## anonymous 5 years ago Suppose X and Y have joint density f(x,y)=1 for 0<x,y<1. Find P(XY<=z)

1. anonymous

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???

2. nowhereman

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) !

3. anonymous

ok. I'll double check the question. thanks

4. anonymous

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 :(

5. nowhereman

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)

6. anonymous

that makes sense...

7. anonymous

so how would this be finished?

8. anonymous

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