A Note on Weaker Conditions for Identifying Restricted Latent Class Models for Binary Responses

Abstract

Restricted latent class models (RLCMs) are an important class of methods that provide researchers and practitioners in the educational, psychological, and behavioral sciences with fine-grained diagnostic information to guide interventions. Recent research established sufficient conditions for identifying RLCM parameters. A current challenge that limits widespread application of RLCMs is that existing identifiability conditions may be too restrictive for some practical settings. In this paper we establish a weaker condition for identifying RLCM parameters for multivariate binary data. Although the new results weaken identifiability conditions for general RLCMs, the new results do not relax existing necessary and sufficient conditions for the simpler DINA/DINO models. Theoretically, we introduce a new form of latent structure completeness, referred to as dyad-completeness, and prove identification by applying Kruskal’s Theorem for the uniqueness of three-way arrays. The new condition is more likely satisfied in applied research, and the results provide researchers and test-developers with guidance for designing diagnostic instruments.

Appendices

Appendix A

1.1 Proof of Proposition 1

Without loss of generality, suppose that items X and $X'$ form a saturated dyad so that ${\varvec{q}}^\top {\varvec{q}}'=2$ and $\Vert {\varvec{q}}\Vert =\Vert {\varvec{q}}'\Vert =2$ and $\mathcal C({\varvec{q}})=\mathcal C({\varvec{q}}')=\{c_0,c_1,c_2,c_3\}$. The $4\times 4$ sub-table of probabilities for the item response patterns and attribute profile configurations for the base columns, $({\varvec{\theta }}{\varvec{*}}{\varvec{\theta }}')_{\mathcal C({\varvec{q}},{\varvec{q}}')}=\varvec{\theta }_{\mathcal C({\varvec{q}})}{\varvec{*}}\varvec{\theta }_{\mathcal C({\varvec{q}}')}'$, is

$$\begin{aligned} \begin{bmatrix} (1-\theta _{c_0})(1-\theta _{c_0}') &{} (1-\theta _{c_1})(1-\theta _{c_1}') &{} (1-\theta _{c_2})(1-\theta _{c_2}') &{} (1-\theta _{c_3})(1-\theta _{c_3}')\\ (1-\theta _{c_0})\theta _{c_0}' &{} (1-\theta _{c_1})\theta _{c_1}' &{} (1-\theta _{c_2})\theta _{c_2}' &{} (1-\theta _{c_3})\theta _{c_3}'\\ \theta _{c_0}(1-\theta _{c_0}') &{} \theta _{c_1}(1-\theta _{c_1}') &{} \theta _{c_2}(1-\theta _{c_2}') &{} \theta _{c_3}(1-\theta _{c_3}')\\ \theta _{c_0}\theta _{c_0}' &{} \theta _{c_1}\theta _{c_1}' &{} \theta _{c_2}\theta _{c_2}' &{} \theta _{c_3}\theta _{c_3}'\\ \end{bmatrix}. \end{aligned}$$

(A1)

We show $({\varvec{\theta }}{\varvec{*}}{\varvec{\theta }}')_{\mathcal C({\varvec{q}},{\varvec{q}}')}$ is full column rank by showing $\det \left( ({\varvec{\theta }}{\varvec{*}}{\varvec{\theta }}')_{\mathcal C({\varvec{q}},{\varvec{q}}')} \right) \ne 0$. In particular, we define

$$\begin{aligned} \mathbf{A} = \mathbf{O}_3 \mathbf{O}_2 \mathbf{O}_1 \left( ({\varvec{\theta }}{\varvec{*}}{\varvec{\theta }}')_{\mathcal C({\varvec{q}},{\varvec{q}}')} \right) \end{aligned}$$

(A2)

and note that, if $\det \left( \mathbf{O}_1\right) =\det \left( \mathbf{O}_2\right) =\det \left( \mathbf{O}_3\right) =1$, then $\det \left( \mathbf{A}\right) =\det \left( \varvec{\theta }_{\mathcal C({\varvec{q}})}{\varvec{*}}\varvec{\theta }_{\mathcal C({\varvec{q}}')}'\right)$. Consequently, $({\varvec{\theta }}{\varvec{*}}{\varvec{\theta }}')_{\mathcal C({\varvec{q}},{\varvec{q}}')}$ is full rank if $\det \left( \mathbf{A}\right) \ne 0$. Let $\mathbf{O}_1$, $\mathbf{O}_2$, and $\mathbf{O}_3$ be defined as

$$\begin{aligned} \mathbf{O}_1= \begin{bmatrix} 1 &{} 1 &{} 1 &{} 1\\ 0 &{} 1 &{} 0 &{} 1\\ 0 &{} 0 &{} 1 &{} 1\\ 0 &{} 0 &{} 0 &{} 1\\ \end{bmatrix}, \mathbf{O}_2= \begin{bmatrix} 1 &{} 0 &{} 0 &{} 0\\ -\theta _{c_0}' &{} 1 &{} 0 &{} 0\\ -\theta _{c_0} &{} 0 &{} 1 &{} 0\\ -\theta _{c_0}\theta _{c_0}' &{} 0 &{} 0 &{} 1\\ \end{bmatrix}, \mathbf{O}_3= \begin{bmatrix} 1 &{} 0 &{} 0 &{} 0\\ 0 &{} 1 &{} 0 &{} 0\\ 0 &{} -\frac{\theta _{c_1}-\theta _{c_0}}{\theta _{c_1}'-\theta _{c_0}'} &{} 1 &{} 0\\ 0 &{} -\theta _{c_1} &{} -\theta _{c_0}' &{} 1\\ \end{bmatrix}. \end{aligned}$$

