currently i'm trying to find the expected value of the mean squares in an ANOVA. However, I think I have a mistake because the latter term is too long and confused me a little. Any help is welcome.
My model is \begin{equation} y_{ij}=\mu+\tau_{i}+\epsilon_{ij}, \qquad i=1,...,k;\quad j=1,...,n_{i} \end{equation} Where
- $\mu$ is the overall mean
- $\tau_{i}$ is the deviation from the mean by the treatment $i$
- $\epsilon_{ij}$ is a random variable with normal distribution and constant variance \begin{equation} \epsilon_{ij}\sim N(0,\sigma^{2}) \end{equation}
The mean squares error for a an unbalanced design the following: \begin{align} \frac{\sum_{i=1}^{k}n_{i}(\bar{y}_{i.}-\bar{y}_{..})^{2}}{k-1} & =\frac{\sum_{i=1}^{k}n_{i}\left(\bar{y}_{i.}^{2}-2\bar{y}_{i.}\bar{y}_{..}+\bar{y}_{..}^{2}\right)}{k-1} \\ & = \frac{\sum_{i=1}^{k}n_{i}\bar{y}^{2}_{i.}-2\bar{y}_{..}\sum_{i=1}^{k}n_{i}\bar{y}_{i.}+n\bar{y}_{..}^{2}}{k-1} \\ & = \frac{\sum_{i=1}^{k}n_{i}\bar{y}^{2}_{i.}-2\bar{y}_{..}\sum_{i=1}^{k}\frac{n_{i}}{n_{i}}\sum_{j=1}^{n_{i}}y_{ij}+n\bar{y}_{..}^{2}}{k-1} \\ & = \frac{\sum_{i=1}^{k}n_{i}\bar{y}^{2}_{i.}-2n\bar{y}^{2}_{..}+n\bar{y}_{..}^{2}}{k-1} \\ & = \frac{\sum_{i=1}^{k}n_{i}\bar{y}^{2}_{i.}-n\bar{y}_{..}^{2}}{k-1} \\ & = \frac{1}{k-1}\left[\sum_{i=1}^{k}\frac{1}{n_{i}}\left(\sum_{j=1}^{n_{i}}y_{ij}\right)^{2}-\frac{1}{n}\left(\sum_{i=1}^{k}\sum_{j=1}^{n_{i}}y_{ij}\right)^{2}\right]\tag{1}\\ & = \frac{1}{k-1}\left[\sum_{i=1}^{k}\frac{1}{n_{i}}\left(\sum_{j=1}^{n_{i}}\mu+\tau_{i}+\epsilon_{ij}\right)^{2}-\frac{1}{n}\left(\sum_{i=1}^{k}\sum_{j=1}^{n_{i}}\mu+\tau_{i}+\epsilon_{ij}\right)^{2}\right]\\ \end{align} Recall that in $(1)$ $y_{ij}=\mu+\tau_{i}+\epsilon_{ij}$
I need the expected value of this