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Thread: pdf minimum of two r.v

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    pdf minimum of two r.v




    Suppose we have a pdf f(x,y) = (1/15) for 0 < x < 3 and 0 < y < 5 we want to calculate the pdf of K = min(x,y)

    for 3<k< 5, => either x or y = 3 and the other => 3
    so the bounds is 0 < x < 3 and 0 < y< 3 => FK(k) = 6/15

    for 0<k<=3
    we get the bounds 0 < x < k and 0 < y < k => fK(k) = k^2 / 15 .

    This obvious is not correct since the cdf of the first is a constant (not equal to 1) and 2nd does not = 1 at k =3

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    Re: pdf minimum of two r.v


    I think the easiest way to do this is with a few tricks

    F_K(k) = P(K \leq k) = P(min(X,Y) \leq k)

    now P(min(X,Y) \leq k) isn't easy to work with directly but we can write...

    P(min(X,Y) \leq k) = 1 - P(min(X,Y) > k)

    Can you think of a way to write P(min(X,Y) > k) in terms of X and Y without the min in there?

    once you do that you can use the fact that X and Y are independent to simplify matters further.
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