Identifiability of Diagnostic Classification Models

Abstract

Diagnostic classification models (DCMs) are important statistical tools in cognitive diagnosis. In this paper, we consider the issue of their identifiability. In particular, we focus on one basic and popular model, the DINA model. We propose sufficient and necessary conditions under which the model parameters are identifiable from the data. The consequences, in terms of the consistency of parameter estimates, of fulfilling or failing to fulfill these conditions are illustrated via simulation. The results can be easily extended to the DINO model through the duality of the DINA and DINO models. Moreover, the proposed theoretical framework could be applied to study the identifiability issue of other DCMs.

Appendix: Proof of Theorems

We begin with two important propositions necessary to prove the main results; their own proofs are postponed to the end of this section.

Recall that identifiability and local identifiability depend on the probability density function $f(x; \theta )$, which, when written as a function of the parameters $\theta$ becomes the likelihood $L(\theta )$.

Under the model specified in Section 2, given the full set of observations $R = \{{\mathbf {R}}^i: i = 1,\ldots , N\}$ and a Q-matrix Q, the likelihood of any set of parameters ${\mathbf {c}},{\mathbf {g}},{\mathbf {p}}$ can be written as

$$\begin{aligned} L({\mathbf {c}},{\mathbf {g}},{\mathbf {p}}; R) = \prod _{i=1}^n P({\mathbf {R}}^i|Q,{\mathbf {c}},{\mathbf {g}},{\mathbf {p}}) = \prod _{{\mathbf {r}}\in \{0,1\}^J} \pi _{\mathbf {r}}^{N_{{\mathbf {r}}}} \end{aligned}$$

(8)

where $N_{{\mathbf {r}}} = |\{i \in \{1,\ldots , N\}: {\mathbf {R}}^i = {\mathbf {r}}\}|$ is the number of observations ${\mathbf {R}}^i$ equal to a particular response vector ${\mathbf {r}}$ and

$$\begin{aligned} \pi _{\mathbf {r}}= P({\mathbf {R}}= {\mathbf {r}}|Q,{\mathbf {c}},{\mathbf {g}},{\mathbf {p}}) = \sum _{\varvec{\alpha }}p_{\varvec{\alpha }}\prod _{j=1}^J P(R_j = r_j|Q,c_j,g_j,{\varvec{\alpha }}) \end{aligned}$$

(9)

is the probability of observing ${\mathbf {r}}$ given $Q,{\mathbf {c}},{\mathbf {g}},{\mathbf {p}}$. The conditional probability $P(R_j = r_j|Q,c_j,g_j,{\varvec{\alpha }})$ may be expressed as

$$\begin{aligned} c_j^{r_j\xi _j(Q,{\varvec{\alpha }})}g_j^{r_j(1-\xi _j(Q,{\varvec{\alpha }}))}(1-c_j)^{(1-r_j) \xi _j(Q,{\varvec{\alpha }})}(1-g_j)^{(1-r_j)(1-\xi _j(Q,{\varvec{\alpha }}))}. \end{aligned}$$

Defining the likelihood leads to the first of the two propositions, which ties the T-matrix to the likelihood:

Proposition 1

For two sets of parameters $(\hat{\mathbf {c}}, \hat{\mathbf {g}}, \hat{\mathbf {p}})$ and $(\bar{\mathbf {c}}, \bar{\mathbf {g}}, \bar{\mathbf {p}})$,

$$\begin{aligned} L(\hat{\mathbf {c}}, \hat{\mathbf {g}}, \hat{\mathbf {p}}; R)=L(\bar{\mathbf {c}}, \bar{\mathbf {g}}, \bar{\mathbf {p}}; R) \end{aligned}$$

for all observation matrices R if and only if the following equation holds:

$$\begin{aligned} T(Q,\hat{\mathbf {c}},\hat{\mathbf {g}})\hat{\mathbf {p}}=T(Q,\bar{\mathbf {c}},\bar{\mathbf {g}})\bar{\mathbf {p}}. \end{aligned}$$

(10)

The second proposition describes the linear relationship between certain pairs of T-matrices.

Proposition 2

There exists a matrix $D({\mathbf {g}}^*)$ depending solely on ${\mathbf {g}}^*=(g_1^*,\ldots ,g_J^*)$, such that for any ${\mathbf {g}}^* \in {\mathbb {R}}^J$,

$$\begin{aligned} T(Q,{\mathbf {c}}-{\mathbf {g}}^*,{\mathbf {g}}-{\mathbf {g}}^*) = D({\mathbf {g}}^*) T(Q,{\mathbf {c}},{\mathbf {g}}). \end{aligned}$$

The matrix $D({\mathbf {g}}^*)$ is always lower triangular with diagonal $\text {diag}(D({\mathbf {g}}^*)) = {\mathbf {1}}$, and thus invertible.

The main idea of the proofs is based on the result of Proposition 1. In particular, to show the identifiability, we only need to show that for two sets of parameters $(\hat{\mathbf {c}},\hat{\mathbf {g}},\hat{\mathbf {p}})$ and $(\bar{\mathbf {c}},\bar{\mathbf {g}},\bar{\mathbf {p}})$ satisfying (10), $(\hat{\mathbf {c}},\hat{\mathbf {g}},\hat{\mathbf {p}})=(\bar{\mathbf {c}},\bar{\mathbf {g}},\bar{\mathbf {p}})$. On the other hand, by the definition, if there exist $(\hat{\mathbf {c}},\hat{\mathbf {g}},\hat{\mathbf {p}})\ne (\bar{\mathbf {c}},\bar{\mathbf {g}},\bar{\mathbf {p}})$ and (10) holds, then the model is unidentifiable. We now prove our main theorems and propositions.

Proof of Theorem 1

The case where ${\mathbf {s}}= {\mathbf {g}}= {\mathbf {0}}$ was shown by (Chiu, 2009). For general known ${\mathbf {s}}$ and ${\mathbf {g}}$, by Proposition 1, ${\mathbf {p}}$ is unidentifiable when $Q,{\mathbf {c}},{\mathbf {g}}$ are known iff there exists $\hat{\mathbf {p}},\bar{\mathbf {p}}\in {\mathbb {R}}_+^{2^K}$ such that

$$\begin{aligned} T(Q,{\mathbf {c}},{\mathbf {g}})\hat{\mathbf {p}}= T(Q,{\mathbf {c}},{\mathbf {g}})\bar{\mathbf {p}}. \end{aligned}$$

This occurs iff $T(Q,{\mathbf {c}},{\mathbf {g}})$ is not a full-rank matrix.

Suppose that the Q-matrix is not complete. WLOG, we assume that the row vector corresponding to the first attribute is missing, i.e., ${\mathbf {e}}_1^\top \not \in \mathcal {R}_Q$. Then, in the T-matrix, the columns corresponding to attribute profiles $\mathbf {0}$ and ${\mathbf {e}}_1$ are both equal to

$$\begin{aligned} \left( 1, g_1, \ldots , g_J , g_1g_2 , \ldots , g_1\cdots g_j \right) ^\top , \end{aligned}$$

and $\text {rank}(T(Q,{\mathbf {c}},{\mathbf {g}}))< 2^K$.

