corrected some pages

Several small mistakes/errors were corrected in several proofs/definitions.

Author: JoramSoch

D/iass.md

@@ -42,4 +42,4 @@ $$ \label{eq:iass}
 \end{split}
 $$
 
-Here, $\bar{y}_{i j \bullet}$ is the mean for the $(i,j)$-th cell (out of $a \times b$ cells), computed from $n_{ij}$ values $y_{ijk}$, $\bar{y}_{i \bullet \bullet}$ and $\bar{y}_{\bullet j \bullet}$ are the level means for the two factors and and $\bar{y}_{\bullet \bullet \bullet}$ is the mean across all values $y_{ijk}$.
+Here, $\bar{y} _{i j \bullet}$ is the mean for the $(i,j)$-th cell (out of $a \times b$ cells), computed from $n_{ij}$ values $y_{ijk}$, $\bar{y} _{i \bullet \bullet}$ and $\bar{y} _{\bullet j \bullet}$ are the level means for the two factors and and $\bar{y} _{\bullet \bullet \bullet}$ is the mean across all values $y_{ijk}$.

D/trss.md

@@ -39,4 +39,4 @@ $$ \label{eq:trss}
 \mathrm{SS}_\mathrm{treat} = \sum_{i=1}^{k} \sum_{j=1}^{n_i} (\bar{y}_i - \bar{y})^2 \; .
 $$
 
-Here, $\bar{y} _i$ is the mean for the $i$-th level of the factor (out of $k$ levels), computed from $n_i$ values $y_{ij}$, and $\bar{y}$ is the mean across all values $y_{ij}$.
+Here, $\bar{y}_i$ is the mean for the $i$-th level of the factor (out of $k$ levels), computed from $n_i$ values $y_{ij}$, and $\bar{y}$ is the mean across all values $y_{ij}$.

P/anova2-pss.md

@@ -54,7 +54,7 @@ $$ \label{eq:anova2-tss}
 \mathrm{SS}_\mathrm{tot} = \sum_{i=1}^{a} \sum_{j=1}^{b} \sum_{k=1}^{n_{ij}} (y_{ijk} - \bar{y}_{\bullet \bullet \bullet})^2
 $$
 
-where $\bar{y}_{\bullet \bullet \bullet}$ is the mean across all values $y_{ijk}$. This can be rewritten as
+where $\bar{y} _{\bullet \bullet \bullet}$ is the mean across all values $y_{ijk}$. This can be rewritten as
 
 $$ \label{eq:anova2-pss-s1}
 \begin{split}

P/mlr-f.md

@@ -171,7 +171,7 @@ F &\overset{\eqref{eq:mlr-f-s1}}{=} \left( \hat{\gamma} -  C^\mathrm{T} \hat{\be
 &\overset{\eqref{eq:mlr-f-h0}}{=} \hat{\gamma}^\mathrm{T} \left( \frac{n-p}{\hat{\varepsilon}^\mathrm{T} \hat{\varepsilon}} Q \right) \hat{\gamma} / q \\
 &\overset{\eqref{eq:fcon}}{=} \hat{\beta}^\mathrm{T} C \left( \frac{n-p}{\hat{\varepsilon}^\mathrm{T} \hat{\varepsilon}} Q \right) C^\mathrm{T} \hat{\beta} / q \\
 &\overset{\eqref{eq:g-est-tau-dist-cond}}{=} \hat{\beta}^\mathrm{T} C \left( \frac{n-p}{\hat{\varepsilon}^\mathrm{T} \hat{\varepsilon}} \left( C^\mathrm{T} (X^\mathrm{T} V^{-1} X)^{-1} C \right)^{-1} \right) C^\mathrm{T} \hat{\beta} / q \\
-&\overset{\eqref{eq:g-est-tau-dist-cond}}{=} \hat{\beta}^\mathrm{T} C \left( \frac{n-p}{(y-X\hat{\beta})^\mathrm{T} V^{-1} (y-X\hat{\beta})} \left( C^\mathrm{T} (X^\mathrm{T} V^{-1} X)^{-1} C \right)^{-1} \right) C^\mathrm{T} \hat{\beta} / q \\
+&\overset{\eqref{eq:mlr-rss}}{=} \hat{\beta}^\mathrm{T} C \left( \frac{n-p}{(y-X\hat{\beta})^\mathrm{T} V^{-1} (y-X\hat{\beta})} \left( C^\mathrm{T} (X^\mathrm{T} V^{-1} X)^{-1} C \right)^{-1} \right) C^\mathrm{T} \hat{\beta} / q \\
 &\overset{\eqref{eq:mlr-est}}{=} \hat{\beta}^\mathrm{T} C \left( \frac{1}{\hat{\sigma}^2} \left( C^\mathrm{T} (X^\mathrm{T} V^{-1} X)^{-1} C \right)^{-1} \right) C^\mathrm{T} \hat{\beta} / q \\
 &= \hat{\beta}^\mathrm{T} C \left( \hat{\sigma}^2 C^\mathrm{T} (X^\mathrm{T} V^{-1} X)^{-1} C \right)^{-1} C^\mathrm{T} \hat{\beta} / q \; .
 \end{split}

