定义以下运动和观测模型:
运动方程: x k = f ( x k − 1 , v k , w k ) , k = 1 , ⋯ , K 观测方程: y k = g ( x k , n k ) , k = 0 , ⋯ , K \begin{aligned}&\text{运动方程:}\boldsymbol{x}_{k}=f\left(\boldsymbol{x}_{k-1}, \boldsymbol{v}_{k},\boldsymbol{w}_{k}\right), \quad k=1, \cdots, K \\&\text{观测方程:}\boldsymbol{y}_{k}=g\left(\boldsymbol{x}_{k}, \boldsymbol{n}_{k}\right), \quad k=0, \cdots, K\end{aligned} 运动方程:xk=f(xk−1,vk,wk),k=1,⋯,K观测方程:yk=g(xk,nk),k=0,⋯,K
变量和符号定义如下:
贝叶斯滤波仅使用过去以及当前的测量,通过构造一个完整的概率密度函数PDF,来计算当前状态 x k \boldsymbol{x}_{k} xk 的置信度:
P ( x k ∣ x ˇ 0 , v 1 : k , y 0 : k ) (1) P\left(\boldsymbol{x}_{k} \mid \check{\boldsymbol{x}}_{0}, \boldsymbol{v}_{1: k}, \boldsymbol{y}_{0: k}\right) \tag{1} P(xk∣xˇ0,v1:k,y0:k)(1)
其中, x ˇ 0 \check{\boldsymbol{x}}_{0} xˇ0 表示 0 时刻系统状态的先验。
贝叶斯滤波器可以用下面公式表示:
p ( x k ∣ x ˇ 0 , v 1 : k , y 0 : k ) ⏟ 后验置信度 = η p ( y k ∣ x k ) ⏟ 利用 g ( ⋅ ) 进行更新 ∫ p ( x k ∣ x k − 1 , v k ) ⏟ 利用 f ( ⋅ ) 进行预测 p ( x k − 1 ∣ x ˇ 0 , v 1 : k − 1 , y 0 : k − 1 ) ⏟ 先验置信度 d x k − 1 (2) \begin{aligned}& \underbrace{p\left(\boldsymbol{x}_{k} \mid \check{\boldsymbol{x}}_{0}, \boldsymbol{v}_{1: k}, \boldsymbol{y}_{0: k}\right)}_{\text {后验置信度 }} \\=& \eta \underbrace{p\left(\boldsymbol{y}_{k} \mid \boldsymbol{x}_{k}\right)}_{\text {利用 } \boldsymbol{g}(\cdot) \text { 进行更新 }} \int \underbrace{p\left(\boldsymbol{x}_{k} \mid \boldsymbol{x}_{k-1}, \boldsymbol{v}_{k}\right)}_{\text {利用 } \boldsymbol{f}(\cdot) \text { 进行预测 }} \underbrace{p\left(\boldsymbol{x}_{k-1} \mid \check{\boldsymbol{x}}_{0}, \boldsymbol{v}_{1: k-1}, \boldsymbol{y}_{0: k-1}\right)}_{\text {先验置信度 }} \mathrm{d} \boldsymbol{x}_{k-1}\end{aligned}\tag{2} =后验置信度 p(xk∣xˇ0,v1:k,y0:k)η利用 g(⋅) 进行更新 p(yk∣xk)∫利用 f(⋅) 进行预测 p(xk∣xk−1,vk)先验置信度 p(xk−1∣xˇ0,v1:k−1,y0:k−1)dxk−1(2)
对(1)式进行贝叶斯展开,有:
p ( x k ∣ x ˇ 0 , v 1 : k , y 0 : k ) = p ( x k , x ˇ 0 , v 1 : k , y 0 : k ) p ( x ˇ 0 , v 1 : k , y 0 : k ) = p ( y k ∣ x k , x ˇ 0 , v 1 : k , y 0 : k − 1 ) p ( x k ∣ x ˇ 0 , v 1 : k , y 0 : k − 1 ) p ( x ˇ 0 , v 1 : k , y 0 : k − 1 ) p ( y k ∣ x ˇ 0 , v 1 : k , y 0 : k − 1 ) p ( x ˇ 0 , v 1 : k , y 0 : k − 1 ) = p ( y k ∣ x k , x ˇ 0 , v 1 : k , y 0 : k − 1 ) p ( x k ∣ x ˇ 0 , v 1 : k , y 0 : k − 1 ) p ( y k ∣ x ˇ 0 , v 1 : k , y 0 : k − 1 ) (3) \begin{aligned}& p\left(\boldsymbol{x}_{k} \mid \check{\boldsymbol{x}}_{0}, \boldsymbol{v}_{1: k}, \boldsymbol{y}_{0: k}\right) \\\\&= \frac{p\left(\boldsymbol{x}_{k}, \check{\boldsymbol{x}}_{0}, \boldsymbol{v}_{1: k}, \boldsymbol{y}_{0: