Analysis of longitudinal health-related quality of life data with terminal events

Abstract

Longitudinal health-related quality of life data arise naturally from studies of progressive and neurodegenerative diseases. In such studies, patients’ mental and physical conditions are measured over their follow-up periods and the resulting data are often complicated by subject-specific measurement times and possible terminal events associated with outcome variables. Motivated by the “Predictor’s Cohort” study on patients with advanced Alzheimer disease, we propose in this paper a semiparametric modeling approach to longitudinal health-related quality of life data. It builds upon and extends some recent developments for longitudinal data with irregular observation times. The new approach handles possibly dependent terminal events. It allows one to examine time-dependent covariate effects on the evolution of outcome variable and to assess nonparametrically change of outcome measurement that is due to factors not incorporated in the covariates. The usual large-sample properties for parameter estimation are established. In particular, it is shown that relevant parameter estimators are asymptotically normal and the asymptotic variances can be estimated consistently by the simple plug-in method. A general procedure for testing a specific parametric form in the nonparametric component is also developed. Simulation studies show that the proposed approach performs well for practical settings. The method is applied to the motivating example.

Appendix

Proof of Theorem 1

(i) Recall that the estimating Eq. (3.9) is used to estimate γ, for convenience, denote

$$U_{n0}(\gamma, \hat{\xi})=\sum_{i=1}^n \int_0^\infty \left(Z_{i1} (t)-\hat{\bar{Z}}_1^*(t;\gamma)\right) \hat{\delta}_i^*(t)dN_i^*(t).$$

The consistency of $\hat{\gamma}$ follows from the almost identical arguments in Appendix A.1 of Lin et al. (2000). Thus, we omit the details.

Now we sketch the proof of $n^{1/2}(\hat{\gamma}-\gamma)=n^{-1/2}\sum_{i=1}^n \psi_i +o_P(1)$. The Taylor series expansion of $U_{n0}(\hat{\gamma}, \hat{\xi})$ at $(\gamma, \hat{\xi})$ and the law of large numbers lead to

$$n^{1/2}(\hat{\gamma}-\gamma)=\Omega_\gamma^{-1} n^{-1/2} U_{n0}(\gamma, \hat{\xi}) +o_P(1).$$

(A.1)

Note that

$$\begin{aligned} U_{n0}(\gamma, \hat{\xi})=&\sum_{i=1}^n \int_0^\infty \left(Z_{i1}(t)-\frac{{R_{z1}^{(1)}(t;\gamma)}} {{R_{z1}^{(0)} (t;\gamma)}}\right) {\delta}_i^*(t)d{\mathcal{M}_i}(t) \\ &+\sum_{i=1}^n \int_0^\infty\left(Z_{i1}(t)-\frac{{R_{z1}^{(1)}(t;\gamma)}} {{R_{z1}^{(0)}(t;\gamma)}}\right) \left(\hat{\delta}_i^*(t) -{\delta}_i^*(t)\right)d{\mathcal{M}_i}(t)+o_P(1).\\ \end{aligned}$$

(A.2)

Following Lin et al. (1994),

$$\begin{aligned} n^{1/2}(\hat{\delta}_i^*(t)-{\delta}_i^*(t))=&{\delta}_i^*(t) n^{-1/2} \sum_{k=1}^n \left[\int_0^t \frac{{e^{\xi^{\rm T} Z_{i2}(u)} d M_k^S(u;\xi)}} {{R_{z2}^{(0)}(u;\xi)}} +\int_0^t e^{\xi^{\rm T} Z_{i2}(u)} \left(Z_{i2}(u)-\frac{{R_{z2}^{(1)}(u;\xi)}} {{R_{z2}^{(0)}(u;\xi)}}\right)^{\rm T}\right.\\ & \left. \times d\Lambda_d(u)\Omega_\xi^{-1} \int_0^\infty \left(Z_{k2}(u)-\frac{{R_{z2}^{(1)}(u;\xi)}} {{R_{z2}^{(0)}(u;\xi)}}\right) d M_k^S(u;\xi)\right] +o_P(1).\end{aligned}$$

(A.3)

Plugging (A.3) and (A.2) into (A.1) and interchanging integrals, we get $n^{1/2}(\hat{\gamma}-\gamma)=n^{-1/2}\sum_{i=1}^n \psi_i +o_P(1)$.