When Q is complete, assume WLOG that $Q_{1:K} = I_K$. The matrix $T(Q,{\mathbf {c}},{\mathbf {g}})$ is full-rank iff $T(Q,{\mathbf {c}}-{\mathbf {g}},{\mathbf {0}})$ is full-rank, since, by Proposition 2, $T(Q,{\mathbf {c}}-{\mathbf {g}},{\mathbf {0}}) = D({\mathbf {g}}) T(Q,{\mathbf {c}},{\mathbf {g}})$ and $D({\mathbf {g}})$ is invertible. Consider the rows of $T(Q,{\mathbf {c}}-{\mathbf {g}},{\mathbf {0}})$ corresponding to combinations of the first K items, i.e., ${\mathbf {r}}\in \{0,1\}^J$ s.t. $r_j = 0$ for all $j > K$. This constitutes an upper-triangular submatrix of size $2^K\times 2^K$ with diagonal entries $\prod _{j:r_j = 1} (c_j-g_j) \ne 0$. Thus, $T(Q,{\mathbf {c}}-{\mathbf {g}},{\mathbf {0}})$ is full-rank, and ${\mathbf {p}}$ is identifiable. $\square$

Proof of Theorem 2

Theorem 2 has been recently proved in Chen et al. (2014). For completeness, we include a proof under the setting of this paper. When ${\mathbf {g}}$ is known, we may combine Propositions 1 and 2 to show that two sets of parameters $(\hat{\mathbf {c}},{\mathbf {g}},\hat{\mathbf {p}})$ and $(\bar{\mathbf {c}},{\mathbf {g}},\bar{\mathbf {p}})$ produce equal likelihoods iff

$$\begin{aligned} T(Q,\hat{\mathbf {c}}-{\mathbf {g}},{\mathbf {0}})\hat{\mathbf {p}}= & {} D({\mathbf {g}})T(Q,\hat{\mathbf {c}},{\mathbf {g}})\hat{\mathbf {p}}\\= & {} D({\mathbf {g}})T(Q,\bar{\mathbf {c}},{\mathbf {g}})\bar{\mathbf {p}}=T(Q,\bar{\mathbf {c}}-{\mathbf {g}},{\mathbf {0}})\bar{\mathbf {p}}. \end{aligned}$$

Note that $c_j \in (g_j,1] \Leftrightarrow c_j- g_j \in (0,1-g_j]$.

Sufficiency For each item $j \in \{1,\ldots ,J\}$, condition C2 implies that there exists some set of items $S^j\subset \{1,\ldots , J\}$ s.t. $j\not \in S^j$ and the attributes required by item j are a subset of the attributes required by the items in $S^j$; then the sets of attributes required by items in $S^j$ and by items in $S^j\cup \{j\}$ are identical. Mathematically, there exists ${\mathbf {r}}^j \in \{0,1\}^J$ s.t. $r^j_j = 0$ and

$$\begin{aligned} T_{{\mathbf {r}}^j}(Q,{\mathbf {1}},{\mathbf {0}}) = T_{{\mathbf {r}}^j+{\mathbf {e}}_j}(Q,{\mathbf {1}},{\mathbf {0}}) . \end{aligned}$$

To find ${\mathbf {r}}^j$ for each item j, first suppose that $j\in \{1,\ldots , K\}$. Then $Q_j = {\mathbf {e}}_j^\top$ and there is some $j' \in \{K+1,\ldots , J\}$ s.t. $q_{j'j}= 1$. Let ${\mathbf {r}}^j = {\mathbf {e}}_{j'}$. Otherwise, when $j \in \{K+1,\ldots , J\}$, let ${\mathbf {r}}^j = \sum _{\{\ell :q_{j\ell }=1\}}{\mathbf {e}}_\ell$.

Then given any two sets of parameters $(\hat{\mathbf {c}},{\mathbf {0}},\hat{\mathbf {p}})$ and $(\bar{\mathbf {c}},{\mathbf {0}},\bar{\mathbf {p}})$ s.t. $T(Q,\hat{\mathbf {c}},{\mathbf {0}})\hat{\mathbf {p}}= T(Q,\bar{\mathbf {c}},{\mathbf {0}})\bar{\mathbf {p}}$,

$$\begin{aligned} \hat{c}_j = \frac{T_{{\mathbf {e}}_j+{\mathbf {r}}^j}(Q,\hat{\mathbf {c}},{\mathbf {0}})\hat{\mathbf {p}}}{T_{{\mathbf {r}}^j} (Q,\hat{\mathbf {c}},{\mathbf {0}})\hat{\mathbf {p}}} = \frac{T_{{\mathbf {e}}_j+{\mathbf {r}}^j}(Q,\bar{\mathbf {c}},{\mathbf {0}})\bar{\mathbf {p}}}{T_{{\mathbf {r}}^j} (Q,\bar{\mathbf {c}},{\mathbf {0}})\bar{\mathbf {p}}} = \bar{c}_j. \end{aligned}$$

Thus, $\hat{\mathbf {c}}= \bar{\mathbf {c}}$; then, by Theorem 1, $\hat{\mathbf {p}}=\bar{\mathbf {p}}$.

Necessity By Theorem 1, condition C1 is necessary. Suppose condition C2 fails to hold. WLOG, it fails to hold for the first attribute and $q_{j1} = 0$ for all $j \ne 1$. Consider any set of parameters $(\hat{\mathbf {c}},\hat{\mathbf {p}})$ s.t. $\hat{c}_j \in (g_j,1]$ for all $j \in \{1,\ldots , J\}$ and $\hat{\mathbf {p}}\in (0,1)^{2^K}$, $\sum _{\varvec{\alpha }}p_{\varvec{\alpha }}= 1$. There exists $\bar{c}_1$ close enough to $\hat{c}_1$ so that $\bar{c}_1 \in (g_1,1]$ and $\bar{p}_{\varvec{\alpha }}\in (0,1)$ for all ${\varvec{\alpha }}\in \{0,1\}^K$, where

$$\begin{aligned} \bar{p}_{\varvec{\alpha }}= {\left\{ \begin{array}{ll}(\hat{c}_1/\bar{c}_1) \hat{p}_{\varvec{\alpha }}&{}\quad \alpha _1 = 1 \\ \hat{p}_{\varvec{\alpha }}+ \hat{p}_{{\varvec{\alpha }}+ {\mathbf {e}}_1}(1-\hat{c}_1/\bar{c}_1) &{}\quad \alpha _1 = 0 \end{array}\right. }. \end{aligned}$$

