 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 $e^{i\mu s-P_0{s}^2/2}\cos (\sqrt{W}s)$, 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, $V>0$, $W\ge 0$, and the given normal filtering law. Conditional on the past, $G{S}_{t-1}+w_t$ is a sum of independent normals, henceThe new observation noise is independent of $S_t$ and the past, so $X_t\mid S_t,{\mathcal{F}}_{t-1}\sim N(F{S}_t,V)$. Write $S_t=m_t+{\eta }_t$ and $X_t=F{m}_t+F{\eta }_t+v_t$. The independent normal pair $({\eta }_t,v_t)$ shows directly thatThe covariance $F{R}_t$ 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 updateThe final expression verifies nonnegativity of the posterior variance.

For $F=G=1$, defineThe recursions become ${\hat{S}}_t=(1-{\alpha }_t){\hat{S}}_{t-1}+{\alpha }_t{X}_t$ and $P_t={\alpha }_{t}V$. If $P_t\to P$, then ${\alpha }_t\to \alpha =P/V$ and, with $c=W/V$,The nonnegative solution is the steady-state local-level Kalman gain: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.