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Math Help - Covariance of two sample means

  1. #1
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    Covariance of two sample means

    Hello all!

    I am looking to calculate VAR(xbar-ybar) where X and Y are not independent.

    Want to try VAR(xbar) +VAR(ybar)+2COV(xbar,ybar).

    I know VAR(xbar) = sigma^2 of X / n1

    I know VAR(ybar) = sigma^2 of Y / n2


    But the COV(xbar,ybar) has me stumped. I wanted to write
    (1/n1)* (1/n2) * COV (SUM(Xi) 1 to n1, SUM(Yi) 1 to n2).


    I have researched the additive rule of COV(X+Y,Z) = COV(X,Y) + COV(Y,Z) but not sure how to apply with a sum in both 'places' for COV(X,Y).

    Anyone help?

    Thanks!
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  2. #2
    MHF Contributor matheagle's Avatar
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    You do need to know the covariance between X_i and Y_j.

     Cov(\bar X,\bar Y)= Cov\biggl({\sum_{i=1}^n X_i\over n},{\sum_{j=1}^m Y_j\over m}\biggr)

    ={1\over nm} Cov\biggl(\sum_{i=1}^n X_i,\sum_{j=1}^m Y_j\biggr)

    ={1\over nm} \sum_{i=1}^n\sum_{j=1}^m Cov(X_i,Y_j) .

    NOW, if they are iid then...

    =Cov(X_1,Y_1) .


    AND VAR(xbar) +VAR(ybar)+2COV(xbar,ybar) is WRONG.
    It's VAR(xbar) +VAR(ybar)-2COV(xbar,ybar).
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  3. #3
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    Thank you Matheagle!

    One point of clairification, when you say if the variables are iid...does this mean the x's and y's are both iid? Could you explain how that then results in

    COV(xbar,ybar) = Cov(X1,Y1)?

    Thanks!~
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  4. #4
    MHF Contributor matheagle's Avatar
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    iid means independent (i) and identically distributed (id)
    Statisticians call that a random sample.
    NOW the sample X_1,\ldots ,X_n are iid
    same for Y_1,\ldots ,Y_m are iid.
    BUT I'm assuming there is a common covariance between any
    X_i and Y_j just like we assume there
    is a common variance of the X_i's and Y_j's.
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  5. #5
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    Oh I see now. Say the covariance between X and Y is 10 (taken from a variance-covariance matrix) and there are 3 X's and 3 Y's:

    (1/9) * (10+10+10+10+10+10+10+10+10) = 10

    Thanks again!
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  6. #6
    MHF Contributor matheagle's Avatar
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    taken from a variance-covariance matrix
    isn't really right.
    There is a variance-covariance matrix for the X's
    and a different one for the Y's.
    The variance-covariance matrix for the X's is \sigma_X^2 times the identity matrix of order n.
    The variance of each X_i is \sigma_X^2
    while all the covariances between the X's are 0.
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