Index: The Book of Statistical ProofsStatistical Models ▷ Univariate normal data ▷ Analysis of variance ▷ Cochran's theorem for two-way ANOVA

Theorem: Assume the two-way analysis of variance model

$\label{eq:anova2} \begin{split} y_{ijk} &= \mu + \alpha_i + \beta_j + \gamma_{ij} + \varepsilon_{ijk} \\ \varepsilon_{ijk} &\overset{\mathrm{i.i.d.}}{\sim} \mathcal{N}(0, \sigma^2), \; i = 1, \ldots, a, \; j = 1, \ldots, b, \; k = 1, \dots, n_{ij} \end{split}$

under the well-known constraints for the model parameters

$\label{eq:anova2-constr} \begin{split} \sum_{i=1}^{a} \frac{n_{ij}}{n} \alpha_i &= 0 \quad \text{for all} \quad j = 1, \ldots, b \\ \sum_{j=1}^{b} \frac{n_{ij}}{n} \beta_j &= 0 \quad \text{for all} \quad i = 1, \ldots, a \\ \sum_{i=1}^{a} \frac{n_{ij}}{n} \gamma_{ij} &= 0 \quad \text{for all} \quad j = 1, \ldots, b \\ \sum_{j=1}^{b} \frac{n_{ij}}{n} \gamma_{ij} &= 0 \quad \text{for all} \quad i = 1, \ldots, a \; . \end{split}$

Then, the following sums of squares are chi-square distributed

$\label{eq:anova2-cochran} \begin{split} \frac{1}{\sigma^2} n (\bar{y}_{\bullet \bullet \bullet} - \mu)^2 = \frac{\mathrm{SS}_M}{\sigma^2} &\sim \chi^2(1) \\ \frac{1}{\sigma^2} \sum_{i=1}^{a} n_{i \bullet} ([\bar{y}_{i \bullet \bullet} - \bar{y}_{\bullet \bullet \bullet}] - \alpha_i)^2 = \frac{\mathrm{SS}_A}{\sigma^2} &\sim \chi^2(a-1) \\ \frac{1}{\sigma^2} \sum_{j=1}^{b} n_{\bullet j} ([\bar{y}_{\bullet j \bullet} - \bar{y}_{\bullet \bullet \bullet}] - \beta_j)^2 = \frac{\mathrm{SS}_B}{\sigma^2} &\sim \chi^2(b-1) \\ \frac{1}{\sigma^2} \sum_{i=1}^{a} \sum_{j=1}^{b} n_{ij} ([\bar{y}_{i j \bullet} - \bar{y}_{i \bullet \bullet} - \bar{y}_{\bullet j \bullet} + \bar{y}_{\bullet \bullet \bullet}] - \gamma_{ij})^2 = \frac{\mathrm{SS}_{A \times B}}{\sigma^2} &\sim \chi^2\left( (a-1)(b-1) \right) \\ \frac{1}{\sigma^2} \sum_{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{n_{ij}} (y_{ijk} - \bar{y}_{i j \bullet})^2 = \frac{\mathrm{SS}_\mathrm{res}}{\sigma^2} &\sim \chi^2(n-ab) \; . \end{split}$

Proof: Denote sample sizes as

$\label{eq:samp-size} \begin{split} n_{ij} &- \text{number of samples in category} \; (i,j) \\ n_{i \bullet} &= \sum_{j=1}^{b} n_{ij} \\ n_{\bullet j} &= \sum_{i=1}^{a} n_{ij} \\ n &= \sum_{i=1}^{a} \sum_{j=1}^{b} n_{ij} \end{split}$

and denote sample means as

$\label{eq:mean-samp} \begin{split} \bar{y}_{\bullet \bullet \bullet} &= \frac{1}{n} \sum_{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{n_{ij}} y_{ijk} \\ \bar{y}_{i \bullet \bullet} &= \frac{1}{n_{i \bullet}} \sum_{j=1}^{b} \sum_{k=1}^{n_{ij}} y_{ijk} \\ \bar{y}_{\bullet j \bullet} &= \frac{1}{n_{\bullet j}} \sum_{i=1}^{a} \sum_{k=1}^{n_{ij}} y_{ijk} \\ \bar{y}_{i j \bullet} &= \frac{1}{n_{ij}} \sum_{k=1}^{n_{ij}} y_{ijk} \; . \end{split}$