(ii) Since $\hat{\gamma}$, $\hat{\xi}$ are consistent, the consistency of $\hat{\beta}$ follows from the expression (3.12) by applying law of large numbers.

Let

$$\begin{aligned} U_{n1}(\beta,\gamma,\hat{\xi})=&\sum_{i=1}^n\int_0^\infty \left\{W(t)\hat{\delta}_i^*(t) \left[X_i(t)-\hat{\bar{X}}^* (t;{\gamma})\right]\right. \\ &\left.\times\left[Y_i(t)-\hat{\bar{Y}}^*(t;{\gamma})-\beta^{\rm T} (X_i(t)-\hat{\bar{X}}^*(t;{\gamma}))\right] \right\}dN_i^*(t), \\ U_{n1}(\beta,\gamma,{\xi})=& \sum_{i=1}^n\int_0^\infty \left\{W(t) {\delta}_i^*(t) \left[X_i(t)-{\bar{X}}^*(t;{\gamma})\right]\right. \\ &\left.\times\left[Y_i(t)-{\bar{Y}}^*(t;{\gamma})-\beta^{\rm T} (X_i(t)-{\bar{X}}^*(t;{\gamma}))\right] \right\}dN_i^*(t).\\ \end{aligned}$$

Clearly, $-\frac{{1}} {{n}} \frac{{\partial U_{n1}(\beta,\gamma,\hat{\xi})}} {{\partial \beta}}$ converges to D as $n\rightarrow \infty$. As in Lin and Ying (2001), $-\frac{{1}}{{n}}\frac{{\partial U_{n1}(\beta,\gamma,\hat{\xi})}} {{\partial \gamma}}$ converges in probability to H. Thus, the Taylor series expansion of $U_{n1}(\hat{\beta},\hat{\gamma},\hat{\xi})$, and the law of large numbers lead to

$$n^{1/2}(\hat{\beta}-\beta)=n^{-1/2} D^{-1}\left[U_{n1}({\beta}, {\gamma},\hat{\xi})-H (\hat{\gamma}-\gamma)\right]+o_P(1).$$

(A.4)

Now

$$\begin{aligned} n^{-1/2}\left[U_{n1}({\beta},{\gamma}, \hat{\xi})-U_{n1}({\beta},{\gamma},{\xi})\right] = & n^{-1/2} \sum_{i=1}^n\int_0^\infty W(t)\left(\hat{\delta}_i^*(t) -{\delta}_i^*(t)\right) \left[X_i(t)-{\bar{X}}^*(t;{\gamma}) \right]\\ & \times\left[Y_i(t)-{\bar{Y}}^*(t;{\gamma})-\beta^{\rm T} (X_i(t)-{\bar{X}}^*(t;{\gamma}))\right] dN_i^*(t) \\ &+o_P(1).\\ \end{aligned}$$

(A.5)

Plugging (A.3) into (A.5) and interchanging integrals with simplification, we get

$$n^{-1/2}U_{n1}({\beta},{\gamma},\hat{\xi})=n^{-1/2}\sum_{i=1}^n (\eta_i+\zeta_i) +o_P(1).$$

(A.6)

Plugging (A.6) and the asymptotic expression of $n^{1/2}(\hat{\gamma}-\gamma)$ in (i) into (A.4), the asymptotic expression of $\hat{\beta}$ follows.

Proof of theorem 2

Note that $n^{1/2}[\hat{\mathcal{A}}(t;\hat{\beta},\hat{\xi},\hat{\gamma}) -\tilde{\mathcal{A}}(t;\hat{\theta},\hat{\xi},\hat{\gamma})]$ can be decomposed as follows:

$$n^{1/2}[\hat{\mathcal{A}}(t;\hat{\beta},\hat{\xi},\hat{\gamma})- \tilde{\mathcal{A}}(t;\hat{\theta},\hat{\xi},\hat{\gamma})] =n^{1/2} (I_1+I_2+I_3+I_4+I_5+I_6+I_7),$$

