# Proof on $SS_{AB} / \sigma^2 \sim \chi^2_{(I-1)(J-1)}$ under the null hypothesis $H_{AB}: \delta_{ij} = 0$ for $i=1,\dotsc,I$ and $j=1,\dotsc,J$

First time to ask a question. I am now reading a textbook "Mathematical Statistics and Data Analysis, 3ed" by Rice. On Page 495, there is a Theorem B for the two-way layout ANOVA.

Let me first give some background. There are $$I$$ levels for one factor, $$J$$ levels for another factor. And $$K$$ independent observations for each of $$I \times J$$ cell. The statistical model for an observation in a cell is $$Y_{ijk} = \mu + \alpha_i + \beta_j + \delta_{ij} + \epsilon_{ijk},$$ with the constraints, $$\sum_{i} \alpha_i = \sum_{j} \beta_j = \sum_{i} \delta_{ij} = \sum_{j} \delta_{ij} = 0.$$ $$\mu$$ is the grand average, $$\alpha_i$$ is $$i^{th}$$ differential effect of one factor, $$\beta_j$$ is $$j^{th}$$ differential effect of another factor, $$\delta_{ij}$$ is the interaction between $$i^{th}$$ and $$j^{th}$$ factor. The error term is normally distributed with mean zero and same variances $$\sigma^2$$.

We know that $$SS_{TOT} = SS_A + SS_B + SS_{AB} + SS_E,$$ where \begin{align*} &SS_A = JK \sum_{i=1}^{I} (\bar{Y}_{i..} - \bar{Y}_{...})^2 \\ &SS_B = IK \sum_{j=1}^{J} (\bar{Y}_{.j.} - \bar{Y}_{...})^2 \\ &SS_{AB} = K \sum_{i=1}^{I} \sum_{j=1}^{J} (\bar{Y}_{ij.} - \bar{Y}_{i..} - \bar{Y}_{.j.} + \bar{Y}_{...})^2 \\ &SS_E = \sum_{i=1}^{I} \sum_{j=1}^{J} \sum_{k=1}^{K} (Y_{ijk} - \bar{Y}_{ij.})^2 \\ &SS_{TOT} = \sum_{i=1}^{I} \sum_{j=1}^{J} \sum_{k=1}^{K} (Y_{ijk} - \bar{Y}_{...})^2 \end{align*} Also, we know that \begin{align*} &E[SS_A] = (I-1)\sigma^2 + JK \sum_{i=1}^{I} \alpha_i^2 \\ &E[SS_B] = (J-1)\sigma^2 + IK \sum_{j=1}^{J} \beta_j^2 \\ &E[SS_{AB}] = (I-1)(J-1) \sigma^2 + K \sum_{i=1}^{I} \sum_{j=1}^{J} \delta_{ij}^2 \\ &E[SS_E] = IJ(K-1) \sigma^2 \end{align*} The following image is the aforesaid Theorem B on which I have questions,

Proofs on a,b and c are straightforward and my question is on the proofs on d and e,

1. How to prove d? (I try to do some algebra and want to relate $$\bar{Y}_{ij.}$$ to $$\bar{Y}_{i..}$$ and $$\bar{Y}_{.j.}$$, but stuck).
2. For e, since $$SS_{A}$$ and $$SS_{B}$$ could be both considered as functions of $$\bar{Y}_{ij.}$$, then it is easy to prove that $$SS_{A}$$ and $$SS_{B}$$ are both independent of $$SS_{E}$$, which allows us to implement F test. But $$SS_{A}$$ and $$SS_{B}$$ seems to be dependent, which contradicts with e. (or the author's meaning is just the dependence between $$SS_{E}$$ and others, but not the dependences among others.)

If you substitute the expressions for $$\overline Y_{ij.},\overline Y_{i..},\overline Y_{.j.}$$ and $$\overline Y_{...}$$ in $$SS_{AB}$$, then you get

$$SS_{AB}=K\sum_{i,j}(\delta_{ij}+\overline\varepsilon_{ij.}-\overline\varepsilon_{i..}-\overline\varepsilon_{.j.}+\overline\varepsilon_{...})^2$$