(A3)

where $\mathbf{O}_3$ assumes that $\theta _{c_1}'\ne \theta _{c_0}'$. We therefore find that $\mathbf{A}$ and $\det \left( ({\varvec{\theta }}{\varvec{*}}{\varvec{\theta }}')_{\mathcal C({\varvec{q}},{\varvec{q}}')} \right)$ are:

$$\begin{aligned} \mathbf{A}= & {} \begin{bmatrix} 1 &{} 1 &{} 1 &{} 1\\ 0 &{} \theta _{c_1}'-\theta _{c_0}' &{} \theta _{c_2}'-\theta _{c_0}' &{} \theta _{c_3}'-\theta _{c_0}'\\ 0 &{} 0&{} (\theta _{c_2}-\theta _{c_0})-\frac{\theta _{c_1}-\theta _{c_0}}{\theta _{c_1}'-\theta _{c_0}'}(\theta _{c_2}'-\theta _{c_0}') &{} (\theta _{c_3}-\theta _{c_0})-\frac{\theta _{c_1}-\theta _{c_0}}{\theta _{c_1}'-\theta _{c_0}'}(\theta _{c_3}'-\theta _{c_0}')\\ 0 &{} 0 &{} (\theta _{c_2}'-\theta _{c_0}')(\theta _{c_2}-\theta _{c_1}) &{} (\theta _{c_3}'-\theta _{c_0}')(\theta _{c_3}-\theta _{c_1}) \end{bmatrix},\nonumber \\ \end{aligned}$$

(A4)

$$\begin{aligned} \det \left( ({\varvec{\theta }}{\varvec{*}}{\varvec{\theta }}')_{\mathcal C({\varvec{q}},{\varvec{q}}')} \right)= & {} \left[ (\theta _{c_1}'-\theta _{c_0}')(\theta _{c_2}-\theta _{c_0})-(\theta _{c_1}-\theta _{c_0})(\theta _{c_2}'-\theta _{c_0}')\right] (\theta _{c_3}'-\theta _{c_0}')(\theta _{c_3}-\theta _{c_1})\nonumber \\&\quad -\left[ (\theta _{c_1}'-\theta _{c_0}')(\theta _{c_3}-\theta _{c_0})-(\theta _{c_1}-\theta _{c_0})(\theta _{c_3}'-\theta _{c_0}')\right] (\theta _{c_2}'-\theta _{c_0}')(\theta _{c_2}-\theta _{c_1}). \end{aligned}$$

(A5)

Appendix B

1.1 Proof of Proposition 3

We prove Proposition 3 by first considering the case where K is even. Let $\mathbb {P}_{jj'}=(\varvec{\theta }_j{\varvec{*}}\varvec{\theta }_{j'})_{\mathcal C({\varvec{q}}_j,{\varvec{q}}_{j'})}$ be the $4\times 4$ matrix of response probabilities by classes for items $X_j$ and $X_{j'}$. The columns of $\mathbb {P}_{jj'}$ correspond with the four possible patterns for the two attributes that load onto $X_j$ and $X_{j'}$. Note that $\mathrm{rank}(\mathbb {P}_{jj'})=4$ given $X_j$ and $X_{j'}$ form a full rank dyad. Consequently, if K is even, $\mathbb {P}_{12},\dots ,\mathbb {P}_{K/2-1,K/2}$ each correspond with distinct pairs of items and attributes, which implies that

(B1)

and $\mathrm{rank}\left( \mathbf{T}_1\right) =2^K$.

If K is odd there are two cases defined by $a\in \{0,1\}$. The $a=0$ case is a direct extension of the even K case. Specifically, $a=0$ implies that $\varvec{\theta }_K$ for $X_K$ has a simple structure pattern so that $\varvec{\theta }_{\mathcal C({\varvec{q}}_K)}$ is a $2\times 2$ matrix with $\det \left( \varvec{\theta }_{\mathcal C({\varvec{q}}_K)}\right) =(\theta _{K,2^{K-1}}-\theta _{K0})$ and $\mathrm{rank}(\varvec{\theta }_{\mathcal C({\varvec{q}}_K)})=2$ if $\theta _{K,2^{K-1}}\ne \theta _{K0}$. The dyad-complete structure for the first $K-1$ items implies that

$$\begin{aligned} \mathbb {P}_{1:(K-1)}=\bigotimes _{d=1}^{(K-1)/2} \mathbb {P}_{2d-1,2d} \end{aligned}$$

