corrected 3 proofs

Author: JoramSoch

P/anova2-fgm.md

@@ -64,177 +64,32 @@ H_1: &\; \mu \neq 0 \; .
 $$
 
 
-**Proof:** Denote sample sizes as
+**Proof:** Applying [Cochran's theorem for two-analysis of variance](/P/anova2-cochran), we find that the following squared sums
 
-$$ \label{eq:samp-size}
+$$ \label{eq:anova2-ss-dist}
 \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}
+\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}_\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}
 $$
 
-and denote sample means as
+are [independent](/D/ind) and [chi-squared distributed](/D/chi2):
 
-$$ \label{eq:mean-samp}
+$$ \label{eq:anova2-cochran-s1}
 \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} \; .
+\frac{\mathrm{SS}_M}{\sigma^2} &\sim \chi^2(1) \\
+\frac{\mathrm{SS}_\mathrm{res}}{\sigma^2} &\sim \chi^2(n-ab) \; .
 \end{split}
 $$
 
-Assume that $\mu$ zero, according to $H_0$ given by \eqref{eq:anova2-h0}. Under this null hypothesis, we have:
-
-$$ \label{eq:yijk-h0}
-y_{ijk} \sim \mathcal{N}(\alpha_i + \beta_j + \gamma_{ij}, \sigma^2) \quad \text{for all} \quad i, j, k \; .
-$$
-
-Thus, the [random variable](/D/rvar) $U_{ijk} = (y_{ijk} - \alpha_i - \beta_j - \gamma_{ij})/\sigma$ [follows a standard normal distribution](/P/norm-snorm)
-
-$$ \label{eq:Uijk-h0}
-U_{ijk} = \frac{y_{ijk} - \alpha_i - \beta_j - \gamma_{ij}}{\sigma} \sim \mathcal{N}(0, 1) \; .
-$$
-
-Now consider the following sum
-
-$$ \label{eq:sum-Uijk-s1}
-\sum_{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{n_{ij}} U_{ijk}^2 = \sum_{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{n_{ij}} \left( \frac{y_{ijk} - \alpha_i - \beta_j - \gamma_{ij}}{\sigma} \right)^2 \\
-$$
-
-which can be rewritten as follows:
-
-$$ \label{eq:sum-Uijk-s2}
-\begin{split}
-\sum_{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{n_{ij}} U_{ijk}^2 = \frac{1}{\sigma^2} \sum_{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{n_{ij}} & \left[ (y_{ijk} - \alpha_i - \beta_j - \gamma_{ij}) - \right. \\
-&\left. [\bar{y}_{\bullet \bullet \bullet} + (\bar{y}_{i \bullet \bullet} - \bar{y}_{\bullet \bullet \bullet}) + (\bar{y}_{\bullet j \bullet} - \bar{y}_{\bullet \bullet \bullet}) + (\bar{y}_{i j \bullet} - \bar{y}_{i \bullet \bullet} - \bar{y}_{\bullet j \bullet} + \bar{y}_{\bullet \bullet \bullet})] \right. + \\
-&\left. [\bar{y}_{\bullet \bullet \bullet} + (\bar{y}_{i \bullet \bullet} - \bar{y}_{\bullet \bullet \bullet}) + (\bar{y}_{\bullet j \bullet} - \bar{y}_{\bullet \bullet \bullet}) + (\bar{y}_{i j \bullet} - \bar{y}_{i \bullet \bullet} - \bar{y}_{\bullet j \bullet} + \bar{y}_{\bullet \bullet \bullet})] \right]^2 \\
-= \frac{1}{\sigma^2}\sum_{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{n_{ij}} & \left[ (y_{ijk} - [\bar{y}_{\bullet \bullet \bullet} + (\bar{y}_{i \bullet \bullet} - \bar{y}_{\bullet \bullet \bullet}) + (\bar{y}_{\bullet j \bullet} - \bar{y}_{\bullet \bullet \bullet}) + (\bar{y}_{i j \bullet} - \bar{y}_{i \bullet \bullet} - \bar{y}_{\bullet j \bullet} + \bar{y}_{\bullet \bullet \bullet})]) + \right. \\
-&\left. (\bar{y}_{\bullet \bullet \bullet}) + ([\bar{y}_{i \bullet \bullet} - \bar{y}_{\bullet \bullet \bullet}] - \alpha_i) + ([\bar{y}_{\bullet j \bullet} - \bar{y}_{\bullet \bullet \bullet}] - \beta_j) \right. + \\
-&\left. ([\bar{y}_{i j \bullet} - \bar{y}_{i \bullet \bullet} - \bar{y}_{\bullet j \bullet} + \bar{y}_{\bullet \bullet \bullet}] - \gamma_{ij}) \right]^2 \\
-= \frac{1}{\sigma^2}\sum_{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{n_{ij}} & \left[ (y_{ijk} - \bar{y}_{i j \bullet}) + (\bar{y}_{\bullet \bullet \bullet}) + (\bar{y}_{i j \bullet} - \bar{y}_{\bullet \bullet \bullet} - \alpha_i - \beta_j - \gamma_{ij}) \right]^2
-\end{split}
-$$
-
-Because the following sum over $k$ is zero for all $(i,j)$
-
-$$ \label{eq:sum-yijk}
-\begin{split}
-\sum_{k=1}^{n_{ij}} (y_{ijk} - \bar{y}_{i j \bullet}) &= \sum_{k=1}^{n_{ij}} y_{ijk} - n_{ij} \bar{y}_{ij \bullet} \\
-&= \sum_{k=1}^{n_{ij}} y_{ijk}  - n_{ij} \cdot \frac{1}{n_{ij}} \sum_{k=1}^{n_{ij}} y_{ijk} \\
-&= 0, \; (i,j) \in \left\lbrace 1, \ldots, a \right\rbrace \times \left\lbrace 1, \ldots, b \right\rbrace \; ,
-\end{split}
-$$
-
-the following sum over $(i,j,k)$ and is also zero
-
-$$ \label{eq:sum-yib}
-\begin{split}
-\sum_{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{n_{ij}} (\bar{y}_{i j \bullet} - \bar{y}_{\bullet \bullet \bullet}) &= \sum_{i=1}^{a} \sum_{j=1}^{b} n_{ij} \bar{y}_{i j \bullet} - \bar{y}_{\bullet \bullet \bullet} \sum_{i=1}^{a} \sum_{j=1}^{b} n_{ij} \\
-&= \sum_{i=1}^{a} \sum_{j=1}^{b} n_{ij} \cdot \frac{1}{n_{ij}} \sum_{k=1}^{n_{ij}} y_{ijk} - n \cdot \frac{1}{n} \sum_{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{n_{ij}} y_{ijk} \\
