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Math Help - Variance of S^2

  1. #1
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    Variance of S^2

    Haven't used this for a while.. not sure if i get the texing.
    I prove first that (this was fine)
      \frac{1}{2n(n-1)} \sum^n_{i=1}\sum^n_{j=1}(X_i-X_j)^2 = S^2

    Then i am to calculate, assuming all the X_i's have finite fourth moment, denoting \theta_1 =EX_i, \theta_j=E(X_i - \theta_1)^j, j=1,2,3, the
    Var(S^2)=\frac{1}{n}\left(\theta_4-\frac{n-3}{n-1}\theta_2^2 \right). I first used Var(S^2)=E(S^4)-E(S^2)^2 and tried to calculate E(S^2) but
    I don't really get how these cross terms work. My first attempt i got \theta_2/(n-1) but i'm only assuming that's wrong.
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  2. #2
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    Re: Variance of S^2

    <br />
Var (S^2) = E(S^4)-E(S^2)^2
    We know that E(S^2) = \theta_1^2 so we just need to calculate E(S^4). Let Y_i = X_i - \theta_2, for i=1,2,...n. Then put
     S^2 =\frac{n \sum_{i=1}^n Y_i^2-(\sum_{i=1}^n Y_i)^2}{n(n-1)}<br />
    So that
    <br />
S^4 = \frac{n^2(\sum Y_i^2)^2 -2n(\sum Y_i^2)(\sum Y_i)^2 +(\sum Y_i)^4}{n^2(n-1)^2}<br />
    and
     E(S^4) = \frac{n^2E(\sum Y_i^2)^2 -2nE[(\sum Y_i^2)(\sum Y_i)^2] +E(\sum Y_i)^4}{n^2(n-1)^2}<br />

    Now we have i \neq j \neq k
    <br />
E(Y_i Y_j) = 0, E(Y_i^3Y_j)=0, E(Y_i^2Y_jY_k)=0,<br />
E(Y_i^2Y_j^2) = \theta_2,  E(Y_i^4)= \theta_4 <br />
    and
    <br />
E(\sum Y_i^2)^2 = n\theta_4 + n(n-1)\theta_2^2 <br />
E[ (\sum Y_i^2)(\sum Y_i)^2] = n\theta_4 + 3n(n-1)\theta_2^2 <br />
E(\sum Y_i)^4 = n\theta_4 + 3n(n-1) \theta_2^2<br />
    Giving us
    <br />
E(S^4) = \frac{n^2(n\theta_4 + n(n-1)\theta_2^2) -2n(n\theta_4 + 3n(n-1)\theta_2^2) +  n\theta_4 + 3n(n-1) \theta_2^2}{n^2(n-1)} <br />
= \frac{(n-1)\theta_4 +(n^2 -2n +3)\theta_2^2}{n(n-1)}<br />
    Which gives
    <br />
Var(S^2) = \frac{(n-1)\theta_4 +(n^2 -2n +3)\theta_2^2}{n(n-1)} - \theta_2^2 \\<br />
 = \frac{1}{n} \left (\theta_4 -\frac{n-3}{n-1} \theta_2^2 \right)<br />

    Not sure why it hates my teXing... anyways its solved
    Last edited by Scopur; November 27th 2012 at 07:19 PM.
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