= Solution
The <state-space model> as printed supplies only uncorrelated <white noise>. That is insufficient for the claimed normal <conditional distributions>. For example, take $S_0$ normal, <independent> process noises $w_t$ taking values $\pm\sqrt W$ equally often, and <independent> Gaussian observation noises. With $G=1$ and no previous observations, $S_1=S_0+w_1$ is a two-component shifted <mixture distribution>, not normal for $W>0$. Its characteristic function is $e^{i\mu s-P_0s^2/2}\cos(\sqrt W\,s)$, which has zeros, whereas a normal characteristic function has none. Similarly a non-Gaussian observation noise prevents the asserted conditional normal law for $X_t$.
The intended <scalar Gaussian Kalman recursion> requires $w_t\sim N(0,W)$ and $v_t\sim N(0,V)$ <independent> of each other and of the previous state and observation history. Assume these Gaussian <independence> conditions, $V>0$, $W\ge0$, and the given normal filtering law. Conditional on the past, $GS_{t-1}+w_t$ is a sum of <independent> normals, hence
$$
\boxed{S_t\mid\mathcal F_{t-1}\sim N(m_t,R_t),\qquad m_t=G\widehat S_{t-1},\quad R_t=G^2P_{t-1}+W.}
$$
The new observation noise is <independent> of $S_t$ and the past, so $X_t\mid S_t,\mathcal F_{t-1}\sim N(FS_t,V)$. Write $S_t=m_t+\eta_t$ and $X_t=Fm_t+F\eta_t+v_t$. The <independent> normal pair $(\eta_t,v_t)$ shows directly that
$$
\binom{X_t}{S_t}\Bigm|\mathcal F_{t-1}\sim N\left(\binom{Fm_t}{m_t},\begin{pmatrix}Q_t&FR_t\\FR_t&R_t\end{pmatrix}\right),\qquad Q_t=F^2R_t+V.
$$
The <covariance> $FR_t$ comes from the common $\eta_t$ term. Applying the <conditional multivariate normal distribution> formula with $X_t$ as the conditioned coordinate gives the <Kalman filter> update
$$
\boxed{\widehat S_t=m_t+\frac{FR_t}{Q_t}(X_t-Fm_t),\qquad P_t=R_t-\frac{F^2R_t^2}{Q_t}=\frac{R_tV}{Q_t}.}
$$
The final expression verifies nonnegativity of the <posterior> <variance>.
For $F=G=1$, define
$$
\boxed{\alpha_t=\frac{P_{t-1}+W}{P_{t-1}+W+V}.}
$$
The recursions become $\widehat S_t=(1-\alpha_t)\widehat S_{t-1}+\alpha_tX_t$ and $P_t=\alpha_tV$. If $P_t\to P$, then $\alpha_t\to\alpha=P/V$ and, with $c=W/V$,
$$
\alpha=\frac{\alpha+c}{\alpha+c+1}\quad\Longrightarrow\quad\alpha^2+c\alpha-c=0.
$$
The nonnegative solution is the <steady-state local-level Kalman gain>:
$$
\boxed{\alpha_\infty=\frac{\sqrt{c^2+4c}-c}{2},\qquad P=\frac{\sqrt{W^2+4WV}-W}{2}.}
$$
For $c=0$ the limit is zero; for $c>0$ it lies strictly between zero and one. These are exact conditional-law results under the additional Gaussian assumptions, not consequences of the weaker printed noise description alone.
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