$$ bel(x_t) = p(x_t|z_{1:t},u_{1:t}) $$
$$ \overline {bel}(x_t) = p(x_t|z_{1:t-1},u_{1:t}) $$
先给出贝叶斯滤波器的递推公式，然后证明。
$$ \overline {bel}(x_t) = \int p(x_t|u_{t},x_{t-1}) \cdot bel(x_{t-1})dx_{t-1} $$
$$ bel(x_t) = \eta \cdot p(z_t|x_{t}) \cdot \overline {bel}(x_t) $$
$p(x_t|u_{t},x_{t-1})$是机器人的状态转移概率
$p(z_t|x_{t})$是测量概率
$\eta$是概率归一化常量
根据这个公式我们可以看出来，只要知道状态转移概率分布，测量概率以及前一时刻的置信概率，就可以递归求得当前机器人状态的置信概率。
\begin{split} bel(x_t) &= p(x_t|z_{1:t},u_{1:t})\\ &= {p(x_t,z_{1:t},u_{1:t}) \over p(z_{1:t},u_{1:t})}条件概率\\ &= {p(z_t|x_t, z_{1:t-1},u_{1:t}) \cdot p(x_t, z_{1:t-1},u_{1:t}) \over p(z_{1:t},u_{1:t})}条件概率\\ &= {p(z_t|x_t, z_{1:t-1},u_{1:t}) \cdot p(x_t|z_{1:t-1},u_{1:t}) \cdot p(z_{1:t-1},u_{1:t}) \over p(z_t|z_{1:t-1},u_{1:t}) \cdot p(z_{1:t-1},u_{1:t})} 条件概率\\ &= {p(z_t|x_t, z_{1:t-1},u_{1:t}) \cdot p(x_t|z_{1:t-1},u_{1:t}) \over p(z_t|z_{1:t-1},u_{1:t})} 约分\\ &= {p(z_t|x_t) \cdot p(x_t|z_{1:t-1},u_{1:t}) \over p(z_t|z_{1:t-1},u_{1:t})} Markov假设\\ &= \eta \cdot p(z_t|x_t) \cdot \overline {bel}(x_t) 替换 \end{split}
\begin{split} \overline {bel}(x_t) &= p(x_t|z_{1:t-1},u_{1:t})\\ &= \int p(x_t|x_{t-1}, z_{1:t-1},u_{1:t}) \cdot p(x_{t-1}|z_{1:t-1},u_{1:t})dx_{t-1} 用全概率计算条件概率\\ &= \int p(x_t|x_{t-1}, u_{t}) \cdot p(x_{t-1}|z_{1:t-1},u_{1:t-1})dx_{t-1} Markov假设并去掉u_t\\ &= \int p(x_t|x_{t-1}, u_{t}) \cdot bel(x_{t-1})dx_{t-1} 替换 \end{split}
这里的$\eta = {p(z_t|z_{1:t-1},u_{1:t})}^{-1}$，可以看作概率的归一化因子。因为它等于分子中两个PDF的乘积的积分：
\begin{split} p(z_t|z_{1:t-1},u_{1:t}) &＝ \int p(z_t|x_t, z_{1:t-1},u_{1:t}) \cdot p(x_t|z_{1:t-1},u_{1:t})dx_t用全概率计算条件概率\\ &= \int p(z_t|x_t) \cdot p(x_t|z_{1:t-1},u_{1:t})dx_t \; Markov假设 \end{split}
Markov假设当前状态是完整的，若已知当前状态$x_t$，则$t$时刻及其之前的测量$z_{1:t}$和控制向量$u_{1:t}$对未来状态没有影响。
Sebastian THRUN, Wolfram BURGARD, Dieter FOX. PROBABILISTIC ROBOTICS. 2005.
M. Sanjeev Arulampalam, Simon Maskell, Neil Gordon, and Tim Clapp. A Tutorial on Particle Filters for Online Nonlinear/Non-Gaussian Bayesian Tracking. IEEE TRANSACTIONS ON SIGNAL PROCESSING, VOL. 50, NO. 2, FEBRUARY 2002.
使用全概率公式计算条件概率. 我的博文，Link.
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