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Math Help - Double Integral Used for For Finding Probability of two Uniform RVs

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    Double Integral Used for For Finding Probability of two Uniform RVs

    Problem: Two points are selected randomly on a line of length L so as to be on opposite sides of the midpoint of the line. [i.e., the two points X and Y are independent random variables such that X is uniformly distributed over (0,L/2) and Y is uniformly distributed over (L/2,L).] Find the probability that the distance between the two points is greater than L/4.

    So we are looking for P(Y-x > L/4) and f(x,y) = 4/L^2
    This probability would be: 4/L^2 * ( \int_{0}^{L/8}dx \int_{L/2}^{L}dy + \int_{L/8}^{L/2}dx \int_{x+L/4}^{L}dy )

    But whenever I do this integration i keep getting (4/L^2)(L^2/4) = 1 and
    The answer should be 14/16.
    Last edited by bjnovak; December 3rd 2011 at 09:02 PM.
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    MHF Contributor harish21's Avatar
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    Re: Double Integral Used for For Finding Probability of two Uniform RVs

    I didnt integrate the entire thing out, but since you're trying to find the probability that the distance between the two points is greater than L/4, then, P(Y-X > L/4) \implies P(Y>X+L/4) so the lower limit of your last integral should be x+L/4
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    Re: Double Integral Used for For Finding Probability of two Uniform RVs

    Yeah sorry about that, edited.
    But I did it with those limits and I'm still not getting it right.
    It's the calculus part I really need help with
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    MHF Contributor harish21's Avatar
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    Re: Double Integral Used for For Finding Probability of two Uniform RVs

    Y-X>\frac{L}{4}
    so, Y>X+\frac{L}{4}
    also, X < y-\frac{L}{4}
    your limits on integration will be:
    therefore,
     0 < X < \frac{L}{2}-\frac{L}{4}=\frac{L}{4} \implies \boxed{0<X<\frac{L}{4}}

    \dfrac{4}{L^2}\bigg[ \int_{L/2}^{L} \int_{0}^{L/4}dx dy\;+\;\int_{L/4}^{L/2} \int_{x+L/4}^{L}dydx\bigg]

    that should give you what you're looking for.
    Last edited by harish21; December 3rd 2011 at 09:30 PM. Reason: typos
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