I am studying this derivation, which makes the following statements:
$$ f(x,y)= \frac{1}{2\pi}(x,y) ^\frac{-1}{2} e^{-\frac{x+y}{2}}$$ $$A= xy $$ $$B=x+y$$ $$ x=\frac{B+\sqrt{B^2-4A}}{2}$$ $$ y=\frac{B-\sqrt{B^2-4A}}{2}$$
What I don't understand is this part:
Given $f(B)=2\times\frac{e^{-\frac{B}{2}}}{2\pi}\int_0^{\frac{B^2}{4}}A^{-\frac{1}{2}}(B^2-4A)^{-\frac{1}{2}}dA$, let $ A=\frac{B^2}{4}\sin^2(t)$.
Then $f(B)=2\times\frac{e^{-\frac{B}{2}}}{2\pi}\int_0^{\frac{\pi}{2}} \, dt$
I can't figure out the exact trigonometric substitution process behind this. How do i prove that $ A=\frac{B^2}{4}\sin^2(t)$ ?
UPDATE -Nov 4
Thanks to J.G. for your help. It's actually the process of deriving $ A=\frac{B^2}{4}\sin^2(t)$ that I am more curious about because this process is not shown on the Wikipedia link, and I am still at loss on how A is computed in such a way using $sin^2$(t).
Update - NOV 5
Thanks again J.G. i have a followup question about using $ A=\frac{B^2}{4}\sin^2(t)$ to get $\int_0^{\frac{\pi}{2}} \, dt$.
I got this outcome instead when I substituted A with $ \frac{B^2}{4}\sin^2(t)$ in $A^{-\frac{1}{2}}(B^2-4A)^{-\frac{1}{2}}\ $:
$$\frac{2}{B^2sin(t)cos(t)}\ $$
Can you please show the correct process so I will end up with $\int_0^{\frac{\pi}{2}} \, dt$ ?
The PDF of $B$ is shown to be $\frac{1}{\pi}e^{-B/2}\int_0^{B^2/4}A^{-1/2}(B^2-4A)^{-1/2}dA$, but then that integral needs to be evaluated. The substitution $A=\frac{B^2}{4}\sin^2t$ was chosen because in view of $\sin^2t+\cos^2t$ it converts the expression to$$\frac{1}{\pi}e^{-B/2}\int_0^{\pi/2}\left(\frac{B^2}{4}\sin^2t\right)^{-1/2}(B^2\cos^2t)^{-1/2}\frac{B^2}{2}\sin t\cos tdt=\frac{1}{\pi}e^{-B/2}\int_0^{\pi/2}dt=\frac12 e^{-B/2}.$$
Let's explore the rationale behind this substitution a bit more. Write $C:=B^2-4A$ so $A+\frac{C}{4}=\frac{B^2}{4}$. that looks a bit like $\sin^2t+\cos^2t=1$, or equivalently $\frac{B^2}{4}\sin^2t+\frac{B^2}{4}\cos^2t=\frac{B^2}{4}$. We may as well identify $A$ with one of the terms on the left-hand side; either way, $t$ can be taken to vary from $0$ to $\pi/2$, so that the substitution is monotonic. Choosing $A=\frac{B^2}{4}\sin^2t$ instead of $A=\frac{B^2}{4}\cos^2t$ has the further advantage that $A$ is an increasing function of $t$, which reduces the risk of sign errors when transforming the integral.