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

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I 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
 5 years ago
Best ResponseYou've already chosen the best response.0Well 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
 5 years ago
Best ResponseYou've already chosen the best response.0ok. I'll double check the question. thanks

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yep. 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
 5 years ago
Best ResponseYou've already chosen the best response.0They 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
 5 years ago
Best ResponseYou've already chosen the best response.0so how would this be finished?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what do you mean by min(1, z/y) and how do i implement that
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