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Math Help - Covariance Question

  1. #1
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    Covariance Question

    Hi all,

    Would just like to confirm something and hopefully hear some of your thoughts.

    Cov(\sum_{i=1}^{N_1}X_i , \sum_{i=1}^{N_2}X_i )

    Here X_i are iid random variables of distribution X~Gamma(a,b) and N_1 and N_2 are dependent negative binomials.

    Would it still be feasible to conclude that the covariance is 0 due to the independence of X_i (even though the N_i are dependent)?

    Cheers
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  2. #2
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    Re: Covariance Question

    Quote Originally Posted by casio415 View Post
    Hi all,

    Would just like to confirm something and hopefully hear some of your thoughts.

    Cov(\sum_{i=1}^{N_1}X_i , \sum_{i=1}^{N_2}X_i )

    Here X_i are iid random variables of distribution X~Gamma(a,b) and N_1 and N_2 are dependent negative binomials.

    Would it still be feasible to conclude that the covariance is 0 due to the independence of X_i (even though the N_i are dependent)?

    Cheers
    Cov\left(\displaystyle{\sum_{i=1}^{N_1}}X_i , \displaystyle{\sum_{j=1}^{N_2}}X_j\right) = \displaystyle{\sum_{i=1}^{N_1}} \displaystyle{\sum_{j=1}^{N_2}}Cov(X_i,X_j) =\begin{array}{ll}&0 : i \neq j \\ &N_1 N_2 Var(X_i) : i=j\end{array}

    so the off diagonal terms of your covariance matrix are still 0 but the diagonal elements do depend on N_1\mbox{ and }N_2
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  3. #3
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    Re: Covariance Question

    Thanks for the reply. One thing though - the LHS covariance presumably is a known constant whereas the RHS is in terms of N_1 N_2 which are random variables. How can this is reconciled?
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  4. #4
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    Re: Covariance Question

    the way to get rid of this dependence is to scale the sum by \frac{1}{\sqrt{N_1 N_2}}
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