Chapter 13.9, Problem 51E

a.

To determine

Derive $E\left(MST\right)$.

Answer to Problem 51E

$E\left(MST\right)={\sigma }^2+\frac{k}{b-1}{\displaystyle \sum _{i=1}^k{\tau }_i^2}$.

Explanation of Solution

It is known that $MST=\frac{b}{k-1}{\displaystyle \sum _{i=1}^k{\left({\overline{Y}}_{i,.}-\overline{Y}\right)}^2}$.

To find $E\left(MST\right)$, it is required to find $E\left({\overline{Y}}_{i,.}^2\right)$.

The model associated with the randomized block design is as follows:

$\begin{array}{l}{Y}_{ij}=\mu +{\tau }_i+{\beta }_j+{\epsilon }_{ij},i=1,2,\mathrm{...}k\text{ and}j=1,2,\mathrm{...}b\\ E\left(Y_{ij}\right)=\mu +{\tau }_i+{\beta }_j\\ V\left(Y_{ij}\right)={\sigma }^2\\ E\left({\epsilon }_{ij}\right)=0\\ V\left({\epsilon }_{ij}\right)={\sigma }^2\\ {\mu }_i=\mu +{\tau }_i\end{array}$.

It is known that ${\displaystyle \sum _{i=1}^k{\tau }_i=0}\text{ and }{\displaystyle \sum _{j=1}^b{\beta }_j=0}$.

${\overline{Y}}_{i.}$ can be rewritten as follows:

$\begin{array}{c}{\overline{Y}}_{i.}=\frac{1}{b}{\displaystyle \sum _{j=1}^b{Y}_{ij}}\\ =\frac{1}{b}{\displaystyle \sum _{j=1}^b\left(\mu +{\tau }_i+{\beta }_j+{\epsilon }_{ij}\right)}\\ =\frac{1}{b}\left(b\mu \right)+\frac{1}{b}{\displaystyle \sum _{j=1}^b\left({\tau }_i\right)}+\frac{1}{b}{\displaystyle \sum _{j=1}^b\left({\beta }_j\right)+}\frac{1}{b}{\displaystyle \sum _{j=1}^b\left({\epsilon }_{ij}\right)}\\ =\mu +{\tau }_i+\frac{1}{b}{\displaystyle \sum _{i=1}^k\left(0\right)}+\frac{1}{b}{\displaystyle \sum _{j=1}^b\left({\epsilon }_{ij}\right)}\\ =\mu +{\tau }_i+\frac{1}{b}{\displaystyle \sum _{j=1}^b\left({\epsilon }_{ij}\right)}\end{array}$

The expected is calculated as follows:

$\begin{array}{c}E\left({\overline{Y}}_{i.}\right)=E\left(\mu \right)+E\left({\tau }_i\right)+\frac{1}{b}{\displaystyle \sum _{j=1}^{b}E\left({\epsilon }_{ij}\right)}\\ =\mu +{\tau }_i\end{array}$

The variance is calculated as follows:

$\begin{array}{c}V\left({\overline{Y}}_{i.}\right)=V\left(\mu \right)+V\left({\tau }_i\right)+\frac{1}{b^2}{\displaystyle \sum _{j=1}^{b}V\left({\epsilon }_{ij}\right)}\\ =0+0+b\frac{{\sigma }^2}{b^2}\\ =\frac{{\sigma }^2}{b}\end{array}$

Now,

$\begin{array}{c}E\left({\overline{Y}}_{i,.}^2\right)=V\left({\overline{Y}}_{i,.}\right)-{\left[E\left({\overline{Y}}_{i,.}\right)\right]}^2\\ =\frac{1}{b}{\sigma }^2+{\left(\mu +{\tau }_i\right)}^2\end{array}$

Also, $\overline{Y}=\frac{1}{bk}{\displaystyle \sum _{i=1}^k{\displaystyle \sum _{j=1}^b{Y}_{i,j}}},E\left(\overline{Y}\right)=\mu ,V\left(\overline{Y}\right)=\frac{1}{bk}{\sigma }^2,E{\left(\overline{Y}\right)}^2=\frac{1}{bk}{\sigma }^2+{\mu }^2$.

