How to derive the solution of $F_S(x)=P \left ({|h|^{2} \le \frac { x \left ({1 + |g|^{2} \rho _{2} }\right)}{\phi \rho _{1}} }\right)$? [EDIT]
I came across a received signal-to-interference-plus-noise-ratio (SINR), $S$, of a wireless communication system as
\begin{align*}
 S = \frac{\phi|h|^2\rho_1}{1+|g|^{2} \rho _{2} }, \tag{1}
\end{align*}
where $\phi$ is the power allocation coefficient, $\rho_1$ is the SNR (signal-to-noise-ratio) of channel 1, $\rho_2$ is the SNR of channel 2, $h \sim\mathcal{CN}(0,\sigma^2)$ is channel 1's gain, and $g \sim\mathcal{CN}(0,\sigma^2)$ is channel 2's gain. Furthermore, $H= \mathbb{E}[|h|^2]$ and $G= \mathbb{E}[|g|^2]$.
The authors state that the CDF of the received SINR is
\begin{align*} F_S(x)=&P \left ({|h|^{2} \le \frac { x \left ({1 + |g|^{2} \rho _{2} }\right)}{\phi \rho _{1}} }\right), \end{align*}
where the RHS is found to be
\begin{align*} P \left ({|h|^{2} \le \frac { x \left ({1 + |g|^{2} \rho _{2} }\right)}{\phi \rho _{1}} }\right) = 1 - \frac {1}{1 + a x} e^{-\frac {x}{\phi H \rho _{1}}},\tag{2} \end{align*}
where  $a=\frac{G \rho_2}{\phi H\rho_1}$.
I understand that $z=h_r^2+h_i^2=\frac{2}{\sigma^2} |h|^2$ is a chi squared distributed random variable with $k=2$ degrees of freedom. Then, the CDF for $k=2$ is
\begin{equation}
F(z;k=2)=1-e^{-z/2}, \tag{3}
\end{equation}
which allows us to obtain the structure of (2).
However, I am unable to figure out how the $\frac {1}{1 + a x}$ in $\frac {1}{1 + a x} e^{-\frac {x}{\phi H \rho _{1}}}$ was derived.
Does the fact that $g$ is a complex random variable also affect the outcome?
Could someone help me understand how the expression in (2) was derived?
Thanks in advance.
 A: The integral turns out to be easier than it looks.
Let's simplify notation a little.  By choosing suitable units for the variables, we may make $\sigma^2=1/2,$ entailing $H=G=1.$  Consequently everything depends on the values of $$\frac{1}{\phi\rho_1} = \beta \gt 0,$$ say, and $$a = \rho_2 \beta \gt 0.$$ $Y=|h|^2$ and $X=|g|^2$ are independent Exponential variables, which means that for any number $\lambda \ge 0,$
$$\Pr(X \le \lambda) = \Pr(Y \le \lambda) = 1 - e^{-\lambda}.$$
The problem is for any $\lambda\ge 0$ to compute
$$\begin{aligned}
\Pr(Y \le \beta \lambda(1 + \rho_2 X)) &= E\left[1 - e^{-( \beta\lambda(1 + \rho_2 X))}\right] \\
&= \int_0^\infty \left(1 - e^{-( \beta\lambda(1 + \rho_2 x))}\right) \,\mathrm{d}\left(1-e^{-x}\right) \\
&= \int_0^\infty e^{-x}\,\mathrm{d}x - e^{-\lambda}\int_0^\infty e^{- \beta\lambda \rho_2 x - x}\,\mathrm{d}x \\
&= 1 - e^{-\beta\lambda} \frac{1}{\beta\lambda\rho_1+1} \\
&= 1 - \frac{1}{1 + a\lambda}e^{-\lambda/(\phi \rho_1)}.
\end{aligned}$$
In the question, "$x$" is used instead of "$\lambda:$" because $H=1,$ this result obviously agrees with the quoted formula in the question, QED.
