# Find the distribution of X and the conditional distribution of Y given X. Are these distributions known? Which?

I have this problem, I am supposed to obtain the distribution of X, for this I know that I must integrate the joint density, but there are many parameters that are not familiar to me, until now by definition I have to:

$$F_X(x)=\int_0^x \frac{\lambda} {\Gamma (\alpha)}(\frac{\lambda}{x})^\alpha y^{\alpha-1}e^{-\lambda (u+\frac{y}{u})} du$$

But I don't know how to continue.

The exercise says:

Given $$\alpha > 0$$ and $$\lambda> 0$$ let $$(X, Y)$$ be an absolutely continuous random vector with joint density:

$$f_{XY}(x,y) = \frac{\lambda} {\Gamma (\alpha)}(\frac{\lambda}{x})^\alpha y^{\alpha-1}e^{-\lambda (x+\frac{y}{x})}$$ $$0

a) Find the distribution of $$X$$ and the conditional distribution of $$Y$$ given $$X$$. Are these distributions.

b) Prove that $$\frac{y}{x}$$ has distribution $${\Gamma (\alpha,\lambda)}$$ and is independent of $$X$$ without appealing to the theorem of change of variables. known? Which?

It looks like your integral limits are wrong. You have to integrate with respect to $$y$$, not $$u$$. For every $$x,y$$ ranges from $$0\to\infty$$ so the integral is\begin{align*}F_X(x)&=\int_0^\color{red}\infty \frac{\lambda} {\Gamma (\alpha)}\left(\frac{\lambda}{x}\right)^\alpha y^{\alpha-1}e^{-\lambda (x+y/x)} dy\\&=\frac{\lambda e^{-\lambda x}}{\Gamma(\alpha)}\int_0^\infty\left(\frac {\lambda y}x\right)^{\alpha-1}e^{-\lambda (y/x)} d(\lambda y/x)\end{align*}since $$x$$ is constant when you integrate with respect to $$y$$. Now keep $$m=\lambda y/x$$, then $$m$$ ranges from $$0\to\infty$$ since $$x,\lambda>0$$. The integral becomes\begin{align*}F_X(x)&=\frac{\lambda e^{-\lambda x}}{\Gamma(\alpha)}\int_0^\infty m^{\alpha-1}e^{-m} dm\\&=\lambda e^{-\lambda x}\end{align*}which is the PDF of exponential distribution.