Hierarchical and multivariate regression models to fit correlated asymmetric positive continuous outcomes

Abstract

In the extant literature, hierarchical models typically assume a flexible distribution for the random-effects. The random-effects approach has been used in the inferential procedure of the generalized linear mixed models . In this paper, we propose a random intercept gamma mixed model to fit correlated asymmetric positive continuous outcomes. The generalized log-gamma (GLG) distribution is assumed as an alternative to the normality assumption for the random intercept. Numerical results demonstrate the impact on the maximum likelihood (ML) estimator when the random-effect distribution is misspecified. The extended inverted Dirichlet (EID) distribution is derived from the random intercept gamma-GLG model that leads to the EID regression model by supposing a particular parameter setting of the hierarchical model. Monte Carlo simulation studies are performed to evaluate the asymptotic behavior of the ML estimators from the proposed models. Analysis of diagnostic methods based on quantile residual and COVARATIO statistic are used to assess departures from the EID regression model and identify atypical subjects. Two applications with real data are presented to illustrate the proposed methodology.

Appendices

Appendix

EIB distribution

Proposition 1

If $z=(\phi \mu /\kappa )y$, then the random variable z follows an inverted-beta distribution with parameters $\phi$ and $\kappa$. In addition, let t be a random variable which follows a beta distribution, $t \sim beta(\phi ,\kappa )$. Then, $v=t/(1-t)$ has an inverted-beta distributions and $(\phi \mu /\kappa )v$ follows a distribution as in (6).

Proof

The proof is direct. $\square$

Proposition 2

For $\kappa >1$ and $\kappa >2$, the mean and variance of $y \sim EIB(\mu , \phi , \kappa )$ are

• (i) $\mathrm{E}(y)=\mu ^{-1}\kappa \varGamma (\kappa -1)/\varGamma (\kappa )$. If $\kappa$ is a positive integer, it follows that $\mathrm{E}(y)=\mu ^{-1}\kappa /(\kappa - 1)$.

• (ii) $\mathrm{Var}(y)= \phi ^{-1}\mu ^{-2}\kappa ^2\left[ (\phi + 1)\left( \varGamma (\kappa - 2)/\varGamma (\kappa )\right) - \varGamma ^{2}(\kappa - 1)/\varGamma ^2(\kappa )\right]$. If $\kappa$ is a positive integer, iy follows that $\mathrm{Var}(y)=\phi ^{-1}\mu ^{-2}\kappa ^{2}(\phi +\kappa -1)/(\kappa -1)^2(\kappa -2),$

where $\varGamma (\cdot )$ is the gamma function.

Observing that when $\kappa \rightarrow \infty$ the mean of E$(y)=\mu ^{-1}$ and the variance increases.

Proof

$$\begin{aligned} \mathrm{E}(y)= & {} \int _{0}^{\infty }yf(y;\varvec{\theta })dy =\frac{1}{B(\phi ,\kappa )}\int _{0}^{\infty }\left( \frac{\phi \mu }{\kappa }y\right) ^{\phi } \left( 1 + \frac{\phi \mu }{\kappa }y\right) ^{-(\phi + \kappa )}dy \nonumber \\= & {} \frac{\mu ^{-1}\phi ^{-1}}{B(\phi ,\kappa )}\int _{0}^{\infty }\frac{( z)^{(\phi + 1)-1}}{(z+ 1)^{((\phi + 1) + (\kappa -1))}}dz =\mu ^{-1}\phi ^{-1}\kappa \frac{B(\phi +1, \kappa -1)}{B(\phi ,\kappa )}. \end{aligned}$$

(9)

The integral in (9) is the inverted-beta pdf with parameters $\phi +1$ and $\kappa -1,$ if $\kappa >1.$ Simplifying (9) leads to the Preposition 2(i). Similarly, we obtain E$(y^2)=\mu ^{-2}\phi ^{-2}\kappa ^2\frac{B(\phi +2, \kappa -2)}{B(\phi ,\kappa )}.$ Thus, the Preposition 2(ii) follows from the expression Var$(y) = \mathrm{E}(y^{2}) - (\mathrm{E}(y))^{2}$. $\square$