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anonymous

  • 5 years ago

I am trying to set up the double integral for: Suppose X and Y have join density f(x,y)=2 for x<y<x<1. Find P(X-Y>z)... any suggestions

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  1. anonymous
    • 5 years ago
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    the answer is supposed to be \[(1-z)^{2}/2\] but I can't get there

  2. nowhereman
    • 5 years ago
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    can you correct the area for x and y? x<y<x<1 makes no sense. are x and y positive?

  3. anonymous
    • 5 years ago
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    sorry - 0<y<x<1

  4. nowhereman
    • 5 years ago
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    The integration itself is of course easy, as the integrand is constant. But you have to find the area you have integrate over. The triangle ⟨(0,0), (1, 0), (1, 1)⟩ is the first restriction (0 < y < x < 1). The other is given by x-y>z or better y < x-z. That gives you a still smaller triangle. Of course you must examine the cases z > 1 and z < 0 separately. For 0 < z < 1 the integral is two times the area of that small triangle. But you can of course write it analytical: \[\int_z^1 \int_0^{x-z}2\; dy\;dx\] As for the supposed answer: if z = 0 the both restrictions are the same and so the overall should be 1 and not 1/2

  5. anonymous
    • 5 years ago
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    excellent! I kept writing my first integral 1 - 0 not 1- z. The z threw me off.

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