--- author: Stéphane Laurent date: '2017-06-06' highlighter: kate output: html_document: keep_md: False md_document: toc: True variant: markdown prettify: True prettifycss: 'twitter-bootstrap' tags: 'maths, statistics' title: 'The geometry of the balanced ANOVA model (with fixed effects)' --- - [Standard normal distribution on a vector space](#standard-normal-distribution-on-a-vector-space) - [The balanced ANOVA model](#the-balanced-anova-model) - [Tensor product](#tensor-product) - [Least-squares estimates](#least-squares-estimates) Most usually, the mathematical treatment of Gaussian linear models starts with the matricial writing $Y=X\beta+\sigma G$, where $Y$ is a random vector modelling the $n$ response values, $X$ is a known matrix, $\beta$ is the vector of unknown parameters, and $G$ has the standard normal distribution on $\mathbb{R}^n$. There are good reasons to use this matricial writing. However it is cleaner to treat the theory with the equivalent vector space notation $Y = \mu + \sigma G$, where $\mu$ is assumed to lie in a linear subspace $W$ of $\mathbb{R}^n$, corresponding to $\text{Im}(X)$ in the matricial notation. For example, denoting by $P_W$ the orthogonal projection on $W$, the least-squares estimate $\hat\mu$ of $\mu$ is simply given by $\hat\mu=P_Wy$, and $P_W^\perp y$ is the vector of residuals, denoting by $P^\perp_W$ the projection on the orthogonal complement of $W$. Thus there is no need to consider $W=\text{Im}(X)$ to derive the general principles of the theory. The balanced one-way ANOVA model, which is the topic of this article, illustrates this approach. Standard normal distribution on a vector space ---------------------------------------------- The main tool used to treat the theory of Gaussian linear models is the standard normal distribution on a linear space.
[Theorem and definition]{#boxtitle style="color:blue;font-size:30px;"}

Let $X$ be a $\mathbb{R}^n$-valued random vector, and $W \subset \mathbb{R}^n$ be a linear space. Say that $X$ has the standard normal distribution on the vector space $W$, and then note $X \sim SN(W)$, if it takes its values in $W$ and its characteristic function is given by $$\mathbb{E} \textrm{e}^{i\langle w, X \rangle} = \textrm{e}^{-\frac12{\Vert w \Vert}^2} \quad \text{for all } w \in W.$$ The three following assertions are equivalent (and this is easy to prove):
1. $X \sim SN(W)$;
2. the coordinates of $X$ in some orthonormal basis of $W$ are i.i.d. standard normal random variables;
3. the coordinates of $X$ in any orthonormal basis of $W$ are i.i.d. standard normal random variables.

Of course we retrieve the standard normal distribution on $\mathbb{R}^n$ when taking $W=\mathbb{R}^n$. From this definition-theorem, the so-called *Cochran's theorem* is an obvious statement. More precisely, if $U \subset W$ is a linear space, and $Z=U^\perp \cap W$ is the orthogonal complement of $U$ in $W$, then the projection $P_UX$ of $X$ on $U$ has the standard normal distribution on $U$, similarly the projection $P_ZX$ of $X$ on $Z$ has the standard normal distribution on $Z$, and moreover $P_UX$ and $P_ZX$ are independent. This is straightforward to see from the definition-theorem of $SN(W)$, and it is also easy to see that ${\Vert P_UX\Vert}^2 \sim \chi^2_{\dim(U)}$. The balanced ANOVA model ------------------------ The balanced ANOVA model is used to model a sample $y=(y_{ij})$ with a tabular structure: $$y=\begin{pmatrix} y_{11} & \ldots & y_{1J} \\ \vdots & y_{ij} & \vdots \\ y_{I1} & \ldots & y_{IJ} \end{pmatrix},$$ $y_{ij}$ denoting the $j$-th measurement in