16.4: Calculation: prediction part
16.4.1: Calculation:${{z}}_s
=
\Phi^{s-1,s}_* (\widetilde{{{z}}}_{s-1})$
in (16.9) (in $\S$16.3)
Furthermore,
the (B$_1$) in Lemma 16.3 and (16.6)
imply that
Thus,
we see, by (B$_2$) in Lemma 16.3, that
We prepare the following lemma.
where the notation "$\approx$" means as follows:
\begin{align}
"f(\omega) \approx g(\omega )"
\Longleftrightarrow
\mbox{"there exists a positive $K$ such that $f(\omega) = K g(\omega )\;\;
(\forall \omega \in \Omega)$"}
\end{align}
$(B_1):$
$
\int_{- \infty}^{\infty}
\frac{1}{\sqrt{2 \pi} A}
\exp[
-\frac{(x -By )^2}{2A^2}
]
\frac{1}{\sqrt{2 \pi} C}
\exp[
-\frac{(y- D)^2}{2C^2}
]
dy
=
\frac{1}{\sqrt{2 \pi} \sqrt{A^2 + B^2C^2}}
\exp[
-\frac{(x- BD)^2}{2(A^2 + B^2C^2)}
]
$
$(B_2):$
$
\exp[
-\frac{(A\omega -B )^2}{2E^2}
]
\exp[
-\frac{(C \omega- D)^2}{2F^2}
]
\approx
\exp
[
-\frac{1}{2}
(
\frac{A^2 F^2+ C^2 E^2}{E^2 F^2}
)
\Big(\omega -
\frac{(
{AB}{F^2}
+
{CD}{E^2}
)}{({A^2}{F^2} + {C^2}{E^2})}
\Big)^2
]
$
Proof
It is easy, thus we omit the proof.
We see, by (16.3) and (A), that
\begin{align}
\widetilde{{{z}}}_0(\omega_0)=&\lim_{\Xi_0 \to x_0} \frac{F(\Xi_0) {{z}}_0}{\int_{\mathbb R} F(\Xi_0) {{z}}_0 d \omega_0}
\nonumber \\
\approx
&
\frac{1}{\sqrt{2 \pi} {q}_0}
\exp[
-\frac{(x_0 -c_0 \omega_0 -d_0 )^2}{2{q}_0^2}
]
\frac{1}{\sqrt{2 \pi} \sigma_0}
\exp[-
\frac{(\omega_0-\mu_0)^2}{2 \sigma_0^2}
]
\nonumber \\
\approx
&
\frac{1}{\sqrt{2 \pi} \widetilde{\sigma}_0}
\exp[-
\frac{(\omega_0-\widetilde{\mu}_0)^2}{2 \widetilde{\sigma}_0^2}
]
\tag{16.10}
\end{align}
where
\begin{align}
\widetilde{\sigma}_0^2=\frac{q_0^2 \sigma_0^2}{q_0^2 + c_0^2 \sigma_0^2},
\quad
\widetilde{\mu}_0
=
\mu_0 + \widetilde{\sigma}_0^2 (
\frac{c_0}{q_0^2})(x_0-d_0- c_0 \mu_0)
\tag{16.11}
\end{align}
Furthermore,
we see, by (B$_1$) in Lemma 16.3, that
\begin{align}
{{z}}_t(\omega_{t})
&
=
[\Phi_*^{{t-1},{t}}
\widetilde{{{z}}}_{t-1}](\omega_{t})
\nonumber \\
&
\approx
\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}
]
\frac{1}{\sqrt{2 \pi} \widetilde{\sigma}_{t-1}}
\exp[-
\frac{(\omega_{t-1}-\widetilde{\mu}_{t-1})^2}{2 \widetilde{\sigma}_{t-1}^2}
]
d \omega_{t-1}
\nonumber \\
&
\approx
\frac{1}{\sqrt{2 \pi} {\sigma}_{t}}
\exp[-
\frac{(\omega_{t}-{\mu}_{t})^2}{2 {\sigma_{t}}^2}
]
\tag{16.16}
\end{align}
where
\begin{align}
{\sigma}_t^2=\ a_t^2 \widetilde{\sigma}_{t-1}^2 + r_t^2,
\quad
{\mu}_t
=
a_t \widetilde{\mu}_{t-1} + b_t
\tag{16.17}
\end{align}
Summing up the above (16.10)--(16.17), we see:
\begin{align}
\overset{\mbox{}}{\underset{\mu_0, \sigma_0}{\fbox{${{z}}_0$}}}
\xrightarrow[(6.11)]{\mbox{$x_0$}}
\underset{\widetilde{\mu}_0, \widetilde{\sigma}_0}{\fbox{$\widetilde{{{z}}}_0$}}
\xrightarrow[(16.13)]{\Phi_*^{0,1}}
\overset{\mbox{}}{\underset{\mu_1, \sigma_1}{\fbox{${{z}}_1$}}}
\xrightarrow[]{\mbox{$x_1$}}
\cdots
\xrightarrow[]{\Phi_*^{t-2,t-1}}
\overset{\mbox{}}{\underset{\mu_{t-1}, \sigma_{t-1}}{\fbox{${{z}}_{t-1}$}}}
\xrightarrow[(16.15)]{\mbox{$x_{t-1}$}}
\underset{\widetilde{\mu}_{t-1}, \widetilde{\sigma}_{t-1}}{\fbox{$\widetilde{{{z}}}_{t-1}$}}
\xrightarrow[(16.17)]{\Phi_*^{t-1,t}}
\overset{\mbox{}}{\underset{\mu_t, \sigma_t}{\fbox{${{z}}_t$}}}
\xrightarrow[]{$x_{t+1}$}
\cdots
\xrightarrow[]{\Phi_*^{s-1,s}}
\overset{\mbox{}}{\underset{\mu_s, \sigma_s}{\fbox{${{z}}_s$}}}
\end{align}
And thus,
we get
\begin{align}
{{z}}_s
=
\Phi^{s-1,s}_* (\widetilde{{{z}}}_{s-1})
\tag{16.18}
\end{align}
in (16.9) (in $\S$16.3).
16.4: Calculation: prediction part of Kalman filter
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)
Lemma 16.3
It holds that