Extended Asymptotic Identifiability of Nonparametric Item Response Models

Abstract

Nonparametric item response models provide a flexible framework in psychological and educational measurements. Douglas (Psychometrika 66(4):531–540, 2001) established asymptotic identifiability for a class of models with nonparametric response functions for long assessments. Nevertheless, the model class examined in Douglas (2001) excludes several popular parametric item response models. This limitation can hinder the applications in which nonparametric and parametric models are compared, such as evaluating model goodness-of-fit. To address this issue, We consider an extended nonparametric model class that encompasses most parametric models and establish asymptotic identifiability. The results bridge the parametric and nonparametric item response models and provide a solid theoretical foundation for the applications of nonparametric item response models for assessments with many items.

Appendix

We present all the lemmas in Section A and provide their proofs in Section B. The proofs of propositions are provided in Section C.

1.1 A. Lemmas

Lemma 1

Consider an integer k such that $\theta _k\in (\alpha ,\beta )$. There exist constants $\tilde{C}_{\alpha \beta }>0$ and $N_{\alpha \beta }$ independent with (n, i) such that when $n \geqslant N_{\alpha \beta }$,

$$\begin{aligned} P\left( \bar{Y}_{n,-i}=\frac{k}{n-1}\right) \geqslant \tilde{C}_{\alpha \beta }n^{-1}. \end{aligned}$$

Lemma 2

Under the conditions of Theorem 2, there exist constants $\tilde{C}_{\alpha \beta ,1}$ and $\tilde{C}_{\alpha \beta ,2}$ independent with (n, i) such that

$$\begin{aligned} \left| 1-P\left( \Theta \in I_{\delta }\mid \mathcal {E}_{n,k} \right) \right| \leqslant \frac{ n\delta \exp (-n\tilde{C}_{\alpha \beta ,1} )}{\tilde{C}_{\alpha \beta ,2}}. \end{aligned}$$

Lemma 3

(Hoeffding inequality of bounded variables) For any (n, i), and $m>0$,

$$\begin{aligned} P(|\bar{Y}_{n,-i}-\bar{P}_{n,-i}(\theta )|> m \mid \Theta =\theta )\leqslant 2\exp [- 2(n-1)m^2]. \end{aligned}$$

1.2 B. Proofs of Lemmas

1.2.1 B1. Proof of Lemma 1

Let $\mu _{\theta }=\sum _{j\ne i} P_{n,j}(\theta )$ and $\sigma ^2_{\theta }=\sum _{j\ne i} P_{n,j}(\theta )(1-P_{n,j}(\theta ))$, which are the mean and variance of $(n-1)\bar{Y}_{n,-i}$ conditioning on $\Theta =\theta$, respectively. By applying a bound on the normal approximation for the distribution of a sum of independent Bernoulli variables (Mikhailov, 1994)

$$\begin{aligned} \left| P\left[ (n-1)\bar{Y}_{n,-i}=k \mid \Theta =\theta \right] -\frac{1}{\sigma _\theta \sqrt{2 \pi }} e^{-\frac{\left( k-\mu _\theta +1 / 2\right) ^2}{2 \sigma _\theta ^2}}\right| \le \frac{c}{\sigma _\theta ^2} \end{aligned}$$

(12)

for some universal constant c.

Define the intervals $I_{n,1/2}=(\theta _k-n^{-1/2}, \theta _k+n^{-1/2})$ and $\tilde{I}_{n,1/2}=I_{n,1/2}\cap (\alpha /2,(1+\beta )/2)$. For $\theta \in \tilde{I}_{n,1/2}$ and $\theta _k \in (\alpha ,\beta )$,

$$\begin{aligned} \frac{|k-\mu _{\theta }|}{n-1}=|\bar{P}_{n,-i}(\theta _k)-\bar{P}_{n,-i}(\theta )|\leqslant {M}_{\alpha \beta ,2} |\theta _k-\theta | < {M}_{\alpha \beta ,2}n^{-1/2}, \end{aligned}$$

