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Thread: Least Square Derivation - Checking

  1. #1
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    Least Square Derivation - Checking

    QUESTION

    I had a 10cm pole which broke in half. The two peices were measured and this resulted in values of $\displaystyle Z_1$ and $\displaystyle Z_2 $ respectively.

    If the true 'correct' length of the first peice is given as $\displaystyle \gamma $ and the standard deviation of the two measurements is $\displaystyle \sigma $

    1) Then find the least-square estimate of $\displaystyle \gamma $
    2) Find the standard error of the estimate in (1) in terms of $\displaystyle \sigma $

    My Answer so Far

    $\displaystyle Z_1 = \gamma + \epsilon_1 $
    $\displaystyle Z_2 = 10 - \gamma + \epsilon_2 $

    By Least-Squares Definition we minimize distance

    $\displaystyle \epsilon_1 = (Z_1 - \gamma)^2 $
    $\displaystyle \epsilon_2 = (Z_2 - 10 + \gamma)^2 $

    Take derivative and set to zero? This is where I am unsure? *Thanks*

    If I do this I get

    $\displaystyle -2(Z_1 - \gamma) + 2(Z_2 - 10 + \gamma) = 0 $

    This solves to

    $\displaystyle \hat\gamma\ = \frac{Z_1 - Z_2 + 10}{2} $

    Is this somewhere close?

    2) $\displaystyle Var(\hat\gamma) = \frac{1}{2^2} Var(Z_1 -Z_2 + 10)$

    Now Im also unsure of this variance property.. but I solve to get

    $\displaystyle Var(\hat\gamma) = \frac{\sigma^2 + \sigma^2}{4} $
    $\displaystyle Var(\hat\gamma) =\frac{\sigma^2}{2} $

    Is it right that the variance of the 10 is zero.. so it just leaves the equation?

    Thanks very much for any help advice
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  2. #2
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    I just realised that (2) should actually say:

    (2) Find the standard error of the estimate in (a) in terms of $\displaystyle \gamma $.. so i think I shouldn't be finding variance.. but standard error?

    Would that just be the sum of errors squared divided by the degrees of freedom (which would be 1?)??
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