16.4.1: 予測部分の計算： $z_s = \Phi^{s-1,s}_* (\widetilde{z}_{s-1})$ in (16.9)(in $\S$16.3)


次の補題を用意しておく.
補題16.3 次が成立する.
$(B_1):$ $ \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 ] $
ここで,　記号"$\approx${\rq\rq}の意味は,以下の通りとする. \begin{align*} "f(\omega) \approx g(\omega )" \Longleftrightarrow \mbox{"$\frac{f(\omega)}{g(\omega )} = $定数" } \end{align*}


証明 証明は,初等的で簡単なので,省略する.



さて, (16.3)と(D)から, 次を得る. \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} ここで, \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}

さらに, 補題16.3の (B$_1$)と(16.6) から, 次が言える. \begin{align} z_1(\omega_1) & = [\Phi_*^{0,1} \widetilde{z}_0](\omega_1) \nonumber \\ & = \int_{- \infty}^{\infty} \frac{1}{\sqrt{2 \pi} r_1} \exp[ -\frac{(\omega_1 - a_1 \omega_{0} -b_1 )^2}{2r_1^2} ] \frac{1}{\sqrt{2 \pi} \widetilde{\sigma}_0} \exp[- \frac{(\omega_0-\widetilde{\mu}_0)^2}{2 \widetilde{\sigma}_0^2} ] d \omega_0 \nonumber \\ & = \frac{1}{\sqrt{2 \pi}{\sigma}_1} \exp[- \frac{(\omega_1-{\mu}_1)^2}{2 {\sigma_1}^2} ] \tag{16.12} \end{align} ここで, \begin{align} {\sigma}_1^2=\ a_1^2 \widetilde{\sigma}_0^2 + r_1^2, \quad {\mu}_1 = a_1 \widetilde{\mu}_0 + b_1 \tag{16.13} \end{align}

したがって, 補題16.3の(B$_2$)から,次を得る. \begin{align} \widetilde{z}_{t-1}(\omega_{t-1}) = & \lim_{\Xi_{t-1} \to x_{t-1}} \frac{F(\Xi_{t-1}) z_{t-1}}{\int_{\mathbb R} F(\Xi_{t-1}) z_{t-1} d \omega_{t-1}} \nonumber \\ \approx & \frac{1}{\sqrt{2 \pi} {q}_{t-1}} \exp[ -\frac{(x_{t-1} -c_{t-1} \omega_{t-1} - d_{t-1})^2}{2{q}_{t-1}^2} ] \frac{1}{\sqrt{2 \pi} \sigma_{t-1}} \exp[- \frac{(\omega_{t-1}-\mu_{t-1})^2}{2 \sigma_{t-1}^2} ] \nonumber \\ \approx & \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} ] \tag{16.14} \end{align} ここで, \begin{align} \widetilde{\sigma}_{t-1}^2 &=\frac{q_{t-1}^2 \sigma_{t-1}^2}{q_{t-1}^2 + c_{t-1}^2 \sigma_{t-1}^2} = \sigma_{t-1}^2 \frac{q_{t-1}^2 + c_{t-1}^2 \sigma_{t-1}^2 +q_{t-1}^2 -q_{t-1}^2 - c_{t-1}^2 \sigma_{t-1}^2}{q_{t-1}^2 + c_{t-1}^2 \sigma_{t-1}^2} \nonumber \\ & = \sigma_{t-1}^2 (1- \frac{ c_{t-1}^2 \sigma_{t-1}^2}{q_{t-1}^2 + c_{t-1}^2 \sigma_{t-1}^2} ) \quad \nonumber \\ \widetilde{\mu}_{t-1} & = \mu_{t-1} + \widetilde{\sigma}_{t-1}^2 ( \frac{c_{t-1}}{q_{t-1}^2})(x_{t-1}- c_{t-1} \mu_{t-1}) \tag{16.15} \end{align}

さらに,補題16.3の(E$_1$)により, \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}

以上の(16.10)--(16.17)をまとめると, \begin{align*} {\small \overset{\mbox{}}{\underset{\mu_0, \sigma_0}{\fbox{$z_0$}}} \xrightarrow[(16.11)]{\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[]{\mbox{$x_{t+1}$}} \cdots \xrightarrow[]{\Phi_*^{s-1,s}} \overset{\mbox{}}{\underset{\mu_s, \sigma_s}{\fbox{$z_s$}}} } \end{align*} したがって,次を得る： \begin{align} z_s = \Phi^{s-1,s}_* (\widetilde{z}_{s-1}) \tag{16.18} \end{align} in (16.9) (in $\S$16.3).