(13)

where in the second inequality, ${M}_{\alpha \beta ,2}$ is a constant that is independent with (n, i) by Condition 3. Moreover, by Condition 4, there exist positive constants $L_{\alpha \beta }$ and $U_{\alpha \beta }$ that are independent with (n, i) such that

$$\begin{aligned} L_{\alpha \beta }<\sigma ^2_{\theta }/(n-1) < U_{\alpha \beta }. \end{aligned}$$

(14)

Therefore, by (12), (13), and (14), for $\theta \in \tilde{I}_{n,1/2}$,

$$\begin{aligned} P\left( \bar{Y}_{n,-i}=\frac{k}{n-1} \ \Big | \ \theta \right)>&~ \frac{1}{\sqrt{2\pi \sigma ^2_{\theta } }} e^{-\frac{\left( k-\mu _\theta +1 / 2\right) ^2}{2\sigma _\theta ^2}} - \frac{c}{\sigma ^2_{\theta }} \\>&~ \frac{1}{\sqrt{2U_{\alpha \beta }(n-1)\pi }} e^{-\frac{\left( {M}_{\alpha \beta ,2}(n-1)n^{-1/2} + \frac{1}{2}\right) ^2}{2 L_{\alpha \beta }(n-1)}} - \frac{c}{L_{\alpha \beta }(n-1)}\\ >&~ \frac{\tilde{C}_{\alpha \beta }}{2n^{1/2}}, \end{aligned}$$

where $\tilde{C}_{\alpha \beta }>0$ is a constant that is independent with (n, i). Therefore,

$$\begin{aligned} P\left( \bar{Y}_{n,-i}=\frac{k}{n-1} \right) \geqslant&~\int _{\theta \in \tilde{I}_{n,1/2}} P\left( \bar{Y}_{n,-i}=\frac{k}{n-1} \ \Big | \ \theta \right) \textrm{d}\theta > \int _{\theta \in \tilde{I}_{n,1/2}}\frac{\tilde{C}_{\alpha \beta }}{2n^{1/2}}\textrm{d}\theta . \end{aligned}$$

(15)

There exists $N_{\alpha \beta }$ independent with i such that when $n\geqslant N_{\alpha \beta }$, $\tilde{I}_{n,1/2}=I_{n,1/2}$ by $\theta _k\in (\alpha ,\beta )$. Then by (15),

$$\begin{aligned} P\left( \bar{Y}_{n,-i}=\frac{k}{n-1} \right) >\int _{\theta \in {I}_{n,1/2}}\frac{\tilde{C}_{\alpha \beta }}{2n^{1/2}}\textrm{d}\theta = \tilde{C}_{\alpha \beta }n^{-1}. \end{aligned}$$

1.2.2 B2. Proof of Lemma 2

To prove Lemma 2, we note that

$$\begin{aligned} 1-P(\Theta \in I_{\delta }\mid \mathcal {E}_{n,k})={P\left( 0<\Theta<\delta \mid \mathcal {E}_{n,k}\right) +P\left( 1-\delta<\Theta <1 \mid \mathcal {E}_{n,k}\right) }. \end{aligned}$$

We next derive an upper bound of $P\left( 0<\Theta <\delta \mid \mathcal {E}_{n,k}\right)$, and a similar upper bound can be obtained for $P\left( 1-\delta<\Theta <1 \mid \mathcal {E}_{n,k}\right)$ following a similar analysis.

By $P(0<\Theta<\delta \mid \mathcal {E}_{n,k})= P(0<\Theta <\delta ,\ \mathcal {E}_{n,k})/P(\mathcal {E}_{n,k})$, and the lower bound of $P(\mathcal {E}_{n,k})$ in Lemma 1, it suffices to derive an upper bound of $P(0<\Theta <\delta ,\ \mathcal {E}_{n,k})$ below. In particular,

$$\begin{aligned} P\left( 0<\Theta<\delta ,\ \mathcal {E}_{n,k}\right) =\int P\biggr (\bar{Y}_{n,-i}=\frac{k}{n-1}\mid \Theta =\theta \biggr ) \, \textrm{I}(0<\theta <\delta )\textrm{d}\theta . \end{aligned}$$

(16)

