Bayesian Semiparametric Longitudinal Inverse-Probit Mixed Models for Category Learning

Abstract

Understanding how the adult human brain learns novel categories is an important problem in neuroscience. Drift-diffusion models are popular in such contexts for their ability to mimic the underlying neural mechanisms. One such model for gradual longitudinal learning was recently developed in Paulon et al. (J Am Stat Assoc 116:1114–1127, 2021). In practice, category response accuracies are often the only reliable measure recorded by behavioral scientists to describe human learning. Category response accuracies are, however, often the only reliable measure recorded by behavioral scientists to describe human learning. To our knowledge, however, drift-diffusion models for such scenarios have never been considered in the literature before. To address this gap, in this article, we build carefully on Paulon et al. (J Am Stat Assoc 116:1114–1127, 2021), but now with latent response times integrated out, to derive a novel biologically interpretable class of ‘inverse-probit’ categorical probability models for observed categories alone. However, this new marginal model presents significant identifiability and inferential challenges not encountered originally for the joint model in Paulon et al. (J Am Stat Assoc 116:1114–1127, 2021). We address these new challenges using a novel projection-based approach with a symmetry-preserving identifiability constraint that allows us to work with conjugate priors in an unconstrained space. We adapt the model for group and individual-level inference in longitudinal settings. Building again on the model’s latent variable representation, we design an efficient Markov chain Monte Carlo algorithm for posterior computation. We evaluate the empirical performance of the method through simulation experiments. The practical efficacy of the method is illustrated in applications to longitudinal tone learning studies.

Appendices

Appendix

Appendix A: Proof of Lemma 1

Proof

It is easy to check that the offset parameters $\delta _{s}$ are not identifiable since

$$\begin{aligned} P(d \mid s,\delta _{s},{\mu }_{1:d_{0},s}, {{\textbf{b}}}_{1:d_{0},s})= & {} \int _{\delta _{s}}^{\infty } g(\tau \mid \delta _{s},\mu _{d,s},b_{d,s}) \prod _{d^{\prime } \ne d} \left\{ 1 - G(\tau \mid \delta _{s},\mu _{d^{\prime },s},b_{d^{\prime },s})\right\} d\tau \\= & {} \int _{0}^{\infty } g(\tau \mid 0,\mu _{d,s},b_{d,s}) \prod _{d^{\prime } \ne d} \left\{ 1 - G(\tau \mid 0,\mu _{d^{\prime },s},b_{d^{\prime },s})\right\} d\tau \\= & {} P(d \mid s, 0,{\mu }_{1:d_{0},s},{{\textbf{b}}}_{1:d_{0},s}). \end{aligned}$$

Next we will show that the drift parameters and decision boundaries are not separately identifiable, even if we fix offset parameters to a constant.

First note that Eq. (3) can also be represented as

$$\begin{aligned} \int _{\delta _{s}}^{\infty } \ldots \int _{\delta _{s}}^{\infty } \prod _{d^{\prime }\ne d} g(\tau _{d^{\prime }} \mid {\theta }_{d^{\prime },s}) \int _{\delta _{s}}^{\wedge _{d\ne d^{\prime }}\tau _{d^{\prime }} } g(\tau _{d} \mid {\theta }_{d,s}) d\tau _{d} \prod _{d^{\prime }\ne d} d\tau _{d^{\prime }}. \end{aligned}$$

(A.1)

First observe that $\tau ^{\star }=\wedge _{d^{\prime }\ne d}\tau _{d^{\prime }} =\tau _{-1}^{\star }\wedge \tau _{1}$, where $\tau _{-1}^{\star }=\wedge _{d^{\prime }\ne \{1,d\}}\tau _{d^{\prime }}$. Thus the integral above can be written as

$$\begin{aligned}&\int _{\delta _{s}}^{\infty } \ldots \int _{\delta _{s}}^{\infty } \prod _{d^{\prime }\ne \{1,d\}} g(\tau _{d^{\prime }} \mid {\theta }_{d^{\prime },s}) \left\{ \int _{\delta _{s}}^{\infty } g(\tau _{1} \mid {\theta }_{1,s}) \int _{\delta _{s}}^{\tau _{-1}^{\star } \wedge \tau _{1} } g(\tau _{d} \mid {\theta }_{d,s}) d\tau _{d}\right\} \prod _{d^{\prime }\ne \{1,d\}} d\tau _{d^{\prime }}\\&\quad =\int _{\delta _{s}}^{\infty } \ldots \int _{\delta _{s}}^{\infty } \prod _{d^{\prime }\ne \{1,d\}} g(\tau _{d^{\prime }} \mid {\theta }_{d^{\prime },s}) \left\{ \int _{\delta _{s}}^{\tau _{-1}^{\star }} g(\tau _{d} \mid {\theta }_{d,s}) \int _{\tau _{d}}^{\infty } g(\tau _{1} \mid {\theta }_{1,s}) d\tau _{1} d\tau _{d} \right\} \prod _{d^{\prime }\ne \{1,d\}} d\tau _{d^{\prime }}. \end{aligned}$$

Proceeding sequentially one can show that the integral above is the same as in (3).

