The joint density of X and Y is given by

f (x,y) = xe^(-x(y+1)); x0,y0
0; otherwise

Question

The joint density of X and Y is given by

f (x,y) = xe^(-x(y+1)); x>0,y>0
             0; otherwise

anonymous · Answer

$$x e^{-x(y+1)}$$

anonymous · Answer

Find the distribution of Z=XY using distribution function method

turingtest · Answer

@Zarkon

turingtest · Answer

@JamesJ I know you're doing ajp's problem, but I wanna know how to do this one as well if you get the chance

zarkon · Answer

Just compute $$P(Z<z)=P(XY<z)$$

integrate over the region in which $XY<z$ is true

anonymous · Answer

How to integrate it when it has two variables?

anonymous · Answer

x y < z
x < z/y
You should compute the integral below to F(z) the CDF and F'(z) would be the pdf

$$
\int _0^{\infty }\int
   _0^{\frac{z}{y}}x e^{-x
   y-x}dxdy
$$
Do you know how to compute this integral?

anonymous · Answer

you compute first the integral below by parts
$$
\int_0^{\frac{z}{y}} x e^{-x
   y-x} \, dx= \frac{1-e^{-\frac{(y+1) z}{y}}
   \left(\frac{z}{y}+z+1\right)
   }{(y+1)^2}
$$
It is messy, who said life is easy.

anonymous · Answer

Then you integrate 
$$
\int_0^{\infty }
   \frac{1-e^{-\frac{(y+1)
   z}{y}}
   \left(\frac{z}{y}+z+1\right)
   }{(y+1)^2} \, dy= 1- e^{-z}
$$
I have to confess that I did not do the above integral by hand and I refuse to do it by hand.

anonymous · Answer

So
$$
F(z) = 1 - e^{-z}\
f(z)=F'(z) = e^{-z}
$$

anonymous · Answer

Thank you for your big help. I will try @eliassaab :)