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Thread: Help understanding Expectation of a sum

  1. #1
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    Help understanding Expectation of a sum



    Hi,

    I would like some help with understanding the properties of the expectation operator and the summation large sigma.

    Denote S as the large sigma sign meaning sum over i=1 to n
    E is the expectation operator
    Ki is a list of constants K1, K2,...,Kn; they are not all the same
    Yi is a list of random variables Y1, Y2,....Yn
    * is the multiplication symbol

    Here is what I have E[S(Ki * Yi)] = S(Ki * E[Yi])

    I can not understand how the E operator is moved inside the summation to give the above result.

    Am I correct in saying that:

    1. S(Ki * Yi) is just a number and that E[a number] = same number; from the rule of the expectation of a constant is a constant

    2. E[Yi] over all Y is just a number

    What other results do I need to be able to be able to accept E[S(Ki * Yi)] = S(Ki * E[Yi])?

    I have tried to do this using i= 1,2,3 as follows

    S(Ki * Yi) = K1Y1 + K2Y2 + K3Y3

    E[S(Ki * Yi)] = E[K1Y1 + K2Y2 + K3Y3] = K1Y1 + K2Y2 + K3Y3 from the expected value of a number is a number. Call this the left hand side.

    Dealing with the right hand side: S(Ki * E[Yi])

    E[Yi] = (Y1 + Y2 + Y3)/3

    S(Ki * (Y1 + Y2 + Y3)/3) =

    (K1 *(Y1 + Y2 + Y3)/3) + (K1 *(Y1 + Y2 + Y3)/3) + (K1 *(Y1 + Y2 + Y3)/3)

    = ((Y1 + Y2 + Y3)/3)) * (K1 + K2 + K3)

    This does not equal the left hand side so where am I making mistakes?

    This is used amongst other results to show that in the case least squares linear regression that the slope of the regressin is an unbiased point estimator of the true slope. It is taken from Applied Linear Statistical Models by Kutner, Nachtsheim, Neter & Li to show the unbiasedness of the point estimator for the slope as stated in the Gauss Markov theorem.

    Many thanks in advance

    Peter
    Last edited by PeterVincent; 10-03-2007 at 03:41 PM. Reason: to include a reference

  2. #2
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    Write down the definition of E(S(kiXi))=E(k1X1+k2X2+...)

    E() is a linear operator so

    E(k1X1+k2X2+...)=E(k1X1)+E(k2X2)+..=SE(kiXi)=S(kiE(Xi))

  3. #3
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    Thank you for the help

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