Then, for any ${\mathbf {r}}\in \{0,1\}^J$ s.t. $r_1 = 0$, $T_{\mathbf {r}}(Q,\hat{\mathbf {c}},{\mathbf {0}}) = T_{\mathbf {r}}(Q,\bar{\mathbf {c}},{\mathbf {0}})$ and

$$\begin{aligned} T_{\mathbf {r}}(Q,\hat{\mathbf {c}},{\mathbf {0}})\hat{\mathbf {p}}= & {} \sum _{\{{\varvec{\alpha }}:\alpha _1=0\}} t_{{\mathbf {r}},{\varvec{\alpha }}}(Q,\hat{\mathbf {c}},{\mathbf {0}})(\hat{p}_{\varvec{\alpha }}+ \hat{p}_{{\varvec{\alpha }}+{\mathbf {e}}_1})\\= & {} \sum _{\{{\varvec{\alpha }}:\alpha _1=0\}} t_{{\mathbf {r}},{\varvec{\alpha }}}(Q,\bar{\mathbf {c}},{\mathbf {0}}))(\bar{p}_{\varvec{\alpha }}+ \bar{p}_{{\varvec{\alpha }}+{\mathbf {e}}_1}) =T_{\mathbf {r}}(Q,\bar{\mathbf {c}},{\mathbf {0}})\bar{\mathbf {p}}. \end{aligned}$$

Otherwise, $r_1 = 1$ and

$$\begin{aligned} T_{\mathbf {r}}(Q,\hat{\mathbf {c}},{\mathbf {0}})\hat{\mathbf {p}}= & {} \sum _{{\varvec{\alpha }}:\alpha _1=1} t_{{\mathbf {r}}-{\mathbf {e}}_1,{\varvec{\alpha }}}(Q,\hat{\mathbf {c}},{\mathbf {0}})\hat{c}_1 \hat{p}_{\varvec{\alpha }}\\= & {} \sum _{{\varvec{\alpha }}:\alpha _1=1} t_{{\mathbf {r}}-{\mathbf {e}}_1,{\varvec{\alpha }}}(Q,\bar{\mathbf {c}},{\mathbf {0}}) \bar{c}_1 \bar{p}_{\varvec{\alpha }}=T_{\mathbf {r}}(Q,\bar{\mathbf {c}},{\mathbf {0}})\bar{\mathbf {p}}. \end{aligned}$$

Thus we have found distinct sets of parameters satisfying (10), and shown that condition C2 is necessary. $\square$

Proof of Theorem 3

Thanks to Theorems 1 and 2, conditions C1 and C2 are necessary for identifiability. We now show the necessity of condition C3. Suppose C3 does not hold, but C1 and C2 do. Then all attributes are required by at least two items and there exists an attribute such that it is only required by two items. WLOG, this is the first attribute.

When both items requiring the first attribute require only the first attribute, the Q-matrix can be written WLOG as

$$\begin{aligned} Q= \left( \begin{array}{c@{\quad }c} 1 &{} {\mathbf {0}}^{\top } \\ 1 &{} {\mathbf {0}}^{\top } \\ {\mathbf {0}} &{} Q' \end{array}\right) . \end{aligned}$$

As was done for $r_1$ in the proof of necessity for Theorem 2, consider each possible value of $(r_1,r_2) \in \{0,1\}^2$ to conclude that, for any distinct sets of parameters $(\hat{\mathbf {c}},\hat{\mathbf {g}},\hat{\mathbf {p}})$ and $(\bar{\mathbf {c}},\bar{\mathbf {g}},\bar{\mathbf {p}})$, $T(Q,\hat{\mathbf {c}},\hat{\mathbf {g}},)\hat{\mathbf {p}}= T(Q,\bar{\mathbf {c}},\bar{\mathbf {g}})\bar{\mathbf {p}}$ if for every ${\varvec{\alpha }}\in \{0,1\}^K$ s.t. $\alpha _1 = 0$,

$$\begin{aligned} {\left\{ \begin{array}{ll} \hat{p}_{\varvec{\alpha }}+ \hat{p}_{{\varvec{\alpha }}+{\mathbf {e}}_1} =\bar{p}_{\varvec{\alpha }}+ \bar{p}_{{\varvec{\alpha }}+{\mathbf {e}}_1} &{}\quad (r_1,r_2) = (0,0)\\ \hat{c}_1 \hat{p}_{{\varvec{\alpha }}+{\mathbf {e}}_1}+\hat{g}_1 \hat{p}_{{\varvec{\alpha }}} =\bar{c}_1 \bar{p}_{{\varvec{\alpha }}+{\mathbf {e}}_1}+\bar{g}_1 \bar{p}_{{\varvec{\alpha }}} &{}\quad (r_1,r_2) = (1,0)\\ \hat{c}_2 \hat{p}_{{\varvec{\alpha }}+{\mathbf {e}}_1}+\hat{g}_2 \hat{p}_{{\varvec{\alpha }}} =\bar{c}_2 \bar{p}_{{\varvec{\alpha }}+{\mathbf {e}}_1}+\bar{g}_2 \bar{p}_{{\varvec{\alpha }}}&{}\quad (r_1,r_2) = (0,1)\\ \hat{c}_1\hat{c}_2 \hat{p}_{{\varvec{\alpha }}+{\mathbf {e}}_1}+\hat{g}_1\hat{g}_2 \hat{p}_{{\varvec{\alpha }}} =\bar{c}_1\bar{c}_2 \bar{p}_{{\varvec{\alpha }}+{\mathbf {e}}_1}+\bar{g}_1\bar{g}_2 \bar{p}_{{\varvec{\alpha }}}&{}\quad (r_1,r_2) = (1,1)\ \end{array}\right. }. \end{aligned}$$

(11)

Otherwise, the Q-matrix can be written WLOG as

$$\begin{aligned} Q= \begin{pmatrix} 1 &{}\quad {\mathbf {0}}^{\top } \\ 1 &{}\quad {\mathbf {v}}^{\top } \\ {\mathbf {0}} &{}\quad Q' \end{pmatrix}, \end{aligned}$$

where ${\mathbf {v}}$ is a $(K-1)$-dimensional nonzero vector. Then $T(Q,\hat{\mathbf {c}},\hat{\mathbf {g}},)\hat{\mathbf {p}}= T(Q,\bar{\mathbf {c}},\bar{\mathbf {g}})\bar{\mathbf {p}}$ if

$$\begin{aligned} {\left\{ \begin{array}{ll} \hat{p}_{\varvec{\alpha }}+ \hat{p}_{{\varvec{\alpha }}+{\mathbf {e}}_1} =\bar{p}_{\varvec{\alpha }}+ \bar{p}_{{\varvec{\alpha }}+{\mathbf {e}}_1} &{} \forall \ {\varvec{\alpha }}: \alpha _1 = 0\\ \hat{c}_1 \hat{p}_{{\varvec{\alpha }}+{\mathbf {e}}_1}+\hat{g}_1 \hat{p}_{{\varvec{\alpha }}} =\bar{c}_1 \bar{p}_{{\varvec{\alpha }}+{\mathbf {e}}_1}+\bar{g}_1 \bar{p}_{{\varvec{\alpha }}} &{} \forall \ {\varvec{\alpha }}: \alpha _1 = 0\\ \hat{c}_2 \hat{p}_{{\varvec{\alpha }}+{\mathbf {e}}_1}+\hat{g}_2 \hat{p}_{{\varvec{\alpha }}} =\bar{c}_2 \bar{p}_{{\varvec{\alpha }}+{\mathbf {e}}_1}+\bar{g}_2 \bar{p}_{{\varvec{\alpha }}} &{} \forall \ {\varvec{\alpha }}: \alpha _1 = 0, {\varvec{\alpha }}\succeq (0\ {\mathbf {v}}^\top )\\ \hat{c}_1\hat{c}_2 \hat{p}_{{\varvec{\alpha }}+{\mathbf {e}}_1}+\hat{g}_1\hat{g}_2 \hat{p}_{{\varvec{\alpha }}} =\bar{c}_1\bar{c}_2 \bar{p}_{{\varvec{\alpha }}+{\mathbf {e}}_1}+\bar{g}_1\bar{g}_2 \bar{p}_{{\varvec{\alpha }}} &{} \forall \ {\varvec{\alpha }}: \alpha _1 = 0,{\varvec{\alpha }}\succeq (0\ {\mathbf {v}}^\top ) \end{array}\right. }. \end{aligned}$$

