I am studying this proof and I don't get why the pdf is doubled.
According to Wikipedia, the pdf of A and B is expressed as below.
\begin{align} f_{A,B}(a,b) &= 2f_{X,Y}(x,y) J((a,b)\rightarrow(x,y)) \\ &= 2\frac{1}{2\pi} \frac{1}{\sqrt{ A }}\left\{ \exp{\left( -\frac{B}{2} \right)} \right\} \left( \frac{1}{\sqrt{ B^2-4A }} \right) \end{align}
And this doubled operation is reasoned like this.
Since the two variable change policies are symmetric, we take the upper one and multiply the result by 2.
What are the two variable change policies and how does that affect this proof?
Thanks, hopeless. I see why I have struggled. My confusion comes from the $\pm$ sign on the Jacobian. It is just a directional difference of vectors, so it doesn't make a significant difference in probability sense. So we must take both cases into account.
This is my part of the derivation in case you are wondering.
$$ \begin{align} J((a,b)\rightarrow(x,y)) &= \begin{vmatrix} \frac{ \partial x }{ \partial a } & \frac{ \partial x }{ \partial b } \\ \frac{ \partial y }{ \partial a } & \frac{ \partial y }{ \partial b } \end{vmatrix} \\ &= \begin{vmatrix} \pm \frac{-4}{4\sqrt{ B^2-4A }} & \frac{1}{2} \pm \frac{2B}{4\sqrt{ B^2-4A }} \\ \mp \frac{-4}{4\sqrt{ B^2-4A }} & \frac{1}{2} \mp \frac{2B}{4\sqrt{ B^2-4A }} \\ \end{vmatrix} \\ &= \begin{vmatrix} \pm \frac{-1}{\sqrt{ B^2-4A }} & \frac{1}{2} \pm \frac{B}{2\sqrt{ B^2-4A }} \\ \mp \frac{-1}{\sqrt{ B^2-4A }} & \frac{1}{2} \mp \frac{B}{2\sqrt{ B^2-4A }} \\ \end{vmatrix} \\ &=\pm \frac{-1}{\sqrt{ B^2-4A }} \left( \frac{1}{2} \mp \frac{B}{2\sqrt{ B^2-4A }} \right) - \left( \frac{1}{2} \pm \frac{B}{2\sqrt{ B^2-4A }} \right)\left( \mp \frac{-1}{\sqrt{ B^2-4A }} \right) \\ &= \pm \frac{-1}{\sqrt{ B^2-4A }}\left( \frac{1}{2} \mp \frac{B}{2\sqrt{ B^2-4A }}+ \frac{1}{2} \pm \frac{B}{2\sqrt{ B^2-4A }}\right) \\ &= \mp \frac{1}{\sqrt{ B^2-4A }} \end{align} $$
Then,
$$ \begin{align} f_{A,B}(a,b) &= 2f_{X,Y}(x,y) \mid J((a,b)\rightarrow(x,y)) \mid \\ &= 2\frac{1}{2\pi} \frac{1}{\sqrt{ A }}\left\{ \exp{\left( -\frac{B}{2} \right)} \right\} \left( \frac{1}{\sqrt{ B^2-4A }} \right) \end{align} $$