Using the above we express the probability in (3) as in (A.1). As the offset parameter $\delta _{s}$ is already shown to be not identifiable, we need to fix the same. Without loss of generality, we fix the offset parameter at 0. The probability density function of inverse Gaussian distribution, with parameters ${\theta }_{d^{\prime },s}=(\mu _{d^{\prime },s},b_{d^{\prime },s})$ evaluated at $\tau _{d^{\prime }}$, $g(\tau _{d^{\prime }}\mid {\theta }_{d^{\prime },s})$ can be obtained from (1) by replacing $\delta _{s}=0$ and $d=d^{\prime }$.

Consider the transformation of $\tau _{d^{\prime }}$ to $\tau _{d^{\prime }}^{\star }$ as $\tau _{d^{\prime }}=c^2\tau _{d^{\prime }}^{\star }$, for some constant $c>0$, and for all $d^{\prime }$. Further, define $b_{d^{\prime },s}^{\star }=b_{d^{\prime },s}/c$ and $\mu _{d^{\prime },s}^{\star }=c\mu _{d^{\prime },s}$, for all $d^{\prime }$. Then observe that

$$\begin{aligned} g(\tau _{d^{\prime }}\mid {\theta }_{d^{\prime },s}) d\tau _{d^{\prime }}&= (2\pi )^{-1/2} b_{d^{\prime },s}^{\star } (\tau _{d^{\prime }}^{\star })^{-3/2}\exp \left\{ -(2\tau _{d^{\prime }}^{\star })^{-1} \left( b_{d^{\prime },s}^{\star } -\mu _{d^{\prime },s}^{\star } \tau _{j}^{\star } \right) ^{2} \right\} d\tau _{d^{\prime }}^{\star }\\&=g(\tau _{d^{\prime }}^{\star }\mid {\theta }_{d^{\prime },s}^{\star } ) d\tau _{d^{\prime }}^{\star }, \end{aligned}$$

where $g(\tau _{d^{\prime }}^{\star }\mid {\theta }_{d^{\prime },s}^{\star } )$ is the pdf of inverse Gaussian distribution with parameters $\mu _{d^{\prime },s}^{\star }$ and $b_{d^{\prime },s}^{\star }$, evaluated at the point $\tau _{d^{\prime }}^{\star }$.

Applying the transformation on $\tau _{d^{\prime }}$ for all $d^{\prime }$ we get that the integral in (A.1) with $\delta _{s}=0$ is same as

$$\begin{aligned} \int _{0}^{\infty } \ldots \int _{0}^{\infty } \prod _{d^{\prime }\ne d} g(\tau _{d^{\prime }}^{\star } \mid {\theta }_{d^{\prime },s}^{\star }) \int _{0}^{\wedge _{d^{\prime }\ne d}\tau _{d^{\prime }}^{\star } } g(\tau _{d}^{\star } \mid {\theta }_{d,s}^{\star }) d\tau _{d}^{\star } \prod _{d^{\prime }\ne d} d\tau _{d^{\prime }}^{\star }. \end{aligned}$$

As c is arbitrary, this shows that the drifts and boundaries are not separately estimable. $\square$

Appendix B: Proof of Theorem 1

Proof

Let ${{\textbf{P}}}({\mu }_{1:d_{0},s})=\{p_{1}({\mu }_{1:d_{0},s}), \dots , p_{d_{0}}({\mu }_{1:d_{0},s}) \}^\textrm{T}$ be the function, given by (4), from $\mathcal{S}_{0,k}$ to unit probability simplex $\Delta ^{d_{0}-1}$. For notational simplicity, we write ${\mu }_{1:d_{0},s} = {\mu }= (\mu _{1}, \dots , \mu _{d_{0}})^\textrm{T}$. We first find the matrix of partial derivative $\nabla {{\textbf{P}}}$ with respect to ${\mu }$.