By $\bar{P}_{n,-i}(\theta _k)=k/(n-1)$,

$$\begin{aligned} \bar{Y}_{n,-i}=\frac{k}{n-1}\quad \Leftrightarrow \quad \bar{Y}_{n,-i}-\bar{P}_{n,-i}(\theta )=[\bar{P}_{n,-i}(\theta _k)-\bar{\kappa }_{-i}]-[\bar{P}_{n,-i}(\theta )-\bar{\kappa }_{-i}], \end{aligned}$$

(17)

where we define $\bar{\kappa }_{-i}=\sum _{j\ne i} \kappa _j/(n-1)$. By $0<\alpha /2<\alpha<\theta _k<\beta$ and Conditions 3 and 4, we have $\bar{P}_{n,-i}(\theta _k) \geqslant \bar{P}_{n,-i}(\alpha )$ and $\bar{P}_{n,-i}(\alpha /2)\geqslant \bar{\kappa }_{-i}.$ Therefore,

$$\begin{aligned} \bar{P}_{n,-i}(\theta _k)-\bar{\kappa }_{-i} \geqslant&~\bar{P}_{n,-i}(\alpha )-\bar{P}_{n,-i}(\alpha /2) = \bar{P}_{n,-i}'(\tilde{\alpha } )\alpha /2\geqslant \tilde{m}_{\alpha } \alpha /2, \end{aligned}$$

(18)

where the second equation is obtained by the intermediate value theorem with $\tilde{\alpha }\in (\alpha /2,\alpha )$, and the last inequality is obtained by Condition 3 with $\tilde{m}_{\alpha }$ being a constant independent with (n, i). Let $\tilde{C}_{\alpha } = \tilde{m}_{\alpha } \alpha /2$. By Condition 4, there exists $\delta _{\alpha }>0$ such that for $\theta <\delta \leqslant \delta _{\alpha }$,

$$\begin{aligned} \bar{P}_{n,-i}(\theta ) -\bar{\kappa }_{-i}< \bar{P}_{n,-i}(\delta ) -\bar{\kappa }_{-i} < \tilde{C}_{\alpha } /2. \end{aligned}$$

(19)

Combining (17), (18), and (19),

$$\begin{aligned}&~P\left( \bar{Y}_{n,-i}=\frac{k}{n-1} \mid \Theta = \theta \right) \\&\quad = ~ P\left( \bar{Y}_{n,-i}-\bar{P}_{n,-i}(\theta )=[\bar{P}_{n,-i}(\theta _k)-\bar{\kappa }_{-i}]-[\bar{P}_{n,-i}(\theta )-\bar{\kappa }_{-i}]\mid \Theta = \theta \right) \\&\quad \leqslant ~P\left( \bar{Y}_{n,-i}-\bar{P}_{n,-i}(\theta )\geqslant \tilde{C}_{\alpha } /2 \mid \Theta =\theta \right) \\&\quad \leqslant ~ 2\exp [-(n-1)\tilde{C}_{\alpha }^2 /2 ], \end{aligned}$$

where the last inequality follows by Lemma 3. By the above inequality and (16),

$$\begin{aligned} P\left( 0<\Theta <\delta \mid \mathcal {E}_{n,k}\right) \leqslant \frac{2\delta \exp [-(n-1)\tilde{C}_{\alpha }^2 /2 ]}{ P(\mathcal {E}_{n,k}) }\leqslant \frac{2n\delta \exp [-(n-1)\tilde{C}_{\alpha }^2 /2 ]}{\tilde{C}_{\alpha \beta }}, \end{aligned}$$

where the second inequality is obtained by Lemma 1. A similar upper bound can be obtained for $P (1-\delta<\Theta <1 \mid \mathcal {E}_{n,k}\ )$ too. Lemma 2 is proved.