(A.7)

where

$$\begin{aligned} I_1=&\hat{\mathcal{A}}(t;\hat{\beta},\hat{\xi},\hat{\gamma}) -\hat{\mathcal{A}}(t;{\beta},\hat{\xi},\hat{\gamma}), \\ I_2=&\hat{\mathcal{A}}(t;{\beta},\hat{\xi},\hat{\gamma})-\hat{\mathcal{A}} (t; \beta,\hat{\xi},\gamma), \\ I_3=&\hat{\mathcal{A}}(t;{\beta}, \hat{\xi},\gamma)-\hat{\mathcal{A}} (t;{\beta},{\xi},\gamma), \\ I_4=&\hat{\mathcal{A}}(t;{\beta}, {\xi},\gamma)-\int_0^t \alpha_0 (s,\theta) d \hat{\Lambda}(s;\xi,\gamma), \\ I_5=&\int_0^t \alpha_0(s,\theta) d \hat{\Lambda} (s;\xi,\gamma)-\int_0^t \alpha_0(s,\theta) d \hat{\Lambda} (s;\hat{\xi},\gamma), \\ I_6=&\int_0^t \alpha_0(s,\theta) d \hat{\Lambda}(s;\hat{\xi},\gamma) -\tilde{\mathcal{A}}(t; \theta,\hat{\xi}, \hat{\gamma}), \\ I_7=&\tilde{\mathcal{A}}(t; \theta,\hat{\xi}, \hat{\gamma}) -\tilde{\mathcal{A}}(t; \hat{\theta},\hat{\xi},\hat{\gamma})].\\ \end{aligned}$$

With the expression (3.13) of $\hat{\mathcal{A}}(t;\beta,\xi,\gamma)$, after some algebra, we get

$$\begin{aligned} I_1=&\sum_{i=1}^n\int_0^t \frac{{X_i^{\rm T}(s) (\beta-\hat{\beta})\hat{\delta}_i^*(s)d N_i^*(s)}} {{\sum_{j=1}^n \hat{\delta}_j^*(s) e^{\hat{\gamma}^{\rm T}Z_{j1}(s)}}}=D_x(t)(\hat{\beta}-\beta)+o_P(\|\hat{\beta}-\beta\|),\\ I_2=&\sum_{i=1}^n\int_0^t \frac{{(Y_i(s)-\beta^{\rm T} X_i(s))\sum_{j=1}^n \hat{\delta}_j^*(s) (e^{{\gamma}^{\rm T}Z_{j1}(s)}-e^{\hat{\gamma}^{\rm T}Z_{j1}(s)})\hat{\delta}_i^*(s)d N_i^*(s)}} {{\sum_{j=1}^n \hat{\delta}_j^*(s) e^{{\gamma}^{\rm T}Z_{j1}(s)} \sum_{k=1}^n \hat{\delta}_k^*(s) e^{\hat{\gamma}^{\rm T}Z_{k1}(s)}}}.\\ \end{aligned}$$

Moreover, by the expression (3.11) of $\hat{\Lambda}$ and Taylor series expansions,

