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Math Help - variance & covariance of orthogonal contrasts

  1. #1
    Senior Member Sambit's Avatar
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    Question variance & covariance of orthogonal contrasts

    Consider an RBD with 5 replications and 4 treatments t_1,t_2,t_3,t_4. We are given the following three treatment-contrasts:
    \frac{t_1-t_2}{\sqrt{2}}
    \frac{t_1+t_2-2t_3}{\sqrt{6}}
    \frac{t_1+t_2+t_3-t_4}{\sqrt{12}}
    Find the covariances of all possible pairs of BLUEs and variances of the BLUEs of the three contrasts.

    What I can see is they are mutually orthogonal. Then?
    Last edited by Sambit; June 12th 2011 at 11:11 PM. Reason: wrong denominator of 1st contrast
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  2. #2
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    Where exactly are you having problems? A precise statement of what the model is would be useful - it isn't clear to me where the blocks are showing up, given the way the question is phrased.

    The anticipation is that the orthogonality is going to cause the contrasts to be uncorrelated. If \hat{\bm t} is the BLUE of t = (t_1, t_2, t_3, t_4)^T and \bm w_1 and \bm w_2 are orthogonal contrasts, you are going to want to see what happens when you calculate

    \mbox{Cov}[\bm w_1 ^T \hat{\bm t}, \bm w_2 ^T \hat{\bm t}] = \bm w_1 ^T \mbox{Var}[\hat {\bm t}] \bm w_2 .

    If \mbox{Var}[\hat t] is proportional to the identity matrix, this is going to give you everything immediately.
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  3. #3
    Senior Member Sambit's Avatar
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    Suppose the BLUE of c_it_i is c_i(\bar{y_{io}}-\bar{y_{oo}}). Then variance of 1st contrast becomes \frac{1}{2}Var(\bar{y_{1o}}-\bar{y_{2o}}) = \frac{1}{2}[Var(\bar{y_{1o}})+Var(\bar{y_{2o}})] = \frac{1}{2}[\frac{\sigma^2}{5}+\frac{\sigma^2}{5}] = \frac{\sigma^2}{5} . But ideally it should be 1...
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  4. #4
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    Quote Originally Posted by Sambit View Post
    Suppose the BLUE of c_it_i is c_i(\bar{y_{io}}-\bar{y_{oo}}). Then variance of 1st contrast becomes \frac{1}{2}Var(\bar{y_{1o}}-\bar{y_{2o}}) = \frac{1}{2}[Var(\bar{y_{1o}})+Var(\bar{y_{2o}})] = \frac{1}{2}[\frac{\sigma^2}{5}+\frac{\sigma^2}{5}] = \frac{\sigma^2}{5} . But ideally it should be 1...
    No it shouldn't. The variance of the contrast (1) needs to depend on \sigma^2 for obvious reasons, and (2) needs to go to 0 as we increase the amount of information we have. The number 1 does neither of these things.
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