f(x,y) = θe-(x+θy) , θ>0 and x>0

find the PDF of XY

thanks :)

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- Oct 7th 2008, 09:27 AMlauren2988finding the PDF!!
f(x,y) = θe-(x+θy) , θ>0 and x>0

find the PDF of XY

thanks :) - Oct 7th 2008, 06:40 PMmr fantastic
Let U = XY.

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

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

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:

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

The calculus stuff is left for you to do.