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Math Help - Order Statistics with a dash of Covariance.

  1. #1
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    Order Statistics with a dash of Covariance.

    Here's the question:

    Let x1,x2,x3 be a random sample from a distribution with a PDF: f(x) = 2x, 0<x<1.

    a) find the probability that the largest of x1,x2,x3 is less than the distribution median.
    b) Find Cov(x(2),x(3))

    Using quantiles, I found the distribution median to be = sqrt(1/2);
    I found the max to be the pdf 6x^5 0<x<1;

    In order to solve this, P(6x^5 < sqrt(1/2)) solved for x. This gave me an integral 2x with bounds from 0 to (((sqrt(.5))/6))^(1/5)) yielding 0.42514

    Did I go through this correctly?

    For the Covariance I am also not confident.

    using the definition that cov(x,y) = E[(x - E(x))(Y - E(y))]

    I found to be something like E[(12x^3-12x^5-24/35)(6x^5-6/7)] which I calculated to be around 78/245.

    Thanks in advance for you help!
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  2. #2
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    Quote Originally Posted by hasbren View Post
    Here's the question:

    Let x1,x2,x3 be a random sample from a distribution with a PDF: f(x) = 2x, 0<x<1.

    a) find the probability that the largest of x1,x2,x3 is less than the distribution median.
    b) Find Cov(x(2),x(3))

    Using quantiles, I found the distribution median to be = sqrt(1/2);
    I found the max to be the pdf 6x^5 0<x<1;

    In order to solve this, P(6x^5 < sqrt(1/2)) solved for x. This gave me an integral 2x with bounds from 0 to (((sqrt(.5))/6))^(1/5)) yielding 0.42514

    Did I go through this correctly?

    For the Covariance I am also not confident.

    using the definition that cov(x,y) = E[(x - E(x))(Y - E(y))]

    I found to be something like E[(12x^3-12x^5-24/35)(6x^5-6/7)] which I calculated to be around 78/245.

    Thanks in advance for you help!
    I don't have time to work through the question and therefore check the details, but your approach looks fine. (All your pdf's integrate to 1 which is a good sign and your means are between 0 and 1 which is another good sign).
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  3. #3
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    Thanks for your approval Mr. Fantastic
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  4. #4
    MHF Contributor matheagle's Avatar
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    To obtain the covariance between two order statistics you need the joint density.

    E(X_{(2)}X_{(3)})=\int_0^1\int_0^{x_3}x_2x_3f_{X_{  (2)}X_{(3)}}(x_2,x_3)dx_2dx_3

    you can obtain E(X_{(2)}) and E(X_{(3)}) via the marginals.
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  5. #5
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    So can I just multiply the two together to get the joint? I don't know if they are independent... oh wait- do I since they are order statistics?
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  6. #6
    MHF Contributor matheagle's Avatar
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    no, they are clearly dependent x(2)<x(3)
    Last edited by matheagle; April 12th 2010 at 04:02 PM.
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  7. #7
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    So, how would I go about getting the joint then?
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  8. #8
    MHF Contributor matheagle's Avatar
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    It's in many books, most likely your textbook.
    I found it here with r=2, n=s=3....
    Joint Distribution of and

    f_{X_{(2)},X_{(3)}}(x_2,x_3)=3![F(x_2)]f(x_2)f(x_3)

    f(x)=2x and F(x)=x^2 on 0<x<1


    so your joint density should be f_{X_{(2)},X_{(3)}}(x_2,x_3)=24x_2^3x_3

    ON the set 0<x_2<x_3<1

    which I just checked is a valid density, it does integrate to one...

    24\int_0^1\int_0^{x_3}x_2^3x_3dx_2dx_3=1
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  9. #9
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    I appreciate the help my friend-
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