$\begin{array}{c}E\left(MST\right)=E\left[\frac{b}{k-1}{\displaystyle \sum _{i=1}^k{\left({\overline{Y}}_{i,.}-\overline{Y}\right)}^2}\right]\\ =\frac{b}{k-1}E\left[{\displaystyle \sum _{i=1}^k{\left({\overline{Y}}_{i,.}-\overline{Y}\right)}^2}\right]\\ =\frac{b}{k-1}\left[{\displaystyle \sum _{i=1}^{k}E\left({\overline{Y}}_{{}_{i,.}}^2\right)-kE\left({\overline{Y}}^2\right)}\right]\\ =\frac{b}{k-1}\left[{\displaystyle \sum _{i=1}^k\left(\frac{1}{b}{\sigma }^2+{\mu }^2+2\mu {\tau }_i+{\tau }_i^2\right)-k\left(\frac{1}{bk}{\sigma }^2+{\mu }^2\right)}\right]\\ =\frac{b}{k-1}\left[\frac{k-1}{b}{\sigma }^2+2\mu {\displaystyle \sum _{i=1}^k{\tau }_i+{\displaystyle \sum _{i=1}^k{\tau }_i^2}}\right]\left[\text{Since,}{\displaystyle \sum _{i=1}^k{\tau }_i=0}\right]\\ ={\sigma }^2+\frac{b}{k-1}{\displaystyle \sum _{i=1}^k{\tau }_i^2}\end{array}$

Thus, $E\left(MST\right)={\sigma }^2+\frac{k}{b-1}{\displaystyle \sum _{i=1}^k{\tau }_i^2}$.

b.

To determine

Derive $E\left(MSB\right)$.

Answer to Problem 51E

$E\left(MSB\right)={\sigma }^2+\frac{k}{b-1}{\displaystyle \sum _{j=1}^b{\beta }_j^2}$.

Explanation of Solution

It is known that $MSB=\frac{k}{b-1}{\displaystyle \sum _{j=1}^b{\left({\overline{Y}}_{.,j}-\overline{Y}\right)}^2}$.

To find $E\left(MST\right)$, it is required to find $E\left({\overline{Y}}_{.,j}^2\right)$.

The model associated with the randomized block design is as follows:

$\begin{array}{l}{Y}_{ij}=\mu +{\tau }_i+{\beta }_j+{\epsilon }_{ij},i=1,2,\mathrm{...}k\text{ and}j=1,2,\mathrm{...}b\\ E\left(Y_{ij}\right)=\mu +{\tau }_i+{\beta }_j\\ V\left(Y_{ij}\right)={\sigma }^2\\ E\left({\epsilon }_{ij}\right)=0\\ V\left({\epsilon }_{ij}\right)={\sigma }^2\\ {\mu }_i=\mu +{\tau }_i\end{array}$.

It is known that $E\left({\overline{Y}}_{.j}\right)=\mu +{\beta }_i$ and $V\left({\overline{Y}}_{.j}\right)=\frac{{\sigma }^2}{k}$.

Now,

$\begin{array}{c}E\left({\overline{Y}}_{j.}^2\right)=V\left({\overline{Y}}_{j,.}\right)-{\left[E\left({\overline{Y}}_{j,.}\right)\right]}^2\\ =\frac{1}{k}{\sigma }^2+{\left(\mu +{\beta }_j\right)}^2\end{array}$

Also, $\overline{Y}=\frac{1}{bk}{\displaystyle \sum _{i=1}^k{\displaystyle \sum _{j=1}^b{Y}_{i,j}}},E\left(\overline{Y}\right)=\mu ,V\left(\overline{Y}\right)=\frac{1}{bk}{\sigma }^2,E{\left(\overline{Y}\right)}^2=\frac{1}{bk}{\sigma }^2+{\mu }^2$.

$\begin{array}{c}E\left(MSB\right)=E\left[\frac{k}{b-1}{\displaystyle \sum _{j=1}^b{\left({\overline{Y}}_{.,j}-\overline{Y}\right)}^2}\right]\\ =\frac{k}{b-1}E\left[{\displaystyle \sum _{j=1}^b{\left({\overline{Y}}_{.,j}-\overline{Y}\right)}^2}\right]\\ =\frac{k}{b-1}\left[{\displaystyle \sum _{j=1}^{b}E\left({\overline{Y}}^2{}_{.,j}\right)-bE\left(\overline{Y}\right)}\right]\\ =\frac{k}{b-1}\left[{\displaystyle \sum _{j=1}^b\left(\frac{1}{k}{\sigma }^2+{\mu }^2+2\mu {\beta }_j+{\beta }_j^2\right)-b\left(\frac{1}{bk}{\sigma }^2+{\mu }^2\right)}\right]\\ =\frac{k}{b-1}\left[\frac{b-1}{k}{\sigma }^2+2\mu {\displaystyle \sum _{j=1}^b{\beta }_j+{\displaystyle \sum _{j=1}^b{\beta }_j^2}}\right]\left[\text{Since,}{\displaystyle \sum _{j=1}^b{\beta }_j=0}\right]\\ ={\sigma }^2+\frac{k}{b-1}{\displaystyle \sum _{j=1}^b{\beta }_j^2}\end{array}$

