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Here is my attempt to show that INDSCAL as a special case of CANDELINC. I am using the following paper as my reference for definitions.

Definitions

INDSCAL Model (Section 5.1 of the source):\begin{equation} \mathcal{X} \approx \langle A, A, C \rangle = \sum_{r=1}^{R} a_r \circ a_r \circ c_r, \end{equation}where:

  1. $A \in \mathbb{R}^{I \times R}$ is the factor matrix shared across the first two modes (symmetry is enforced).
  2. $C \in \mathbb{R}^{K \times R}$ represents the third mode's variation.

The symmetry implies$x_{ijk} = x_{jik}$ for all$i$,$j$, and$k$.

CANDELINC Model (Section 5.3 of the source):\begin{equation} \mathcal{X} \approx \langle \Phi_A A, \Phi_B B, \Phi_C C \rangle, \end{equation}where:

  1. $\Phi_A, \Phi_B, \Phi_C$ are constraint matrices.
  2. For example,$\Phi_A A$ represents that the factor matrix$A$ is constrained to lie in the subspace defined by$\Phi_A$.

Mapping INDSCAL onto CANDELINC

  1. In the INDSCAL model, the symmetry between the first two modes enforces that$B = A$.
  2. This can be written as alinear constraint in the CANDELINC framework:$\Phi_B = \Phi_A = I, \quad B = A$.
  3. Substituting these constraints into the CANDELINC model gives:\begin{equation} \mathcal{X} \approx \langle \Phi_A A, \Phi_A A, \Phi_C C \rangle. \end{equation}

Linear Constraints in CANDELINCIn CANDELINC, the linear constraints are applied via matrices$\Phi_A, \Phi_B, \Phi_C$:

  1. Setting$\Phi_B = \Phi_A = I$ ensures that$A$ is shared across the first two modes.
  2. The matrix$\Phi_C$ allows$C$ to remain unconstrained, capturing the individual differences specific to the third mode.

Specialization of CANDELINCCANDELINC accommodates a broader range of linear constraints through$\Phi_A, \Phi_B, \Phi_C$. By specifying:$\Phi_B = \Phi_A = I \quad \text{and enforcing symmetry on } A$,the INDSCAL model is obtained. Thus, INDSCAL is a specialized case of CANDELINC with symmetry and shared structure constraints.

Would love to know if there is any issue with this mapping.

askedJan 14 at 19:55
Omar Shehab's user avatar
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