批量估计问题

最大后验估计MAP
$$\hat{x}=\arg \underset{x}{\max }p\left(x|u,y\right)$$
我们希望在给定先验信息和所有时刻的输入$u$和观测$y$，推断出所有时刻的最优状态$\hat{x}$。为此我们定义几个宏观变量。
$$\begin{array}{rl}x& =x_{0:K}=\left(x_0,?,x_K\right)\\ u& =({\check{x}}_0,u_{1:k})=\left({\check{x}}_0,u_1,?,u_K\right)\\ y& =y_{0:K}=\left(y_0,?,y_K\right)\end{array}$$
用贝叶斯公式重写MAP估计：
$$\hat{x}=\arg \underset{x}{\max }p\left(x|u,y\right)=\arg \underset{x}{\max }\frac{p\left(y|x,u\right)p\left(x|u\right)}{p\left(y|u\right)}=\arg \underset{x}{\max }p\left(y|x\right)p\left(x|u\right)$$
这里我们吧分母略去，因为它与$x$无关。同时省略$p(y|x,u)$中的$u$，因为如果$x$已知，它不会影响观测数据(观测方程与它无关)。
接下来我们做出一个重要假设：对于所有时刻$k=0,?,K$，所有噪声项$w_k$和$n_k$之间是无关的，即$y_k$只与$x_k$有关，则可以用乘法公式对$p(y|x)$进行因子分解：
$$\begin{array}{rl}p\left(y|x\right)& =p\left(y_0|x_{0:K}\right)p\left(y_1|x_{0:K},y_0\right)?p\left(y_k|x_{0:K},y_{0:K?1}\right)\\ & =\prod _{k=0}^{K}p\left(y_k|x_k\right)\end{array}$$
同理，$x_k$只与$x_{k?1},u_k$ 有关，可得：
$$\begin{array}{rl}p\left(x|u\right)& =p\left(x_0|u\right)p\left(x_1|u,x_0\right)\dots p\left(x_K|u,x_{0:K?1}\right)\\ & =p\left(x_0|{\check{x}}_0\right)\prod _{k=1}^{K}p\left(x_k|x_{k?1},u_k\right)\end{array}$$

$$p\left(x_0|{\check{x}}_0\right)=\frac{1}{\sqrt{{\left(2\pi \right)}^N\det {\check{P}}_0}}\times \exp \left(?\frac{1}{2}{\left(x_0?{\check{x}}_0\right)}^T{\check{P}}_0^{?1}\left(x_0?{\check{x}}_0\right)\right)$$

$$p\left(x_k|x_{k?1},u_k\right)=\frac{1}{\sqrt{{\left(2\pi \right)}^N\det Q_k}}\times \exp \left(?\frac{1}{2}{\left(x_k?f\left(x_{k?1},u_k,0\right)\right)}^T{Q}_k^{?1}\left(x_k?f\left(x_{k?1},u_k,0\right)\right)\right)$$

$$p\left(y_k|x_k\right)=\frac{1}{\sqrt{{\left(2\pi \right)}^M\det R_k}}\times \exp \left(?\frac{1}{2}{\left(y_k?h\left(x_k,0\right)\right)}^T{R}_k^{?1}\left(y_k?h\left(x_k,0\right)\right)\right)$$
对等式两侧取对数：
$$\ln \left(p\left(y|x\right)p\left(x|u\right)\right)=\ln p\left(x_0|{\check{x}}_0\right)+\sum _{k=1}^k\ln p\left(x_k|x_{k?1},u_k\right)+\sum _{k=0}^k\ln p\left(y_k|x_k\right)$$

$$\ln p\left(x_0|{\check{x}}_0\right)=?\frac{1}{2}{\left(x_0?{\check{x}}_0\right)}^T{P}_0^{?1}\left(x_0?{\check{x}}_0\right)?\underset{与x无关}{\underbrace{\frac{1}{2}\ln \left({\left(2\pi \right)}^N\det {\check{P}}_0\right)}}$$
$$\ln p\left(x_k|x_{k?1},u_k\right)=?\frac{1}{2}{\left(x_k?f\left(x_{k?1},u_k,0\right)\right)}^T{Q}_k^{?1}\left(x_k?f\left(x_{k?1},u_k,0\right)\right)?\underset{与x无关}{\underbrace{\frac{1}{2}\ln \left({\left(2\pi \right)}^N\det Q_k\right)}}$$
$$\ln p\left(y_k|x_k\right)=?\frac{1}{2}{\left(y_k?h\left(x_k,0\right)\right)}^{\pi }{R}_k^{?1}\left(y_k?h\left(x_k,0\right)\right)?\underset{与x无关}{\underbrace{\frac{1}{2}\ln \left({\left(2\pi \right)}^M\det R_k\right)}}$$