最大后验估计MAP
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\hat{x}=\arg \max _{x}p\left( x|u,y\right)
x^=argxmax?p(x∣u,y)
我们希望在给定先验信息和所有时刻的输入
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\begin{aligned} x&=x_{0:K}=\left( x_{0},\cdots ,x_{K}\right) \\ u&=(\check{x}_{0},u_{1:k}) =\left( \check{x}_{0},u_{1},\cdots ,u_{K}\right)\\ y&=y_{0:K}=\left( y_{0},\cdots ,y_{K}\right) \end{aligned}
xuy?=x0:K?=(x0?,?,xK?)=(xˇ0?,u1:k?)=(xˇ0?,u1?,?,uK?)=y0:K?=(y0?,?,yK?)?
用贝叶斯公式重写MAP估计：
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\hat{x}=\arg \max _{x}p\left( x| u,y\right) =\arg \max _{x}\dfrac{p\left( y|x,u\right) p\left( x|u\right) }{p\left( y| u\right) }=\arg \max _{x}p\left( y| x\right) p\left( x| u\right)
x^=argxmax?p(x∣u,y)=argxmax?p(y∣u)p(y∣x,u)p(x∣u)?=argxmax?p(y∣x)p(x∣u)
这里我们吧分母略去，因为它与
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p(y∣x,u)中的
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接下来我们做出一个重要假设：对于所有时刻
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p(y∣x)进行因子分解：
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\begin{aligned} p\left( y| x\right) &=p\left( y_{0}| x_{0:K}\right) p\left( y_{1}| x_{0:K},y_{0}\right) \cdots p\left( y_{k}| x_{0:K},y_{0:K-1}\right) \\ &=\prod ^{K}_{k=0}p\left( y_{k}| x_{k}\right) \end{aligned}
p(y∣x)?=p(y0?∣x0:K?)p(y1?∣x0:K?,y0?)?p(yk?∣x0:K?,y0:K?1?)=k=0∏K?p(yk?∣xk?)?
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\begin{aligned} p\left( x| u\right) &=p\left( x_{0}| u\right) p\left( x_{1}| u,x_{0}\right) \ldots p\left( x_{K}| u,x_{0:K-1}\right) \\ &=p\left( x_{0}| \check{x}_{0}\right) \prod ^{K}_{k=1}p\left( x_{k}| x_{k-1},u_{k}\right) \end{aligned}
p(x∣u)?=p(x0?∣u)p(x1?∣u,x0?)…p(xK?∣u,x0:K?1?)=p(x0?∣xˇ0?)k=1∏K?p(xk?∣xk?1?,uk?)?
p ( x 0 ∣ x ˇ 0 ) = 1 ( 2 π ) N det ? P ˇ 0 × exp ? ( ? 1 2 ( x 0 ? x ˇ 0 ) T P ˇ 0 ? 1 ( x 0 ? x ˇ 0 ) ) p\left( x_{0}| \check{x}_{0}\right) =\dfrac{1}{\sqrt{\left( 2\pi \right) ^{N}\det \check{P}_{0}}}\times \exp \left( -\dfrac{1}{2}\left( x_{0}-\check{x}_{0}\right) ^{T}\check{P}_{0}^{-1}\left( x_{0}-\check{x}_{0}\right) \right) p(x0?∣xˇ0?)=(2π)NdetPˇ0??1?×exp(?21?(x0??xˇ0?)TPˇ0?1?(x0??xˇ0?))
p ( x k ∣ x k ? 1 , u k ) = 1 ( 2 π ) N det ? Q k × exp ? ( ? 1 2 ( x k ? f ( x k ? 1 , u k , 0 ) ) T Q k ? 1 ( x k ? f ( x k ? 1 , u k , 0 ) ) ) p\left( x_{k}| x_{k-1},u_{k}\right) =\dfrac{1}{\sqrt{\left( 2\pi \right) ^{N}\det Q_{k}}}\times \exp \left( -\dfrac{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(xk?∣xk?1?,uk?)=(2π)NdetQk??1?×exp(?21?(xk??f(xk?1?,uk?,0))TQk?1?(xk??f(xk?1?,uk?,0)))
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p\left( y_{k}| x_{k}\right) =\dfrac{1}{\sqrt{\left( 2\pi \right) ^{M}\det R_{k}}}\times \exp \left( -\dfrac{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)
p(yk?∣xk?)=(2π)MdetRk??1?×exp(?21?(yk??h(xk?,0))TRk?1?(yk??h(xk?,0)))
对等式两侧取对数：
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\ln \left( p\left( y| x\right) p\left( x|u\right) \right) =\ln p\left( x_{0}| \check{x}_{0}\right) +\sum ^{k}_{k=1}\ln p\left( x_{k}| x_{k-1},u_{k}\right) +\sum ^{k}_{k=0}\ln p\left( y_{k}| x_{k}\right)
ln(p(y∣x)p(x∣u))=lnp(x0?∣xˇ0?)+k=1∑k?lnp(xk?∣xk?1?,uk?)+k=0∑k?lnp(yk?∣xk?)
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\ln p\left( x_{0}| \check{x}_{0}\right) =-\dfrac{1}{2}\left( x_{0}-\check{x}_{0}\right) ^{T}P_{0}^{-1}\left( x_{0}-\check{x}_{0}\right) -\underbrace{\dfrac{1}{2}\ln \left( \left( 2\pi \right) ^{N}\det \check{P}_{0}\right)}_{与x无关}
lnp(x0?∣xˇ0?)=?21?(x0??xˇ0?)TP0?1?(x0??xˇ0?)?与x无关
21?ln((2π)NdetPˇ0?)??
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\ln p\left( x_{k}| x_{k-1},u_{k}\right) =-\dfrac{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) -\underbrace{\dfrac{1}{2}\ln \left( \left( 2\pi \right) ^{N}\det Q_{k}\right)}_{与x无关}
lnp(xk?∣xk?1?,uk?)=?21?(xk??f(xk?1?,uk?,0))TQk?1?(xk??f(xk?1?,uk?,0))?与x无关
21?ln((2π)NdetQk?)??
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\ln p\left( y _{k}| x_{k}\right) =-\dfrac{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) -\underbrace{\dfrac{1}{2}\ln \left( \left( 2\pi \right) ^{M}\det R_{k}\right)}_{与x无关}
lnp(yk?∣xk?)=?21?(yk??h(xk?,0))πRk?1?(yk??h(xk?,0))?与x无关
21?ln((2π)MdetRk?)??