Theorem and definition
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
$$
{\Vert P_Z \mu \Vert}^2 = J \sum_{i=1}^I {(\mu_i - \bar\mu_\bullet)}^2 =
J \sum_{i=1}^I \alpha_i^2.
$$
```{r, echo=FALSE}
knitr::knit_exit()
```
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effect size