k}\right)}{p\left(\check{\boldsymbol{x}}_{0}, \boldsymbol{v}_{1: k}, \boldsymbol{y}_{0: k}\right) } \\\\&= \frac{p\left(\boldsymbol{y}_{ k} \mid \boldsymbol{x}_{k}, \check{\boldsymbol{x}}_{0}, \boldsymbol{v}_{1: k}, \boldsymbol{y}_{0: k-1}\right) p\left(\boldsymbol{x}_{k} \mid \check{\boldsymbol{x}}_{0}, \boldsymbol{v}_{1: k}, \boldsymbol{y}_{0: k-1}\right)p\left( \check{\boldsymbol{x}}_{0}, \boldsymbol{v}_{1: k}, \boldsymbol{y}_{0: k-1}\right)}{p\left(\boldsymbol{y}_{ k} \mid \check{\boldsymbol{x}}_{0}, \boldsymbol{v}_{1: k}, \boldsymbol{y}_{0: k-1}\right) p\left( \check{\boldsymbol{x}}_{0}, \boldsymbol{v}_{1: k}, \boldsymbol{y}_{0: k-1}\right)} \\\\&=\frac{p\left(\boldsymbol{y}_{ k} \mid \boldsymbol{x}_{k}, \check{\boldsymbol{x}}_{0}, \boldsymbol{v}_{1: k}, \boldsymbol{y}_{0: k-1}\right) p\left(\boldsymbol{x}_{k} \mid \check{\boldsymbol{x}}_{0}, \boldsymbol{v}_{1: k}, \boldsymbol{y}_{0: k-1}\right)}{p\left(\boldsymbol{y}_{ k} \mid \check{\boldsymbol{x}}_{0}, \boldsymbol{v}_{1: k}, \boldsymbol{y}_{0: k-1}\right) }\end{aligned}\tag{3} p(xk∣xˇ0,v1:k,y0:k)=p(xˇ0,v1:k,y0:k)p(xk,xˇ0,v1:k,y0:k)=p(yk∣xˇ0,v1:k,y0:k−1)p(xˇ0,v1:k,y0:k−1)p(yk∣xk,xˇ0,v1:k,y0:k−1)p(xk∣xˇ0,v1:k,y0:k−1)p(xˇ0,v1:k,y0:k−1)=p(yk∣xˇ0,v1:k,y0:k−1)p(yk∣xk,xˇ0,v1:k,y0:k−1)p(xk∣xˇ0,v1:k,y0:k−1)(3)
由于 y k \boldsymbol{y}_k yk 的状态只与 x k \boldsymbol{x}_k xk 相关,所以上式可化简为:
p ( x k ∣ x ˇ 0 , v 1 : k , y 0 : k ) = p ( y k ∣ x k ) p ( x k ∣ x ˇ 0 , v 1 : k , y 0 : k − 1 ) p ( y k ) (4) \begin{aligned}& p\left(\boldsymbol{x}_{k} \mid \check{\boldsymbol{x}}_{0}, \boldsymbol{v}_{1: k}, \boldsymbol{y}_{0: k}\right) \\\\&= \frac{p\left(\boldsymbol{y}_{ k} \mid \boldsymbol{x}_{k}\right) p\left(\boldsymbol{x}_{k} \mid \check{\boldsymbol{x}}_{0}, \boldsymbol{v}_{1: k}, \boldsymbol{y}_{0: k-1}\right)}{p\left(\boldsymbol{y}_k\right) }\end{aligned}\tag{4} p(xk∣xˇ0,v1:k,y0:k)=p(yk)p(yk∣xk)p(xk∣xˇ0,v1:k,y0:k−1)(4)
令 p − 1 ( y k ) = η p^{-1}\left(\boldsymbol{y}_k\right) = \eta p−1(yk)=η,则有:
p ( x k ∣ x ˇ 0 , v 1 : k , y 0 : k ) = η p ( y k ∣ x k ) p ( x k ∣ x ˇ 0 , v 1 : k , y 0 : k − 1 ) (5) \begin{aligned}& p\left(\boldsymbol{x}_{k} \mid \check{\boldsymbol{x}}_{0}, \boldsymbol{v}_{1: k}, \boldsymbol{y}_{0: k}\right) = \eta p\left(\boldsymbol{y}_{ k} \mid \boldsymbol{x}_{k}\right) p\left(\boldsymbol{x}_{k} \mid \check{\boldsymbol{x}}_{0}, \boldsymbol{v}_{1: k}, \boldsymbol{y}_{0: k-1}\right) \end{aligned}\tag{5} p(xk∣xˇ0,v1:k,y0:k)=ηp(yk∣xk)p(xk∣xˇ0,v1:k,y0:k−1)(5)
由全概率公式和贝叶斯公式有:
