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Math Help - Find the Coefficient of Correlation. (Need help understanding the posted solution)

  1. #1
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    Find the Coefficient of Correlation. (Need help understanding the posted solution)

    \\\text{Given }n\text{ random variables }X_1,X_2,...,X_n\text{ each having the same variance of }\sigma^2\text{, and defining}\\\\U=2X_1+X_2+...+X_{n-1}\text{ and }V=X_2+X_3+...+2X_n\text{, find the coefficient of correlation between U and V.}

    Solution:

    \\\rho_{UV}=\frac{Cov[U,V]}{\sigma_U\sigma_V}\text{; }\sigma^2_U=(4+1+1+1+...+1)\sigma^2=(n+2)\sigma^2=  \sigma^2_V\text{.}\\\\\text{Since the X's are independent, if }i\neq j\text{ then }Cov[X_i,X_j]=0\text{. Then, noting that }\\\\Cov[W,W]=Var[W]\text{,}\\\\\text{we have }Cov[U,V]=Cov[2X_1,X_2]+Cov[2X_1,X_3]+...+Cov[X_{n-1},2X_n]=\\\\Var[X_2]+Var[X_3]+...+Var[X_{n-1}]=(n-2)\sigma^2\text{.}\\\\\text{Then, }\rho_{UV}=\frac{(n-2)\sigma^2}{(n+2)\sigma^2}=\frac{(n-2)}{(n+2)}


    I understand the first line. The rest is confusing. I don't understand how  Cov[U,V] is found. Any help would be appreciated. Thanks

    Edit: I guess it helps to take the time to type everything out and look at it. I have an understanding of the solution now, but I wouldn't be able to come up with a solution on my own. I'd love any insights for this sort of problem. Thanks
    Last edited by downthesun01; August 8th 2013 at 12:24 AM.
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  2. #2
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    Re: Find the Coefficient of Correlation. (Need help understanding the posted solution

    Hey downthesun01.

    The definition of covariance is Cov[X,Y] = E[XY] - E[X]E[Y].

    If X and Y are independent then P(A and B) = P(A)P(B) and from that you can prove that E[XY] = E[X]E[Y] which would prove that Cov[X,Y] = 0 for this particular case of X and Y being independent.

    The rest of the argument expands out Cov[U,V] in terms of the Xi terms and note that Cov[Xi,Xj] = 0 for i != j (which is what they used above for independent random variables) and when i = j then you calculate E[Xi^2] which can be calculated in the normal way if you know the probability density function.

    Thats the basic idea behind the independence and the algebra that is used.
    Thanks from downthesun01
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