According to the model given by \eqref{eq:anova2}, the observations are distributed as:

$\label{eq:yijk-h0} y_{ijk} \sim \mathcal{N}(\mu + \alpha_i + \beta_j + \gamma_{ij}, \sigma^2) \quad \text{for all} \quad i, j, k \; .$

Thus, the random variable $U_{ijk} = (y_{ijk} - \mu - \alpha_i - \beta_j - \gamma_{ij})/\sigma$ follows a standard normal distribution

$\label{eq:Uijk-h0} U_{ijk} = \frac{y_{ijk} - \mu - \alpha_i - \beta_j - \gamma_{ij}}{\sigma} \sim \mathcal{N}(0, 1) \; .$

Now consider the following sum

$\label{eq:sum-Uijk-s1} \sum_{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{n_{ij}} U_{ijk}^2 = \sum_{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{n_{ij}} \left( \frac{y_{ijk} - \mu - \alpha_i - \beta_j - \gamma_{ij}}{\sigma} \right)^2 \\$

which can be rewritten as follows:

$\label{eq:sum-Uijk-s2} \begin{split} \sum_{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{n_{ij}} U_{ijk}^2 = \frac{1}{\sigma^2} \sum_{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{n_{ij}} & \left[ (y_{ijk} - \mu - \alpha_i - \beta_j - \gamma_{ij}) - \right. \\ &\left. [\bar{y}_{\bullet \bullet \bullet} + (\bar{y}_{i \bullet \bullet} - \bar{y}_{\bullet \bullet \bullet}) + (\bar{y}_{\bullet j \bullet} - \bar{y}_{\bullet \bullet \bullet}) + (\bar{y}_{i j \bullet} - \bar{y}_{i \bullet \bullet} - \bar{y}_{\bullet j \bullet} + \bar{y}_{\bullet \bullet \bullet})] \right. + \\ &\left. [\bar{y}_{\bullet \bullet \bullet} + (\bar{y}_{i \bullet \bullet} - \bar{y}_{\bullet \bullet \bullet}) + (\bar{y}_{\bullet j \bullet} - \bar{y}_{\bullet \bullet \bullet}) + (\bar{y}_{i j \bullet} - \bar{y}_{i \bullet \bullet} - \bar{y}_{\bullet j \bullet} + \bar{y}_{\bullet \bullet \bullet})] \right]^2 \\ = \frac{1}{\sigma^2} \sum_{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{n_{ij}} & \left[ (y_{ijk} - [\bar{y}_{\bullet \bullet \bullet} + (\bar{y}_{i \bullet \bullet} - \bar{y}_{\bullet \bullet \bullet}) + (\bar{y}_{\bullet j \bullet} - \bar{y}_{\bullet \bullet \bullet}) + (\bar{y}_{i j \bullet} - \bar{y}_{i \bullet \bullet} - \bar{y}_{\bullet j \bullet} + \bar{y}_{\bullet \bullet \bullet})]) + \right. \\ &\left. (\bar{y}_{\bullet \bullet \bullet} - \mu) + ([\bar{y}_{i \bullet \bullet} - \bar{y}_{\bullet \bullet \bullet}] - \alpha_i) + ([\bar{y}_{\bullet j \bullet} - \bar{y}_{\bullet \bullet \bullet}] - \beta_j) \right. + \\ &\left. ([\bar{y}_{i j \bullet} - \bar{y}_{i \bullet \bullet} - \bar{y}_{\bullet j \bullet} + \bar{y}_{\bullet \bullet \bullet}] - \gamma_{ij}) \right]^2 \\ = \frac{1}{\sigma^2} \sum_{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{n_{ij}} & \left[ (y_{ijk} - \bar{y}_{i j \bullet}) + (\bar{y}_{\bullet \bullet \bullet} - \mu) + ([\bar{y}_{i \bullet \bullet} - \bar{y}_{\bullet \bullet \bullet}] - \alpha_i) + \right. \\ &\left. ([\bar{y}_{\bullet j \bullet} - \bar{y}_{\bullet \bullet \bullet}] - \beta_j) + ([\bar{y}_{i j \bullet} - \bar{y}_{i \bullet \bullet} - \bar{y}_{\bullet j \bullet} + \bar{y}_{\bullet \bullet \bullet}] - \gamma_{ij}) \right]^2 \; . \end{split}$