-&= 0
-\end{split}
-$$
-
-and the term $\bar{y}_{\bullet \bullet \bullet}$ does not depend on $i$, $j$ and $k$
-
-$$ \label{eq:yb-const}
-\bar{y}_{\bullet \bullet \bullet} = \text{const.} \; ,
-$$
-
-non-square products in \eqref{eq:sum-Uijk-s2} disappear and the sum reduces to
-
-$$ \label{eq:sum-Uijk-s3}
-\begin{split}
-\sum_{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{n_{ij}} U_{ijk}^2 = \frac{1}{\sigma^2} & \sum_{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{n_{ij}} \left[ (y_{ijk} - \bar{y}_{i j \bullet})^2 + (\bar{y}_{\bullet \bullet \bullet})^2 + (\bar{y}_{i j \bullet} - \bar{y}_{\bullet \bullet \bullet} - \alpha_i - \beta_j - \gamma_{ij})^2 + \right] \\
-= \frac{1}{\sigma^2} & \left[ \sum_{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{n_{ij}} \right. (y_{ijk} - \bar{y}_{i j \bullet})^2 + \sum_{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{n_{ij}} (\bar{y}_{\bullet \bullet \bullet})^2 + \\
-& \sum_{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{n_{ij}} \left. (\bar{y}_{i j \bullet} - \bar{y}_{\bullet \bullet \bullet} - \alpha_i - \beta_j - \gamma_{ij})^2 \right]
-\end{split}
-$$
-
-[Cochran's theorem](/P/snorm-cochran) states that, if a sum of squared [standard normal](/D/snorm) [random variables](/D/rvar) 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](/D/ind) and each term $Q_j$ follows a [chi-squared distribution](/D/chi2) 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 $U$:
-
-$$ \label{eq:U}
-U = \left[ \begin{matrix} u_{1 \bullet} \\ \vdots \\ u_{a \bullet} \end{matrix} \right] \quad \text{where} \quad u_{i \bullet} = \left[ \begin{matrix} u_{i1} \\ \vdots \\ u_{ib} \end{matrix} \right] \quad \text{where} \quad u_{ij} = \left[ \begin{matrix} (y_{i,j,1} - \alpha_i - \beta_j - \gamma_{ij})/\sigma \\ \vdots \\ (y_{i,j,n_{ij}} - \mu - \alpha_i - \beta_j)/\sigma \end{matrix} \right] \; .
-$$
-
-Next, we specify the $n \times n$ matrices $B$
-
-$$ \label{eq:B}
-\begin{split}
-B^{(1)} &= I_n - \mathrm{diag}\left[ \mathrm{diag}\left( \frac{1}{n_{11}} J_{n_{11}}, \; \ldots, \; \frac{1}{n_{1b}} J_{n_{1b}} \right), \; \ldots, \; \mathrm{diag}\left( \frac{1}{n_{a1}} J_{n_{a1}}, \; \ldots, \; \frac{1}{n_{ab}} J_{n_{ab}} \right) \right] \\
-B^{(2)} &= \frac{1}{n} J_n \\
-B^{(3)} &= \mathrm{diag}\left[ \mathrm{diag}\left( \frac{1}{n_{11}} J_{n_{11}}, \; \ldots, \; \frac{1}{n_{1b}} J_{n_{1b}} \right), \; \ldots, \; \mathrm{diag}\left( \frac{1}{n_{a1}} J_{n_{a1}}, \; \ldots, \; \frac{1}{n_{ab}} J_{n_{ab}} \right) \right] - \frac{1}{n} J_n
-\end{split}
-$$
-
-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
-
-$$ \label{eq:U-Q-B}
-\sum_{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{n_{ij}} U_{ijk]^2 = Q_1 + Q_2 + Q_3 = U^\mathrm{T} B^{(1)} U + U^\mathrm{T} B^{(2)} U + U^\mathrm{T} B^{(3)} U
-$$
-
-as well as
-
-$$ \label{eq:B-In}
-B^{(1)} + B^{(2)} + B^{(3)} = I_n
-$$
-
-and their ranks are
-
-$$ \label{eq:B-rk}
-\begin{split}
-\mathrm{rank}\left( B^{(1)} \right) &= n - a \cdot b \\
-\mathrm{rank}\left( B^{(2)} \right) &= 1 \\
-\mathrm{rank}\left( B^{(3)} \right) &= n - (n-ab) - 1 = a \cdot b - 1 \; .
-\end{split}
-$$
-
-Let's write down the [explained sum of squares](/D/ess) and the [residual sum of squares](/D/rss) for [two-way analysis of variance](/D/anova2) as
-
-$$ \label{eq:ess-rss}
-\begin{split}
-\mathrm{ESS} &= \sum_{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{n_{ij}} (\bar{y}_{\bullet \bullet \bullet})^2 \\
-\mathrm{RSS} &= \sum_{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{n_{ij}} (y_{ijk} - \bar{y}_{i j \bullet})^2 \; .
-\end{split}
-$$
-
-Then, using \eqref{eq:sum-Uijk-s3}, \eqref{eq:cochran-p1}, \eqref{eq:cochran-p2}, \eqref{eq:B} and \eqref{eq:B-rk}, we find that
-
-$$ \label{eq:ess-rss-dist}
-\begin{split}
-\frac{\mathrm{ESS}}{\sigma^2} = \sum_{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{n_{ij}} \left( \frac{\bar{y}_{\bullet \bullet \bullet}}{\sigma} \right)^2 &= Q_2 = U^\mathrm{T} B^{(2)} U \sim \chi^2(1) \\
-\frac{\mathrm{RSS}}{\sigma^2} = \sum_{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{n_{ij}} \left( \frac{y_{ijk} - \bar{y}_{i j \bullet}}{\sigma} \right)^2 &= Q_1 = U^\mathrm{T} B^{(1)} U \sim \chi^2(n-ab) \; .
-\end{split}
-$$
-
-Because $\mathrm{ESS}/\sigma^2$ and $\mathrm{RSS}/\sigma^2$ are also independent by \eqref{eq:cochran-p2}, the F-statistic from \eqref{eq:anova2-fgm} is equal to the ratio of two independent [chi-squared distributed](/D/chi2) [random variables](/D/rvar) divided by their degrees of freedom
+Thus, the F-statistic from \eqref{eq:anova2-fgm} is equal to the ratio of two [independent](/D/ind) [chi-squared distributed](/D/chi2) [random variables](/D/rvar) divided by their degrees of freedom
 
