Variance of an Addative Model

**statmajor** · March 25th 2010, 04:16 PM

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?

**matheagle** · March 25th 2010, 10:21 PM

I n=bc instead, but I don't follow your comment on multiplying by a, which may then agree with my computation.

Originally Posted by statmajor

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?

**statmajor** · March 26th 2010, 01:40 PM

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 $Var(\hat{\alpha_i})$

**Random Variable** · March 26th 2010, 02:16 PM

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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