I need your help with the following formula. If we have:
\begin{equation} \gamma=\frac{E_s}{N_0} Y^2 \end{equation}
And:
\begin{equation} \bar{\gamma}=\frac{E_s}{N_0}\prod_{i=1}^N \Omega_i \end{equation}
How did they come up with the following expression: \begin{eqnarray} F_\gamma (\gamma )= {F_Y}\left(\sqrt{\frac{\gamma \prod_{i=1}^N \Omega_i}{\bar{\gamma}} }\right) \end{eqnarray}
Where $F$ is the CDF.
Thank you.
This is rather straight forward: $$\frac{\gamma}{\bar \gamma} = \frac{Y^2}{\prod_{i=1}^N \Omega_i} \Rightarrow Y = \pm \sqrt{\frac{\gamma\prod_{i=1}^N \Omega_i}{\bar \gamma}}$$ if we further assume $Y \ge 0$ then we have $Y = \sqrt{\frac{\gamma\prod_{i=1}^N \Omega_i}{\bar \gamma}}$.
Now $F_\gamma (x) = \text{Pr}(\gamma \le x)$ and $F_Y (y) = \text{Pr}(Y \le y)$.
So if $\gamma \le x \Rightarrow Y = \sqrt{\frac{\gamma\prod_{i=1}^N \Omega_i}{\bar \gamma}} \le \sqrt{\frac{x\prod_{i=1}^N \Omega_i}{\bar \gamma}} \Rightarrow F_\gamma (x )= {F_Y}\left(\sqrt{\frac{x \prod_{i=1}^N \Omega_i}{\bar{\gamma}} }\right)$
I considered $Y \ge 0$ because i think $E_s$ = symbol Energy and $N_0 = $ noise variance.