A Marginalized Overdispersed Location Scale Model for Clustered Ordinal Data

Abstract

Overdispersion and intra cluster correlation are two important issues in clustered categorical/ordinal data and failure to account for them can result in misleading inferences. Generalized estimating equations and mixed effects models are two common frameworks for analyzing clustered data which are recently combined and extended to the marginalized random effects model. The location scale models are a different extension of the mixed effects models that furthermore allow the variance to vary as a function of covariates. In this paper, we extend a marginalized location scale model for longitudinal ordinal responses by allowing a log-linear model for variance components that facilitates both population-averaged and subject-specific interpretations. We then extend the marginalized location scale model by incorporating an additional random term into the model to handle the overdispersion aspect of the data. We conduct extensive simulation studies to investigate the statistical properties of the maximum likelihood estimators of the model parameters. We illustrate this methodology using a dataset on a children’s growth failure.

Appendices

Appendix

We provide a proof of Theorem 2 below. Since, Theorem 1 can be obtained by essentially replacing 𝜃 by the constant 1, a separate proof of Theorem 1 is not provided.

The following lemma is need for the proof of Theorem 2.

Lemma 1.

For normally distributed random variable b _i , $\int {\Phi } \left (\frac {{\Delta }_{ijk}+b_{i}}{\sigma _{\varepsilon _{ij}}}\right ) f(b_{i})$ $db_{i} =\sigma _{\varepsilon _{ij}} {\Phi } \left (\frac {{\Delta }_{ijk}}{\sqrt {\sigma _{\varepsilon _{ij}}^{2}+\sigma _{b}^{2}}} \right )$ .

Proof of Lemma 1.

$\int {{\Phi }{(\frac {{\Delta }_{ijk}+b_{i}}{\sigma _{\varepsilon _{ij}}}\mathrm ) } f{(b_{i})} db_{i}} \mathrm {=}\sigma _{\varepsilon _{ij}}{\Phi } (\frac {{{\Delta }}_{ijk}}{\sqrt {\sigma _{\varepsilon _{ij}}^{2}+{\sigma _{b}^{2}}}} \mathrm )$. □

By definition,

$${\Phi} (x)=\int\limits_{-\infty}^{x} {\frac{1}{\sqrt{2\pi} } exp {\left( \frac{-t^{2}}{2}\right)}dt.} $$

Hence

$$\begin{array}{@{}rcl@{}} \int {{\Phi} \left( \frac{{\Delta}_{ijk}+b_{i}}{\sigma_{\varepsilon_{ij}}}\right) f(b_{i}) db_{i}} &=&\int\limits_{-\infty}^{+\infty}{\int\limits_{-\infty }^{\frac{{\Delta}_{ijk}+b_{i}}{\sigma _{\varepsilon_{ij}}}} {\frac{1}{\sqrt {2\pi} } {exp}\left( \frac{-t^{2}}{2}\right) . \frac{1}{\sqrt {2\pi}s}{exp}\left( \frac{-{b_{i}^{2}}}{2\sigma_{b}^{2}}\right)dt db_{i},}}\\ &=&\int\limits_{-\infty}^{+\infty} {\int\limits_{-\infty}^{\frac{{\Delta}_{ijk}+b_{i}}{\sigma_{\varepsilon_{ij}}}} {{\left( \frac{1}{\sqrt{2\pi} }\right)}^{2}\frac{1}{\sigma_{b}} {exp}\left\{\frac{-1}{2} [{b_{i}^{2}} \sigma_{b}^{-2}+(t^{2})]\right\}} dt} db_{i}. \end{array} $$

Here we make a change of variable $t=\frac {z+b_{i}}{\sigma _{\varepsilon _{ij}}}$ to conclude that the above expression equals

$$\int\limits_{-\infty}^{+\infty} {\int\limits_{-\infty}^{{\Delta}_{ijk}} {{\left( \frac{1}{\sqrt{2\pi} } \right)}^{2}\frac{1}{\sigma_{b}} {exp}\left\{\frac{-1}{2} \left[{b_{i}^{2}}\sigma_{b}^{-2}+ {\left( \frac{z+b_{i}}{\sigma_{\varepsilon_{ij}}}\right)}^{2}\right]\right\}} \frac{1}{\sigma_{\varepsilon_{ij}}}dz} db_{i}. $$

Furthermore, letting $\acute {z}=\frac {z}{\sigma _{\varepsilon _{ij}}}$ and $\acute {b}_{i}=\frac {b_{i}}{\sigma _{\varepsilon _{ij}}}$ we see that the above expression further equals

$$\int\limits_{-\infty}^{+\infty}{\int\limits_{-\infty}^{{\Delta}_{ijk}} {{\left( \frac{1}{\sqrt {2\pi} } \right)}^{2}\frac{1}{\sigma_{b}} {exp}\left\{\frac{-1}{2} \left[\acute{b}_{i}^{2} \sigma _{\varepsilon_{ij}}^{2} \sigma_{b}^{-2}+ {(\acute{z}+\acute{b}_{i})}^{2}\right]\right\}} \frac{1}{\sigma_{\varepsilon_{ij}}}\sigma_{\varepsilon_{ij}}d\acute{z}} \sigma_{\varepsilon _{ij}}d\acute{b}_{i}. $$