Note that the following sums are all zero:

$\label{eq:sum-Uijk-s3a} \begin{split} \sum_{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{n_{ij}} (y_{ijk} - \bar{y}_{i j \bullet}) &= \sum_{i=1}^{a} \sum_{j=1}^{b} \left[ \sum_{k=1}^{n_{ij}} y_{ijk} - n_{ij} \cdot \bar{y}_{i j \bullet} \right] \\ &\overset{\eqref{eq:mean-samp}}{=} \sum_{i=1}^{a} \sum_{j=1}^{b} \left[ \sum_{k=1}^{n_{ij}} y_{ijk} - n_{ij} \cdot \frac{1}{n_{ij}} \sum_{k=1}^{n_{ij}} y_{ijk} \right] \\ &= \sum_{i=1}^{a} \sum_{j=1}^{b} 0 = 0 \end{split}$ $\label{eq:sum-Uijk-s3b} \begin{split} \sum_{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{n_{ij}} ([\bar{y}_{i \bullet \bullet} - \bar{y}_{\bullet \bullet \bullet}] - \alpha_i) &= \sum_{i=1}^{a} n_{i \bullet} \cdot (\bar{y}_{i \bullet \bullet} - \bar{y}_{\bullet \bullet \bullet} - \alpha_i) \\ &= \sum_{i=1}^{a} n_{i \bullet} \cdot \bar{y}_{i \bullet \bullet} - \bar{y}_{\bullet \bullet \bullet} \sum_{i=1}^{a} n_{i \bullet} - \sum_{i=1}^{a} n_{i \bullet} \alpha_i \\ &\overset{\eqref{eq:mean-samp}}{=} \sum_{i=1}^{a} n_{i \bullet} \cdot \frac{1}{n_{i \bullet}} \sum_{j=1}^{b} \sum_{k=1}^{n_{ij}} y_{ijk} - n \cdot \frac{1}{n} \sum_{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{n_{ij}} y_{ijk} - \sum_{i=1}^{a} n_{i \bullet} \alpha_i \\ &= - \sum_{i=1}^{a} n_{i \bullet} \alpha_i \overset{\eqref{eq:samp-size}}{=} - n \sum_{i=1}^{a} \sum_{j=1}^{b} \frac{n_{ij}}{n} \alpha_i \overset{\eqref{eq:anova2-constr}}{=} 0 \end{split}$ $\label{eq:sum-Uijk-s3c} \begin{split} \sum_{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{n_{ij}} ([\bar{y}_{\bullet j \bullet} - \bar{y}_{\bullet \bullet \bullet}] - \beta_j) &= \sum_{j=1}^{b} n_{\bullet j} \cdot (\bar{y}_{\bullet j \bullet} - \bar{y}_{\bullet \bullet \bullet} - \beta_j) \\ &= \sum_{j=1}^{b} n_{\bullet j} \cdot \bar{y}_{\bullet j \bullet} - \bar{y}_{\bullet \bullet \bullet} \sum_{j=1}^{b} n_{\bullet j} - \sum_{j=1}^{b} n_{\bullet j} \beta_j \\ &\overset{\eqref{eq:mean-samp}}{=} \sum_{j=1}^{b} n_{\bullet j} \cdot \frac{1}{n_{\bullet j}} \sum_{i=1}^{a} \sum_{k=1}^{n_{ij}} y_{ijk} - n \cdot \frac{1}{n} \sum_{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{n_{ij}} y_{ijk} - \sum_{j=1}^{b} n_{\bullet j} \beta_j \\ &= - \sum_{j=1}^{b} n_{\bullet j} \beta_j \overset{\eqref{eq:samp-size}}{=} - n \sum_{j=1}^{b} \sum_{i=1}^{a} \frac{n_{ij}}{n} \beta_j \overset{\eqref{eq:anova2-constr}}{=} 0 \end{split}$ $\label{eq:sum-Uijk-s3d} \begin{split} &\hphantom{=} \sum_{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{n_{ij}} ([\bar{y}_{i j \bullet} - \bar{y}_{i \bullet \bullet} - \bar{y}_{\bullet j \bullet} + \bar{y}_{\bullet \bullet \bullet}] - \gamma_{ij}) \\ &= \sum_{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{n_{ij}} \left[ (\bar{y}_{i j \bullet} - \bar{y}_{\bullet \bullet \bullet}) - (\bar{y}_{i \bullet \bullet} - \bar{y}_{\bullet \bullet \bullet}) - (\bar{y}_{\bullet j \bullet} - \bar{y}_{\bullet \bullet \bullet}) - \gamma_{ij} \right] \\ &= \sum_{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{n_{ij}} (\bar{y}_{i j \bullet} - \bar{y}_{\bullet \bullet \bullet} - \gamma_{ij}) - \sum_{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{n_{ij}} (\bar{y}_{i \bullet \bullet} - \bar{y}_{\bullet \bullet \bullet}) - \sum_{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{n_{ij}} (\bar{y}_{\bullet j \bullet} - \bar{y}_{\bullet \bullet \bullet}) \\ &\overset{\eqref{eq:sum-Uijk-s3c}}{=} \sum_{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{n_{ij}} (\bar{y}_{i j \bullet} - \bar{y}_{\bullet \bullet \bullet} - \gamma_{ij}) - \sum_{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{n_{ij}} (\bar{y}_{i \bullet \bullet} - \bar{y}_{\bullet \bullet \bullet}) \\ &\overset{\eqref{eq:sum-Uijk-s3b}}{=} \sum_{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{n_{ij}} (\bar{y}_{i j \bullet} - \bar{y}_{\bullet \bullet \bullet} - \gamma_{ij}) \\ &= \sum_{i=1}^{a} \sum_{j=1}^{b} n_{ij} \bar{y}_{i j \bullet} - \bar{y}_{\bullet \bullet \bullet} \sum_{i=1}^{a} \sum_{j=1}^{b} n_{ij} - \sum_{i=1}^{a} \sum_{j=1}^{b} n_{ij} \gamma_{ij} \\ &\overset{\eqref{eq:mean-samp}}{=} \sum_{i=1}^{a} \sum_{j=1}^{b} n_{ij} \cdot \frac{1}{n_{ij}} \sum_{k=1}^{n_{ij}} y_{ijk} - n \cdot \frac{1}{n} \sum_{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{n_{ij}} y_{ijk} - \sum_{i=1}^{a} \sum_{j=1}^{b} n_{ij} \gamma_{ij} \\ &= - \sum_{i=1}^{a} \sum_{j=1}^{b} n_{ij} \gamma_{ij} = - \frac{1}{n} \sum_{i=1}^{a} \sum_{j=1}^{b} \frac{n_{ij}}{n} \gamma_{ij} \overset{\eqref{eq:anova2-constr}}{=} 0 \; . \end{split}$

Note further that $\bar{y}_{\bullet \bullet \bullet}$ and $\mu$ are not dependent on $i$, $j$ and $k$:

$\label{eq:yb-mu-const} \bar{y}_{\bullet \bullet \bullet} = \text{const.} \quad \text{and} \quad \mu = \text{const.}$

Thus, all the non-square products in \eqref{eq:sum-Uijk-s2} disappear and the sum reduces to

