Question

The joint density of X and Y is given by

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

Answer 1

\[x e^{-x(y+1)}\]

Answer 2

Find the distribution of Z=XY using distribution function method

Answer 3

@Zarkon

Answer 4

@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

Answer 5

Just compute \[P(Z<z)=P(XY<z)\]

integrate over the region in which \(XY<z\) is true

Answer 6

How to integrate it when it has two variables?

Answer 7

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?

Answer 8

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.

Answer 9

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.

Answer 10

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

Answer 11

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