P/mlr-rssdist.md

@@ -46,7 +46,7 @@ $$ \label{eq:mlr}
 y = X\beta + \varepsilon, \; \varepsilon \sim \mathcal{N}(0, \sigma^2 V)
 $$
 
-and consider estimation using [weighted least squares](/P/mlr-wls). Then, the [residual sum of squares](/D/rss) $\hat{\varepsilon}^\mathrm{T} \hat{\varepsilon}$, divided by the true [error variance](/D/mlr) $\sigma^2$, follows a [chi-squared distribution] with $n-p$ [degrees of freedom](/D/dof)
+and consider estimation using [weighted least squares](/P/mlr-wls). Then, the [residual sum of squares](/D/rss) $\hat{\varepsilon}^\mathrm{T} \hat{\varepsilon}$, divided by the true [error variance](/D/mlr) $\sigma^2$, follows a [chi-squared distribution](/D/chi2) with $n-p$ [degrees of freedom](/D/dof)
 
 $$ \label{eq:mlr-rss-dist}
 \frac{\hat{\varepsilon}^\mathrm{T} \hat{\varepsilon}}{\sigma^2} \sim \chi^2(n-p)
@@ -79,7 +79,7 @@ $$ \label{eq:y-tilde-dist}
 \tilde{y} \sim \mathcal{N}(\tilde{X} \beta, \sigma^2 I_n) \; .
 $$
 
-With that, we have obtained a [linear regression model](/D/mlr) with independent observations. [Cochran's theorem for multivariate normal variables](/P/mvn-cochran) states that, for an $n \times 1$ [normal random vector] whose [covariance matrix](/D/covmat) is a scalar multiple of the identity matrix, a specific squared form will follow a [non-central chi-squared distribution](/D/ncchi2) where the degrees of freedom and the non-centrality paramter depend on the matrix in the quadratic form:
+With that, we have obtained a [linear regression model](/D/mlr) with independent observations. [Cochran's theorem for multivariate normal variables](/P/mvn-cochran) states that, for an $n \times 1$ [normal random vector](/D/mvn) whose [covariance matrix](/D/covmat) is a scalar multiple of the identity matrix, a specific squared form will follow a [non-central chi-squared distribution](/D/ncchi2) where the degrees of freedom and the non-centrality paramter depend on the matrix in the quadratic form:
 
 $$ \label{eq:mvn-cochran}
 x \sim \mathcal{N}(\mu, \sigma^2 I_n) \quad \Rightarrow \quad y = x^\mathrm{T} A x /\sigma^2 \sim \chi^2\left( \mathrm{tr}(A), \mu^\mathrm{T} A \mu \right) \; .

P/mvn-indprod.md

@@ -48,8 +48,8 @@ $$
 
 Then, due to the [linear transformation theorem](/P/mvn-ltt), we have
 
-$$ \label{eq:CZ}
-CZ = \left[ \begin{array}{c} AZ \\ BZ \end{array} \right] \sim \mathcal{N}\left( \left[ \begin{array}{c} A\mu \\ B\mu \end{array} \right], C \Sigma C^\mathrm{T} \right)
+$$ \label{eq:CX}
+CZ = \left[ \begin{array}{c} AX \\ BX \end{array} \right] \sim \mathcal{N}\left( \left[ \begin{array}{c} A\mu \\ B\mu \end{array} \right], C \Sigma C^\mathrm{T} \right)
 $$
 
 with the combined [covariance matrix](/D/covmat)
@@ -58,7 +58,7 @@ $$ \label{eq:CSC}
 C \Sigma C^\mathrm{T} = \left[ \begin{array}{cc} A \Sigma A^\mathrm{T} & A \Sigma B^\mathrm{T} \\ B \Sigma A^\mathrm{T} & B \Sigma B^\mathrm{T} \end{array} \right] \; .
 $$
 
-We know that [the necessary and sufficient condition for two components of a multivariate normal random vector to be independent is that their entries in the covariance matrix are zero](/P/mvn-ind). Thus, $AZ$ and $BZ$ are [independent](/D/ind), if and only if
+We know that [the necessary and sufficient condition for two components of a multivariate normal random vector to be independent is that their entries in the covariance matrix are zero](/P/mvn-ind). Thus, $AX$ and $BX$ are [independent](/D/ind), if and only if
 
 $$ \label{eq:mvn-indprod-qed}
 A \Sigma B^\mathrm{T} = (B \Sigma A^\mathrm{T})^\mathrm{T} = 0_{kl}