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Math Help - Variance of Sample mean (not independent)

  1. #1
    lpd
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    Variance of Sample mean (not independent)

    Hi I have this question.

    Suppose that the random variables X_1, X_2, ... X_n are identically distributed, with mean \mu and variance \sigma^2, but not independent. Assume the correlation between any pair is equal to \rho. i.e. corr(X_i,X_j)=\rho for i \neq j.

    1) Derive Var(\bar{X}) for this situation.
    2) What is Var(\bar{X}) when \rho=0? Explain.
    3) What is Var(\bar{X}) when \rho=1? Explain.
    4) Use the rsult you have derived to comment on how small \rho can be in this situation. Explain.


    To me, this looks hard. I can do it for the case if the random variables are independent and identically distributed (because it is such a well-known result).

    But how do you do it if the random variables are not independent? I'm a bit confused. Can someone lend me a hand and direct me to a suitable source for this question?

    Thanks!!

    Regards,

    Lpd
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  2. #2
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    Re: Variance of Sample mean (not independent)

    Assuming Xbar is the sum of all Xi I think u'll need these relations. var(Xi) = sig^2:

    p = corr(Xi,Xj) = cov(Xi,Xj)/[sqrt(var(Xi)var(Xj)]
    cov(Xi,Xj) = E(XiXj) - E(Xi)E(Xj)
    Var(Xi+Xj) = Var(Xi) + Var(Xj) + Cov(Xi,Xj)
    Last edited by TheCountador; September 16th 2012 at 08:25 PM.
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  3. #3
    MHF Contributor harish21's Avatar
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    Re: Variance of Sample mean (not independent)

    For correlated variables, you can use the fact that the variance of the sum of correlated variables is equal to the sum of the covariances.

    Var(\bar{X_n})= Var(\dfrac{1}{n}\sum_{i=1}^n X_i)=\dfrac{1}{n^2}Var(\sum_{i=1}^n X_i)=\dfrac{1}{n^2} \sum_{i=1}^n \sum_{j=1}^n Cov(X_i,X_j)

    reading this link http://sites.stat.psu.edu/~dhunter/a...ures/asymp.pdf section 2.2.3 on page 54 might help


    Variance - Wikipedia, the free encyclopedia

    \rho = 0 means the variables are independent and the variance will be given by \dfrac{\sigma^2}{n}
    Last edited by harish21; September 17th 2012 at 09:56 AM.
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