前節の問題16.2を再掲しておく。
さて,問題16.2を解こう.
このためには,
次の
$z_s \in L^1_{+1} (\Omega_s )$
を計算すればよい.
\begin{align}
\small{
\lim_{ \Xi_t \to x_t \;\;(t\in T)}
\frac{\int_{\Omega_0} [{\widehat F}_{0}(({{{\times}}}_{t=0}^n \Xi_{t}) \times \Gamma_s)](\omega_0) \;\;
z_0 (\omega_0 ) d\omega_0}{\int_{\Omega_0} [{\widehat F}_{0}({{{\times}}}_{t=0}^n \Xi_{t}) ](\omega_0)\;\;
z_0 (\omega_0 ) d\omega_0}
=
\int_{\Omega_s} [G_s(\Gamma_s )](\omega_s)
\;\;
z_s (\omega_s ) d \omega_s
\quad (\forall \Gamma_s \in {\mathcal F}_s )
}
\end{align}
以下に,
$z_s= [B_{\widehat{\mathsf{O}}_{0} }^s ({{{\times}}}_{t \in T} \{x_t \})]({z}_0)$
を計算しよう.
\begin{align}
&
\int_{\Omega_0} [{\widehat F}_{0}(({{{\times}}}_{t=0}^n \Xi_{t}) \times \Gamma_s)](\omega_0) \;\;
z_0 (\omega_0 ) d\omega_0
\nonumber \\
=
&
{}_{{}_{L^1(\Omega_0)}} \langle z_0,
{\widehat F}_{0}(({{{\times}}}_{t=0}^n \Xi_{t}) \times \Gamma_s)
\rangle {}_{{}_{L^\infty(\Omega_0)}}
\nonumber \\
=
&
{}_{{}_{L^1(\Omega_1)}} \langle \Phi^{0,1}_* (F_0(\Xi_0) z_0),
{\widehat F}_{1}(({{{\times}}}_{t=1}^n \Xi_{t}) \times \Gamma_s)
\rangle {}_{{}_{L^\infty(\Omega_1)}}
\tag{16.7}
\end{align}
\begin{align}
(16.7)
=
&
{}_{{}_{L^1(\Omega_1)}} \langle \Phi^{0,1}_* (\widetilde{z}_0),
{\widehat F}_{1}(({{{\times}}}_{t=1}^n \Xi_{t}) \times \Gamma_s)
\rangle {}_{{}_{L^\infty(\Omega_1)}}
\nonumber \\
=
&
{}_{{}_{L^1(\Omega_2)}} \langle \Phi^{1,2}_* ( \widetilde{z}_1),
{\widehat F}_{2}(({{{\times}}}_{t=2}^n \Xi_{t}) \times \Gamma_s)
\rangle {}_{{}_{L^\infty(\Omega_2)}}
\nonumber \\
&
\cdots \cdots
\nonumber \\
=
&
{}_{{}_{L^1(\Omega_{s+1})}} \langle
\Phi^{s,s+1}_* ( \widetilde{z}_{s}),
{\widehat F}_{s+1}(({{{\times}}}_{t=s+1}^n \Xi_{t}) \times \Gamma_s)
\rangle {}_{{}_{L^\infty(\Omega_{s+1})}}
\nonumber \\
=
&
{}_{{}_{L^1(\Omega_{s})}} \langle
\Phi^{s-1,s}_* ( \widetilde{z}_{s-1}),
{\widehat F}_{s}(({{{\times}}}_{t=s}^n \Xi_{t}) \times \Gamma_s)
\rangle {}_{{}_{L^\infty(\Omega_{s})}}
\nonumber \\
=
&
{}_{{}_{L^1(\Omega_{s})}} \langle
\Phi^{s-1,s}_* (\widetilde{z}_{s-1}),
F_{s} (\Xi_{s})G_{s} (\Gamma_{s})
\Phi^{s,s+1}{\widehat F}_{s+1}({{{\times}}}_{t=s+1}^n \Xi_{t})
\rangle {}_{{}_{L^\infty(\Omega_{s})}}
\nonumber \\
=
&
{}_{{}_{L^1(\Omega_{s})}} \langle
\Big(
F_{s} (\Xi_{s})
\Phi^{s,s+1}{\widehat F}_{s+1}({{{\times}}}_{t=s+1}^n \Xi_{t})
\Big)
\Big(
\Phi^{s-1,s}_*
(\widetilde{z}_{s-1})
\Big),
G_{s} (\Gamma_{s})
\rangle {}_{{}_{L^\infty(\Omega_{s})}}
\tag{16.8}
\end{align}
となる.
