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Math Help - Calculate some weird RV's variance

  1. #1
    Junior Member
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    Calculate some weird RV's variance

    Can you help me with something? I get stuck everytime i try to solve the following problem.

    Let Xi, 1<=i<=10 be independent random variables such that EXi = 10 and Var Xi = 6. Let Y=X_1X_2+X_2X_3+...+X_{10}X_{11}
    Calculate Var Y.

    I tried to solve it using the fact that Var Y = Cov(Y,Y). That leads to:

    Var Y = Cov(Y,Y) = Cov(\sum_{i=1}^{10}X_iX_{i+1},\sum_{j=1}^{10}X_jX_  {j+1})=
    =\sum_{i,j}Cov(X_iX_{i+1},X_jX_{j+1})=
    =\sum_{i=j}Cov(X_iX_{i+1},X_jX_{j+1})+2\sum_{i<j}C  ov(X_iX_{i+1},X_jX_{j+1})

    The second term can be decomposed in two types:
    1. Cov(X_iX_{i+1},X_{i+1}X_{i+2})
    2. Cov(X_iX_{i+1},X_{j}X_{j+1}), j>i+1
    The second type is zero because the variables are independent. The first type can be calculated:

    Cov(X_iX_{i+1},X_{i+1}X_{i+2})=E(X_iX_{i+1}^2X_{i+  2})-E(X_i)E(X_{i+1})^2E(X_{i+2})
    =EX_iEX_{i+2}(E(X_{i+1}^2)-E(X_{i+1})^2)=EX_iEX_{i+2}Var(X_{i+1})

    Ok. But here's the problem: look what i get when I calculate Cov(XiXi+1,XiXi+1):

    Cov(X_iX_{i+1},X_iX_{i+1})=\mathbb{E}(X_i^2X_{i+1}  ^2)-\mathbb{E}(X_i)^2\mathbb{E}(X_{i+1})^2=
    E(X_i^2)E(X_{i+1}^2)-E(X_i)^2E(X_{i+1})^2=
    =E(X_i^2)E(X_{i+1}^2)-E(X_i)^2E(X_{i+1})^2+E(X_i)^2E(X_{i+1}^2)-E(X_i)^2E(X_{i+1}^2)=
    =E(X_{i})^2Var(X_{i+1})+E(X_{i+1}^2)Var(X_i)

    And I can't get rid of this E(X_{i+1)^2)!

    I appreciate any help. Thanks
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  2. #2
    MHF Contributor
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    i haven't not checked your working at all, but:

    its always true that:
     V(X) = E(X^2) - (E(X))^2)

    so
     6 = E(X^2) - 100
    E(X^2) = 106
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  3. #3
    Junior Member
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    Jeepers! How could i not realize it was that simple?

    Thank you so much!
    Last edited by mr fantastic; July 11th 2010 at 06:48 PM.
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