16.1: Bayes=Kalman method (in $L^\infty(\Omega, m)$)

Recall Theorem 9.11 (Bayes' theorem), which will be generalized as Bayes=Kalman operator as follows.

Let $t_0$ be the root of a tree $T$. For each $t \in T$, consider the classical basic structure:

\begin{align} [C_0(\Omega_t) \subseteq L^\infty ( \Omega_t, m_t) \subseteq B(L^2(\Omega_t, m_t ) ) ] \end{align}

Let $[{\mathbb O}_T{}]$ $=$ $[{} \{ {\mathsf{O}}_t (\equiv (X_t ,$ ${\cal F}_{t} , {F}_t )) \}_{ t \in T} , \{ \Phi^{t_1,t_2}{}:$ $L^\infty (\Omega_{t_2}) \to L^\infty (\Omega_{t_1}) \}_{(t_1,t_2) \in T^2_\le }$ $]$ be a sequential causal observable with the realization $\widehat{\mathsf{O}}_{t_0}$ $\equiv$ $(\times_{t \in T } X_t ,$ $\boxtimes_{t \in T } {\cal G}_t,$ ${\widehat F}_{t_0})$ in $L^\infty (\Omega_{t_0})$.

For example, For each $t \in T$, consider another observable ${\mathsf O}_t'=( Y_t, {\mathcal G}_t, G_t)$ in $L^\infty(\Omega_t, m_t )$, and the simultaneous observable ${\mathsf O} \times {\mathsf O}_t'=( X_t \times Y_t, {\mathcal F}_t \boxtimes {\mathcal G}_t, F_t \times G_t)$ in $L^\infty(\Omega_t, m_t )$. And let $[{\mathbb O}_T^\times{}]$ $=$ $[{} \{ {\mathsf{O}}_t^\times (\equiv (X_t \times Y_t ,$ ${\cal F}_{t} \boxtimes {\cal G}_t , {F}_t \times G_t )) \}_{ t \in T} , \{ \Phi^{t_1,t_2}{}:$ $L^\infty (\Omega_{t_2}) \to L^\infty (\Omega_{t_1}) \}_{(t_1,t_2) \in T^2_\le }$ $]$ be a sequential causal observable with the realization $\widehat{\mathsf{O}}_{t_0}^\times$ $\equiv$ $(\times_{t \in T } (X_t \times Y_t) ,$ $\boxtimes_{t \in T } ({\cal F}_{t} \boxtimes {\cal G}_t),$ ${\widehat H}_{t_0})$ in $L^\infty (\Omega_{t_0})$.

For example, Thus we have the mixed measurement ${\mathsf M}_{L^\infty (\Omega_{t_0})} (\widehat{\mathsf{O}}_{t_0}^\times, \overline{S}_{[\ast]}({{z}}_0 ) )$, where ${{z}}_0 \in L^1_{+1} (\Omega_{t_0} )$. Assume that we know that the measured value $(x,y)$ $( = ( (x_t)_{t\in T}, (y_t)_{t\in T}, ) \in (\times_{t \in T} X_t) \times (\times_{t \in T} Y_t) )$ obtained by the measurement ${\mathsf M}_{L^\infty (\Omega_{t_0})} (\widehat{\mathsf{O}}_{t_0}^\times, \overline{S}_{[\ast]}({{z}}_0 ) )$ belongs to $(\times_{t \in T} \Xi_t) \times$ $(\times_{t \in T}Y_t) \;$ $(\in (\boxtimes_{t \in T}{\mathcal F}_t) \boxtimes (\boxtimes_{t \in T} {\mathcal G}_t) )$. Then, by Axiom${}^{{ (m)}}$ 1($\S$9.1), we can infer that

 $(A):$ the probability $P_{\times_{t \in T} \Xi_t} ( (G_t (\Gamma_t))_{t \in T} )$ that $y$ belongs to $\times_{t \in T} \Gamma_t (\in \boxtimes_{t \in T} {\cal G}_t)$ is given by \begin{align} & P_{\times_{t \in T} \Xi_t} ( (G_t (\Gamma_t))_{t \in T} ) \nonumber \\ = & \frac{\int_{\Omega_0} [ {\widehat H}_{t_0} ( (\times_{t \in T} \Xi_t) \times (\times_{t \in T} \Gamma_t) ) ] (\omega_0) \; {{z}}_0 (\omega_0 ) \; m_0 (d \omega_0) }{ \int_{\Omega_0} [ {\widehat H}_{t_0} (\times_{t \in T} \Xi_t) \times (\times_{t \in T}Y_t) ](\omega_0) \; {{z}}_0 (\omega_0 ) \; m_0(d \omega_0 )} \tag{16.1} \\ & \quad (\forall \Gamma_t \in {\cal G}_t, t \in T). \nonumber \end{align}
Let $s \in T$ be fixed. Assume that \begin{align} \Gamma_t = Y_t \quad (\forall t \in T \mbox{ such that $t \not= s$}) \end{align}

Thus, putting ${\widehat P}_{\times_{t \in T} \Xi_t} ( G_s (\Gamma_s) )=P_{\times_{t \in T} \Xi_t} ( (G_t (\Gamma_t))_{t \in T} )$, we see that ${\widehat P}_{\times_{t \in T} \Xi_t} \in L^1_{+1}(\Omega_s, m_s )$. That is, there uniquely exists ${{z}}_s^a \in L^1_{+1} (\Omega_s , m_s)$ such that

\begin{align} & {\widehat P}_{\times_{t \in T} \Xi_t} ( (G_s (\Gamma_s)) = {}_{{}_{L^1(\Omega_s)}} \langle {{z}}_s^a, G_s (\Gamma_s) \rangle {}_{{}_{L^\infty(\Omega_s)}} = \int_{\Omega_s} [G_s (\Gamma_s)](\omega_s ) {{z}}_s^a(\omega_s ) m_s ( d \omega_s ) \nonumber \end{align}

for any observable $(Y_s, {\cal G}_s, G_s)$ in $L^\infty (\Omega_s)$. That is because the linear functional ${\widehat P}_{\times_{t \in T} \Xi_t} : L^\infty(\Omega_s ) \to {\mathbb C}$ (complex numbers) is weak$^*$ continuous. After all,

 $(B):$ we can define the Bayes-Kalman operator $[B_{\widehat{\mathsf{O}}_{t_0} }^s(\times_{t \in T} \Xi_t)]: L^1_{+1}(\Omega_{t_0})$ $\to L^1_{+1}( \Omega_s)$ such that \begin{align} \quad \overset{\mbox{(pretest state)}}{ \underset{ (\in L^1_{+1}(\Omega_{t_0}))} {\fbox{${{{z}}_0}$ }} } \xrightarrow[\scriptsize{\mbox{Bayes-Kalman operator}}]{\mbox{ $\qquad [B_{\widehat{\mathsf{O}}_{t_0} }^s(\times_{t \in T} \Xi_t)] \qquad$}} \overset{\mbox{(posttest state)}}{ \underset{( \in L^1_{+1}(\Omega_s))} {\fbox{${{{z}}_s^a}$ }} } \tag{16.2} \end{align}

which is the generalization of the Bayes operator (9.5).

Remark 16.1 We have frequently discussed the Bayes=Kalman filter, for example, in ref.  in $\S$0.0. However, these arguments are too theoretical. In this chapter, we devote ourselves to the numerical aspect of the Kalman filter.