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    1. Kalman Filter Tutorial (Cont'd...)

    1.3.1 Derivation of Kalman filter (cont'd...)

    Now, we have to expand $P_{k|k}$, (last expression in earlier page) and regrouped linear and quadratic term. i.e. \begin{equation}\label{3n} P_{k|k}=P_{k|k-1}-\underbrace{(K_{k}H_{k}P_{k|k-1}+P_{k|k-1}H_{k}^{T}K_{k}^{T})}_{Linear \ in\ k}+\underbrace{K_{k}(H_{k}P_{k|k-1}H_{k}^{T}+R_{k})K_{k}^{T}}_{Quadratic \ in\ k}, \end{equation} where, $P_{k|k}$ is quadratic only when $(H_{k}P_{k|k-1}H_{k}^{T}+R_{k})$ is symmetric and positive definite. Let us assume \begin{equation}\label{3o} A_{k}=H_{k}P_{k|k-1}H_{k}^{T}+R_{k}. \end{equation} To substitute the value of $A_{k}$ in the last expression of $P_{k|k}$, we obtain \begin{equation}\label{3p} P_{k|k}=P_{k|k-1}-K_{k}H_{k}P_{k|k-1}-P_{k|k-1}H_{k}^{T}K_{k}^{T}+K_{k}A_{k}^{T}K_{k}^{T}. \end{equation} Now, we have to minimize $P_{k|k}$, which may be called as cost function. So, we differentiate $P_{k|k}$ with $K_{k}$ and equate it with zero. \begin{equation*} \dfrac{\partial P_{k|k}}{\partial K_{k}}=-(H_{k}P_{k|k-1})^{T}-P_{k|k-1}H_{k}^{T}+2K_{k}A_{k}=0, \end{equation*} where, $P_{k|k-1}$ is symmetric matrix. Now, from the above equation, we could write $K_{k}$ as: \begin{equation}\label{3q} K_{k}=P_{k|k-1}H_{k}^{T}(H_{k}P_{k|k-1}H_{k}^{T}+R_{k})^{-1}. \end{equation} This is the expression of Kalman gain matrix. Now, substituting the value of Kalman gain matrix to the expression of $P_{k|k}$, we reach \begin{equation}\label{3r} P_{k|k}=(I-K_{k}H_{k})P_{k|k-1}. \end{equation} As the variance is minimized, we call it as minimum variance estimator.

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