(12)

Since the equations in (12) are a subset of the equations in (11), finding sets of parameters $(\hat{\mathbf {c}},\hat{\mathbf {g}},\hat{\mathbf {p}})$ and $(\bar{\mathbf {c}},\bar{\mathbf {g}},\bar{\mathbf {p}})$ fulfilling (11) completes the proof for both types of Q-matrices.

Choose a valid set of parameters $(\hat{\mathbf {c}},\hat{\mathbf {g}},\hat{\mathbf {p}})$ s.t. $\hat{p}_{\varvec{\alpha }}/ \hat{p}_{{\varvec{\alpha }}+{\mathbf {e}}_1} = \rho$ is constant over all ${\varvec{\alpha }}\in \{0,1\}^K$ s.t. $\alpha _1 =0$. Then, for any $\bar{\mathbf {g}}\in {\mathbb {R}}^J$, setting

$$\begin{aligned} \bar{c}_j = {\left\{ \begin{array}{ll} \bar{g}_1 + \frac{(\hat{c}_1-\bar{g}_1) (\hat{c}_2-\bar{g}_2) + \rho (\hat{g}_1-\bar{g}_1) (\hat{g}_2-\bar{g}_2)}{(\hat{c}_2-\bar{g}_2) + \rho (\hat{g}_2-\bar{g}_2)}, &{} j = 1\\ \bar{g}_2 + \frac{(\hat{c}_1-\bar{g}_1) (\hat{c}_2-\bar{g}_2) + \rho (\hat{g}_1-\bar{g}_1) (\hat{g}_2-\bar{g}_2)}{(\hat{c}_1-\bar{g}_1) + \rho (\hat{g}_1-\bar{g}_1)}, &{} j = 2\\ \hat{c}_j, &{} j =3, \ldots , J \end{array}\right. } \end{aligned}$$

and setting

$$\begin{aligned} \bar{p}_{{\varvec{\alpha }}+{\mathbf {e}}_1}= & {} \frac{((\hat{c}_1-\bar{g}_1) + \rho (\hat{g}_1-\bar{g}_1))((\hat{c}_2-\bar{g}_2) + \rho (\hat{g}_2 - \bar{g}_2) )}{(\hat{c}_1-\bar{g}_1)( \hat{c}_2-\bar{g}_2) + \rho (\hat{g}_1-\bar{g}_1)( \hat{g}_2-\bar{g}_2)} \hat{p}_{{\varvec{\alpha }}+{\mathbf {e}}_1},\\ \bar{p}_{\varvec{\alpha }}= & {} \hat{p}_{\varvec{\alpha }}+ \hat{p}_{{\varvec{\alpha }}+{\mathbf {e}}_1} - \bar{p}_{{\varvec{\alpha }}+{\mathbf {e}}_1} \end{aligned}$$

for every ${\varvec{\alpha }}\in \{0,1\}^K$ s.t. $\alpha _1=0$ results in a solution to (11). By continuity, there is $\bar{\mathbf {g}}$ sufficiently close to $\hat{\mathbf {g}}$ so that $\bar{\mathbf {c}},\bar{\mathbf {g}}\in [0,1]^J$, ${\mathbf {c}}\succ {\mathbf {g}}$, and $\bar{\mathbf {p}}\succ {\mathbf {0}}$. Thus, the model is non-identifiable when condition C3 fails, making it a necessary condition. $\square$

Proof of Theorem 4

Suppose that conditions C1 and C3 hold, and let $(\hat{\mathbf {c}},\hat{\mathbf {g}},\hat{\mathbf {p}})$ and $(\bar{\mathbf {c}},\bar{\mathbf {g}},\bar{\mathbf {p}})$ be two sets of parameters solving Equation (10). According to condition C1, there is an item requiring solely the kth attribute for each $k \in \{1,\ldots , K\}$. Moreover, by condition C3, there are also least two additional items requiring the kth attribute. We begin the proof of sufficiency by showing that for every k, there exists an item j requiring the kth attribute s.t. $\hat{g}_j = \bar{g}_j$. The case where all these items require solely the kth attribute and the case where at least one requires multiple attributes are treated separately.

Case 1 :

All items requiring the kth attribute require solely the kth attribute. WLOG, $k=1$ and the first three rows of Q are as follows:

$$\begin{aligned} Q_{1:3}= \left( \begin{array}{c@{\quad }c} 1 &{} {\mathbf {0}}^{\top } \\ 1 &{} {\mathbf {0}}^{\top } \\ 1 &{} {\mathbf {0}}^{\top } \\ \end{array}\right) . \end{aligned}$$

By Proposition 2, $T(Q,\hat{\mathbf {c}},\hat{\mathbf {g}})\hat{\mathbf {p}}= T(Q,\bar{\mathbf {c}},\bar{\mathbf {g}})\bar{\mathbf {p}}$ iff

$$\begin{aligned} T(Q,\hat{\mathbf {c}}-\hat{\mathbf {g}},{\mathbf {0}})\hat{\mathbf {p}}= T(Q,\bar{\mathbf {c}}-\hat{\mathbf {g}},\bar{\mathbf {g}}-\hat{\mathbf {g}})\bar{\mathbf {p}}. \end{aligned}$$

Then, since

$$\begin{aligned} \frac{T_{{\mathbf {e}}_1+{\mathbf {e}}_3}(\hat{\mathbf {c}}-\hat{\mathbf {g}},{\mathbf {0}})\hat{\mathbf {p}}}{T_{{\mathbf {e}}_1} (\hat{\mathbf {c}}-\hat{\mathbf {g}},{\mathbf {0}})\hat{\mathbf {p}}} = \hat{c}_3 - \hat{g}_3 =\frac{T_{{\mathbf {e}}_1+{\mathbf {e}}_2+{\mathbf {e}}_3}(\hat{\mathbf {c}}-\hat{\mathbf {g}},{\mathbf {0}})\hat{\mathbf {p}}}{T_{{\mathbf {e}}_1+{\mathbf {e}}_2}(\hat{\mathbf {c}}-\hat{\mathbf {g}},{\mathbf {0}})\hat{\mathbf {p}}} , \end{aligned}$$

we may conclude that

$$\begin{aligned} \frac{T_{{\mathbf {e}}_1+{\mathbf {e}}_3}(\bar{\mathbf {c}}-\hat{\mathbf {g}},\bar{\mathbf {g}}-\hat{\mathbf {g}})\bar{\mathbf {p}}}{T_{{\mathbf {e}}_1} (\bar{\mathbf {c}}-\hat{\mathbf {g}},\bar{\mathbf {g}}-\hat{\mathbf {g}})\bar{\mathbf {p}}} =\frac{T_{{\mathbf {e}}_1+{\mathbf {e}}_2+{\mathbf {e}}_3}(\bar{\mathbf {c}}-\hat{\mathbf {g}},\bar{\mathbf {g}}-\hat{\mathbf {g}})\bar{\mathbf {p}}}{T_{{\mathbf {e}}_1+{\mathbf {e}}_2}(\bar{\mathbf {c}}-\hat{\mathbf {g}},\bar{\mathbf {g}}-\hat{\mathbf {g}})\bar{\mathbf {p}}}. \end{aligned}$$

