# Thread: variance & covariance of orthogonal contrasts

1. ## 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?

2. 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.

3. 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...

4. Originally Posted by Sambit
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.