Robust estimation in joint mean–covariance regression model for longitudinal data

Abstract

In this paper, we develop robust estimation for the mean and covariance jointly for the regression model of longitudinal data within the framework of generalized estimating equations (GEE). The proposed approach integrates the robust method and joint mean–covariance regression modeling. Robust generalized estimating equations using bounded scores and leverage-based weights are employed for the mean and covariance to achieve robustness against outliers. The resulting estimators are shown to be consistent and asymptotically normally distributed. Simulation studies are conducted to investigate the effectiveness of the proposed method. As expected, the robust method outperforms its non-robust version under contaminations. Finally, we illustrate by analyzing a hormone data set. By downweighing the potential outliers, the proposed method not only shifts the estimation in the mean model, but also shrinks the range of the innovation variance, leading to a more reliable estimation in the covariance matrix.

Appendix

1.1 Proofs

Regularity conditions:

• A1. We assume that the dimensions $p,\,q$ and $d$ of the covariates $x_{ij}$ $z_{ij}$ and $z_{ijk}$ are fixed and that $\{n_i\}$ is a bounded sequence of positive integers. The first four moments of $y_{ij}$ exist.

• A2. The parameter space of $(\beta ^{\prime },\gamma ^{\prime },\lambda ^{\prime })^{\prime },\,\varTheta$, is a compact subset of $R^{p+q+d}$, and the true parameter value $(\beta ^{\prime }_{0},\gamma ^{\prime }_{0},\lambda ^{\prime }_{0})^{\prime }$ is in the interior of the parameter space $\varTheta$.

• A3. The covariates $z_{ijk}$ and $z_{ij}$, the matrices $W_{i}^{-1}$ are all bounded, meaning that all the elements of the vectors are bounded. The function $\dot{g}^{-1}(\cdot )$ has bounded second derivatives.

Proof of Theorem 1

For illustration we only give the proof that $\hat{\beta }_m\rightarrow \beta _0$ almost surely. The proofs for $\hat{\gamma }_ m$ and $\hat{\lambda }_m$ are similar. According to McCullagh (1983), we have

$$\begin{aligned} \hat{\beta }_m-\beta _0&= \bigg \{ \frac{1}{m}\sum ^m_{i=1}X_i^{\prime }\varDelta _i (V^{\beta }_i)^{-1}\varGamma _i^{\beta }\varDelta ^{\prime }_iX_i\bigg \}^{-1} _{\beta =\beta _0}\\&\times \bigg \{\sum ^m_{i=1}X_i^{\prime }\varDelta _i(V^{\beta }_i) ^{-1}h^{\beta }_i(\mu _i(\beta ))\bigg \}_{\beta =\beta _0}+o_p(m^{-1/2}). \end{aligned}$$

On the other hand, the expectation and variance matrix of $U_{1i}=X_i^{\prime }\varDelta _i(V^{\beta }_i)^{-1}h^{\beta }_i(\mu _i(\beta ))$ at $\beta =\beta _0$ are given by $E_0(U_{1i})=0$ and

$$\begin{aligned} \text{ var}_0(U_{1i})=\bigg \{X_i^{\prime }\varDelta _i(V^{\beta }_i)^{-1}\varGamma _ i^{\beta }\varDelta ^{\prime }_iX_i\bigg \}_{\beta _0}=(G_i^0X_i^{\prime }X_i)^{\prime } (V^{\beta }_i)^{-1}\varGamma ^{\beta }_i(G^{0}_iX_i^{\prime }X_i), \end{aligned}$$

where $G_i^{0}=\text{ diag}\{\dot{g}^{-1}(x_{i1}^{\prime }\beta _0),\dots , \dot{g}^{-1}(x^{\prime }_{in_i}\beta _0)\}$ is an $n_i\times n_i$ diagonal matrix.

Since $V_i^{\beta }=A_i^{-1/2}\varSigma _i$ and $\varSigma _i^{-1}=\varPhi ^{\prime }_iD^{-1}_i\varPhi _i$, the variance can be further written as $\text{ var}_0(U_{1i})=(G^0_iX_i^{\prime }X_i)^{\prime }\varPhi _i(D_i^{-1}A_i^{-1/2})\varPhi ^{\prime } _i\varGamma ^{\beta }_i(G^0_iX_i^{\prime }X_i)$. Condition A3 above implies that there exists a constant $\kappa _0$ such that $\text{ var}_0(U_{1i})\le \kappa _01_{p\times p}$ for all $i$ and all $\theta \in \varTheta$, where $1_{p\times p}$ is the $p\times p$ matrix with all elements being 1’s, meaning that all elements of $\text{ var}_0(U_{1i})$ are bounded by $\kappa _0$. Thus $\sum ^{\infty }_{i=1}\text{ var}_0(U_{1i})/i^2<\infty$. By Kolmogorov’s strong law of large numbers we know that

