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Math Help - Find the first order statistic

  1. #1
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    Find the first order statistic

    Suppose Y_1,..., Y_n is a random sample where the density of each random variable Y_i is f(y) = 2*x^2*y^(-3), y >= 0 for some parameter x > 1. Find the first order statistic, Y_(1).
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  2. #2
    MHF Contributor matheagle's Avatar
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    use the complement, let M be the min

    F_M(y)=P(M\le y)=1-P(M>y)
    now if the min exceeds the value, y, they all do...

    =1-P(Y_1>y,\ldots ,Y_n>y)

    using independence

    =1-P(Y_1>y)\cdots P(Y_n>y)

    =1-\bigl[P(Y_1>y)\bigr]^n

    =1-\bigl[1-F_Y(y)\bigr]^n

    now differentiate and you have

    f_M(y)=P(M\le y)=n\bigl[1-F_Y(y)\bigr]^{n-1}f_Y(y)

    finally plug in your density/distribution.
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  3. #3
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    Thanks, that's what I had but I'm still getting the wrong answer. When computing the pdf and I'm integrating the cdf given what should my limits of integration be?
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  4. #4
    MHF Contributor matheagle's Avatar
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    show your work and we can better answer your question
    y goes from zero to infinity, x is a fixed constant.
    NEVERMIND
    this is NOT a valid density
    MY guess is that y is bounded below by x, clearly it is not bounded below by 0.
    Last edited by matheagle; March 21st 2011 at 04:28 PM.
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  5. #5
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    Quote Originally Posted by matheagle View Post
    show your work and we can better answer your question
    y goes from zero to infinity, x is a fixed constant.
    NEVERMIND
    this is NOT a valid density
    MY guess is that y is bounded below by x, clearly it is not bounded below by 0.
    Ok, so when I integrate (2*x^2*y^(-3)) with respect to y I get -(x^2/y^2) and plugging in the bounds of x to infinity, I get 0 - (-1) = 1. But this can't be right because then the entire pdf for the order statistic ends up equaling 0. Would I just use the indefinite integral of the pdf -(x^2/y^2) as my cdf and plug that in for the cdf found in the equation for the first order statistic that you listed above?
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  6. #6
    MHF Contributor matheagle's Avatar
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    IT is right
    ONE means it is a valid density
    that's how I figured out what your density must be
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  7. #7
    MHF Contributor matheagle's Avatar
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    F_Y(y)=2x^2\int_x^yt^{-3}dt

    x is not a variable, it's just a fixed constant.
    BUT you should look up the problem again, having y bounded below by zero, seems off.

    2x^2\int_x^{\infty}y^{-3}dy

    must be one.
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  8. #8
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    Thanks for all the help, I got it
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  9. #9
    MHF Contributor matheagle's Avatar
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    but is that the correct density, where y>x?
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  10. #10
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    Quote Originally Posted by matheagle View Post
    F_Y(y)=2x^2\int_x^yt^{-3}dt

    x is not a variable, it's just a fixed constant.
    BUT you should look up the problem again, having y bounded below by zero, seems off.

    2x^2\int_x^{\infty}y^{-3}dy

    must be one.
    Yes, you're right. y >= x, not y >= 0 as I typed in the original problem statement. My mistake
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