Let $\tilde{\mathbf {c}}= \bar{\mathbf {c}}-\hat{\mathbf {g}}$ and let $\tilde{\mathbf {g}}= \bar{\mathbf {g}}-\hat{\mathbf {g}}$. In addition, let $\bar{p}_i = \sum _{{\varvec{\alpha }}: \alpha _1=i} p_{\varvec{\alpha }}$ for $i = 0,1$. Then the previous equation may be written as

$$\begin{aligned} \frac{\tilde{g}_1 \tilde{g}_3 \bar{p}_0 + \tilde{c}_1 \tilde{c}_3 \bar{p}_1}{\tilde{g}_1 \bar{p}_0 + \tilde{c}_1 \bar{p}_1} =\frac{\tilde{g}_1 \tilde{g}_2 \tilde{g}_3 \bar{p}_0 + \tilde{c}_1\tilde{c}_2 \tilde{c}_3 \bar{p}_1}{\tilde{g}_1 \tilde{g}_2 \bar{p}_0 + \tilde{c}_1 \tilde{c}_2\bar{p}_1} \end{aligned}$$

and

$$\begin{aligned} \tilde{g}_1 \tilde{c}_1 (\tilde{c}_2 - \tilde{g}_2) (\tilde{c}_3 - \tilde{g}_3) \bar{p}_0 \bar{p}_1 =0. \end{aligned}$$

By assumption, $\bar{\mathbf {p}}\succ {\mathbf {0}}$, $\tilde{\mathbf {c}}\succ \tilde{\mathbf {g}}$, so $\hat{g}_1 = \bar{g}_1 \text { or } \bar{c}_1$. By symmetry, $\bar{g}_1 = \hat{g}_1 \text { or } \hat{c}_1$. If $\hat{g}_1\ne \bar{g}_1$, then $\hat{c}_1=\bar{g}_1$ and $\bar{c}_1=\hat{g}_1$. This contradicts the assumption that $\hat{\mathbf {c}}\succ \hat{\mathbf {g}}$ and $\bar{\mathbf {c}}\succ \bar{\mathbf {g}}$. Thus $\hat{g}_1= \bar{g}_1$.

Case 2 :

At least one item requiring the kth attribute requires multiple attributes. WLOG, $k = 1$ and

$$\begin{aligned} Q_{1:3}= \left( \begin{array}{c@{\quad }c@{\quad }c} 1 &{} 0 &{}{\mathbf {0}}^{\top } \\ 1 &{} 1 &{} {\mathbf {v}}^{\top } \\ 0 &{} 1 &{} {\mathbf {0}}^{\top } \\ \end{array}\right) , \end{aligned}$$

for some vector ${\mathbf {v}}\in \{0,1\}^{K-2}$. We will show that $\hat{g}_2 = \bar{g}_2$. Since

$$\begin{aligned} \frac{T_{{\mathbf {e}}_1+{\mathbf {e}}_2}(\hat{\mathbf {c}}-\hat{\mathbf {g}},{\mathbf {0}})\bar{\mathbf {p}}}{T_{{\mathbf {e}}_2}(\hat{\mathbf {c}}-\hat{\mathbf {g}},{\mathbf {0}})\bar{\mathbf {p}}}= \hat{c}_1 - \hat{g}_1 = \frac{T_{{\mathbf {e}}_1+{\mathbf {e}}_2+{\mathbf {e}}_3}(\hat{\mathbf {c}}-\hat{\mathbf {g}},{\mathbf {0}})\bar{\mathbf {p}}}{T_{{\mathbf {e}}_2+{\mathbf {e}}_3}(\hat{\mathbf {c}}-\hat{\mathbf {g}},{\mathbf {0}})\bar{\mathbf {p}}}, \end{aligned}$$

we know that

$$\begin{aligned} \frac{T_{{\mathbf {e}}_1+{\mathbf {e}}_2}(\bar{\mathbf {c}}-\hat{\mathbf {g}},\bar{\mathbf {g}}-\hat{\mathbf {g}})\bar{\mathbf {p}}}{T_{{\mathbf {e}}_2} (\bar{\mathbf {c}}-\hat{\mathbf {g}},\bar{\mathbf {g}}-\hat{\mathbf {g}})\bar{\mathbf {p}}} = \frac{T_{{\mathbf {e}}_1+{\mathbf {e}}_2+{\mathbf {e}}_3}(\bar{\mathbf {c}}-\hat{\mathbf {g}},\bar{\mathbf {g}}-\hat{\mathbf {g}})\bar{\mathbf {p}}}{T_{{\mathbf {e}}_2+{\mathbf {e}}_3}(\bar{\mathbf {c}}-\hat{\mathbf {g}},\bar{\mathbf {g}}-\hat{\mathbf {g}})\bar{\mathbf {p}}}. \end{aligned}$$

Thus,

$$\begin{aligned}&\frac{\tilde{g}_1\tilde{g}_2 \bar{p}_{0,0} +\tilde{c}_1\tilde{g}_2 \bar{p}_{1,0} +\tilde{g}_1\tilde{g}_2 \bar{p}_{0,1} +\tilde{c}_1\tilde{c}_2 \bar{p}_{1,1}}{\tilde{g}_2 \bar{p}_{0,0} +\tilde{g}_2 \bar{p}_{1,0} +\tilde{g}_2 \bar{p}_{0,1} +\tilde{c}_2 \bar{p}_{1,1}}\\&\quad = \frac{\tilde{g}_1\tilde{g}_3\tilde{g}_2 \bar{p}_{0,0} +\tilde{c}_1\tilde{g}_3\tilde{g}_2 \bar{p}_{1,0} +\tilde{g}_1\tilde{c}_3\tilde{g}_2 \bar{p}_{0,1} +\tilde{c}_1\tilde{c}_3\tilde{c}_2 \bar{p}_{1,1}}{\tilde{g}_3\tilde{g}_2 \bar{p}_{0,0} +\tilde{g}_3\tilde{g}_2 \bar{p}_{1,0} +\tilde{c}_3\tilde{g}_2 \bar{p}_{0,1} +\tilde{c}_3\tilde{c}_2 \bar{p}_{1,1}}, \end{aligned}$$

where $\bar{p}_{i,j}=\sum _{{\varvec{\alpha }}: (\alpha _1,\alpha _2) = (i,j)} \bar{p}_{\varvec{\alpha }}$ for $(i,j)\in \{0,1\}^2$, $\tilde{g}_j=\bar{g}_j-\hat{g}_j$ for $j=1,2,3$, $\tilde{c}_j=\bar{c}_j-\hat{c}_j$ for $j=1,3$, and

$$\begin{aligned} \tilde{c}_2=\frac{(\bar{c}_2-\hat{c}_2)\sum _{{\varvec{\alpha }}: {\varvec{\alpha }}\succeq Q_2}p_{\varvec{\alpha }}+(\bar{g}_2-\hat{g}_2)\sum _{{\varvec{\alpha }}:\alpha _1=\alpha _2=1,{\varvec{\alpha }}\not \succeq Q_2} \bar{p}_{\varvec{\alpha }}}{\bar{p}_{1,1}}. \end{aligned}$$