$\label{eq:sum-Uijk-s4} \begin{split} \sum_{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{n_{ij}} U_{ijk}^2 = \frac{1}{\sigma^2} \sum_{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{n_{ij}} & \left[ (y_{ijk} - \bar{y}_{i j \bullet})^2 + (\bar{y}_{\bullet \bullet \bullet} - \mu)^2 + ([\bar{y}_{i \bullet \bullet} - \bar{y}_{\bullet \bullet \bullet}] - \alpha_i)^2 + \right. \\ &\left. ([\bar{y}_{\bullet j \bullet} - \bar{y}_{\bullet \bullet \bullet}] - \beta_j)^2 + ([\bar{y}_{i j \bullet} - \bar{y}_{i \bullet \bullet} - \bar{y}_{\bullet j \bullet} + \bar{y}_{\bullet \bullet \bullet}] - \gamma_{ij})^2 \right] \\ = \frac{1}{\sigma^2} \left[ \sum_{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{n_{ij}} \right. & (y_{ijk} - \bar{y}_{i j \bullet})^2 + \sum_{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{n_{ij}} (\bar{y}_{\bullet \bullet \bullet} - \mu)^2 + \\ \sum_{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{n_{ij}} &\left. ([\bar{y}_{i \bullet \bullet} - \bar{y}_{\bullet \bullet \bullet}] - \alpha_i)^2 + \sum_{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{n_{ij}} ([\bar{y}_{\bullet j \bullet} - \bar{y}_{\bullet \bullet \bullet}] - \beta_j)^2 + \right. \\ \sum_{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{n_{ij}} &\left. ([\bar{y}_{i j \bullet} - \bar{y}_{i \bullet \bullet} - \bar{y}_{\bullet j \bullet} + \bar{y}_{\bullet \bullet \bullet}] - \gamma_{ij})^2 \right] \; . \end{split}$

Cochran’s theorem states that, if a sum of squared standard normal random variables can be written as a sum of squared forms

$\label{eq:cochran-p1} \begin{split} \sum_{i=1}^{n} U_i^2 = \sum_{j=1}^{m} Q_j \quad &\text{where} \quad Q_j = U^\mathrm{T} B^{(j)} U \\ &\text{with} \quad \sum_{j=1}^{m} B^{(j)} = I_n \\ &\text{and} \quad r_j = \mathrm{rank}(B^{(j)}) \; , \end{split}$

then the terms $Q_j$ are independent and each term $Q_j$ follows a chi-squared distribution with $r_j$ degrees of freedom:

$\label{eq:cochran-p2} Q_j \sim \chi^2(r_j), \; j = 1, \ldots, m \; .$

First, we define the $n \times 1$ vector $U$:

$\label{eq:U} U = \left[ \begin{matrix} u_{1 \bullet} \\ \vdots \\ u_{a \bullet} \end{matrix} \right] \quad \text{where} \quad u_{i \bullet} = \left[ \begin{matrix} u_{i1} \\ \vdots \\ u_{ib} \end{matrix} \right] \quad \text{where} \quad u_{ij} = \left[ \begin{matrix} (y_{i,j,1} - \mu - \alpha_i - \beta_j - \gamma_{ij})/\sigma \\ \vdots \\ (y_{i,j,n_{ij}} - \mu - \alpha_i - \beta_j - \gamma_{ij})/\sigma \end{matrix} \right] \; .$