$$\begin{aligned} I_3=& \sum_{i=1}^n \int_0^t\left(Y_i(s)-\beta^{\rm T} X_i (s)\right) \left[\frac{{\hat{\delta}_i^*(t)}} {{\sum_{j=1}^n \hat{\delta}_j^*(s) e^{\hat{\gamma}^{\rm T}Z_{j1}(s)}}}-\frac{{\delta_i^*(t)}} {{\sum_{j=1}^n {\delta}_j^*(s) e^{\hat{\gamma}^{\rm T}Z_{j1}(s)}}}\right]dN_i^*(t), \\ I_4=&\sum_{i=1}^n \int_0^t \frac{{\delta_i^*(s)\epsilon_i(s)dN_i^*(s)}} {{\sum_{j=1}^n {\delta}_j^*(s) e^{{\gamma}^{\rm T}Z_{j1}(s)}}} =\sum_{i=1}^n \int_0^t \frac{{\delta_i^*(s)\left[dM_i(s)-\alpha_0(s)d\mathcal{M}_i(s)\right]}} {{\sum_{j=1}^n {\delta}_j^*(s) e^{{\gamma}^{\rm T}Z_{j1}(s)}}}, \\ I_5=&\sum_{i=1}^n \int_0^t \alpha_0(s,\theta)\left[\frac{{\delta_i^*(t)}} {{\sum_{j=1}^n {\delta}_j^*(s) e^{\hat{\gamma}^{\rm T}Z_{j1}(s)}}}-\frac{{\hat{\delta}_i^*(t)}} {{\sum_{j=1}^n \hat{\delta}_j^*(s) e^{\hat{\gamma}^{\rm T}Z_{j1}(s)}}}\right]dN_i^*(t), \\ I_6=&\sum_{i=1}^n \int_0^t \alpha_0(s,\theta)\hat{\delta}_i^*(s) \left[\frac{{1}}{{\sum_{j=1}^n \hat{\delta}_j^*(s) e^{{\gamma}^{\rm T}Z_{j1}(s)}}} -\frac{{1}}{{\sum_{j=1}^n \hat{\delta}_j^*(s) e^{\hat{\gamma}^{\rm T}Z_{j1}(s)}}}\right]dN_i^*(t), \\ I_7=&\sum_{i=1}^n\int_0^t\frac{{[\alpha_0(s,\theta)-\alpha_0 (s,\hat{\theta})]\hat{\delta}_i^*(s)dN_i^*(s)}} {{\sum_{j=1}^n \hat{\delta}_j^*(s) e^{\hat{\gamma}^{\rm T}Z_{j1}(s)}}}.\\ \end{aligned}$$

The consistency of $\hat{\gamma}$ and $\hat{\xi}$ and Taylor series expansion, we can get

$$\begin{aligned} I_7=&\left[-\sum_{i=1}^n\int_0^t \frac{{[\frac{{\partial{\alpha_0(s,\theta)}}} {{\partial{\theta}}}\hat{\delta}_i^*(s)dN_i^*(s)}} {{\sum_{j=1}^n \hat{\delta}_j^*(s) e^{\hat{\gamma}^{\rm T}Z_{j1}(s)}}}\right] (\hat{\theta}-\theta)+o_P(\|\hat{\theta}-\theta\|),\\ I_2+I_6=&\left[-\int_0^t\frac{{\hat{\delta}_i^*(s)\epsilon_i(s) \sum_{j=1}^n\hat{\delta}_j^*(s) e^{{\gamma}^{\rm T}Z_{j1}(s)}Z_{j1}(s) dN_i^*(s)}} {{\left(\sum_{j=1}^n \hat{\delta}_j^*(s) e^{{\gamma}^{\rm T}Z_{j1}(s)}\right)^2}}\right](\hat{\gamma}-\gamma)+o_P (\|\hat{\gamma}-\gamma\|) \\ =&H(t)(\hat{\gamma}-\gamma)+o_P(\|\hat{\gamma}-\gamma\|).\\ \end{aligned}$$

In addition,

$$\begin{aligned} I_3+I_5=&\sum_{i=1}^n \int_0^t \epsilon_i(t) \left[\frac{{\delta_i^*(t)}} {{\sum_{j=1}^n {\delta}_j^*(s) e^{\hat{\gamma}^{\rm T}Z_{j1}(s)}}}-\frac{{\hat{\delta}_i^*(t)}} {{\sum_{j=1}^n \hat{\delta}_j^*(s) e^{\hat{\gamma}^{\rm T}Z_{j1}(s)}}}\right]dN_i^*(t) \\ =&\sum_{i=1}^n \int_0^t \frac{{[\hat{\delta}_i^*(s)-\delta_i^*(s)]\left[dM_i(s)-\alpha_0(s) d\mathcal{M}_i(s)\right]}} {{nr_{z2}^{(0)}(s)}}+o_P(1).\\ \end{aligned}$$