p ( x k ∣ x ˇ 0 , v 1 : k , y 0 : k − 1 ) = ∫ p ( x k , x k − 1 ∣ x ˇ 0 , v 1 : k , y 0 : k − 1 ) d x k − 1 = ∫ p ( x k ∣ x k − 1 , x ˇ 0 , v 1 : k , y 0 : k − 1 ) p ( x k − 1 ∣ x ˇ 0 , v 1 : k , y 0 : k − 1 ) d x k − 1 (6) \begin{aligned}& p\left(\boldsymbol{x}_{k} \mid \check{\boldsymbol{x}}_{0}, \boldsymbol{v}_{1: k}, \boldsymbol{y}_{0: k-1}\right)\\\\=&\int p\left(\boldsymbol{x}_{k}, \boldsymbol{x}_{k-1} \mid \check{\boldsymbol{x}}_{0}, \boldsymbol{v}_{1: k}, \boldsymbol{y}_{0: k-1}\right) \mathrm{d} \boldsymbol{x}_{k-1} \\\\=& \int p\left(\boldsymbol{x}_{k} \mid \boldsymbol{x}_{k-1}, \check{\boldsymbol{x}}_{0}, \boldsymbol{v}_{1: k}, \boldsymbol{y}_{0: k-1}\right) p\left(\boldsymbol{x}_{k-1} \mid \check{\boldsymbol{x}}_{0}, \boldsymbol{v}_{1: k}, \boldsymbol{y}_{0: k-1}\right) \mathrm{d} \boldsymbol{x}_{k-1}\end{aligned}\tag{6} ==p(xk∣xˇ0,v1:k,y0:k−1)∫p(xk,xk−1∣xˇ0,v1:k,y0:k−1)dxk−1∫p(xk∣xk−1,xˇ0,v1:k,y0:k−1)p(xk−1∣xˇ0,v1:k,y0:k−1)dxk−1(6)
然后,由一阶马尔可夫性假设( k 时刻的状态只和 k - 1 时刻状态有关)有:
p ( x k ∣ x k − 1 , x ˇ 0 , v 1 : k , y 0 : k − 1 ) = p ( x k ∣ x k − 1 , v k ) (7) p\left(\boldsymbol{x}_{k} \mid \boldsymbol{x}_{k-1}, \check{\boldsymbol{x}}_{0}, \boldsymbol{v}_{1: k}, \boldsymbol{y}_{0: k-1}\right)=p\left(\boldsymbol{x}_{k} \mid \boldsymbol{x}_{k-1},\boldsymbol{v}_{k}\right)\tag{7} p(xk∣xk−1,xˇ0,v1:k,y0:k−1)=p(xk∣xk−1,vk)(7)
并且 k - 1 时刻的状态与 k 时刻的输入无关,即:
p ( x k − 1 ∣ x ˇ 0 , v 1 : k , y 0 : k − 1 ) = p ( x k − 1 ∣ x ˇ 0 , v 1 : k − 1 , y 0 : k − 1 ) (8) p\left(\boldsymbol{x}_{k-1} \mid \check{\boldsymbol{x}}_{0}, \boldsymbol{v}_{1: k}, \boldsymbol{y}_{0: k-1}\right) = p\left(\boldsymbol{x}_{k-1} \mid \check{\boldsymbol{x}}_{0}, \boldsymbol{v}_{1: k-1}, \boldsymbol{y}_{0: k-1}\right) \tag{8} p(xk−1∣xˇ0,v1:k,y0:k−1)=p(xk−1∣xˇ0,v1:k−1,y0:k−1)(8)
结合 (5)、(6)、(7)、(8) 式,即可得到:
p ( x k ∣ x ˇ 0 , v 1 : k , y 0 : k ) = η p ( y k ∣ x k ) ∫ p ( x k ∣ x k − 1 , v k ) p ( x k − 1 ∣ x ˇ 0 , v 1 : k − 1 , y 0 : k − 1 ) d x k − 1 (9) \begin{aligned}& p\left(\boldsymbol{x}_{k} \mid \check{\boldsymbol{x}}_{0}, \boldsymbol{v}_{1: k}, \boldsymbol{y}_{0: k}\right) \\=& \eta p\left(\boldsymbol{y}_{ k} \mid \boldsymbol{x}_{k}\right) \int p\left(\boldsymbol{x}_{k} \mid \boldsymbol{x}_{k-1},\boldsymbol{v}_{k}\right)p\left(\boldsymbol{x}_{k-1} \mid \check{\boldsymbol{x}}_{0}, \boldsymbol{v}_{1: k-1}, \boldsymbol{y}_{0: k-1}\right) \mathrm{d} \boldsymbol{x}_{k-1}\end{aligned}\tag{9} =p(xk∣xˇ0,v1:k,y0:k)ηp(yk∣xk)∫p(xk∣xk−1,vk)p(xk−1∣xˇ0,v1:k−1,y0:k−1)dxk−1(9)
即证。