OpenStudy (anonymous):

The joint density of X and Y is given by f (x,y) = xe^(-x(y+1)); x>0,y>0 0; otherwise

6 years ago
OpenStudy (anonymous):

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

6 years ago
OpenStudy (anonymous):

Find the distribution of Z=XY using distribution function method

6 years ago
OpenStudy (turingtest):

@Zarkon

6 years ago
OpenStudy (turingtest):

@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

6 years ago
OpenStudy (zarkon):

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

6 years ago
OpenStudy (anonymous):

How to integrate it when it has two variables?

6 years ago
OpenStudy (anonymous):

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?

6 years ago
OpenStudy (anonymous):

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.

6 years ago
OpenStudy (anonymous):

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.

6 years ago
OpenStudy (anonymous):

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

6 years ago
OpenStudy (anonymous):

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

6 years ago