Next, we specify the $n \times n$ matrices $B$

$\label{eq:B} \begin{split} B^{(1)} &= I_n - \mathrm{diag}\left[ \mathrm{diag}\left( \frac{1}{n_{11}} J_{n_{11}}, \; \ldots, \; \frac{1}{n_{1b}} J_{n_{1b}} \right), \; \ldots, \; \mathrm{diag}\left( \frac{1}{n_{a1}} J_{n_{a1}}, \; \ldots, \; \frac{1}{n_{ab}} J_{n_{ab}} \right) \right] \\ B^{(2)} &= \frac{1}{n} J_n \\ B^{(3)} &= \mathrm{diag}\left( \frac{1}{n_{1 \bullet}} J_{n_{1 \bullet}}, \; \ldots, \; \frac{1}{n_{a \bullet}} J_{n_{a \bullet}} \right) - \frac{1}{n} J_n \\ B^{(4)} &= M_B - \frac{1}{n} J_n \\ B^{(5)} &= \mathrm{diag}\left[ \mathrm{diag}\left( \frac{1}{n_{11}} J_{n_{11}}, \; \ldots, \; \frac{1}{n_{1b}} J_{n_{1b}} \right), \; \ldots, \; \mathrm{diag}\left( \frac{1}{n_{a1}} J_{n_{a1}}, \; \ldots, \; \frac{1}{n_{ab}} J_{n_{ab}} \right) \right] \\ &- \mathrm{diag}\left( \frac{1}{n_{1 \bullet}} J_{n_{1 \bullet}}, \; \ldots, \; \frac{1}{n_{a \bullet}} J_{n_{a \bullet}} \right) - M_B + \frac{1}{n} J_n \end{split}$

with the factor B matrix $M_B$ given by

$\label{eq:MB} M_B = \left[ \begin{matrix} \mathrm{diag}\left( \frac{1}{n_{\bullet 1}} J_{n_{11},n_{11}}, \; \ldots, \; \frac{1}{n_{\bullet b}} J_{n_{1b},n_{1b}} \right) & \cdots & \mathrm{diag}\left( \frac{1}{n_{\bullet 1}} J_{n_{11},n_{a1}}, \; \ldots, \; \frac{1}{n_{\bullet b}} J_{n_{1b},n_{ab}} \right) \\ \vdots & \ddots & \vdots \\ \mathrm{diag}\left( \frac{1}{n_{\bullet 1}} J_{n_{a1},n_{11}}, \; \ldots, \; \frac{1}{n_{\bullet b}} J_{n_{ab},n_{1b}} \right) & \cdots & \mathrm{diag}\left( \frac{1}{n_{\bullet 1}} J_{n_{a1},n_{a1}}, \; \ldots, \; \frac{1}{n_{\bullet b}} J_{n_{ab},n_{ab}} \right) \end{matrix} \right] \; .$

where $J_n$ is an $n \times n$ matrix of ones, $J_{n,m}$ is an $n \times m$ matrix of ones and $\mathrm{diag}\left( A_1, \ldots, A_n \right)$ denotes a block-diagonal matrix composed of $A_1, \ldots, A_n$. We observe that those matrices satisfy

$\label{eq:U-Q-B} \sum_{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{n_{ij}} U_{ijk}^2 = \sum_{l=1}^{5} Q_l = \sum_{l=1}^{5} U^\mathrm{T} B^{(l)} U$

as well as

$\label{eq:B-In} \sum_{l=1}^{5} B^{(l)} = I_n$

and their ranks are

$\label{eq:B-rk} \begin{split} \mathrm{rank}\left( B^{(1)} \right) &= n - a b \\ \mathrm{rank}\left( B^{(2)} \right) &= 1 \\ \mathrm{rank}\left( B^{(3)} \right) &= a - 1 \\ \mathrm{rank}\left( B^{(4)} \right) &= b - 1 \\ \mathrm{rank}\left( B^{(5)} \right) &= (a-1)(b-1) \; . \end{split}$

Thus, the conditions for applying Cochran’s theorem given by \eqref{eq:cochran-p1} are fulfilled and we can use \eqref{eq:sum-Uijk-s4}, \eqref{eq:cochran-p2}, \eqref{eq:B} and \eqref{eq:B-rk} to conclude that

