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. ", style.title=list(color="blue", "font-size"="30px"))` 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 should 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 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 $${(x \otimes y)}_{ij}=x_iy_j.$$ Indeed, one has $$\mu = (\mu_1, \mu_2) \otimes (1,1,1),$$ and then the linear space $W$ in which $\mu$ is assumed to lie is $$W = \mathbb{R}^I\otimes{\bf 1}_J.$$ 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.