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Math Help - Variance calc with weighted coefficients

  1. #1
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    Variance calc with weighted coefficients

    I want to understand the calculation of the variance of a sum of rvs where each variable is weighted using a constant coefficient. I post the full question and answer here and I highlight the areas that I am struggling to follow.


    Question.

    Y_1,... Y_n is a sample from a population with mean \mu and variance \sigma.

    The sample is not random, Cov(Y_i,Y_j)={\rho}\sigma^2 for i\neq{j}. Let U=\Sigma_{i=1}^na_iY_i.

    (a) Give the condition on a constants a_1,...a_n for U to be unbiased estimator of \mu.
    (b) Under this condition, calculate MSE(U).


    Answer.

    (a) is not a problem, just need to calculate a if E(U)=\mu which gives

    E(U)=\Sigma_{i=1}^na_iE(Y_i)=\mu\Sigma_{i=1}^na_i and \Sigma_{i=1}^na_i=1 is the condition.

    (b) given (a) is met, then MSE(U)=Var(U):

    Var(U)=Var(\Sigma_{i=1}^na_iY_i)=E[\Sigma_{i=1}^na_i(Y_i-\mu)\Sigma_{j=1}^na_j(Y_j-\mu)] (1)

    and starting from (2) I find it tricky to follow the solution:

    =\Sigma_{i=1}^n\Sigma_{j=1}^na_ia_jCov(Y_i, Y_j) (2)


    =\Sigma_{i=1}^na_i^2Var(Y_i)+\Sigma_{j=1}^n\Sigma_  {i\neq{j}}Cov(Y_i, Y_j) (3)


    It looks to me like a sum of all entries of the variance matrix of an nx1 vector... but I cannot yet move beyond that.


    Then it gets OK again, a straightforward substitution

    =\sigma^2\Sigma_{i=1}^na_i^2+\rho\sigma^2\Sigma_{j  =1}^n\Sigma_{i\neq{j}a_ia_j} (4).

    I'd appreciate a bit of clarification on manipulations in (2) and (3)...
    Last edited by Volga; March 17th 2011 at 08:41 PM.
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  2. #2
    MHF Contributor matheagle's Avatar
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    I'm confused as to the covariance of Y's being \sigma^2
    that means the correlation is 1.
    The variance is just \sigma^2\sum_{i=1}^n\sum_{j=1}^n a_ia_j
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  3. #3
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    Quote Originally Posted by matheagle View Post
    I'm confused as to the covariance of Y's being \sigma^2
    that means the correlation is 1.
    I misspelt rho as ro! Latext didn't recognise it. Sorry! I edited. It's \rho\sigma^2
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  4. #4
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    I've just realised I messed up the final answer too, it should read:

    \sigma^2\Sigma_{i=1}^na_i^2+\rho\sigma^2\Sigma_{j\  neq{i}}a_ia_j

    so apologies, my bad.


    Anyways, I had a bright idea to look up "variance of a sum of correlated variables" and I've figure it out

    Var(\Sigma_{i=1}^na_iY_i)=\Sigma_{i=1}^n\Sigma_{j=  1}^nCov(a_iY_i,a_jY_j) (per Wiki),

    which also includes Var(a_iY_i)=Cov(a_iY_i,a_iY_i) when i=j:

    =\Sigma_{i=1}^nVar(a_iY_i)+\Sigma_{i=1}^n\Sigma_{j  \neq{i}}Cov(a_iY_i,a_iY_i)

    =\Sigma_{i=1}^na_i^2Var(Y_i)+\Sigma_{i=1}^n\Sigma_  {j\neq{i}}a_ia_jCov(Y_i,Y_i) (we can take constants outside)

    =\Sigma_{i=1}^na_i^2\sigma^2+\Sigma_{i=1}^n\Sigma_  {j\neq{i}}a_ia_j\rho\sigma^2 (substibute info given in the question)

    Finally, taking common factors outside the summation sign

    =\sigma^2\Sigma_{i=1}^na_i^2+\rho\sigma^2\Sigma_{i  =1}^n\Sigma_{j\neq{i}}a_ia_j as per the final book answer.


    Strange though that I didn't use the fact that from (a) that \Sigma_{i=1}^na_i=1
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  5. #5
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    Quote Originally Posted by Volga View Post
    Strange though that I didn't use the fact that from (a) that \Sigma_{i=1}^na_i=1
    If \sum a_i \ne 1 then you would have to tack on the square of the bias, so you are using it implicitly when you jump straight into calculating the variance.
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