$\label{eq:anova2-cochran-s1} \begin{split} \frac{1}{\sigma^2} \sum_{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{n_{ij}} (\bar{y}_{\bullet \bullet \bullet} - \mu)^2 &= Q_2 = U^\mathrm{T} B^{(2)} U \sim \chi^2(1) \\ \frac{1}{\sigma^2} \sum_{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{n_{ij}} ([\bar{y}_{i \bullet \bullet} - \bar{y}_{\bullet \bullet \bullet}] - \alpha_i)^2 &= Q_3 = U^\mathrm{T} B^{(3)} U \sim \chi^2(a-1) \\ \frac{1}{\sigma^2} \sum_{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{n_{ij}} ([\bar{y}_{\bullet j \bullet} - \bar{y}_{\bullet \bullet \bullet}] - \beta_j)^2 &= Q_4 = U^\mathrm{T} B^{(4)} U \sim \chi^2(b-1) \\ \frac{1}{\sigma^2} \sum_{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{n_{ij}} ([\bar{y}_{i j \bullet} - \bar{y}_{i \bullet \bullet} - \bar{y}_{\bullet j \bullet} + \bar{y}_{\bullet \bullet \bullet}] - \gamma_{ij})^2 &= Q_5 = U^\mathrm{T} B^{(5)} U \sim \chi^2\left( (a-1)(b-1) \right) \\ \frac{1}{\sigma^2} \sum_{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{n_{ij}} (y_{ijk} - \bar{y}_{i j \bullet})^2 &= Q_1 = U^\mathrm{T} B^{(1)} U \sim \chi^2(n-ab) \; . \end{split}$

Finally, we identify the terms $Q$ with sums of squares in two-way ANOVA and simplify them to reach the expressions given by \eqref{eq:anova2-cochran}:

$\label{eq:anova2-cochran-s2} \begin{split} \frac{\mathrm{SS}_M}{\sigma^2} &= \frac{1}{\sigma^2} \sum_{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{n_{ij}} (\bar{y}_{\bullet \bullet \bullet} - \mu)^2 = \frac{1}{\sigma^2} n (\bar{y}_{\bullet \bullet \bullet} - \mu)^2 \\ \frac{\mathrm{SS}_A}{\sigma^2} &= \frac{1}{\sigma^2} \sum_{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{n_{ij}} ([\bar{y}_{i \bullet \bullet} - \bar{y}_{\bullet \bullet \bullet}] - \alpha_i)^2 = \frac{1}{\sigma^2} \sum_{i=1}^{a} n_{i \bullet} ([\bar{y}_{i \bullet \bullet} - \bar{y}_{\bullet \bullet \bullet}] - \alpha_i)^2 \\ \frac{\mathrm{SS}_B}{\sigma^2} &= \frac{1}{\sigma^2} \sum_{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{n_{ij}} ([\bar{y}_{\bullet j \bullet} - \bar{y}_{\bullet \bullet \bullet}] - \beta_j)^2 = \frac{1}{\sigma^2} \sum_{j=1}^{b} n_{\bullet j} ([\bar{y}_{\bullet j \bullet} - \bar{y}_{\bullet \bullet \bullet}] - \beta_j)^2 \\ \frac{\mathrm{SS}_{A \times B}}{\sigma^2} &= \frac{1}{\sigma^2} \sum_{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{n_{ij}} ([\bar{y}_{i j \bullet} - \bar{y}_{i \bullet \bullet} - \bar{y}_{\bullet j \bullet} + \bar{y}_{\bullet \bullet \bullet}] - \gamma_{ij})^2 = \frac{1}{\sigma^2} \sum_{i=1}^{a} \sum_{j=1}^{b} n_{ij} ([\bar{y}_{i j \bullet} - \bar{y}_{i \bullet \bullet} - \bar{y}_{\bullet j \bullet} + \bar{y}_{\bullet \bullet \bullet}] - \gamma_{ij})^2 \\ \frac{\mathrm{SS}_\mathrm{res}}{\sigma^2} &= \frac{1}{\sigma^2} \sum_{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{n_{ij}} (y_{ijk} - \bar{y}_{i j \bullet})^2 = \frac{1}{\sigma^2} \sum_{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{n_{ij}} (y_{ijk} - \bar{y}_{i j \bullet})^2 \; . \end{split}$
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Metadata: ID: P378 | shortcut: anova2-cochran | author: JoramSoch | date: 2022-11-16, 15:15.