group $i$. It is assumed that the $y_{ij}$ are independent and the population mean depends on the group index $i$. More precisely, the $y_{ij}$ are modelled by random variables $Y_{ij} \sim_{\text{iid}} {\cal N}(\mu_i, \sigma^2)$. So, how to write this model as $Y=\mu + \sigma G$ where $G \sim SN(\mathbb{R}^n)$ and $\mu$ lies in a linear space $W \subset \mathbb{R}^n$ ? Tensor product -------------- Here $n=IJ$ and one could consider $Y$ as the vector obtained by stacking the $Y_{ij}$. For example if $I=2$ and $J=3$, we should write $$Y={(Y_{11}, Y_{12}, Y_{13}, Y_{21}, Y_{22}, Y_{23})}'.$$ Actually this is not a good idea to loose the tabular structure. The appropriate approach for writing the balanced ANOVA model involves the *tensor product*. We keep the tabular structure of the data: $$Y = \begin{pmatrix} Y_{11} & Y_{12} & Y_{13} \\ Y_{21} & Y_{22} & Y_{23} \end{pmatrix}$$ and we take $$G \sim SN(\mathbb{R}^I\otimes\mathbb{R}^J)$$ where the *tensor poduct* $\mathbb{R}^I\otimes\mathbb{R}^J$ of $\mathbb{R}^I$ and $\mathbb{R}^J$ is nothing but the space of matrices with $I$ rows and $J$ columns. Here $$\mu = \begin{pmatrix} \mu_1 & \mu_1 & \mu_1 \\ \mu_2 & \mu_2 & \mu_2 \end{pmatrix},$$ lies, as we will see, in a linear space $W \subset \mathbb{R}^I\otimes\mathbb{R}^J$ which is convenient to define with the help of the *tensor product* $x \otimes y$ of two vectors $x \in \mathbb{R}^I$ and $y \in \mathbb{R}^J$, defined as the element of $\mathbb{R}^I\otimes\mathbb{R}^J$ given by $${(x \otimes y)}_{ij}=x_iy_j.$$ Not all vectors of $\mathbb{R}^I\otimes\mathbb{R}^J$ can be written $x \otimes y$, but the vectors $x \otimes y$ span $\mathbb{R}^I\otimes\mathbb{R}^J$. In consistence with this notation, the tensor product $U \otimes V$ of two vector spaces $U \subset \mathbb{R}^I$ and $V \subset \mathbb{R}^J$ is defined as the vector space spanned by the vectors of the form $x \otimes y$, $x \in U$, $y \in V$. Then\ $$\mu = (\mu_1, \mu_2) \otimes (1,1,1),$$ and $\mu$ is assumed to lie in the linear space $$W = \mathbb{R}^I \otimes [{\bf 1}_J].$$ Moreover, there is a nice orthogonal decomposition of $W$ corresponding to the usual other parameterization of the model: $$\boxed{\mu_i = m + \alpha_i} \quad \text{with } \sum_{i=1}^I\alpha_i=0.$$ Indeed, writing $\mathbb{R}^I=[{\bf 1}_I] \oplus {[{\bf 1}_I]}^\perp$ yields the following decomposition of $\mu$: \begin{align*} \mu = (\mu_1, \ldots, \mu_I) \otimes {\bf 1}_J & = \begin{pmatrix} m & m & m \\ m & m & m \end{pmatrix} + \begin{pmatrix} \alpha_1 & \alpha_1 & \alpha_1 \\ \alpha_2 & \alpha_2 & \alpha_2 \end{pmatrix} \\ & = \underset{\in \bigl([{\bf 1}_I]\otimes[{\bf 1}_J]\bigr)}{\underbrace{m({\bf 1}_I\otimes{\bf 1}_J)}} + \underset{\in \bigl([{\bf 1}_I]^{\perp}\otimes[{\bf 1}_J] \bigr)}{\underbrace{(\alpha_1,\ldots,\alpha_I)\otimes{\bf 1}_J}} \end{align*} Least-squares estimates ----------------------- With the theory introduced above, the least-squares estimates of $m$ and the $\alpha_i$ are given by $\hat m({\bf 1}_I\otimes{\bf 1}_J) = P_U y$ and $\hat\alpha\otimes{\bf 1}_J = P_Zy$ where $U = [{\bf 1}_I]\otimes[{\bf 1}_J]$ and $Z = {[{\bf 1}_I]}^{\perp}\otimes[{\bf 1}_J] = U^\perp \cap W$, and we also know that $\hat m$ and the $\hat\alpha_i$ are independent. The least-squares estimates of the $\mu_i$ are given by $\hat\mu_i=\hat m +\hat\alpha_i$. Deriving the expression of these estimates and their distribution is left as an exercise to the reader. As another exercise, check that $${\Vert P_Z \mu \Vert}^2 = J \sum_{i=1}^I {(\mu_i - \bar\mu_\bullet)}^2 = J \sum_{i=1}^I \alpha_i^2.$$