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Math Help - Bias estimators - please help

  1. #1
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    Bias estimators - please help

    Hi, i posted this in the urgent maths help forum, but i thought it would be better here, any sort of help would be greatly appreciated, thanks.
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    Quote Originally Posted by skamoni View Post
    Hi, i posted this in the urgent maths help forum, but i thought it would be better here, any sort of help would be greatly appreciated, thanks.
    There is a problem with the formula given in Q1. The right hand side reduces to c(X_n - X_1) and I can't see the expected value of this ever being equal to \sigma^2 .....
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    Quote Originally Posted by skamoni View Post
    Hi, i posted this in the urgent maths help forum, but i thought it would be better here, any sort of help would be greatly appreciated, thanks.
    An estimator \Phi of a parameter \theta is unbiased iff:

    E(\Phi)=\theta

    You first problem is to determine the value of c so that:

    \overline{\sigma^2}=c\sum_{i=1}^{n-1}(X_{i+1}-X_i)

    is an unbiased estimator for the variance.

    This is imposible unless the variance is 0 as the expectation of the right hand side is 0.

    Now if instead it was meant to be:

    \overline{\sigma^2}=c\sum_{i=1}^{n-1}(X_{i+1}-X_i)^2

    we have:

    E(\overline{\sigma^2})=c\sum_{i=1}^{n-1}E[(X_{i+1}-X_i)^2]=c (n-1)E(X_{2}^2-2X_{2}X_1+X_1^2)

    This last bit of simplification because the X_i are independently identically distributed.

    So:

    E(\overline{\sigma^2})=c (n-1)(\overline{X^2}-2\bar{X}^2+\overline{X^2})=2c(n-1)\sigma^2

    so can you now tell us what value of c will make  \overline{\sigma^2} an unbiased estimator?


    RonL
    Last edited by CaptainBlack; April 25th 2008 at 05:37 AM.
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    thanks

    yeh sorry i got the question wrong, you had to include the squared, thanks alot for your help
    Last edited by CaptainBlack; April 25th 2008 at 06:21 AM. Reason: "mate" deleted as it can be offensive
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