Y(t)=Y_0(t)+\theta^T\Lambda(t)+\sigma_bB(t)\\
Y_0(t)\sim N(\mu_0,\sigma_0)
Y_{i,j}-Y_{i,j-1}=\sigma_b[B(t_{i,j}-t_{i,j-1})]
其似然公式为
L_d(\mu_0,\sigma_0,\theta,\sigma_b;D)=\prod_{i=1}^{n}\bigg[\frac{1}{\sqrt{2\pi\sigma_0^2}}exp\bigg(-\frac{(\Delta y_{i,0}-\mu_0)^2}{2\sigma_0^2}\bigg)\times\\\prod_{j=1}^{m_i}\frac{1}{\sqrt{2\pi (t_{i,j}-t_{i,j-1})\sigma_b^2}}exp\bigg(-\frac{(\Delta y_{i,j}-\theta^T(\Lambda(t_{i,j})-\Lambda(t_{i,j-1})))^2}{2(t_{i,j}-t_{i,j-1})\sigma_b^2}\bigg]
log(L_d(\mu_0,\sigma_0,\theta,\sigma_b;D))=\sum_{i=1}^{n}\big(-\frac{1}{2}log2\pi-log(\sigma)-\frac{(\Delta y_{i,0}-\mu_0)^2}{2\sigma_0^2}\big)\\ +\sum_{i=1}^{n}\sum_{j=1}^{m_i}\big(-\frac{1}{2}log(2\pi(t_{i,j}-t_{i,j-1})-log(\sigma_b)\\-\frac{(\Delta y_{i,j}-\theta^T(\Lambda(t_{i,j})-\Lambda(t_{i,j-1})))^2}{2(t_{i,j}-t_{i,j-1})\sigma_b^2}\big)
上式分别对\mu_0,\sigma_0,\theta,\sigma_b求导
1、
\frac{\partial L}{\partial \mu_0}=\sum_{i=1}^{n}\bigg(-\frac{2(\Delta y_{i,0}-\mu_0)}{2\sigma_0^2}\bigg)
令上式等于0可得
\sum_{i=1}^{n}(\Delta y_{i,0}-\mu_0)=0
\\ \mu_0=\frac{\sum_{i=1}^{n}\Delta y_{i,0}}{n}
2、
\frac{\partial L}{\partial \sigma_0}=\sum_{i=1}^{n}\Bigg(-\frac{1}{\sigma_0}+\frac{(\Delta y_{i,0}-\mu_0)^2}{\sigma_0^3}\Bigg)\\=\sum_{i=1}^{n}\bigg(\frac{(\Delta y_{i,0}-\mu_0)^2-\sigma_0^2}{\sigma_0^3}\bigg)
令上式等于0可得
\sum_{i=1}^{n}\bigg(\frac{(\Delta y_{i,0}-\mu_0)^2-\sigma_b^2}{\sigma_b^3}\bigg)=0
\\ \sigma_0=\sqrt\frac{\sum_{i=1}^{n}(\Delta y_{i,0}-\mu_0)^2}{n}
3、
\frac{\partial L}{\partial \theta^T}=\sum_{i=1}^{n}\bigg(-\frac{2\Big(\Delta y_{i,j}-\theta^T\big(\Lambda(t_{i,j})-\Lambda(t_{i,j-1})\big)\Big)\times\Big(\Lambda(t_{i,j})-\Lambda(t_{i,j-1})\Big)}{2\sigma_b^2}\bigg)
令上式等于0可得
\sum_{i=1}^{n}\bigg(\Delta y_{i,j}-\theta^T\Big(\Lambda(t_{i,j})-\Lambda(t_{i,j-1})\Big)\bigg)=0
\\ \sum_{i=1}^{n}\bigg(\Big(\Lambda(t_{i,j})-\Lambda(t_{i,j-1})\Big)\bigg)\theta^T=\sum_{i=1}^{n}(\Delta y_{i,j})
4、
\frac{\partial L}{\partial \sigma_b}=\sum_{i=1}^{n}\sum_{j=1}^{m_i}\Bigg(-\frac{1}{\sigma_b}+-\frac{\Big(\Delta y_{i,j}-\theta^T\big(\Lambda(t_{i,j})-\Lambda(t_{i,j-1})\big)\Big)^2}{(t_{i,j}-t_{i,j-1})\sigma_b^3}\Bigg)\\=\sum_{i=1}^{n}\sum_{j=1}^{m_i}\bigg(\frac{\Big(\Delta y_{i,j}-\theta^T\big(\Lambda(t_{i,j})-\Lambda(t_{i,j-1})\big)\Big)^2-(t_{i,j}-t_{i,j-1})\sigma_b^2}{\sigma_b^3}\bigg)
令上式等于0可得
\sum_{i=1}^{n}\sum_{j=1}^{m_i}\Big(\big(\Delta y_{i,j}-\theta^T(\Lambda(t_{i,j})-\Lambda(t_{i,j-1}))\big)^2-(t_{i,j}-t_{i,j-1})\sigma_b^2\Big)=0
\\ \sigma_b^2=\frac{\sum_{i=1}^{n}\sum_{j=1}^{m_i}\big(\Delta y_{i,j}-\theta^T(\Lambda(t_{i,j})-\Lambda(t_{i,j-1}))\big)^2}{\sum_{i=1}^{n}\sum_{j=1}^{m_i}(t_{i,j}-t_{i,j-1})} 