Thus, $E\left(MSB\right)={\sigma }^2+\frac{k}{b-1}{\displaystyle \sum _{j=1}^b{\beta }_j^2}$.

c.

To determine

Derive $E\left(MSE\right)$.

Answer to Problem 51E

$E\left(MSE\right)={\sigma }^2$.

Explanation of Solution

It is known that $SSE=TSS-SST-SSB$.

From Parts (a) and (b), $E\left(MST\right)={\sigma }^2+\frac{k}{b-1}{\displaystyle \sum _{i=1}^k{\tau }_i^2}$ and $E\left(MSB\right)={\sigma }^2+\frac{k}{b-1}{\displaystyle \sum _{j=1}^b{\beta }_j^2}$.

It is required to find $E\left(TSS\right)$.

$TSS={\displaystyle \sum _{i=1}^k{\displaystyle \sum _{j=1}^b{Y}_{i,j}^2}}-bk{\overline{Y}}^2$

$\begin{array}{c}E\left(Y_{i,j}^2\right)=V\left(Y_{i,j}\right)+{\left(E\left(Y_{i,j}\right)\right)}^2\\ ={\sigma }^2+{\left(\mu +{\tau }_i+{\beta }_j\right)}^2\end{array}$

$\begin{array}{c}E\left(TSS\right)=E\left[{\displaystyle \sum _{i=1}^k{\displaystyle \sum _{j=1}^b{Y}_{i,j}^2}}-bk{\overline{Y}}^2\right]\\ =E\left({\displaystyle \sum _{i=1}^k{\displaystyle \sum _{j=1}^b{Y}_{i,j}^2}}\right)-bkE\left({\overline{Y}}^2\right)\\ ={\displaystyle \sum _{i=1}^k{\displaystyle \sum _{j=1}^b\left({\sigma }^2+{\mu }^2+{\tau }_i{}^2+{\beta }_j{}^2\right)}}-kb\left(\frac{1}{bk}{\sigma }^2+{\mu }^2\right)\\ =\left(bk-1\right){\sigma }^2+b{\displaystyle \sum _{i=1}^k{\tau }_i{}^2+k}{\displaystyle \sum _{j=1}^b{\beta }_j{}^2}\end{array}$

From Part (a), $E\left(SST\right)=E\left(MST\right)\left(n-k-b+1\right)$, and from Part (b), $E\left(SSB\right)=E\left(MSB\right)\left(b-1\right)$.

Now,

$\begin{array}{c}E\left(SSE\right)=E\left(TSS\right)-E\left(SST\right)-E\left(SSB\right)\\ =\left[\begin{array}{l}\left(bk-1\right){\sigma }^2+b{\displaystyle \sum _{i=1}^k{\tau }_i{}^2+k}{\displaystyle \sum _{j=1}^b{\beta }_j{}^2}-\\ E\left(MST\right)\left(n-k-b+1\right)-\\ E\left(MSB\right)\left(b-1\right)\end{array}\right]\\ =\left[\begin{array}{l}\left(bk-1\right){\sigma }^2+b{\displaystyle \sum _{i=1}^k{\tau }_i{}^2+k}{\displaystyle \sum _{j=1}^b{\beta }_j{}^2}-\\ {\sigma }^2+\frac{k}{b-1}{\displaystyle \sum _{i=1}^k{\tau }_i^2}\left(n-k-b+1\right)-\\ {\sigma }^2+\frac{k}{b-1}{\displaystyle \sum _{j=1}^b{\beta }_j^2}\left(b-1\right)\end{array}\right]\\ =\left(bk-k-b+1\right){\sigma }^2\\ E\left(MSE\right)=\frac{E\left(SSE\right)}{bk-k-b+1}\\ =\frac{bk-k-b+1\left({\sigma }^2\right)}{bk-k-b+1}\\ ={\sigma }^2\end{array}$

Thus, $E\left(MSE\right)={\sigma }^2$.