For ${\mu }\in \mathcal{S}_{0,k}$, $\textbf{1}^{T} {\mu }=k$, and hence the probability reduces to

$$\begin{aligned} p_{d}\left( {\mu }\right) = \frac{\left( be^{b}\right) ^{d_{0}}}{ (2 \pi )^{d_{0}/2}}\int _{0}^{\infty } \int _{\tau _{d}}^{\infty } \cdots \int _{\tau _{d}}^{\infty } |{\tau }|^{-3/2} \exp \left\{ -\frac{1}{2} \left( \textbf{1}^{T} {\tau }^{-1} \textbf{1} + {\mu }^{T} {\tau }{{\mu }} \right) \right\} d{\tau }_{-d} d\tau _{d}, \end{aligned}$$

for $d=1,\ldots , d_{0}$, where ${\tau }=\textrm{diag}(\tau _{1}, \ldots , \tau _{d_{0}})$, and ${\tau }_{-d}$ is the sub-vector of ${\tau }$ excluding the d-th element. Next, differentiating $p_{d}\left( {\mu }\right)$ with respect to ${\mu }$, we get

$$\begin{aligned} \frac{\partial {p_{d}\left( {\mu }\right) }}{\partial {\mu }}= & {} \displaystyle \frac{\left( be^{b}\right) ^{d_{0}}}{ (2 \pi )^{d_{0}/2}}\int _{0}^{\infty } \int _{\tau _{d}}^{\infty } \cdots \int _{\tau _{d}}^{\infty } |{\tau }|^{-3/2} \left( -{\tau }{\mu }\right) \exp \left\{ -\frac{1}{2} \left( \textbf{1}^{T} {\tau }^{-1} \textbf{1} + {\mu }^{T} {\tau }{{\mu }} \right) \right\} d{\tau }_{-d} d\tau _{d}, \\= & {} \begin{bmatrix} \mu _{1} \eta _{2}&\quad \cdots&\quad \mu _{d-1} \eta _{2}&\quad \mu _{d} \eta _{1}&\quad \mu _{d+1} \eta _{2}&\quad \cdots&\quad \mu _{d_{0}} \eta _{2} \end{bmatrix}^{T}, \end{aligned}$$

where $\eta _{1} = - E\left\{ \tau _{1} {\mathbb {I}}\left( \tau _{2}> \tau _{1}, \ldots , \tau _{d_{0}}>\tau _{1}\right) \left| {\mu }\right. \right\}$, and $\eta _{2} = -E\left\{ \tau _{2} {\mathbb {I}}\left( \tau _{2}> \tau _{1}, \ldots , \tau _{d_{0}}>\tau _{1}\right) \left| {\mu }\right. \right\}$, and ${\mathbb {I}}(A)$ is the indicator function of the event A. Here the expectation is considered under the joint distribution of $\left( \tau _{1}, \ldots , \tau _{d}\right)$, which is independent inverse Gaussian. Clearly $\eta _{1}>\eta _{2}>0$.

From the above derivation, it is easy to obtain that

$$\begin{aligned} \nabla {{\textbf{P}}}\left( {\mu }\right) = \begin{bmatrix} \mu _{1} \eta _{1} &{}\quad \mu _{2} \eta _{2} &{}\quad \cdots &{}\quad \mu _{d_{0}} \eta _{2} \\ \mu _{1} \eta _{2} &{}\quad \mu _{2} \eta _{1} &{}\quad \cdots &{}\quad \mu _{d_{0}} \eta _{2} \\ \vdots &{}\quad \vdots &{}\quad \cdots &{}\quad \vdots \\ \mu _{1} \eta _{2} &{}\quad \mu _{2} \eta _{2} &{}\quad \cdots &{}\quad \mu _{d_{0}} \eta _{1} \end{bmatrix} = {{\textbf{M}}}\left\{ \left( \eta _{1}-\eta _{2}\right) I+\eta _{2} \textbf{1} \textbf{1}^{T} \right\} , \end{aligned}$$

where ${{\textbf{M}}}=\textrm{diag}\left( \mu _{1}, \ldots , \mu _{d_{0}}\right)$.

Now, suppose there exists ${\mu }$ and ${\nu }$ in $\mathcal{S}_{k}$ such that ${\mu }\ne {\nu }$ and ${{\textbf{P}}}\left( {\mu }\right) = {{\textbf{P}}}\left( {\nu }\right)$. Define ${\gamma }: [0,1] \rightarrow {\mathbb {R}}^{d_{0}}$ such that ${\gamma }(t)= {\mu }+ t \left( {\nu }- {\mu }\right)$, $t\in [0,1]$. Further, define $h(t)= \langle {{\textbf{P}}}\left( {\gamma }(t) \right) - {{\textbf{P}}}\left( {\mu }\right) , {\nu }- {\mu }\rangle$, as the cross-product of ${{\textbf{P}}}\left( {\gamma }(t) \right) - {{\textbf{P}}}\left( {\mu }\right)$ and ${\nu }-{\mu }$. Then $h(1)=h(0)=0$ under the proposition that ${{\textbf{P}}}\left( {\mu }\right) = {{\textbf{P}}}\left( {\nu }\right)$. Therefore, by the Mean Value Theorem, as ${\mu }\ne {\nu }$, there exists some point $c\in (0,1)$ such that $\left. \partial h(t) / \partial t \right| _{t=c} =0$. Now,