(B2)

and $\mathrm{rank}\left( \mathbb {P}_{1:(K-1)}\right) =2^{K-1}$. Note that $\mathbb {P}_{1:(K-1)}$ describes how $X_1,\dots , X_{K-1}$ relate to attributes $\alpha _2,\dots ,\alpha _K$ and $X_1,\dots , X_{K-1}$ are unrelated to $\alpha _1$. The fact that $X_K$ is the only item loading onto $\alpha _1$ implies that

$$\begin{aligned} \mathbf{T}_1=\varvec{\theta }_{\mathcal C({\varvec{q}}_K)}\otimes \left( \bigotimes _{d=1}^{(K-1)/2} \mathbb {P}_{2d-1,2d}\right) \end{aligned}$$

(B3)

and $\mathrm{rank}\left( \mathbf{T}_1\right) =2\cdot 2^{K-1}=2^K$.

For $a=1$, $X_K$ loads onto both $\alpha _1$ and $\alpha _2$, so, as shown in Example 2, the $2\times 2^K$ matrix $\varvec{\theta }_K$ has four unique values so that

$$\begin{aligned} \varvec{\theta }_K=\left( \varvec{\theta }_{K0},\varvec{\theta }_{K,2^{K-2}},\varvec{\theta }_{K,2^{K-1}},\varvec{\theta }_{K,3\cdot 2^{K-2}}\right) \otimes {\varvec{1}}_{2^{K-2}}^\top . \end{aligned}$$

(B4)

Note that we can partition the attribute profile as ${\varvec{\alpha }}=(\alpha _1,{\varvec{\alpha }}_{2:K})$ so that $(0,{\varvec{\alpha }}_{2:K})$ corresponds with $2^{K-1}$ patterns for ${\varvec{\alpha }}_{2:K}$ when $\alpha _1=0$ and $(1,{\varvec{\alpha }}_{2:K})$ similarly includes $2^{K-1}$ patterns for ${\varvec{\alpha }}_{2:K}$ with $\alpha _1=1$. The fact that $X_1,\dots ,X_{K-1}$ are unrelated to $\alpha _1$ implies that the probability of response patterns for $X_1,\dots ,X_{K-1}$ given $(0,{\varvec{\alpha }}_{2:K})$ is $\mathbb {P}_{1:(K-1)}$ as is the probability of $X_1,\dots ,X_{K-1}$ given $(1,{\varvec{\alpha }}_{2:K})$. Notice that $(0,{\varvec{\alpha }}_{2:K}^\top ){\varvec{v}}\in \{0,\dots ,2^{K-1}-1\}$ and $(1,{\varvec{\alpha }}_{2:K}^\top ){\varvec{v}}\in \{2^{K-1},\dots ,2^K-1\}$. Therefore, the $2^{K-1}\times 2^K$ matrix $\varvec{\theta }_{1:(K-1)}$ for $X_1,\dots , X_{K-1}$ is defined as

$$\begin{aligned} \varvec{\theta }_{1:(K-1)} = {\varvec{1}}_{2}^\top \otimes \mathbb {P}_{1:(K-1)}. \end{aligned}$$

(B5)

So, $\varvec{\theta }_{1:(K-1)}$ is a block matrix with the first and last $2^{K-1}$ columns equal to $\mathbb {P}_{1:(K-1)}$. Therefore, we can write $\mathbf{T}_1$ as

(B6)

where

$$\begin{aligned} \mathbf{D}_{11}&= \begin{bmatrix} 1-\theta _{K0}&{}0\\ 0&{} 1-\theta _{K,2^{K-2}} \end{bmatrix}\otimes \mathbf{I}_{2^{K-2}} \end{aligned}$$

(B7)

$$\begin{aligned} \mathbf{D}_{12}&= \begin{bmatrix} 1-\theta _{K,2^{K-1}}&{}0\\ 0&{} 1-\theta _{K,3\cdot 2^{K-2}} \end{bmatrix} \otimes \mathbf{I}_{2^{K-2}} \end{aligned}$$

(B8)

$$\begin{aligned} \mathbf{D}_{21}&= \begin{bmatrix} \theta _{K0}&{}0\\ 0&{} \theta _{K,2^{K-2}} \end{bmatrix} \otimes \mathbf{I}_{2^{K-2}} \end{aligned}$$

(B9)

$$\begin{aligned} \mathbf{D}_{22}&= \begin{bmatrix} \theta _{K,2^{K-1}}&{}0\\ 0&{} \theta _{K,3\cdot 2^{K-2}} \end{bmatrix} \otimes \mathbf{I}_{2^{K-2}}. \end{aligned}$$

(B10)

Recall the determinant of a block matrix

$$\begin{aligned} \mathbf{Y} = \begin{bmatrix} \mathbf{A}&{}\mathbf{B}\\ \mathbf{C}&{}\mathbf{D} \end{bmatrix} \end{aligned}$$

(B11)

is $\det (\mathbf{Y})=\det (\mathbf{A})\det (\mathbf{D}-\mathbf{C}\mathbf{A}^{-1}\mathbf{B})$. Therefore,

(B12)

which is nonzero if the conditions for $a=1$ of Proposition 3 are satisfied.