$$\begin{aligned} \bigg \{\frac{1}{m}\sum ^m_{i=1}X_i^{\prime }\varDelta _i(V^{\beta }_i)^{-1}h^ {\beta }_i(\mu _i(\beta ))\bigg \}_{\beta =\beta _0}\rightarrow 0 \end{aligned}$$

almost surely as $m\rightarrow \infty$. In the same manner it can be shown that

$$\begin{aligned} \bigg \{\frac{1}{m}\sum ^m_{i=1}X_i^{\prime }\varDelta _i(V^{\beta }_i)^{-1} \varGamma _i^{\beta }\varDelta ^{\prime }_iX_i\bigg \}_{\beta =\beta _0} \end{aligned}$$

is a bounded matrix. This leads to $\hat{\beta }_m-\beta _0\rightarrow 0$ almost surely as $m\rightarrow \infty$. The proof is complete. $\square$

Proof of Theorem 2

First we give some notations. Define

$$\begin{aligned}&H_m=\sum ^m_{i=1} X_i^{\prime } \varDelta _i(V^\beta _i)^{-1}\varGamma _i^\beta \varDelta _iX_i, \nonumber \\&B_m=\sum ^m_{i=1} T_i^{\prime }(V_i^\gamma )^{-1}\varGamma ^\gamma _i T_i, \nonumber \\&C_m=\sum ^m_{i=1} Z_i^{\prime } D_i (V_i^\lambda )^{-1}\varGamma _i^\lambda D_iZ_i. \nonumber \\&\tilde{U}_1(\beta )=\sum ^m_{i=1}X_i^{\prime } \varDelta _{0i} (V_{0i}^\beta )^{-1} h^\beta _{0i}(\mu _{0i}(\beta )),\end{aligned}$$

(11)

$$\begin{aligned}&\tilde{U}_2(\gamma )=\sum ^m_{i=1} T_i^{\prime }(V^\gamma _{0i})^{-1} h^\gamma _{0i}(\hat{r}_{0i}(\gamma )),\end{aligned}$$

(12)

$$\begin{aligned}&\tilde{U}_3(\lambda )=\sum ^m_{i=1} Z_i^{\prime } D_{0i} (V^\lambda _{0i})^{-1} h^\lambda _{0i} (\sigma ^2_{0i}(\lambda )). \end{aligned}$$

(13)

$$\begin{aligned}&\xi =H_m^{1/2}(\beta -\beta _0),\ \hat{\xi }=\xi (\hat{\beta }_m)=H_m^{1/2}(\hat{\beta }_m -\beta _0),\ \tilde{\xi }=H_m^{1/2}\tilde{U}_1;\\&\eta =B_m^{1/2}(\gamma -\gamma _0),\ \hat{\eta }=\eta (\hat{\gamma }_m)=B_m^{1/2}(\hat{\gamma }_m-\gamma _0), \ \tilde{\eta }=B_m^{1/2}\tilde{U}_2;\\&\zeta =C_m^{1/2}(\lambda -\lambda _0),\ \hat{\zeta }=\zeta (\hat{\lambda }_m)=C_m^{1/2}(\hat{\lambda }_m-\lambda _0),\ \tilde{\zeta }=C_m^{1/2}\tilde{U}_3. \end{aligned}$$

Next we prove the following Lemma:

Lemma

Under condition (A1)–(A3),

$$\begin{aligned}&||\hat{\xi }-\tilde{\xi }||=o_p(1),\end{aligned}$$

(14)

$$\begin{aligned}&||\hat{\eta }-\tilde{\eta }||=o_p(1),\end{aligned}$$

(15)

$$\begin{aligned}&||\hat{\zeta }-\tilde{\zeta }||=o_p(1). \end{aligned}$$

(16)

Define

$$\begin{aligned}&\Psi (\xi )=H_m^{1/2}U_1(\beta )=H_m^{1/2}U_1(\xi ),\end{aligned}$$

(17)

$$\begin{aligned}&\varPhi (\xi )=H_m^{1/2}\tilde{U}_1(\xi )-\xi . \end{aligned}$$

(18)

