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Covariance of sums - pls check solution

  1. Aug 1, 2009 #1
    1. The problem statement, all variables and given/known data

    Suppose X1 , X2 , X3 , and X4 are independent with a common mean 1 and common variance 2. Compute Cov( X1 + X2 , X2 + X3).

    2. Relevant equations

    Cov (X,Y) = E[(X-u)(Y-v)] = E[XY] - uv, where u and v are the means of X and Y
    E[XY] = E[X]E[Y]
    E[X+Y] = E[X] + E[Y]

    3. The attempt at a solution

    Since there is a common mean=1, Cov(X,Y) = E[XY] -1 = E[X]E[Y] - 1
    Substituting for E[XY]= E[X1+X2]E[X2+X3]
    Substituting for E[X+Y]= (E[X1]+E[X2])(E[X2]+E[X3])
    Expanding (E[X1]+E[X2])(E[X2]+E[X3]) = E[X1]E[X2] + E[X1]E[X3] + E[X2]^2+ E[X2]E[X3]
    Substituting for mean=1: (1)(1) + (1)(1) + 1^2 +(1)(1) = 4

    Cov (X,Y) = E[XY] - 1 = 4 -1 = 3

    Can someone check this for me? I think it is wrong. If the variables are independent with identical means and variances, shouldn't the covariance equal the variance?
     
    Last edited: Aug 1, 2009
  2. jcsd
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