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Math Help - Correlation coeff. of tetrahedron

  1. #1
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    Correlation coeff. of tetrahedron

    A tetrahedron with 4 sides (1,2,3,4) is tossed two times. X=min of two observations, Y=max of two observations.
    I am trying to find the correlation coefficient of X and Y.

    I think I am suppose to make a table for sample space for tosses, a table for the values of X and Y for each outcome, and the table for the pmf.

    but i am just having trouble trying to start on the first table...
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  2. #2
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    Quote Originally Posted by mightydog78 View Post
    A tetrahedron with 4 sides (1,2,3,4) is tossed two times. X=min of two observations, Y=max of two observations.
    I am trying to find the correlation coefficient of X and Y.

    I think I am suppose to make a table for sample space for tosses, a table for the values of X and Y for each outcome, and the table for the pmf.

    but i am just having trouble trying to start on the first table...
    Suppose that the tetrahedra are distinguishable, then the equally likely outcomes of tossing the two are:

    (1,1),(1,2),(1,3),(1,4)
    (2,1),(2,2),(2,3),(2,4)
    (3,1),(3,2),(3,3),(3,4)
    (4,1),(4,2),(4,3),(4,4)

    Now your actual experiment has outcomes (a,b) where a is the smaller of the two throws and b the larger. So we can list these and calculate their probabilities using the above table of equally likely outcomes of the underlying experiment.

    (1,1) 1/16
    (1,2) 2/16
    (1,3) 2/16
    (1,4) 2/16
    (2,2) 1/16
    (2,3) 2/16
    (2,4) 2/16
    (3,3) 1/16
    (3,4) 2/16
    (4,4) 1/16

    CB
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  3. #3
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    so now i would just need to find the standard deviation of X and Y as well as the covariance, then correlation coeff=Covariance/(stand.dev(X)*stand.dev(Y))

    thanks for the help!
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  4. #4
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    i noticed on those probabilities you found that you skipped (2,1),(3,1),(4,1),(3,2),(4,2),(4,3)

    is that because X=min of the observations and Y=max of two observations?
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  5. #5
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    Quote Originally Posted by mightydog78 View Post
    i noticed on those probabilities you found that you skipped (2,1),(3,1),(4,1),(3,2),(4,2),(4,3)

    is that because X=min of the observations and Y=max of two observations?
    For the final calculation you will see that I enumerate them as (a,b) where a<=b, so (2,1) et al are not in the set of outrcomes that are possible when labeled this way.

    CB
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  6. #6
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    the first list is the sample space for the two tosses and I am assuming the second list of data is a table for the values of X and Y correct? which is why some are left blank since X is the max(a,b) and Y is the min(a,b)

    just trying to clarify here
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  7. #7
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    Quote Originally Posted by mightydog78 View Post
    the first list is the sample space for the two tosses and I am assuming the second list of data is a table for the values of X and Y correct? which is why some are left blank since X is the max(a,b) and Y is the min(a,b)

    just trying to clarify here
    What is left blank?

    CB
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  8. #8
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    i have the table for X vs. Y and found the marginal probabilities of PY and PX
    my question now is how do i find the covariance

    the equation for covariance that i have is (E(X-uy)(Y-uy)) / ((stand. dev. of x)*(stand. dev. of y))

    how do i find the standard deviation of X or Y when there are 4 different columns for each? should I use the marginal probabilities?
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