よって,次を得る.
\begin{align}
\small{
[B_{\widehat{\mathsf{O}}_{0} }^s ({{{\times}}}_{t \in T} \{x_t \})](z_0)
=
\lim_{ \Xi_t \to x_t \;\;(t\in T)}
\frac{
\Big(
F_{s} (\Xi_{s})
\Phi^{s,s+1}{\widehat F}_{s+1}({{{\times}}}_{t=s+1}^n \Xi_{t})
\Big)
\times
\Big(
\Phi^{s-1,s}_* \widetilde{z}_{s-1})
\Big)
}
{\int_{\Omega_0} [{\widehat F}_{0}({{{\times}}}_{t=0}^n \Xi_{t}) ](\omega_0)\;\;
z_0 (\omega_0 ) d\omega_0}
}
\tag{16.9}
\end{align}
$(A):$
ここで,
$\widetilde{z}_0=F_0(\Xi_0) z_0$
(もっと正確には,その正規形
$\widetilde{z}_0=\lim_{\Xi_0 \to x_0}
\frac{F_0(\Xi_0) z_0}{\int_{\Omega_0}{F_0(\Xi_0) z_0} d\omega_0}$)
,
$\widetilde{z}_1=F_1(\Xi_1) \Phi^{0,1}_* (\widetilde{z}_0)$,
$\widetilde{z}_2=F_2(\Xi_2) \Phi^{1,2}_* ( \widetilde{z}_1)$,
$\cdots$ ,
$\widetilde{z}_{s-1}=F_{s-1}(\Xi_{s-1}) \Phi^{s-2,s-1}_* (\widetilde{z}_{s-2})$
とおいて,さらに計算を進めて,
16.3: ベイズ-カルマン作用素 $B_{\widehat{\mathsf{O}}_{0} }^s ({{{\times}}}_{t \in T} \{x_t \})$
This web-site is the html version of "Linguistic Copehagen interpretation of quantum mechanics; Quantum language [Ver. 4]" (by Shiro Ishikawa; [home page] )
PDF download : KSTS/RR-18/002 (Research Report in Dept. Math, Keio Univ. 2018, 464 pages)
問題16.2 [カルマンフィルタ; 計算]
測定
${\mathsf M}_{L^\infty (\Omega_0)}$
$(\widehat{\mathsf O}_{0},$
$
\overline{S}_{[\ast]}(z_0 )
)$
によって,測定値
$(x_0, x_2, \cdots, x_n )$
$(\in {{{\times}}}_{t=0}^n X_t)$
が得られたとしよう.
$s(\in T)$を固定する.
このとき,(16.2)式で定めた
ベイズ-カルマン作用素
$[B_{\widehat{\mathsf{O}}_{0} }^s({{{\times}}}_{t \in T} \{x_t\})]
( z_0 )$
を計算せよ.すなわち,
\begin{align*}
[B_{\widehat{\mathsf{O}}_{0} }^s({{{\times}}}_{t \in T} \{x_t\})]
( z_0 )
=
\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*}
を計算せよ.
すなわち,
\begin{align*}
L^1_{+1}(\Omega_0)
\ni
{{z}}_0
\xrightarrow[B_{\widehat{\mathsf{O}}_{0} }^s({{{\times}}}_{t \in T} \{x_t\})]{\mbox{測定値 :}(x_0,x_1,...,x_n)}
{{z}}_s^a
\in
L^1_{+1}(\Omega_s)
\end{align*}
を計算せよ。