1.3 C. Proofs of Propositions

1.3.1 C1. Proof of Proposition 1

Let $x=\Phi ^{-1}(\theta )$. We can equivalently write $P'_{n,i}(\theta )=h_{n,i}(x)$, where

$$\begin{aligned} h_{n,i}(x)=\frac{a_{n,i} \phi [a_{n,i}(x-b_{n,i})] }{ \phi (x)} = a_{n,i}\exp \left[ -\frac{1}{2}(a_{n,i}^2-1)x^2+a_{n,i}^2b_{n,i}\left( x-\frac{1}{2}\right) \right] . \end{aligned}$$

Note that $\theta \downarrow 0$ and $\theta \uparrow 0$ correspond to $x \rightarrow -\infty$ and $x\rightarrow +\infty$, respectively. Let $p_{+}=\lim _{\theta \rightarrow 1}P_{n,i}'(\theta )=\lim _{x\rightarrow +\infty } h_{n,i}(x)$ and $p_{-}=\lim _{\theta \rightarrow 0}P_{n,i}'(\theta )=\lim _{x\rightarrow -\infty } h_{n,i}(x)$. As $h_{n,i}(x)=0$ when $a_{n,i}=0$, it suffices to consider $a_{n,i}\ne 0$ below. In particular,

$$\begin{aligned} (p_{-}, p_{+})={\left\{ \begin{array}{ll} (+\infty , +\infty ) &{} \text { when } 0<a_{n,i}^2<1 \\ (0, 0) &{} \text { when } a_{n,i}^2>1\\ (0, +\infty ) &{} \text { when } a_{n,i}^2=1, b_{n,i}>0\\ ( +\infty , 0) &{} \text { when } a_{n,i}^2=1, b_{n,i}<0\\ (1, 1) &{} \text { when } a_{n,i}^2=1, b_{n,i}=0. \end{array}\right. } \end{aligned}$$

This suggests that $P'_{n,i}(\theta )$ cannot be uniformly bounded on (0, 1) when $a_{n,i}^2\ne 1$ or $b_{n,i}\ne 0.$

On the other hand, when $\theta \in [\alpha , \beta ]$, $x\in [\Phi ^{-1}(\alpha ), \Phi ^{-1}(\beta )]$ with the two end points bounded away from $-\infty$ and $+\infty$. Therefore, when $a_{n,i}\in [1/C, C]$ and $|b_{n,i}|\in [1/C', C']$, there exist $0< m_{\alpha \beta }< M_{\alpha \beta } <\infty$ such that $P'_{n,i}(\theta ) = h_{n,i}(x) \in (m_{\alpha \beta }, M_{\alpha \beta })$. Thus Condition 3 is satisfied.

We next prove that Condition 4 is also satisfied. When $\epsilon \in (0,1/2)$, $\Phi ^{-1}(\epsilon )<0$, and then we set $l_{\epsilon }=\Phi \left[ C_a\Phi ^{-1}(\epsilon ) - C_b\right] .$ When $\theta \in [0,l_{\epsilon }]$,

$$\begin{aligned} P_{n,i}(\theta )\leqslant&~ P_{ni}(l_{\epsilon }) =\Phi \big [a_{ni}(\Phi ^{-1}(l_{\epsilon }) - b_{ni})\big ] \leqslant \Phi \{ a_{ni}[C_a\Phi ^{-1}(\epsilon )-C_b+ |b_{ni}| ] \}\\ \leqslant&~ \Phi \{ a_{ni}[C_a\Phi ^{-1}(\epsilon )-C_b+C_b] \} \leqslant \Phi \left[ \frac{1}{C_a} C_a \Phi ^{-1}(\epsilon )\right] = \epsilon . \end{aligned}$$

When $\epsilon \in (1/2,1)$, $\Phi ^{-1}(\epsilon )>0$, and we set $l_{\epsilon }=\Phi \left[ C_a^{-1}\Phi ^{-1}(\epsilon ) - C_b\right] .$ Then we can obtain $P_{n,i}(\theta )\leqslant \epsilon$ similarly to the above analysis. Following similar analysis, we can construct $u_{\epsilon }$ so that $1-P_{n,i}(\theta )\leqslant \epsilon$ for $\theta \in [u_{\epsilon },1]$. In brief, Condition 4 is satisfied.

