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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)

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  1. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    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
    • 5 years ago
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    ok. I'll double check the question. thanks

  4. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    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
    • 5 years ago
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    that makes sense...

  7. anonymous
    • 5 years ago
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    so how would this be finished?

  8. anonymous
    • 5 years ago
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    what do you mean by min(1, z/y) and how do i implement that

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