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:

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

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?

I appreciate your ideas.


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.


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