16.2: Problem establishment (concrete calculation)

In the previous section, we study the general theory of Kalman filter. In this section, we devote ourselves to the calculation of Kalman filter in the case of a linear ordered tree $T=\{0,1,2, \cdots, n\}$ such that the parent map $\pi : T\setminus \{0\} \to T$ is defined by $\pi(k) = k-1$:

\begin{align} 0 \xleftarrow[]{\quad \pi \quad} 1 \xleftarrow[]{\quad \pi \quad} 2 \xleftarrow[]{\quad \pi \quad} \cdots \xleftarrow[]{\quad \pi \quad} n-1 \xleftarrow[]{\quad \pi \quad} n \end{align}
Figure 16.3: Linear ordered tree

For each $k \in T$, consider the classical basic structure: \begin{align} [ C_0(\Omega_k ) \subseteq L^\infty ( \Omega_k, m_k) \subseteq B(L^\infty ( \Omega_k, m_k)) ] \Big( = [ C_0 ({\mathbb R}) \subseteq L^\infty ( {\mathbb R}, d \omega ) \subseteq B(L^2 ( {\mathbb R}, d \omega ))] \Big) \end{align}

where $d \omega$ is the Lebesgue measure on ${\mathbb R}$.

Consider the {sequential causal observable} $[{\mathbb O}_T] $ $=$ $[ \{{\mathsf O}_t \}_{t \in T}$, $\{ \Phi^{t-1, t} : L^\infty (\Omega_t) \to L^\infty (\Omega_{t-1}) \}_{T=1,2, \cdots, n} $ $]$, and assume the initial state ${{z}}_0 \in L^1_{+1} (\Omega_0 , m_0 )$.

Thus, we have the following situation: \begin{align} \overset{\mbox{initial state ${{z}}_0$}}{\underset{{\mathsf O}_0=(X_0, {\mathcal F}_0 F_0)}{\fbox{$L^\infty(\Omega_0, m_0 )$}}} \xleftarrow[]{\Phi^{0,1}} \underset{{\mathsf O}_1=(X_1, {\mathcal F}_1 F_1)}{\fbox{$L^\infty(\Omega_1, m_1 )$}} \xleftarrow[]{\Phi^{1,2}} \cdots \xleftarrow[]{\Phi^{s-1,s}} \underset{{\mathsf O}_s=(X_s, {\mathcal F}_s F_s)}{\fbox{$L^\infty(\Omega_s, m_s )$}} \xleftarrow[]{\Phi^{s,s+1}} \cdots \xleftarrow[]{\Phi^{n-1,n}} \underset{{\mathsf O}_n=(X_n, {\mathcal F}_n F_n)}{\fbox{$L^\infty(\Omega_n, m_n )$}} \end{align} or, equivalently, \begin{align} \overset{\mbox{initial state ${{z}}_0$}}{\underset{{\mathsf O}_0=(X_0, {\mathcal F}_0, F_0)}{\fbox{$L^1(\Omega_0, m_0 )$}}} \xrightarrow[]{\Phi_*^{0,1}} \underset{{\mathsf O}_1=(X_1, {\mathcal F}_1, F_1)}{\fbox{$L^1(\Omega_1, m_1 )$}} \xrightarrow[]{\Phi_*^{1,2}} \cdots \xrightarrow[]{\Phi_*^{s-1,s}} \underset{{\mathsf O}_s=(X_s, {\mathcal F}_s, F_s)}{\fbox{$L^1(\Omega_s, m_s )$}} \xrightarrow[]{\Phi_*^{s,s+1}} \cdots \xrightarrow[]{\Phi_*^{n-1,n}} \underset{{\mathsf O}_n=(X_n, {\mathcal F}_n, F_n)}{\fbox{$L^1(\Omega_n, m_n )$}} \end{align}
In the above, the initial state ${{z}}_0 (\in L^1_{+1} (\Omega_0 , m_0 ))$ is defined by \begin{align} {{z}}_0(\omega_0) = \frac{1}{\sqrt{2 \pi} \sigma_0} \exp[- \frac{(\omega_0-\mu_0)^2}{2 \sigma_0^2} ] \qquad( \forall \omega_0 \in \Omega_0) \tag{16.3} \end{align}

where it is assumed that $\mu_0$ and $\sigma_0$ are known.

