FCA-$3$modes and FCA-$k$modes

As for PTA-$k$modes one will present only the case $k=3$, the framework for $k > 3$ being the same. With similar notations for a three-way table $I \times J \times K$ , one performs the PTA- 3modes of the quadruple:

$$((D_I^{-1}\otimes D_J^{-1}\otimes D_K^{-1})P, \quad D_I, \quad D_J, \quad D_K)$$ (26)

This has similar properties as for FCA-2modes moreover if one notes:

$$\Pi_{ijk}=\Pi_{.jk}+\Pi_{i.k}+\Pi_{ij.}+\Delta_{ijk}$$

for

$$(\frac{p_{ijk}-p_{i..}p_{.j.}p_{..k}}{ p_{i..}p_{.j.}p_{..k... ..._{.j.}})+(\frac{p_{ijk}-\delta_{ijk}}{ p_{i..}p_{.j.}p_{..k}}),$$

where $\delta_{ijk}=p_{ij.}p_{..k}+p_{i.k}p_{.j.}+p_{.jk}p_{i..}-2p_{i..}p_{.j.}p_{..k}$, one has the following property:

$$\left\Vert \Pi_{ijk} \right\Vert ^2=\left\Vert \Pi_{.jk} \... ...ij.} \right\Vert ^2 +\left\Vert \Delta_{ijk} \right\Vert ^2,$$ (27)

where $\left\Vert \right\Vert$ is the norm on the tensor space, i.e. using the metric $D_I \otimes D_J \otimes D_K$. This result dating from Lancaster(1951, 1980) was reported recently in [1] where a particular generalisation of correspondence analysis based on [9]'s book was derived. Equation (27) means that deviation from three-way independence can be orthogonally decomposed into deviations from independence for the two-way margins of the three-way table, and a three-way interaction term. Each two-way margins deviation from independence is reminiscent of (simple) correspondence analysis. To be convinced of this point just rewrite equation (27) as below wherein terms as in equation (22) can be identified:

$$\frac{\chi^2}{N} = \sum_{ijk}p_{i..}p_{.j.}p_{..k} (\frac{p_{ijk}-p_{i..}p_{.j.}p_{..k}}{ p_{i..}p_{.j.}p_{..k}})^2$$ (28)

$$= \sum_{jk}p_{.j.}p_{..k}(\frac{p_{.jk}-p_{.j.}p_{..k}}{ p_{.j.}p_{... ...}})^2+\sum_{ij}p_{i..}p_{.j.}(\frac{p_{ij.}-p_{i..}p_{.j.}}{ p_{i..}p_{.j.}})^2$$

$$+ \sum_{ijk}p_{i..}p_{.j.}p_{..k}(\frac{p_{ijk}-\delta_{ijk}}{ p_{i..}p_{.j.}p_{..k}})^2.$$

When performing the PTA-$3$modes (26) one retrieves simply and naturally these lack of marginal independence. The inertia or sum of squares is :

$$\sum_{ijk}p_{i..}p_{.j.}p_{..k} (\frac{p_{ijk}}{ p_{i..}p_{.... ...s=0}^{r}\sigma_s=1+\sum_{s=1}^{r}\sigma_s =1+\frac{\chi^2}{N};$$

the first ($s=0$) principal tensor being ${\it 1}\!\!I_I \otimes{\it 1}\!\!I_J \otimes{\it 1}\!\!I_K$ with $\sigma_0=1$, its associated principal tensors relate to two-way margins decompositions, i.e. each term of the second row of equation (28). One can write a reconstruction formula similar to expressions (23) or (24):

$$\widehat{P} = (D_I\otimes D_J\otimes D_K) (1 {\it 1}\!\!I_I \otimes{\it 1}\!\!I... ...times{\it 1}\!\!I_K + \sum_{s=1}^r\sigma_s\psi_s\otimes\varphi_s \otimes\phi_s)$$ (29)

$$= (P_{I..}\otimes P_{.J.}\otimes P_{..K}) + (D_I\otimes D_J\otimes D_K)(\sum_{s=1}^r\sigma_s\psi_s\otimes\varphi_s \otimes\phi_s)$$

and also achieve the full decomposition (or reconstruction). Though no explicit expression of the maximal rank $r$ can be calculated beforehand and is a subject of research in multiway analysis.