finding the PDF!!

**lauren2988** · October 7th 2008, 09:27 AM

f(x,y) = θe-(x+θy) , θ>0 and x>0

find the PDF of XY

thanks

**mr fantastic** · October 7th 2008, 06:40 PM

Originally Posted by lauren2988

f(x,y) = θe-(x+θy) , θ>0 and x>0

find the PDF of XY

thanks

Let U = XY.

Calculate the cumulative density function F(u). Then the pdf of U is $g(u) = \frac{dF}{du}$ .

$F(u) = \Pr(U < u) = \Pr(XY < u) = \Pr\left( Y < \frac{u}{X}\right) = \int_{x = -\infty}^{+\infty} \int_{y = -\infty}^{+\infty} f(x, y) \, dy \, dx$

Are you given that the joint pdf has the exponential rule for y > 0 as well as x > 0? If not, then the next line won't be correct:

$= \int_{x=0}^{x = +\infty} \int_{y=0}^{y = u/x} \theta \, e^{-(x + \theta y)} \, dy \, dx$ .

The calculus stuff is left for you to do.

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