Suppose X and Y have joint density f(x,y)=1 for 0

Question

Suppose X and Y have joint density f(x,y)=1 for 0<x,y<1. Find P(XY<=z)

anonymous · Answer

I set this up like: $$\int\limits_{0}^{1} \int\limits_{0}^{z/y} dx dy$$ -  I end up with a ln y which takes everything to zero.  is there something I'm missing becuase of the XY is less than or equal to z???

nowhereman · Answer

Well that integral does not exist. And neither does the original density. A probability density of 1 over an unbounded area is wrong.
Or maybe the initial condition was $$0<x<1 ∧ 0<y<1$$ that would make sense. But then you should not integrate from 0 to z/y but from 0 to min(1, z/y) !

anonymous · Answer

ok.  I'll double check the question.  thanks

anonymous · Answer

Yep.  I've got the question right... I hope this isn't one of those partial derivative things.  It's been too long since I've done these :(

nowhereman · Answer

They probably mean that both x and y are restricted to [0, 1] because the integral of the density over that domain must be 1. So just replace z/y by min(1, z/y)

anonymous · Answer

that makes sense...

anonymous · Answer

so how would this be finished?

anonymous · Answer

what do you mean by min(1, z/y) and how do i implement that