I am trying to understand why $V+W$ is independent from $\frac{V}{V+W}$. With $V \sim \chi^2_{(1)}$ and $W \sim \chi^2_{(2)}$. I do not see how this comes about. Note: $V$ and $W$ are independent
I tried deriving the density of $\frac{V}{V+W}$, which to me seems to be some form of Beta, given that cannot be more than $1$ or less than $0$.
$$F_{\frac{V}{V+W}}(\xi)=\int_0^\infty \int_0^{\frac{v-\xi}{1-\xi}}\frac{v^{-1/2}}{2^{1/2}\Gamma(1/2)} \frac{e^{-w^2 /2}}{2} dv dw $$ $$=\frac{1}{2^{1/2}\Gamma(1/2)}[\int_0^\infty v^{-1/2} e^{-1/2}] - \int_0^\infty v^{-1/2}e^{\frac{-v}{2(1-\xi)}}dv$$ $$= 1- \frac{1}{2^{1/2}\Gamma(1/2)}\int_0^\infty v^{-1/2}e^{\frac{-v}{2(1-\xi)}}dv$$ Which now I do not think it is right. I mainly want to be able to understand why $V+W$ is independent of $\frac{V}{V+W}$
Due to independence of $V$ and $W$, joint density of $(V,W)$ is
\begin{align} f_{V,W}(v,w)&=f_V(v)f_W(w) \\\\&=\frac{e^{-v/2}}{\sqrt{2\pi v}}\mathbf1_{v>0}\frac{e^{-w/2}}{2}\mathbf1_{w>0} \\\\&=\frac{1}{2\sqrt{2\pi}}\frac{e^{-(v+w)/2}}{\sqrt{v}}\mathbf1_{v>0,w>0} \end{align}
Transform $(V,W)\to(X,Y)$ such that $$X=\frac{V}{V+W}\quad,\quad Y=V+W$$
Then, $$v=xy\quad,\quad w=y(1-x)$$
Clearly, $$v>0\,,\,w>0\implies 0<x<1\,,\,y>0$$
The Jacobian determinant is $$J=\det\begin{bmatrix}\dfrac{\partial v}{\partial x}&\dfrac{\partial v}{\partial y}\\\dfrac{\partial w}{\partial x}&\dfrac{\partial w}{\partial y}\end{bmatrix}=\cdots$$
So the joint density of $(X,Y)$ is of the form
$$f_{X,Y}(x,y)=f_{V,W}(xy,y(1-x))|J|=\cdots$$
Just from the above expression, establish that $f_{X,Y}(x,y)$ can be factored as the product of two densities. Those are the marginal densities of $X$ and $Y$. Hence the independence of $X$ and $Y$ would be proved.