,x_2)\cdot \left|J(x_1,x_2)\right|^{-1}=\frac12 e^{-\frac{y_1+y_2}{2}} e^{-\frac{y_2-y_1}{2}}=\frac12 e^{-y_2}.$
Is this answer correct? I am afraid because of the fact that integral from $\frac12 e^{-y_2}$ over whole space is not equal to $1$.
We don't have $y_1$ in the formula of density so maybe the answer $\frac12$ is suitable, but I wanted to make sure.