Calculate the marginal distribution of $X$ given the joint distribution of $(X,Y)$

Question

Given,

$$\ f(x,y)={y\over(\alpha e^{-\beta x}+\sigma)^2} exp\left({-y\over(\alpha e^{-\beta x})+\sigma)} -x\right), \ x,y>0$$

I am trying to find,

$$f_{Y|X}(y|x)=\frac{f_{X,Y}(x,y)}{f_X(x)}$$

In the above equation,

$$f_X(x)=\int\limits_{-\infty}^{\infty}f_{X,Y}(x,y)dy$$

I am unable to compute the above integral. Even Wolfram is unable to compute the integral. Anyone have some ideas what to do? Thanks for answer.

Answer 1

If $Z \sim gamma(a,b)$, then its pdf is,

$$f_{Z}(z; a,b) = \frac{z^{a-1}e^{-x/b}}{\Gamma(a)b^a}, \, z>0$$

$$\int_{0}^{\infty}f_{Z}(z;a,b)dz = 1$$

$$f_{X}(x) = \int_{0}^{\infty}{y\over(\alpha e^{-\beta x}+\sigma)^2} e^{-\left({y\over(\alpha e^{-\beta x}+\sigma)} +x\right)}dy\\ = e^{-x}\int_{0}^{\infty}{y\over(\alpha e^{-\beta x}+\sigma)^2} e^{-\left({y\over(\alpha e^{-\beta x}+\sigma)}\right)}dy\\ = e^{-x}\int_{0}^{\infty}f_{Y}(y; 2,\alpha e^{-\beta x}+\sigma)dy= e^{-x}\cdot1 = e^{-x}$$