Cross-multiply and cancel to obtain that

$$\begin{aligned} \bar{p}_{0,1} \bar{p}_{1,0} (\tilde{c}_1- \tilde{g}_1) \tilde{g}_2^2(\tilde{c}_3- \tilde{g}_3)= \bar{p}_{0,0} \bar{p}_{1,1}(\tilde{c}_1 - \tilde{g}_1) \tilde{c}_2 \tilde{g}_2 (\tilde{c}_3 - \tilde{g}_3) \end{aligned}$$

Now suppose that $\hat{g}_2 \ne \bar{g}_2$. Since $\tilde{c}_j > \tilde{g}_j$ for $j = 1,2,3$,

$$\begin{aligned} \bar{p}_{1,0}\bar{p}_{0,1}(\bar{g}_2 - \hat{g}_2)= \bar{p}_{0,0}\bar{p}_{1,1}(\bar{c}_2-\hat{g}_2). \end{aligned}$$

(13)

In addition, by symmetry,

$$\begin{aligned} \hat{p}_{1,0}\hat{p}_{0,1} (\hat{g}_2 - \bar{g}_2)= \hat{p}_{0,0}\hat{p}_{1,1}(\hat{c}_2 - \bar{g}_2), \end{aligned}$$

(14)

where $\hat{p}_{i,j}=\sum _{{\varvec{\alpha }}: (\alpha _1,\alpha _2) = (i,j)} \hat{p}_{\varvec{\alpha }}$ for $(i,j)\in \{0,1\}^2$. Taken together, (13) and (14) imply that either $\hat{c}_3 > \hat{g}_3> \bar{c}_3>\bar{g}_3$ or $\bar{c}_3 > \bar{g}_3> \hat{c}_3>\hat{g}_3$. However, since $T_{{\mathbf {e}}_2}(\hat{\mathbf {c}},\hat{\mathbf {g}})\hat{\mathbf {p}}= T_{{\mathbf {e}}_2}(\bar{\mathbf {c}},\bar{\mathbf {g}})\bar{\mathbf {p}}$,

$$\begin{aligned} \hat{g}_2(\hat{p}_{0,0}+\hat{p}_{1,0}+\hat{p}_{0,1})+\hat{c}_2 p_{1,1} = \bar{g}_2(\bar{p}_{0,0}+\bar{p}_{1,0}+\bar{p}_{0,1})+\bar{c}_2 p_{1,1}. \end{aligned}$$

This is a contradiction; thus $\hat{g}_2 = \bar{g}_2$.

WLOG, the Q-matrix can be written as

$$\begin{aligned} Q = \begin{pmatrix} {\mathcal {I}}_K \\ Q' \end{pmatrix}. \end{aligned}$$

We have shown that for each $k \in \{1,\ldots , K\}$, there exists some item $j_k >K$ requiring the kth attribute s.t. $\hat{g}_{j_k} = \bar{g}_{j_k}$. For each item $j >K$, let ${\mathbf {r}}^j = \begin{pmatrix}Q_j^\top&{\mathbf {0}}\end{pmatrix}$ be the response vector selecting those among the first K items requiring attributes required by the jth item. Then ${\mathbf {r}}^j$ and ${\mathbf {r}}^j+{\mathbf {e}}_j$ denote distinct sets of items with identical attribute requirements and

$$\begin{aligned} \hat{c}_j -\hat{g}_j = \frac{T_{{\mathbf {r}}^j+{\mathbf {e}}_j}(Q,\hat{\mathbf {c}}-\hat{\mathbf {g}},{\mathbf {0}})\hat{\mathbf {p}}}{T_{{\mathbf {r}}^j}(Q,\hat{\mathbf {c}}-\hat{\mathbf {g}},{\mathbf {0}})\hat{\mathbf {p}}} = \frac{T_{{\mathbf {r}}^j+{\mathbf {e}}_j}(Q,\bar{\mathbf {c}}-\hat{\mathbf {g}},\bar{\mathbf {g}}-\hat{\mathbf {g}})\bar{\mathbf {p}}}{T_{{\mathbf {r}}^j}(Q,\bar{\mathbf {c}}-\hat{\mathbf {g}},\bar{\mathbf {g}}-\hat{\mathbf {g}})\bar{\mathbf {p}}}. =\bar{c}_j - \hat{g}_j \end{aligned}$$

Thus, $\hat{c}_j = \bar{c}_j$ if $\hat{g}_j = \bar{g}_j$; by the proof of Case 2, this includes all items j requiring multiple attributes. Otherwise, $Q_j = {\mathbf {e}}_k$ for some $k \in \{1,\ldots ,K\}$, and the response vectors ${\mathbf {e}}_j + {\mathbf {e}}_{j_k}$ and ${\mathbf {e}}_{j_k}$ represent distinct combinations of items with identical attribute requirements, so that

$$\begin{aligned} \hat{c}_j= & {} \frac{T_{{\mathbf {e}}_j+{\mathbf {e}}_{j_k}}(Q,\hat{\mathbf {c}}-\hat{g}_{j_k}{\mathbf {e}}_{j_k},\hat{\mathbf {g}}- \hat{g}_{j_k}{\mathbf {e}}_{j_k})\hat{\mathbf {p}}}{T_{{\mathbf {e}}_{j_k}}(Q,\hat{\mathbf {c}}-\hat{g}_{j_k} {\mathbf {e}}_{j_k},\hat{\mathbf {g}}- \hat{g}_{j_k}{\mathbf {e}}_{j_k})\hat{\mathbf {p}}}\\= & {} \frac{T_{{\mathbf {e}}_j + {\mathbf {e}}_{j_k}}(Q,\bar{\mathbf {c}}-\hat{g}_{j_k}{\mathbf {e}}_{j_k}, \bar{\mathbf {g}}-\hat{g}_{j_k}{\mathbf {e}}_{j_k})\bar{\mathbf {p}}}{T_{{\mathbf {e}}_{j_k}}(Q,\bar{\mathbf {c}}- \hat{g}_{j_k}{\mathbf {e}}_{j_k},\bar{\mathbf {g}}-\hat{g}_{j_k}{\mathbf {e}}_{j_k})\bar{\mathbf {p}}} =\bar{c}_j. \end{aligned}$$

Thus, $\hat{c}_j = \bar{c}_j$ for every $j\in \{1,\ldots , J\}$, i.e., $\hat{\mathbf {c}}= \bar{\mathbf {c}}$.

