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Math Help - Variance-covariance matrix of random vector

  1. #1
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    Variance-covariance matrix of random vector

    Notation:
    Var(Y) is the variance-covariance matrix of a random vector Y
    B' is the tranpose of the matrix B.

    1) Let A be a m x n matrix of constants, and Y be a n x 1 random vector. Then Var(AY) = A Var(Y) A'

    Proof:
    Var(AY)
    = E[(AY-A E(Y)) (AY-A E(Y))' ]
    = E[A(Y-E(Y)) (Y-E(Y))' A' ]
    = A E[(Y-E(Y)) (Y-E(Y))'] A'
    = A Var(Y) A'

    Now, I don't understand the step in red. What theorem is that step using?
    I remember a theorem that says if B is a m x n matrix of constants, and X is a n x 1 random vector, then BX is a m x 1 matrix and E(BX) = B E(X), but this theorem doesn't even apply here since it requries X to be a column vector, not a matrix of any dimension.


    2) Theorem: Let Y be a n x 1 random vector, and B be a n x 1 vector of constants(nonrandom), then Var(B+Y) = Var(Y).

    I don't see why this is true. How can we prove this?
    Is it also true that Var(Y+B) = Var(Y) ?


    Any help is greatly appreciated!

    note: also under discussion in Talk Stats forum
    Last edited by kingwinner; June 15th 2009 at 09:32 PM.
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  2. #2
    MHF Contributor matheagle's Avatar
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    Constants can be pulled out of ANY expectation.
    Just like in calculus, constants are not part of the integration,
    nor the summations in the discrete case.
    Last edited by matheagle; June 15th 2009 at 11:33 PM.
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  3. #3
    MHF Contributor matheagle's Avatar
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    Of course Var(Y+B) = Var(Y), since Y+B=B+Y.

    You only need the dimensions to be the same, of Y and B.
    And that is assumed when you are asked to add two vectors.
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