递归最小二乘估计
作者：互联网
@[TOC](递归最小二乘估计(Recursive Least Square Estimation))
递归最小二乘估计(Recursive Least Square Estimation)
随着测量次数增加，最小二乘计算量计算量会快速增加，递归最小二乘给出了在线实时计算的递推方程。
矩阵的迹
可参考链接
导数性质
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\begin{aligned} \frac{\partial{Tr(AX^T)}}{\partial{X}} = A \end{aligned}
∂X∂Tr(AXT)=A
推导
x ^ k − 1 , y k \hat x_{k-1},y_k x^k−1,yk分别为k-1时的估计值及k时测量值，线性递归估计可以表示为：
y k = H k x + v k y_k=H_kx+v_k yk=Hkx+vk
x ^ k = x ^ k − 1 + K k ( y k − H k x ^ k − 1 ) \hat x_k = \hat x_{k-1}+K_k(y_k-H_k\hat x_{k-1}) x^k=x^k−1+Kk(yk−Hkx^k−1)
在已知估计量
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k
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1
\hat x_{k-1}
x^k−1和当前测量值
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y_k
yk基础上计算
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\hat x_{k}
x^k。
K
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K_k
Kk为增益矩阵，
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(y_k-H_k\hat x_{k-1})
(yk−Hkx^k−1)为矫正项。如果
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(y_k-H_k\hat x_{k-1})=0
(yk−Hkx^k−1)=0，则
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\hat x_k = \hat x_{k-1}
x^k=x^k−1。
估计量误差均值可以计算如下：
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\begin{aligned} E(\epsilon_{x,k})&= E(x-\hat x) \\ &=E(x-\hat x_{k-1}-K_k(y_k-H_k\hat x_{k-1}))\\ &=E(\epsilon_{x,k-1}-K_k(H_kx+v_k -H_k\hat x_{k-1}))\\ &=E(\epsilon_{x,k-1}-K_kH_k\epsilon_{x,k-1}-K_kv_k)\\ &=(I-K_kH_k)E(\epsilon_{x,k-1})-K_k E(v_k) \end{aligned}
E(ϵx,k)=E(x−x^)=E(x−x^k−1−Kk(yk−Hkx^k−1))=E(ϵx,k−1−Kk(Hkx+vk−Hkx^k−1))=E(ϵx,k−1−KkHkϵx,k−1−Kkvk)=(I−KkHk)E(ϵx,k−1)−KkE(vk)
显然，如果
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E(\epsilon_{x,k-1})=0
E(ϵx,k−1)=0和
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E(v_k)=0
E(vk)=0，则
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(\epsilon_{x,k})=0
(ϵx,k)=0。
但如果测量噪声
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vk均值不为0，估计量的初值设为估计值的期望
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\hat x_0=E(x))
x^0=E(x))，如此对任意k,有
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x_k=E(\hat x_k
xk=E(x^k)，因而最小二乘估计为无偏估计。
求增益矩阵 K k K_k Kk
优化的目标选取最小化k时估计量的方差，即代价函数
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\begin{aligned} J_k&=E[(x_1-\hat x_1)^2+(x_2-\hat x_2)^2+\dots+(x_n-\hat x_n)^2]\\ &=E[\epsilon_{x1,k}^2+\epsilon_{x2,k}^2+\dots+\epsilon_{xn,k}^2]\\ &=E[\epsilon_{x,k}^T\epsilon_{x,k}]\\ &=E[Tr(\epsilon_{x,k}\epsilon_{x,k}^T)]\\ &=Tr(P_k) \end{aligned}