Further let us to define:

$$\left\{ {\begin{array}{*{20}c} {\begin{array}{*{20}l} A = 1+\frac{\sigma_{\varepsilon_{ij}}^{2}}{{\sigma_{b}^{2}}} \\ B=-\acute{z} {\sigma_{b}^{2}} \end{array}} \\ {\begin{array}{*{20}l} C=\acute{z}^{2}E \\ E=\frac{1}{\sigma_{\varepsilon_{ij}}^{2}+{\sigma_{b}^{2}}} \end{array}} \end{array}} \right. $$

hence the above expression equals

$$\int\limits_{-\infty}^{+\infty}{\int\limits_{-\infty}^{{\Delta}_{ijk}} {{\left( \frac{1}{\sqrt {2\pi} } \right)}^{2}\frac{\sigma_{\varepsilon_{ij}}}{\sigma_{b}} {exp}\left\{\frac{-1}{2} [\acute{b}_{i}^{2}A + 2\acute{z}\acute{b}_{i}+\acute{z}^{2}]\right\}} d\acute{z}} d\acute{b}_{i}. $$

Consider that $\acute {b}_{i}^{2} \sigma _{\varepsilon _{ij}}^{2} \sigma _{b}^{-2}+ {(\acute {z}+\acute {b}_{i})}^{2}\mathrm {=}\acute {b}_{i}^{2}A + 2\acute {z}\acute {b}_{i}+\acute {z}^{2}{=}{{(\acute {b}_{i}{-}B)}}^{2}A\mathrm {+}C$, the above expression equals

$$\int\limits_{-\infty}^{+\infty} {\int\limits_{-\infty}^{{\Delta}_{ijk}} {{\left( \frac{1}{\sqrt {2\pi} } \right)}^{2}\frac{\sigma_{\varepsilon_{ij}}}{\sigma_{b}} {exp}\left\{\frac{-1}{2} \frac{{(\acute{b}_{i}-B)}^{2}}{\frac{1}{A}}\right\} . {exp}\left\{\frac{-1}{2}\frac{\acute{z}^{2}}{\frac{1}{E}}\right\}} d\acute{z}} d\acute{b}_{i}. $$

By multiplying the above formula by $\frac {1}{A^{\frac {1}{2}}A^{\frac {\mathrm {-1}}{2}}\quad E^{\frac {1}{2}}{E}^{-\frac {1}{2}}}$, it further equals

$$\int\limits_{-\infty}^{+\infty} \int\limits_{-\infty}^{{\Delta}_{ijk}}{\frac{1}{A^{\frac{1}{2}}{A}^{\frac{\mathrm{-1}}{2}}E^{\frac{1}{2}}{E}^{-\frac{1}{2}}}.\left( \frac{1}{\sqrt{2\pi}}\right)}^{2}\frac{\sigma_{\varepsilon_{ij}}}{\sigma_{b}}{exp}\left\{\frac{-1}{2} \frac{{(\acute{b}-B)}^{2}}{\frac{1}{A}}\right\}.{exp}\left\{\frac{-1}{2}\frac{\acute{z}^{2}}{\frac{1}{E}}\right\} d\acute{z} d\acute{b}_{i}. $$

With a slight rearrangement the above equation equals

$$\sigma_{\varepsilon_{ij}}\times \int\limits_{-\infty}^{+\infty} \frac{1}{\sqrt {2\pi } { A}^{\frac{\mathrm{-1}}{2}}}{exp}\left\{\frac{-1}{2} \frac{{(\acute{b}-B)}^{2}}{\frac{1}{A}}\right\}d\acute{b}_{i} \int\limits_{-\infty}^{{\Delta}_{\boldsymbol{ijk}}}{\frac{1}{\sqrt{2\pi} E^{\frac{\mathrm{-1}}{2}}}{exp}\left\{\frac{-1}{2} \frac{\acute{z}^{2}}{\frac{1}{E}}\right\}} d\acute{z}. $$

Since the first integral equals one, then the above formula equals

$$\sigma_{\varepsilon_{ij}}\times 1\times \int\limits_{-\infty}^{{\Delta}_{ijk}} {\frac{1}{\sqrt {2\pi} E^{\frac{\mathrm{-1}}{2}}}{exp}\left\{\frac{-1}{2} \frac{\acute{z}^{2}}{\frac{1}{E}}\right\}} d. $$

Here we make a change of variable $q= \acute {z}\sqrt E$ to conclude that the above expression equals

$$\begin{array}{@{}rcl@{}} \sigma_{\varepsilon_{ij}}\times \int\limits_{-\infty}^{\frac{{\Delta}_{ijk}}{\sqrt{\sigma_{\varepsilon_{ij}}^{2}+{\sigma_{b}^{2}}}}} \frac{1}{\sqrt {2\pi} E^{\frac{\mathrm{-1}}{2}}}{exp}\left\{\frac{-1}{2} q^{2}\right\} {E}^{\frac{\mathrm{-1}}{2}} dq,\\ =\sigma_{\varepsilon_{ij}}{\Phi} \left( \frac{{\Delta}_{ijk}}{\sqrt {\sigma_{\varepsilon_{ij}}^{2}+\sigma_{b}^{2}}} \right). \end{array} $$

Proof of the Theorem 2.