Taking $$u_{ij}=\delta_{ij}+\overline\varepsilon_{ij.}$$, this can be rewritten as

\begin{align} SS_{AB}&=K\sum _{i,j}(u_{ij}-\overline u_{i.}-\overline u_{.j}+\overline u_{..})^2 \\&=K\sum _{i,j}\left\{(u_{ij}-\overline u_{i.})-(\overline u_{.j}-\overline u_{..})\right\}^2 \\&=K\sum_{i,j}(u_{ij}-\overline u_{i.})^2-IK\sum_j (\overline u_{.j}-\overline u_{..})^2 \tag{1} \end{align}

Now, the $$u_{ij}$$'s are independent normal variables, i.e. for every $$i,j$$,

$$u_{ij}\stackrel{\text{ind.}}\sim N\left(\delta_{ij},\frac{\sigma^2}{K}\right)$$

So under $$H_{AB}:\delta_{ij}=0$$,

$$\sum_j \frac{\left(u_{ij}-\overline u_{i.}\right)^2}{\sigma^2/K}\stackrel{\text{ind.}}\sim \chi^2_{J-1} \quad,\forall\, i$$

This implies

$$\sum_{i,j}\frac{\left(u_{ij}-\overline u_{i.}\right)^2}{\sigma^2/K}\sim \chi^2_{I(J-1)} \tag{2}$$

Similarly,

$$\overline u_{.j}\stackrel{\text{ind.}}\sim N\left(0,\frac{\sigma^2}{IK}\right)\,,$$

so that

$$\sum_j\frac{\left(\overline u_{.j}-\overline u_{..}\right)^2}{\sigma^2/IK}\sim \chi^2_{J-1} \tag{3}$$

If you could argue that $$\sum_j \left(\overline u_{.j}-\overline u_{..}\right)^2$$ and $$\sum_{i,j}(u_{ij}-\overline u_{i.}-\overline u_{.j}+\overline u_{..})^2$$ are independent, then $$(1),(2)$$ and $$(3)$$ would give you the desired distribution of $$SS_{AB}/\sigma^2$$ under $$H_{AB}$$. Or you could use general theory from distribution of quadratic forms to establish this. It also might be of interest to refer to the Fisher-Cochran theorem on quadratic forms.

I am not sure of part (e) of the theorem in your post. But it is certainly true that $$SS_E$$ is independent of each of $$SS_A,SS_B$$ and $$SS_{AB}$$.

• Thanks and sorry for the late reply due to being busy with other stuffs. I follow the approach (but keep $\epsilon$ notation) and also get $$\frac{1}{\sigma^2/K} \sum_{i,j} (\bar{\epsilon}_{ij.} - \bar{\epsilon}_{i..})^2 \sim \chi^2_{I(J-1)}$$ and $$\frac{1}{\sigma^2/IK} \sum_{j} (\bar{\epsilon}_{.j.} - \bar{\epsilon}_{...})^2 \sim \chi^2_{J-1}.$$ Then, I try to prove $\sum_{i,j} (\bar{\epsilon}_{ij.} - \bar{\epsilon}_{i..})^2$ and $\sum_{j} (\bar{\epsilon}_{.j.} - \bar{\epsilon}_{...})^2$ are independent. Commented Oct 12, 2021 at 15:30
• My idea is as follows: The first term is function of the vector $U = (\bar{\epsilon}_{11.} - \bar{\epsilon}_{1..}, \bar{\epsilon}_{11.} - \bar{\epsilon}_{1..}, \dotsc)$, the second term is function of the vector $V=(\bar{\epsilon}_{.1.} - \bar{\epsilon}_{...}, \bar{\epsilon}_{.1.} - \bar{\epsilon}_{...}, \dotsc)$. Since $X = (U, V)$ is multivariate normal, thus if we can prove every element of $U$ is uncorrelated with every element of $V$, then $U$ and $V$ are independent, which completes the proof. Commented Oct 12, 2021 at 15:41
• Re-label element in $U$ and $V$ as $\bar{\epsilon}_{ij_{1}.} - \bar{\epsilon}_{i..}$ and $\bar{\epsilon}_{.j_{2}.} - \bar{\epsilon}_{...}$. For $j_1 \neq j_2$, they are independent. But for $j_1 = j_2 = j$, I try to compute $$\text{Cov}(\bar{\epsilon}_{ij.} - \bar{\epsilon}_{i..}, \,\bar{\epsilon}_{.j.} - \bar{\epsilon}_{...} )$$ but get nonzero result. I don't know where i am wrong. Commented Oct 12, 2021 at 15:46