Jk=E[(x1−x^1)2+(x2−x^2)2+⋯+(xn−x^n)2]=E[ϵx1,k2+ϵx2,k2+⋯+ϵxn,k2]=E[ϵx,kTϵx,k]=E[Tr(ϵx,kϵx,kT)]=Tr(Pk)
即协方差矩阵
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P_k=E[\epsilon_{x,k}\epsilon_{x,k}^T]
Pk=E[ϵx,kϵx,kT]，代入得
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\begin{aligned} P_k&=E[\epsilon_{x,k}\epsilon_{x,k}^T]\\ &=E[((I-K_kH_k)E(\epsilon_{x,k-1})-K_k E(v_k))((I-K_kH_k)E(\epsilon_{x,k-1})-K_k E(v_k))^T]\\ &=(I-K_kH_k)E(\epsilon_{x,k-1})E(\epsilon_{x,k-1})^T(I-K_kH_k)^T-(I-K_kH_k)E(\epsilon_{x,k-1})E(v_k)^TK_k^T\\&-K_k E(v_k)E(\epsilon_{x,k-1})^T(I-K_kH_k)^T+K_k E(v_k)E(v_k)^TK_k^T\\ &=(I-K_kH_k)E(\epsilon_{x,k-1}\epsilon_{x,k-1}^T)(I-K_kH_k)^T+K_k E(v_kv_k^T)K_k^T\\ \end{aligned}
Pk=E[ϵx,kϵx,kT]=E[((I−KkHk)E(ϵx,k−1)−KkE(vk))((I−KkHk)E(ϵx,k−1)−KkE(vk))T]=(I−KkHk)E(ϵx,k−1)E(ϵx,k−1)T(I−KkHk)T−(I−KkHk)E(ϵx,k−1)E(vk)TKkT−KkE(vk)E(ϵx,k−1)T(I−KkHk)T+KkE(vk)E(vk)TKkT=(I−KkHk)E(ϵx,k−1ϵx,k−1T)(I−KkHk)T+KkE(vkvkT)KkT
因为
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E(v_k)E(\epsilon_{x,k-1})^T=0
E(vk)E(ϵx,k−1)T=0，所以
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P_k=(I-K_kH_k)P_{k-1}(I-K_kH_k)^T+K_k R_kK_k^T
Pk=(I−KkHk)Pk−1(I−KkHk)T+KkRkKkT
对代价函数关于增益矩阵求偏导，得
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\begin{aligned} {\partial J_k \over \partial K_k}=2(I-K_kH_k)P_{k-1}(-H_k^T)+2K_k R_k \end{aligned}
∂Kk∂Jk=2(I−KkHk)Pk−1(−HkT)+2KkRk
令
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{\partial J_k \over \partial K_k}=0
∂Kk∂Jk=0，得
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\begin{aligned} 2(I-K_kH_k)P_{k-1}(-H_k^T)+2K_k R_k = 0\\ K_k=P_{k-1}H_k^T(H_kP_{k-1}H_k^T+R_k)^{-1} \end{aligned}
2(I−KkHk)Pk−1(−HkT)+2KkRk=0Kk=Pk−1HkT(HkPk−1HkT+Rk)−1
递归最小二乘估计流程
1.初始化
x ^ 0 = E ( x ) \hat x_0=E(x) x^0=E(x)
P 0 = E [ ( x − x ^ 0 ) ( x − x ^ 0 ) T ] P_0=E[(x-\hat x_0)(x - \hat x_0)^T] P0=E[(x−x^0)(x−x^0)T]
2.for k = 1,2,
…
\dots
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K_k=P_{k-1}H_k^T(H_kP_{k-1}H_k^T+R_k)^{-1}
Kk=Pk−1HkT(HkPk−1HkT+Rk)−1
x ^ k = x ^ k − 1 + K k ( y k − H k x ^ k − 1 ) \hat x_k = \hat x_{k-1}+K_k(y_k-H_k\hat x_{k-1}) x^k=x^k−1+Kk(yk−Hkx^k−1)
P k = ( I − K k H k ) P k − 1 ( I − K k H k ) T + K k R k K k T P_k=(I-K_kH_k)P_{k-1}(I-K_kH_k)^T+K_k R_kK_k^T Pk=(I−KkHk)Pk−1(I−KkHk)T+KkRkKkT
替代形式
标签：vk,递归,Kk,epsilon,kH,最小,Hk,hat,二乘 来源： https://blog.csdn.net/weixin_42170603/article/details/106504553