$$\begin{aligned} \frac{\partial h(t) }{ \partial t}= & {} \sum _{d^{\prime }=1}^{d_{0}} \left( \nu _{d^{\prime }} - \mu _{d^{\prime }} \right) \frac{\partial }{\partial t}\left[ p_{d^{\prime }} \left\{ {\gamma }(t) \right\} - p_{d^{\prime }} \left( {\mu }\right) \right] \\= & {} \sum _{d^{\prime }=1}^{d_{0}} \left( \nu _{d^{\prime }} - \mu _{d^{\prime }} \right) \left\{ \frac{\partial }{\partial {\gamma }} p_{d^{\prime }} \left( {\gamma }\right) \right\} ^{T} \frac{\partial {\gamma }(t)}{\partial t}\\= & {} \left( {\nu }-{\mu }\right) ^{T} \nabla {{\textbf{P}}}\{{\gamma }(t)\} \left( {\nu }-{\mu }\right) \\= & {} \left( \eta _{1}-\eta _{2}\right) \left( {\nu }-{\mu }\right) ^{T} {\Gamma }(t) \left( {\nu }-{\mu }\right) + \eta _{2} \left( {\nu }-{\mu }\right) ^{T} M \textbf{1} \textbf{1}^{T} \left( {\nu }-{\mu }\right) \\= & {} \left( \eta _{1}-\eta _{2}\right) \left( {\nu }-{\mu }\right) ^{T} {\Gamma }(t) \left( {\nu }-{\mu }\right) , \end{aligned}$$

as $\textbf{1}^{T} \left( {\nu }-{\mu }\right) =0$, where ${\Gamma }(t)=\textrm{diag}\{ {\gamma }(t) \}$.

As every component of ${\mu }$ and ${\nu }$ is positive, for any $c\in (0,1)$, the matrix ${\Gamma }(c)$ is positive definite. Further, as $\eta _{1}>\eta _{2}$, $\left. \partial h(t) / \partial t \right| _{t=c} =0$ only if ${\mu }={\nu }$, which contradicts the proposition. $\square$

Appendix C: Algorithm for Minimal Distance Mapping

The problem of finding projection of a point ${\mu }$ onto the space $\mathcal{S}_{k,\varepsilon }$ is equivalent to the following nonlinear optimization problem:

$$\begin{aligned} \textrm{minimize}_{{{\textbf{w}}}} \Vert {{{\textbf{w}}}} -{\mu }\Vert ^2 \quad \text{ such } \text{ that } \quad \sum _{i=1}^{d_{0}} w_{i}=k,\quad w_{i} \ge \varepsilon . \end{aligned}$$

Duchi et al. (2008, Algorithm 1) provides a solution to the problem of projection of a given point ${\mu }$ onto the space $\mathcal{S}_{k,\varepsilon }$ for $\varepsilon =0$, which is modified for any given $\varepsilon$ below.

Appendix D: Proof of Lemma 3

Proof

We consider the unconditional distribution of $\tau _{1:d_{0}}$, given the parameters $\mu _{1:d_{0}}$ as the proposal distribution, g. Clearly, the proposal distribution g and the target conditional joint distribution f satisfies $f(\tau _{1:d_{0}}|\mu _{1:d_{0}})/g(\tau _{1:d_{0}}|\mu _{1:d_{0}})\le M$, where $M^{-1}=P\left( \tau _{d} \le \tau _{1:d_{0}}\right)$. Therefore, for any random sample $U\sim U(0,1)$, $f(\tau _{1:d_{0}}|\mu _{1:d_{0}})\ge M U g(\tau _{1:d_{0}}|\mu _{1:d_{0}})$ if the sample satisfies the condition $\tau _{d} \le \tau _{1:d_{0}}$, and $f(\tau _{1:d_{0}}|\mu _{1:d_{0}})< M U g(\tau _{1:d_{0}}|\mu _{1:d_{0}})$ otherwise. Hence, by Lemma 2.3.1 of Robert and Casella (2004), algorithm above produces samples from the target distribution. $\square$