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Math Help - Variance of an Addative Model

  1. #1
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    Variance of an Addative Model

    Let X_{ijk} be normal distributed with mean \mu_{ijk} and common variance \sigma^2. i = 1,...,a ; j = 1,...,b;k = 1,...,c

    Find the variance of \hat{\alpha_i} = \overline{X}_{i..} - \overline{X}_{...}

    Var(\hat{\alpha_1}) = Var(\overline{X}_{1..} - \frac{\overline{X}_{1..} + ... + \overline{X}_{a..}}{a})=Var(\frac{(a-1)\overline{X}_{1..} - ... - \overline{X}_{a..}}{a}) =  a^{-2}[(a-1)^2 \frac{\sigma^2}{n} + (a-1) \frac{\sigma^2}{n}]

    where n = abc. Then I would multiply it by 'a' to find Var(\hat{\alpha_i}) Have I made a mistake somewhere?
    Last edited by statmajor; March 25th 2010 at 07:40 PM.
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  2. #2
    MHF Contributor matheagle's Avatar
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    I n=bc instead, but I don't follow your comment on multiplying by a, which may then agree with my computation.

    Quote Originally Posted by statmajor View Post
    Let X_{ijk} be normal distributed with mean \mu_{ijk} and common variance \sigma^2. i = 1,...,a ; j = 1,...,b;k = 1,...,c

    Find the variance of \hat{\alpha_i} = \overline{X}_{i..} - \overline{X}_{...}

    Var(\hat{\alpha_1}) = Var(\overline{X}_{1..} - \frac{\overline{X}_{1..} + ... + \overline{X}_{a..}}{a})=Var(\frac{(a-1)\overline{X}_{1..} - ... - \overline{X}_{a..}}{a}) =  a^{-2}[(a-1)^2 \frac{\sigma^2}{n} + (a-1) \frac{\sigma^2}{n}]

    where n = abc. Then I would multiply it by 'a' to find Var(\hat{\alpha_i}) Have I made a mistake somewhere?
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  3. #3
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    I calculated Var(\hat{\alpha_1}), and not Var(\hat{\alpha_i}), so I'll need to sum up Var(\hat{\alpha_1}) from i= 1,...,a to get Var(\hat{\alpha_i})
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  4. #4
    Super Member Random Variable's Avatar
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    Are the observations independent?

    If they are, then  V(\hat{\alpha}_{i}) = \sigma^{2}\Big(\frac{1}{bc} + \frac{1}{abc} \Big)
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