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Can we find the CDF of the random variable where $$\gamma = \frac{a_1\, g}{a_2 \, g +1}+ \frac{b_1\, h}{b_2 \, h +1},$$

where $g$ and $h$ are independent exponential random variables with different parameters and $a_1, a_2,b_1,b_2$ are positive numbers less than $1$? Also: $a_1 + a_2 =1$ and $b_1 + b_2 =1$.

In other words, I need to find:

$$F_{\gamma}(x) = P(\gamma <x)$$

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  • $\begingroup$ Are $g$ and $h$ independent? Are they characterized by the same parameter or not? $\endgroup$ – rubik May 12 '18 at 16:16
  • $\begingroup$ I updated the question. $\endgroup$ – Basem Elhalawany May 12 '18 at 16:22
  • $\begingroup$ I guess you need a numerical resource. $\endgroup$ – Felix Marin May 12 '18 at 19:36
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Let $$\begin{align} G &\sim \mathsf{Exp}(\lambda)\\ H &\sim \mathsf{Exp}(\mu)\\ \gamma &\sim \frac{a_1 G}{a_2 G +1}+ \frac{b_1 H}{b_2 H +1}, \end{align}$$ with $\lambda,\mu > 0$, $a_\star, b_\star > 0$ and $a_1 + a_2 = b_1 + b_2 = 1$.

Then $$F_\gamma(x) = \mathsf P(\gamma \le x) = \int_0^{+\infty} \mathsf P(\gamma \le x \mid H = h) f_H(h)\,\mathrm dh,$$ where $f_H$ is the PDF of $H$, that is $$f_H(h) = \mu e^{-\mu h}.$$

Through simple algebra, we find that $$\mathsf P(\gamma \le x \mid H = h) = \mathsf P\left(G \le \frac{y}{a_1 - a_2 y}\right) = F_G\left(\frac{y}{a_1 - a_2 y}\right),$$ where $$y = x - \frac{b_1 h}{b_2 h + 1}$$ and $F_G$ is the CDF of $G$.

Therefore, $$\begin{align} F_\gamma(x) &= \int_0^{+\infty} F_G\left(\frac{y}{a_1 - a_2y}\right)f_H(h)\,\mathrm dh =\\[1ex] &= \int_0^{+\infty} \left[1 - \exp\left(-\frac{y\lambda}{a_1 - a_2y}\right)\right]\mu e^{-\mu h}\,\mathrm dh =\\ &= 1 -\mu \int_0^{+\infty} \exp\left(-\frac{y\lambda}{a_1 - a_2y} - \mu h\right)\mathrm dh \end{align}$$

Maybe this integral has a closed form, but I wasn't able to find it. I asked a new question here in case someone has an idea.

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