NOTE: This is in further extension to my quest of understanding the paper discussion in question here (I will really appreciate any assistance, though I am myself trying to find the solution as well)
$$E = K\Omega\Phi\gamma \ \ \ (1)$$
Where $\Omega$ and $\Phi$ are two random variables with pdf as follows:
$$ f_\Omega(w) = \frac{1}{(1+w)^2}$$
Authors are interested in finding a Test Statistic $E$ and based on eq. (1), they claim that $E$ has the same distribution as $\Omega$ for a given vaule of $\Phi$. The conditional PDF is hence found by authors as :
$$f_E(\varepsilon|\Phi) = \frac{K\Phi\gamma}{(K\Phi\gamma + \varepsilon)^2}$$
Where $\Omega = h_1^2/h_2^2$ and $h_1^2 \sim \exp(u)$ and $h_2^2 \sim \exp(u)$
My Questions are as following (in order)
(i) How is eq. 2 obtained.
(ii) Why is $E$ has same distribution as $\Omega$
Given $\Phi$, we have the probability: \begin{align} P(E\leq \xi|\Phi) &= P(K\Phi\gamma\Omega \leq \xi|\Phi) \\ &= P\left(\left. \Omega\leq \frac{\xi}{K\Phi\gamma}\right|\Phi\right) \\ &= \frac{\frac{\xi}{K\Phi\gamma}}{1+\frac{\xi}{K\Phi\gamma}}\\ &=\frac{\xi}{K\Phi\gamma + \xi} \end{align} We differentiate this to obtain the PDF: \begin{align} f_E(\xi|\Phi)&=\frac{1}{K\Phi\gamma + \xi}-\frac{\xi}{(K\Phi\gamma + \xi)^2}\\ &=\frac{K\Phi\gamma}{(K\Phi\gamma + \xi)^2} \end{align}