## Appendix A  
**Proof of Proposition 1**. According to the statistical properties of Brownian motion, we have \(\mathrm{E}(B(t_{i}))=0\) and \(\mathrm{D}(B(t_{i}))=t_{i}\), where \(\mathrm{E}(\cdot)\) and \(\mathrm{D}(\cdot)\) denote the mean and variance, respectively. In this way, the mean of the health indicators at \(t_{i}\) is clearly derived as  
\[\mathrm{E}(y_{i})=\mathrm{E}\!\left(\lambda\int_{0}^{t_{i}}\varphi(\tau;\theta )d\tau\right)+\mathrm{E}\!\left(\sqrt{m\lambda}B(t_{i})\right)+\mathrm{E}( \varepsilon)=\mu_{\lambda}\int_{0}^{t_{i}}\varphi(\tau;\theta)d\tau, \tag{10}\]  
and then  
\[\mathrm{E}\!\left(\boldsymbol{y}_{\boldsymbol{c}}\right)=[\mathrm{E}(y_{1}), \cdots,\mathrm{E}(y_{c})]^{\prime}=\mu_{\lambda}\boldsymbol{t}_{\boldsymbol{c}}. \tag{11}\]  
For the covariance between \(y_{i}\) and \(y_{j}\), we calculate it by using  
\[\begin{array}{rl}\mathrm{Cov}\!\left(y_{i},y_{j}\right)&=\mathrm{Cov}\! \left(\lambda\int_{0}^{t_{i}}\varphi(\tau;\theta)d\tau,\lambda\int_{0}^{t_{j} }\varphi(\tau;\theta)d\tau\right)\\ &\quad+\mathrm{Cov}\!\left(\sqrt{m\lambda}B(t_{i}),\sqrt{m\lambda}B\!\left(t_ {j}\right)\right)\\ &\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad+ \mathrm{Cov}\!\left(\varepsilon,\varepsilon\right)\end{array} \tag{12}\]  
where each item can be given as  
\[\begin{array}{c}\mathrm{Cov}\!\left(\lambda\int_{0}^{t_{i}}\varphi(\tau; \theta)d\tau,\lambda\int_{0}^{t_{j}}\varphi(\tau;\theta)d\tau\right)\\ =\mathrm{E}\!\left(\lambda\int_{0}^{t_{i}}\varphi(\tau;\theta)d\tau\cdot\lambda \int_{0}^{t_{j}}\varphi(\tau;\theta)d\tau\right)-\mathrm{E}\!\left(\lambda \int_{0}^{t_{i}}\varphi(\tau;\theta)d\tau\right)\cdot\mathrm{E}\!\left(\lambda \int_{0}^{t_{j}}\varphi(\tau;\theta)d\tau\right)\\ =\int_{0}^{t_{i}}\varphi(\tau;\theta)d\tau\cdot\int_{0}^{t_{j}}\varphi(\tau; \theta)d\tau\cdot\left(\mathrm{E}\!\left(\lambda^{2}\right)-\mathrm{E}^{2}( \lambda)\right)\\ =\sigma_{\lambda}^{2}\cdot\int_{0}^{t_{i}}\varphi(\tau;\theta)d\tau\cdot\int_{0 }^{t_{j}}\varphi(\tau;\theta)d\tau\end{array} \tag{13}\]  
\[\begin{array}{c}\mathrm{Cov}\!\left(\sqrt{m\lambda}B(t_{i}),\sqrt{m\lambda}B \!\left(t_{j}\right)\right)\\ =E\!\left(\sqrt{m\lambda}B(t_{i})\cdot\sqrt{m\lambda}B\!\left(t_{j}\right) \right)-E\!\left(\sqrt{m\lambda}B(t_{i})\right)\cdot E\!\left(\sqrt{m\lambda}B \!\left(t_{j}\right)\right)\\ =\left\{\begin{array}{l}E\!\left(\sqrt{m\lambda}B\!\left(t_{i}\right)\!\left( \sqrt{m\lambda}B\!\left(t_{j}\right)\!-\sqrt{m\lambda}B\!\left(t_{j}\right) \right)+m\lambda B\!\left(t_{j}\right)^{2}\right)=E\!\left(m\lambda B\!\left(t _{i}\right)^{2}\right)=m\mu_{\lambda}t_{i},\quad i\leq j\\ E\!\left(\sqrt{m\lambda}B\!\left(t_{j}\right)\!\left(\sqrt{m\lambda}B\!\left(t_ {i}\right)-\sqrt{m\lambda}B\!\left(t_{j}\right)\right)+m\lambda B\!\left(t_{j }\right)^{2}\right)=E\!\left(m\lambda B\!\left(t_{j}\right)^{2}\right)=m\mu_{ \lambda}t_{j},\quad i>j\end{array}\right.\end{array} \tag{14}\]  
Denote the results in matrix and we derive  
\[\mathrm{D}\!\left(\boldsymbol{y}_{\boldsymbol{c}}\right)=\sigma_{2}^{\lambda}t _{\boldsymbol{c}}t_{\boldsymbol{c}}^{\prime}+m\mu_{\lambda}\mathbf{Q}_{ \boldsymbol{c}}\sigma_{2}^{m}\mathbf{E}_{\boldsymbol{c}}. \tag{15}\]  
\(\square\)