- The probability density functions of two independent random variables $X$ & $Y$ is known.
- To be more specific $X$ & $Y$ have Rayleigh and Rician distribution respectively.
- I want to find the following probability: $P\left(\frac{XY}{X + Y + 1} < c\right)$, where $c$ is a constant.
After slight reframing the problem, I have put it as $$ P\left[\left(\frac{1}{X} + \frac{1}{Y} + \frac{1}{XY}\right) > \frac{1}{c}\right] $$ Can anyone suggest a simple way to arrive at the final solution $?$.
Sketch for a solution: let $Z:= \frac{XY}{X+Y+1}$ where $X$ and $Y$ are independent and have density, then you have that
$$ \Pr [Z\leqslant c]=\int_{\mathbb{R}^2}\mathbf{1}_{A_c}(x,y)f_X(x)f_Y(y)\mathop{}\!d (x,y) $$
for $A_c:=\{(x,y)\in \mathbb{R}^2:xy/(1+x+y)\leqslant c\}$. Then when $(1+x+y)(c-x)>0$ we have that $$ \frac{xy}{1+x+y}\leqslant c \iff y\geqslant c\frac{1+x}{x-c}\\ \text{ and }(1+x+y)(c-x)>0 \iff c+(c-1-y)x-x^2>0 $$ Then, solving the last inequality with the roots, we can divide $A_c$ in a finite number of disjoint sets with the form $\{(x,y)\in \mathbb{R}^2: y\geqslant f(x)\}$ for some functions $f$, then we can apply Fubini's theorem easily in each of these sets to compute each integral (the set where $(1+x+y)(c-x)=0$ have Lebesgue measure zero so we can ignore it).