Recall that D ₁ is the limit of $(-1/n)[(\partial{U}_n(\theta;\beta))/(\partial {\theta})]$, D ₂ is the limit of $(-1/n)[(\partial{U}_n(\theta;\beta))/(\partial {\beta})]$, D _x (t) is the limit of $-\frac{{1}} {{n}}\sum_{i=1}^n\int_0^t \frac{{X_i(s)\delta_i^*(s) d N_i^*(s)}} {{R_{z1}(s;\hat{\gamma})}}$. With Taylor series expansion, notice that

$$0=n^{-1/2}U_n(\hat{\theta};\hat{\beta},\hat{\xi})=n^{-1/2} \left(U_n({\theta};{\beta},\hat{\xi}) -D_1 n^{1/2}(\hat{\theta}- \theta) - D_2 n^{1/2}(\hat{\beta}-\beta)\right)+o_P(1).$$

which leads to,

$$\begin{aligned} n^{1/2}(\hat{\theta}-\theta) =&D_1^{-1}\left(n^{-1/2} U_n({\theta};{\beta},\hat{\xi})-D_2 n^{1/2}(\hat{\beta}-\beta)\right) +o_P(1) \\ =&n^{-1/2}D_1^{-1}\left(U_n({\theta};{\beta},{\xi}) +U_n({\theta}; {\beta},\hat{\xi})-U_n({\theta};{\beta},{\xi})\right)-D_1^{-1}D_2 n^{1/2}(\hat{\beta}-\beta)+o_P(1).\\ \end{aligned}$$

Use the asymptotic representation (A.3) of $\hat{\delta}_i^*(t)-\delta_i^*(t)$ and the fact that

$$U_n({\theta};{\beta},{\xi})=\sum_{i=1}^n \int_0^\infty \frac{{\partial{\alpha_0(t,\theta)}}} {{\partial{\theta}}} \left[d M_i(t)-\alpha_0(t)d \mathcal{M}_i(t)\right]$$

$$U_n({\theta};{\beta},\hat{\xi})-U_n({\theta};{\beta},{\xi}) =\sum_{i=1}^n \int_0^\infty \left[K (Z_{i2}-\bar{Z}_{i2}(t))+ \frac{{q_e(t)}} {{r_{z2}^{(0)}(t)}} \right] dM_i^S(t) +o_P(n\|\hat{\xi} -\xi\|), $$

we can get ${n}^{1/2} (\hat{\mathcal{A}} (t;\hat{\beta},\hat{\xi},\hat{\gamma}) - \tilde{\mathcal{A}}(t;\hat{\theta},\hat{\xi},\hat{\gamma})) =n^{-1/2}\sum_{i=1}^n \Gamma_i(t) +o_p(1)$, where

$$\begin{aligned} \Gamma_i(t)=&\int_0^t\frac{{\epsilon_i(s)[dM_i^*(s;\mathcal{A}, \beta,\gamma,\xi)-\alpha_0(s)d\mathcal{M}_i^*(s;\gamma,\xi)]}} {{r_{z1}^0(s;\gamma)}}-H(t)\psi_i \\ &+\int_0^\infty C(t)\left[Z_{i2}-\frac{{r_{z2}^{(1)}(s;\xi)}} {{r_{z2}^{(0)}(s;\xi)}}\right]dM_i^S(s)+\frac{{q_v(t,s)}} {{r_{z2}^{(0)}(s;\xi)}}dM_i^S(s) \\ &+\left[D_x^{T}(t)+\int_0^t \left(\frac{{\partial{\alpha_0} (s,\theta)}} {{\partial \theta}}\right)^{ T} d {\Lambda}(s) D_1^{-1} D_2\right]\phi_i \\ &-\left[\int_0^t \left(\frac{{\partial{\alpha_0}(s,\theta)}} {{\partial \theta}}\right)^{ T} d {\Lambda}(s)\right] D_1^{-1}\int_0^\infty \left(\frac{{\partial{\alpha_0}(s,\theta)}} {{\partial \theta}}\right)^{ T}\left[dM_i^*(s;\mathcal{A},\beta,\gamma,\xi)-\alpha_0(s)d\mathcal{M}_i^* (s;\gamma,\xi)\right] \\ &-\left[\int_0^t \left(\frac{{\partial{\alpha_0}(s,\theta)}} {{\partial \theta}}\right)^{ T} d {\Lambda}(s)\right] D_1^{-1}\int_0^\infty \left\{K\left[Z_{i2}-\frac{{r_{z2}^{(1)}(s;\xi)}} {{r_{z2}^{(0)}(s;\xi)}}\right]+\frac{{q_e(s)}} {{r_{z2}^{(0)}(s;\xi)}}\right\}dM_i^S(s).\\ \end{aligned}$$

This completes the proof. □