# pdf of product of two function of the Exponential variables

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}}

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 $$W_{1}W_{2}$$)