If $X:=(X_1,X_2)$ has density $e^{-(x_1+x_2)}1_{\mathbb{R}_{\ge 0}^2}(x_1,x_2)$, then $X_1+X_2$ and $X_1/X_2$ are independent Let $X_1$ and $X_2$ be real-valued random variables, such that $X:=(X_1,X_2)$ has the density $$f_X(x_1,x_2)=\begin{cases}e^{-(x_1+x_2)}&\text{, if }x_1,x_2\ge 0\\0&\text{, otherwise}\end{cases}\tag{1}$$
Then


*

*$X_1+X_2$ and $X_1/X_2$ are independent

*$X_1+X_2$ and $X_1/(X_1+X_2)$ are independent



I know the following fact: If $X_i$ has a continuous density $f_{X_i}$, then $X_1$ and $X_2$ are independent iff $$f_X(x_1,x_2)=f(x_1)f(x_2)\tag{2}$$
Since $(1)$ is obviously continuous, this fact seems to be useful here. However, I'm unsure what exactly I really know about $X_1$,$X_2$,$X_1+X_2$,$X_1/X_2$ and $X_1/(X_1+X_2)$ if the only thing given is the density of $X$.
 A: Let $U=X_1+X_2$ and $V=\frac{X_1}{X_2}$. First, observe that $U,V \in (0,+\infty)$ and that $X_1, X_2$ are independent since $$f_{X_1,X_2}(x_1,x_2)=e^{-x_1}1_{\{x_1\ge0\}}\cdot e^{-x_2}1_{\{x_2\ge0\}}=f_{X_1}(x_1)f_{X_2}(x_2)$$ Second $$\begin{cases}U=X_1+X_2\\V=\frac{X_1}{X_2}\end{cases}\implies\begin{cases}X_1=\dfrac{UV}{V+1}\\X_2=\dfrac{U}{V+1}\end{cases}$$ which gives you $$det(J)=\begin{vmatrix}\frac{\partial X_1}{\partial U}&\frac{\partial X_1}{\partial V}\\\frac{\partial X_2}{\partial U}&\frac{\partial X_2}{\partial V}\end{vmatrix}=\begin{vmatrix}\frac{V}{V+1}&\frac{U}{(V+1)^2}\\\frac{1}{V+1}&-\frac{U}{(V+1)^2}\end{vmatrix}=-\frac{U}{(V+1)^2}$$ and therefore $$\begin{align*}f_{U,V}(u,v)&=f_{X_1,X_2}\left(\frac{uv}{v+1},\frac{u}{v+1}\right)\cdot|det(J)|=\exp\left(-\frac{uv}{v+1}-\frac{u}{v+1}\right)\cdot\frac{u}{(v+1)^2}\\&=ue^{-u}1_{\{u\ge0\}}\cdot\frac{1}{(v+1)^2}1_{\{v\ge0\}}=g(u)\cdot h(v)\end{align*}$$ which shows that $U,V$ are independent since $f_{U,V}$ is of "separating variables" with respect to $U,V$. Note, however that you should multiply (and divide) $g(u), h(v)$ with appropriate constants to obtain the marginal densities of $U,V$.

Work similarly for the second case, using the result of the first.