We now consider the identifiability of the remaining $g_j$. For each $j >K$ s.t. $Q_j = {\mathbf {e}}_k$ for some $k \in \{1,\ldots ,K\}$, let ${\mathbf {c}}^* = \hat{c}_k {\mathbf {e}}_k + \hat{c}_j {\mathbf {e}}_j$. Then

$$\begin{aligned} \hat{g}_k - \hat{c}_k =\frac{T_{{\mathbf {e}}_k + {\mathbf {e}}_j} (\hat{\mathbf {c}}-{\mathbf {c}}^*, \hat{\mathbf {g}}- {\mathbf {c}}^*)\hat{\mathbf {p}}}{T_{{\mathbf {e}}_k} (\hat{\mathbf {c}}-{\mathbf {c}}^*, \hat{\mathbf {g}}- {\mathbf {c}}^*)\hat{\mathbf {p}}} =\frac{T_{{\mathbf {e}}_k + {\mathbf {e}}_j} (\bar{\mathbf {c}}-{\mathbf {c}}^*, \bar{\mathbf {g}}- {\mathbf {c}}^*)\bar{\mathbf {p}}}{T_{{\mathbf {e}}_k} (\bar{\mathbf {c}}-{\mathbf {c}}^*, \bar{\mathbf {g}}- {\mathbf {c}}^*)\bar{\mathbf {p}}} =\bar{g}_k - \hat{c}_k \end{aligned}$$

and $\hat{g}_k = \bar{g}_k$. Thus $g_j$ is identifiable for all $j >K$.

To show the identifiability of $g_1,\ldots , g_K$, for each $k \le K$ let

$$\begin{aligned} {\mathbf {r}}^k = \sum _{j=K+1}^J {\mathbf {e}}_j(1-q_{jk}) \end{aligned}$$

represent the set of items in $Q'$ not requiring the kth attribute. When condition C4 holds, there is some item $\ell >K$ requiring the kth attribute and no other attributes not required by the set of items denoted by ${\mathbf {r}}^k$. Let ${\mathbf {g}}^* = (\hat{c}_1,\ldots , \hat{c}_k, \hat{g}_{K+1},\ldots , \hat{g}_J)^\top$. Then, for any set of parameters $({\mathbf {c}},{\mathbf {g}},{\mathbf {p}})$ s.t. $g_j = \hat{g}_j$ for all $j>K$,

$$\begin{aligned} T_{{\mathbf {r}}}(Q,{\mathbf {c}}- {\mathbf {g}}^*, {\mathbf {g}}- {\mathbf {g}}^*) {\mathbf {p}}=\left( \prod _{j=K+1}^J ( c_j-\hat{g}_j)^{r_j}\right) \sum _{{\varvec{\alpha }}\in \{0,1\}^K} p_{\varvec{\alpha }}t_{{\mathbf {r}},{\varvec{\alpha }}}(Q) \end{aligned}$$

for all response vectors ${\mathbf {r}}$ s.t. $r_j = 0$ for all $j\le K$. Since $\hat{\mathbf {c}}=\bar{\mathbf {c}}$ and $\bar{g}_j = \hat{g}_j$ for all $j>K$, this implies that

$$\begin{aligned} \sum _{{\varvec{\alpha }}\in \{0,1\}^K} \hat{p}_{\varvec{\alpha }}t_{{\mathbf {r}},{\varvec{\alpha }}}(Q) = \sum _{{\varvec{\alpha }}\in \{0,1\}^K} \bar{p}_{\varvec{\alpha }}t_{{\mathbf {r}},{\varvec{\alpha }}}(Q) \end{aligned}$$

(15)

for all such ${\mathbf {r}}$. Consider the row of $T(Q,{\mathbf {c}}-{\mathbf {g}}^*,{\mathbf {g}}-{\mathbf {g}}^*)$ corresponding to the combination of the kth item with all the items denoted by ${\mathbf {r}}^k$. The entries of this row-vector are non-zero only for attribute profiles denoting mastery of the skills required by ${\mathbf {r}}^k$ and non-mastery of the kth attribute. Thus,

$$\begin{aligned}&T_{{\mathbf {e}}_k + {\mathbf {r}}^k}(Q,{\mathbf {c}}-{\mathbf {g}}^*,{\mathbf {g}}-{\mathbf {g}}^*){\mathbf {p}}\\&\quad = (g_k - \hat{c}_k)\left( \prod _{j=K+1}^J ( c_j-\hat{g}_j)^{r_j^k}\right) \sum _{{\varvec{\alpha }}\in \{0,1\}^K} p_{\varvec{\alpha }}(t_{{\mathbf {r}}^k,{\varvec{\alpha }}}(Q) - t_{{\mathbf {e}}_k + {\mathbf {r}}^k,{\varvec{\alpha }}}(Q)). \end{aligned}$$

When condition C4 holds, there is some ${\mathbf {r}}$ s.t. $r_j = 0$ for all $j \le K$ and $T_{{\mathbf {e}}_k + {\mathbf {r}}^k}(Q) = T_{{\mathbf {r}}+ {\mathbf {r}}^k}(Q)$. Then, by (15)

$$\begin{aligned} \sum _{{\varvec{\alpha }}\in \{0,1\}^K} \hat{p}_{\varvec{\alpha }}(t_{{\mathbf {r}}^k}(Q) - t_{{\mathbf {e}}_k + {\mathbf {r}}^k,{\varvec{\alpha }}}(Q)) = \sum _{{\varvec{\alpha }}\in \{0,1\}^K} \bar{p}_{\varvec{\alpha }}(t_{{\mathbf {r}}^k}(Q) - t_{{\mathbf {e}}_k + {\mathbf {r}}^k,{\varvec{\alpha }}}(Q)). \end{aligned}$$

Since $T_{{\mathbf {e}}_k + {\mathbf {r}}^k}(Q,\hat{\mathbf {c}}-{\mathbf {g}}^*,\hat{\mathbf {g}}-{\mathbf {g}}^*)\hat{\mathbf {p}}=T_{{\mathbf {e}}_k + {\mathbf {r}}^k}(Q,\bar{\mathbf {c}}-{\mathbf {g}}^*,\bar{\mathbf {g}}-{\mathbf {g}}^*)\bar{\mathbf {p}}$, it must be true that $\hat{g}_k = \bar{g}_k$. Thus, ${\mathbf {g}}$ is fully identifiable and by Theorem 1 so is ${\mathbf {p}}$. $\square$