1.3.2 C2. Proof of Proposition 2

Let $x=\Phi ^{-1}(\theta )$. We can equivalently write $P'_{n,i}(\theta )=h_{n,i}(x)$, where

$$\begin{aligned} h_{n,i}(x)=&~(d_{n,i}-c_{n,i})a_{n,i}\frac{ g'\left[ a_{n,i}(x-b_{n,i})\right] }{\phi (x)} \\ =&~ (d_{n,i}-c_{n,i})a_{n,i} \frac{ e^{x^2/2}/[e^{a_{n,i}(x-b_{n,i})}+e^{-a_{n,i}(x-b_{n,i})}]}{1+2/[e^{a_{n,i}(x-b_{n,i})}+e^{-a_{n,i}(x-b_{n,i})}]} \end{aligned}$$

which is obtained by

$$\begin{aligned} g'(x)= \frac{1/(e^x+e^{-x})}{1+2/(e^x+e^{-x})}. \end{aligned}$$

Note that $\theta \downarrow 0$ and $\theta \uparrow 0$ correspond to $x \rightarrow -\infty$ and $x\rightarrow +\infty$, respectively. When $a_{n,i}=0$, $h_{n,i}(x)=0$. When $a_{n,i}\ne 0$, $e^{a_{n,i}(x-b_{n,i})}+e^{-a_{n,i}(x-b_{n,i})}\rightarrow +\infty$ and $e^{x^2/2}$ as $|x|\rightarrow +\infty$. Moreover, as the quadratic term diverges faster than the linear term, we know $e^{x^2/2}/[e^{a_{n,i}(x-b_{n,i})}+e^{-a_{n,i}(x-b_{n,i})}]\rightarrow \infty$. Thus, $h_{n,i}(x)\rightarrow +\infty$.

On the other hand, when $\theta \in [\alpha , \beta ]$, $x\in [\Phi ^{-1}(\alpha ), \Phi ^{-1}(\beta )]$ with the two end points bounded away from $-\infty$ and $+\infty$. Therefore, under the conditions in (ii) of Proposition 2, there exist $0< m_{\alpha \beta }< M_{\alpha \beta } <\infty$ such that $P'_{n,i}(\theta ) = h_{n,i}(x) \in (m_{\alpha \beta }, M_{\alpha \beta })$. Thus, Condition 3 is satisfied.

We next prove that Condition 4 is also satisfied. When $\epsilon$ is small such that $g^{-1}(\epsilon /C_{c,d})<0$, we set $l_{\epsilon }=\Phi \left[ C_ag^{-1}(\epsilon /C_{c,d}) - C_b\right] .$ Then for $\theta \in [0,l_{\epsilon }]$,

$$\begin{aligned} P_{n,i}(\theta )-c_{n,i} \leqslant P_{n,i}(l_{\epsilon })-c_{n,i} =&~ \left( d_{n,i}-c_{n,i}\right) g\left( a_{n,i}[C_ag^{-1}(\epsilon /C_{c,d}) - C_b - b_{n,i} ] \right) \\ \leqslant&~ \left( d_{n,i}-c_{n,i}\right) g\left( a_{n,i}[C_ag^{-1}(\epsilon /C_{c,d}) - C_b + |b_{n,i}|] \right) \\ \overset{(i1)}{\leqslant }\&~ \left( d_{n,i}-c_{n,i}\right) g[g^{-1}(\epsilon /C_{c,d})] \leqslant \epsilon , \end{aligned}$$

where the inequality (i1) above is obtained by $a_{n,i}C_a\geqslant 1 > 0$ and $|b_{n,i}|-C_b\leqslant 0$. When $\epsilon$ is large such that $g^{-1}(\epsilon /C_{c,d})>0$, we set $l_{\epsilon }=\Phi \left[ C_a^{-1}g^{-1}(\epsilon /C_{c,d}) - C_b\right] \!,$ and then $P_{n,i}(\theta )-c_{n,i}\leqslant \epsilon$ can be obtained similarly. Following similar analysis, we can construct $u_{\epsilon }$ so that $d_{n,i}-P_{n,i}(\theta ) \leqslant \epsilon$ for $\theta \in [u_{\epsilon },1]$. In brief, Condition 4 is satisfied.