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I would like to find this probability: $Pr(Z<2^{2R})$ for $R>0$.

So, I try several ways and finally decide to find pdf of Z. If $X$ and $Y$ are independent and exponentially distributed with parameter $\lambda_1$ and $\lambda_2$, respectively, and $a_i>0$ for $i=1,...,6$, which is the pdf of $Z$? Where $Z$ is given by

$$Z=\frac{(a_1X+a_2)(a_3Y+a_5)}{(a_3X+a_4)(a_1Y+a_6)} $$

any idea?

First,considering $Z=W_{1}W_{2}$, I calculate pdf of $W1=\frac{(a_1X+a_2)}{(a_3X+a_4)}=\frac{W_{11}}{W_{12}}$ as:

$f_{W_{11}}(w_{11})=\frac{1}{a_{1}}f_{X}(\frac{w_{11}-a_{2}}{a_{1}})$ for $w_{11}>a_{2}$

$f_{W_{12}}(w_{12})=\frac{1}{a_{3}}f_{X}(\frac{w_{12}-a_{4}}{a_{3}})$ for $w_{12}>a_{4}$

$F_{W_{1}}(W_{1})=Pr(W1=\frac{W_{11}}{W_{12}}<w_{1})=\int_{a_{4}}^{\infty}\int_{a_{2}}^{ww_{2}} f_{w_{11}}(w_{11})f_{w_{12}}(w_{12})\, dw_{11}dw_{12}=1-\frac{{{e}^{{{\lambda }_{1}}(\frac{{{a}_{2}}}{{{a}_{1}}}+\frac{{{a}_{4}}}{{{a}_{3}}})}}{{e}^{-{{\lambda }_{1}}(\frac{{{a}_{4}}}{{{a}_{3}}}+\frac{{{a}_{4}}{{w}_{1}}}{{{a}_{1}}})}}}{(1+\frac{{{a}_{3}}}{{{a}_{1}}}{{w}_{1}})}$

If we derive from the above, pdf of $W_{1}$ is achieved as:

$\frac{a_{3}{{e}^{{{\lambda }_{1}}(\frac{{{a}_{2}}}{{{a}_{1}}}+\frac{{{a}_{4}}}{{{a}_{3}}})}}{{e}^{-{{\lambda }_{1}}(\frac{{{a}_{4}}}{{{a}_{3}}}+\frac{{{a}_{4}}{{w}_{1}}}{{{a}_{1}}})}}}{a_{1}(1+\frac{{{a}_{3}}}{{{a}_{1}}}{{w}_{1}})^{2}}+\frac{\lambda_{1} a_{4}{{e}^{{{\lambda }_{1}}(\frac{{{a}_{2}}}{{{a}_{1}}}+\frac{{{a}_{4}}}{{{a}_{3}}})}}{{e}^{-{{\lambda }_{1}}(\frac{{{a}_{4}}}{{{a}_{3}}}+\frac{{{a}_{4}}{{w}_{1}}}{{{a}_{1}}})}}}{a_{1}(1+\frac{{{a}_{3}}}{{{a}_{1}}}{{w}_{1}})}$

But, I have a problem: I know that integral of $f_{W_{11}}(W_{11})$ over whole interval equal to $1$. What are the integral bounds?(I need these bounds for calculating pdf of product of $W_{1}W_{2}$)

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