 $$ \label{eq:anova2-fgm-ess-tss}
 \begin{split}
-F_M &= \frac{(\mathrm{ESS}/\sigma^2)/(1)}{(\mathrm{RSS}/\sigma^2)/(n-ab)} \\
-&= \frac{\mathrm{ESS}/(1)}{\mathrm{RSS}/(n-ab)} \\
-&= \frac{\sum_{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{n_{ij}} (\bar{y}_{\bullet \bullet \bullet})^2}{\frac{1}{n-ab} \sum_{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{n_{ij}} (y_{ijk} - \bar{y}_{i j \bullet})^2} \\
-&= \frac{(\bar{y}_{\bullet \bullet \bullet})^2 \sum_{i=1}^{a} \sum_{j=1}^{b} n_{ij}}{\frac{1}{n-ab} \sum_{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{n_{ij}} (y_{ijk} - \bar{y}_{i j \bullet})^2} \\
-&= \frac{n (\bar{y}_{\bullet \bullet \bullet})^2}{\frac{1}{n-ab} \sum_{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{n_{ij}} (y_{ijk} - \bar{y}_{i j \bullet})^2}
+F_M &= \frac{(\mathrm{SS}_M/\sigma^2)/(1)}{(\mathrm{SS}_\mathrm{res}/\sigma^2)/(n-ab)} \\
+&= \frac{\mathrm{SS}_M/(1)}{\mathrm{SS}_\mathrm{res}/(n-ab)} \\
+&\overset{\eqref{eq:anova2-ss-dist}}{=} \frac{n (\bar{y}_{\bullet \bullet \bullet} - \mu)^2}{\frac{1}{n-ab} \sum_{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{n_{ij}} (y_{ijk} - \bar{y}_{i j \bullet})^2} \\
+&\overset{\eqref{eq:anova2-fgm}}{=} \frac{n (\bar{y}_{\bullet \bullet \bullet})^2}{\frac{1}{n-ab} \sum_{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{n_{ij}} (y_{ijk} - \bar{y}_{i j \bullet})^2}
 \end{split}
 $$