Also, for each $t \in T =\{0,1, \cdots, n \}$, consider the observable ${\mathsf O}_t=(X_t, {\mathcal F}_t, F_t)$ $=({\mathbb R}, {\mathcal B}_{{\mathbb R}}, F_t )$ in $L^\infty ( \Omega_t, m_t )$ such that

\begin{align} & [F_t (\Xi_t )](\omega_t) = \int_{\Xi_t} \frac{1}{\sqrt{2 \pi} {q}_t} \exp[ -\frac{(x_t -c_t \omega_t -d_t)^2}{2{q}_t^2} ] dx_t \equiv \int_{\Xi_t} f_{x_t}(\omega_t) dx_t \quad(\forall \Xi_t \in {\mathcal F}_t, \;\; \forall \omega_t \in \Omega_t ) \\ & {\quad} \tag{16.4} \end{align}

where it is assumed that $c_t$, $d_t$ and $q_t$ are known $(t \in T)$.

And further, the causal operator $\Phi^{t-1,t}:L^\infty (\Omega_t) \to L^\infty ( \Omega_{t-1})$ is defined by

\begin{align} & [\Phi^{t-1,t} \widetilde{f}_{x_{t}}](\omega_{t-1} ) = \int_{- \infty}^{\infty} \frac{1}{\sqrt{2 \pi} r_t} \exp[ -\frac{(\omega_t - a_t \omega_{t-1} -b_t )^2}{2r_t^2} ] \widetilde{f}_{x_{t}} ) d \omega_{t} \equiv f_{t-1}(\omega_{t-1}) \tag{16.5} \\ & \qquad \qquad \qquad (\forall \widetilde{f}_{x_{t}} \in L^\infty(\Omega_{t}, m_{t}), \;\; \forall \omega_{t-1} \in \Omega_{t-1}) \nonumber \end{align}

where it is assumed that $a_t$, $b_t$ and $r_t$ are known $(t \in T)$.

Or, equivalently, the pre-dual causal operator $\Phi_*^{t-1,t}:L^1_{+1} (\Omega_{t-1}) \to L^1_{+1} ( \Omega_{t})$ is defined by

\begin{align} & [\Phi^{t-1,t}_* \widetilde{{{z}}}_{t-1}](\omega_t ) = \int_{- \infty}^{\infty} \frac{1}{\sqrt{2 \pi} r_t} \exp[ -\frac{(\omega_t - a_t \omega_{t-1} -b_t )^2}{2r_t^2} ] \widetilde{{{z}}}_{t-1}(\omega_{t-1}) d\omega_{t-1} \tag{16.6} \\ & \qquad \qquad \qquad (\forall \widetilde{{{z}}}_{t-1} \in L^1_{+1}(\Omega_{t-1}, m_{t-1}), \; \forall \omega_t \in \Omega_t) \nonumber \end{align}

Now we have the sequential causal observable

\begin{align} [{\mathbb O}_T] = [ \{{\mathsf O}_t \}_{t \in T}, \{ \Phi^{t-1, t} : L^\infty (\Omega_{t}) \to L^\infty (\Omega_{t-1}) \}_{T=1,2, \cdots, n} \end{align}

Let $\widehat{\mathsf O}_{0}$ $(\times_{t=0}^n X_t, \boxtimes_{t=0}^n {\mathcal F}_t, {\widehat F} )$ be its realization. Then we have the following problem:

Problem16.2 [Kalman filter; calculation]
$\quad$ Assume that a measured value $(x_0, x_2, \cdots, x_n )$ $(\in \times_{t=0}^n X_t)$ is obtained by the measurement ${\mathsf M}_{L^\infty (\Omega_0)}$ $(\widehat{\mathsf O}_{0},$ $ \overline{S}_{[\ast]}({{z}}_0 ) )$. Let $s(\in T)$ be fixed. Then, calculate the Bayes-Kalman operator $[B_{\widehat{\mathsf{O}}_{0} }^s(\times_{t \in T} \{x_t\})] ( {{z}}_0 )$ in (6.2), where \begin{align} [B_{\widehat{\mathsf{O}}_{0} }^s(\times_{t \in T} \{x_t\})] ( {{z}}_0 ) = {{z}}_s^a = \lim_{\Xi_t \to x_t \;(t\in T)} [B_{\widehat{\mathsf{O}}_{0} }^s(\times_{t \in T} \Xi_t)]({{z}}_0) \end{align} That is, \begin{align} L^1_{+1}(\Omega_0) \ni {{z}}_0 \xrightarrow[B_{\widehat{\mathsf{O}}_{0} }^s(\times_{t \in T} \{x_t\})]{\mbox{ measured value:}(x_0,x_1,...,x_n)} {{z}}_s^a \in L^1_{+1}(\Omega_s) \end{align}