By condition $($A1$)$–$($A3$)$, $\Psi (\xi )$ and $U_1$ give the same root for $\xi$. The solution of $\varPhi$ is $\tilde{\xi }$. Following the proof of Theorem $1$ in He et al. (2005), we immediately obtain that

$$\begin{aligned} \text{ sup}_{||\xi ||\le L}||\Psi (\xi )-\varPhi (\xi )||=o_p (1), \ ||\xi ||=O_p (1), \end{aligned}$$

where $L$ is a sufficiently large number. By Brouwer’s fixed-point theorem, (11) is verified. We can prove (12) and (13) similarly. $\square$

By Lemma, we only need to show the asymptotic normality of $(\tilde{\xi }^{\prime },\ \tilde{\eta }^{\prime },\ \tilde{\zeta }^{\prime })^{\prime }/\sqrt{m}$. This is equivalent to the asymptotic normality of $(\tilde{U}_{1}^{\prime },\ \tilde{U}_{2}^{\prime },\ \tilde{U}_{3}^{\prime })/\sqrt{m}$. Note that Conditions (A1)–(A3) imply that

$$\begin{aligned} \text{ E}_{0}[\varsigma ^{\prime }\{X_{i}^{\prime }\varDelta _{0i}(V_{i}^{\beta })^{-1}h _{i}^{\beta }\}+\omega ^{\prime }\{T_{i}^{\prime }(V_{i}^{\gamma })^{-1}h_{i}^{\gamma }\} +\phi ^{\prime }\{Z_{i}^{\prime }D_{0i}(V_{i}^{\rho })^{-1}h_{i}^{\lambda } \} ]^{3}<\kappa , \end{aligned}$$

for any $\varsigma \in R^{p+K},\ \omega \in R^{q}\, \text{ and}\, \phi \in R^{d+K^{\prime }}$, where $\kappa$ is a constant independent of $i$.

Furthermore, we have

$$\begin{aligned}&\frac{1}{m}\sum _{i=1}^{m}V[\varsigma ^{\prime }\{X_{i}^{\prime }\varDelta _{0i}(V_{i} ^{\theta })^{-1}h_{i}^{\theta }\}+\omega ^{\prime }\{T_{i}^{\prime }(V_{i}^{\gamma }) ^{-1}h_{i}^{\gamma }\}+\phi ^{\prime }\{Z_{i}^{\prime }D_{0i}(V_{i}^{\rho })^{-1}h_{i}^{\rho } \} ]\\&\qquad =(\varsigma ^{\prime },\omega ^{\prime },\phi ^{\prime })\frac{1}{m}V_m(\varsigma ^{\prime },\omega ^{\prime },\phi ^{\prime })^{\prime } \rightarrow (\varsigma ^{\prime },\omega ^{\prime },\phi ^{\prime })^{\prime }V(\varsigma ^{\prime },\omega ^{\prime },\phi ^{\prime })^{\prime }>0. \end{aligned}$$

Therefore, the asymptotic normality of $(\tilde{U}_{1}^{\prime },\ \tilde{U}_{2}^{\prime },\ \tilde{U}_{3}^{\prime })/\sqrt{m}$ is easily proved by multivariate Liapounov central limit theorem. Therefore,

$$\begin{aligned} \sqrt{m}\left( \begin{array}{c} \hat{\beta }_m-\beta _0 \\ \hat{\gamma }_m-\gamma _0 \\ \hat{\lambda }_m-\lambda _0 \\ \end{array} \right)=\left( \begin{array}{c@{\quad }c@{\quad }c} (H_m/m)^{-1}&0&0 \\ 0&(B_m/m)^{-1}&0 \\ 0&0&(C_m/m)^{-1} \\ \end{array} \right) \left( \begin{array}{c} \tilde{U}_1/\sqrt{m} \\ \tilde{U}_2/\sqrt{m} \\ \tilde{U}_3/\sqrt{m} \\ \end{array} \right)\qquad \quad \end{aligned}$$

(19)

$$\begin{aligned} \rightarrow N \left\{ 0,\ \left( \begin{array}{c@{\quad }c@{\quad }c} v^{11}&0&0 \\ 0&v^{22}&0 \\ 0&0&v^{33} \\ \end{array} \right)^{-1}\left( \begin{array}{c@{\quad }c@{\quad }c} v^{11}&v^{12}&v^{13} \\ v^{21}&v^{22}&v^{23} \\ v^{31}&v^{32}&v^{33} \\ \end{array} \right)\left( \begin{array}{c@{\quad }c@{\quad }c} v^{11}&0&0 \\ 0&v^{22}&0 \\ 0&0&v^{33} \\ \end{array} \right)^{-1} \right\} \qquad \quad \end{aligned}$$

(20)

The proof of Theorem 2 is completed. $\square$