Proof of Proposition 1

The observations follow a multinomial distribution over the set of possible responses ${\mathbf {r}}\in \{0,1\}^J$, with probabilities $\pi _{\mathbf {r}}$ as defined in (9). For a particular $(\hat{\mathbf {c}},\hat{\mathbf {g}},\hat{\mathbf {p}})$ and $(\bar{\mathbf {c}},\bar{\mathbf {g}},\bar{\mathbf {p}})$,

$$\begin{aligned} L(\hat{\mathbf {c}},\hat{\mathbf {g}},\hat{\mathbf {p}}; R)= & {} \prod _{{\mathbf {r}}\in \{0,1\}^J} \pi _{\mathbf {r}}(Q,\hat{\mathbf {c}},\hat{\mathbf {g}},\hat{\mathbf {p}})^{N_{\mathbf {r}}}\\= & {} \prod _{{\mathbf {r}}\in \{0,1\}^J} \pi _{\mathbf {r}}(Q,\bar{\mathbf {c}},\bar{\mathbf {g}},\bar{\mathbf {p}})^{N_{\mathbf {r}}} =L(\bar{\mathbf {c}},\bar{\mathbf {g}}, \bar{\mathbf {p}}; R) \end{aligned}$$

for all observation matrices R iff $\pi _{\mathbf {r}}(Q,\hat{\mathbf {c}},\hat{\mathbf {g}},\hat{\mathbf {p}})= \pi _{\mathbf {r}}(Q,\bar{\mathbf {c}},\bar{\mathbf {g}},\bar{\mathbf {p}})$ for all ${\mathbf {r}}\in \{0,1\}^J$. Consider a P-matrix $P(Q,{\mathbf {c}},{\mathbf {g}})$ indexed like the T-matrix by item subsets ${\mathbf {r}}\in \{0,1\}^J$ and attribute profiles ${\varvec{\alpha }}\in \{0,1\}^K$. The entries of $P(Q,{\mathbf {c}},{\mathbf {g}})$ are denoted by the quantities

$$\begin{aligned} p_{{\mathbf {r}},{\varvec{\alpha }}}(Q,{\mathbf {c}},{\mathbf {g}}) = P({\mathbf {R}}= {\mathbf {r}}|Q,{\mathbf {c}},{\mathbf {g}}). \end{aligned}$$

Then the statement $\pi _{\mathbf {r}}(Q,\hat{\mathbf {c}},\hat{\mathbf {g}},\hat{\mathbf {p}})= \pi _{\mathbf {r}}(Q,\bar{\mathbf {c}},\bar{\mathbf {g}},\bar{\mathbf {p}})$ for all ${\mathbf {r}}\in \{0,1\}^J$ can be written in matrix notation as $P(Q,\hat{\mathbf {c}},\hat{\mathbf {g}})\hat{\mathbf {p}}= P(Q,\bar{\mathbf {c}},\bar{\mathbf {g}})\bar{\mathbf {p}}$. Since

$$\begin{aligned} t_{{\mathbf {r}},{\varvec{\alpha }}}(Q,{\mathbf {c}},{\mathbf {g}}) = P({\mathbf {R}}\succeq {\mathbf {r}}|Q,{\mathbf {c}},{\mathbf {g}}) = \sum _{{\mathbf {r}}\succeq {\mathbf {r}}} p_{{\mathbf {r}},{\varvec{\alpha }}}(Q,{\mathbf {c}},{\mathbf {g}}) \end{aligned}$$

there is a one-to-one linear transformation between $P(Q,{\mathbf {c}},{\mathbf {g}})$ and $T(Q,{\mathbf {c}},{\mathbf {g}})$ that is not dependent on $Q,{\mathbf {c}},{\mathbf {g}}$, and

$$\begin{aligned} P(Q,\hat{\mathbf {c}},\hat{\mathbf {g}})\hat{\mathbf {p}}= P(Q,\bar{\mathbf {c}},\bar{\mathbf {g}})\bar{\mathbf {p}}\Leftrightarrow T(Q,\hat{\mathbf {c}},\hat{\mathbf {g}})\hat{\mathbf {p}}= T(Q,\bar{\mathbf {c}},\bar{\mathbf {g}})\bar{\mathbf {p}}. \end{aligned}$$

$\square$

Proof of the Proposition 2

In what follows, we construct such a D matrix satisfying the conditions in the proposition, i.e., $D({\mathbf {g}}^*)$ is a matrix only depending on ${\mathbf {g}}^*$ such that $D_{{\mathbf {g}}^*} T(Q,{\mathbf {c}},{\mathbf {g}}) =T_{{\mathbf {c}}-{\mathbf {g}}^*,{\mathbf {g}}-{\mathbf {g}}^*}(Q)$ for any $Q,{\mathbf {c}},{\mathbf {g}}$. Recall that for any $Q,{\mathbf {c}},{\mathbf {g}}$,

$$\begin{aligned} t_{{\mathbf {r}},{\varvec{\alpha }}}(Q,{\mathbf {c}},{\mathbf {g}}) =\prod _{j\in S} c_j ^{\xi _j(Q,{\varvec{\alpha }})}g_j^{1-\xi _j(Q,{\varvec{\alpha }})}\ \forall \ {\mathbf {r}}\in \{0,1\}^J,\quad {\varvec{\alpha }}\in \{0,1\}^K. \end{aligned}$$

We may extend this definition to include ${\mathbf {c}},{\mathbf {g}}\not \in [0,1]^M$, though in such cases the $t_{{\mathbf {r}},{\varvec{\alpha }}}$ will no longer correspond to probabilities. Then for any ${\mathbf {g}}^* \in {\mathbb {R}}$,

$$\begin{aligned} t_{{\mathbf {r}},{\varvec{\alpha }}}(Q,{\mathbf {c}}-{\mathbf {g}}^*,{\mathbf {g}}-{\mathbf {g}}^*) =\prod _{j:r_j=1} (c_j-g_j^*) ^{\xi _j(Q,{\varvec{\alpha }})}(g_j-g_j^*)^{1-\xi _j(Q,{\varvec{\alpha }})} = \prod _{j:r_j=1} (b_j-g_j^*), \end{aligned}$$

where $b_j = c_j^{\xi _j(Q,{\varvec{\alpha }})}g_j^{1-\xi _j(Q,{\varvec{\alpha }})} = t_{{\mathbf {e}}_j,{\varvec{\alpha }}}(Q,{\mathbf {c}},{\mathbf {g}})$. By polynomial expansion,

$$\begin{aligned} t_{{\mathbf {r}},{\varvec{\alpha }}}(Q,{\mathbf {c}}-{\mathbf {g}}^*,{\mathbf {g}}-{\mathbf {g}}^*) = \sum _{{\mathbf {r}}' \preceq {\mathbf {r}}} (-1)^{\sum _{j=1}^J r_j-r_j'}\prod _{j: r_j - r_j' = 1} g_j^* \prod _{k: r_k' = 1} b_k. \end{aligned}$$

Define the entries $d_{{\mathbf {r}},{\mathbf {r}}'}({\mathbf {g}}^*)$ of $D({\mathbf {g}}^*)$ as

Then

$$\begin{aligned} T(Q,{\mathbf {c}}-{\mathbf {g}}^*,{\mathbf {g}}-{\mathbf {g}}^*) = D({\mathbf {g}}^*)T(Q,{\mathbf {c}},{\mathbf {g}}), \end{aligned}$$

where $D({\mathbf {g}}^*)$ is a lower triangular matrix depending solely on ${\mathbf {g}}^*$ with eigenvalues equal to its diagonal. Since $\text {diag}(D({\mathbf {g}}^*)) = {\mathbf {1}}$, $D({\mathbf {g}}^*)$ is invertible. $\square$