${\Delta }_{ijk} = {\Phi }^{-1}\,\left \{\!\frac {1}{E\left (\theta \right ) \sigma _{e_{ij}}}\,expit\,(\,\!\alpha _{0k}\,+\, \boldsymbol {x}_{\boldsymbol {ij}}^{\boldsymbol {T}} \boldsymbol {\beta }\,)\!\right \} \times$$\left \{\sqrt {\sigma _{e_{ij}}^{2}+\sigma _{b}^{2}}\right \}.$□

Knowing the relationship between the marginal and conditional probabilities as follows:

$$F_{ijk}=\int F_{ijk} (b_{i}\theta ) f(b_{i}) db_{i}, $$

we can write:

$$\begin{array}{@{}rcl@{}} \beta_{0k}+\boldsymbol{x}_{\boldsymbol{ij}}^{\boldsymbol{T}} \boldsymbol{\beta} =Logit\left[\int {E(\theta ) {\Phi} \left( \frac{{\Delta}_{ijk}+b_{i}}{\sigma_{\varepsilon_{ij}}}\right) f(b_{i}) db_{i}} \right],\\ \beta_{0k}+\boldsymbol{x}_{\boldsymbol{ij}}^{\boldsymbol{T}} \boldsymbol{\beta} =Logit\left[E(\theta )\int {{\Phi} \left( \frac{{\Delta}_{ijk}+b_{i}}{\sigma_{\varepsilon_{ij}}}\right) f(b_{i}) db_{i}} \right], \end{array} $$

(A1)

where $\int {{\Phi } (\frac {{\Delta }_{ijk}+b_{i}}{\sigma _{\varepsilon _{ij}}}) f(b_{i}) db_{i}}$ simplifies to $\sigma _{\varepsilon _{ij}} {\Phi } (\frac {{\Delta }_{ijk}}{\sqrt {\sigma _{\varepsilon _{ij}}^{2}+{\sigma _{b}^{2}}}} )$ by Lemma 1.

Therefore,

$$\beta_{0k}+\boldsymbol{x}_{\boldsymbol{ij}}^{\boldsymbol{T}} \boldsymbol{\beta} =Logit\left[E(\theta ) \sigma_{\varepsilon_{ij}} {\Phi} \left( \frac{{\Delta}_{ijk}}{\sqrt {\sigma_{\varepsilon_{ij}}^{2}+{\sigma_{b}^{2}}}} \right)\right],$$

and hence

$$expit[\beta_{0k}+\boldsymbol{x}_{\boldsymbol{ij}}^{\boldsymbol{T}} \boldsymbol{\beta} ]=E(\theta ) \sigma_{\varepsilon_{ij}} {\Phi} \left( \frac{{\Delta}_{ijk}}{\sqrt {\sigma_{\varepsilon_{ij}}^{2}+\sigma_{b}^{2}}} \right). $$

A slight rearrangement shows

$$\frac{1}{\sigma_{\varepsilon_{ij}} E(\theta )}expit[\beta_{0k}+\boldsymbol{x}_{\boldsymbol{ij}}^{\boldsymbol{T}} \boldsymbol{\beta} ]={\Phi} \left( \frac{{\Delta}_{ijk}}{\sqrt {\sigma_{\varepsilon_{ij}}^{2}+\sigma_{b}^{2}}} \right), $$

implying

$${\Phi}^{-1}\left[\frac{1}{\sigma_{\varepsilon_{ij}} E(\theta )} expit \{\beta_{0k}+ \boldsymbol{x}_{\boldsymbol{ij}}^{\boldsymbol{T}} \boldsymbol{\beta }\}\right]=\frac{{\Delta}_{ijk}}{\sqrt {\sigma_{\varepsilon_{ij}}^{2}+\sigma_{b}^{2}}}, $$

and finally

$${{\Delta}_{ijk}={\Phi}}^{-1}\left[\frac{1}{\sigma_{\varepsilon_{ij}} E(\theta )} expit \{\beta_{0k}+ \boldsymbol{x}_{\boldsymbol{ij}}^{\boldsymbol{T}} \boldsymbol{\beta}\} \right]\times \sqrt {\sigma_{\varepsilon_{ij}}^{2}+{\sigma_{b}^{2}}}. $$

Proof of the Theorem 3.

For the OMLS model in equation (2.7), if α₀₁ < ⋯ < α_0K− 1 then Δ_ij1 < ⋯ < Δ_ijK− 1. □

The proof follows along the same lines as in Vahabi et al. (2017).

Web Supplements

SAS Code for Generating Simulated Data Based on the Proposed OMLS Model in Section 2.3

SAS Code for the LS Model Reviewed in Section 2.1

SAS Code for the MLS Model Introduced in Section 2.2

SAS Code for the OMLS Model Introduced in Section 2.3