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With a fixed gain $0<K<1$, exponential smoothing recursively combines the newest observation with the preceding estimate. Iteration assigns geometric weights $K(1-K{)}^j$ to older observations, together with a diminishing weight on the initial estimate. The steady-state local-level Kalman gain gives a probabilistic interpretation of this recursion.

Consider a local-level state-space model $x_t=x_{t-1}+w_t$, $y_t=x_t+v_t$, with mutually independent centered Gaussian noises of variances $Q=1/4$ and $R=1$. Start with $x_0\sim N(10,4)$, independently of these noises. Before the first observation, the Kalman prediction step gives $m_1^{-}=10$, $P_1^{-}=17/4$. If $y_1=14$, the Kalman gain is $17/21$, and the posterior mean and posterior variance becomeThe next Kalman prediction step keeps the mean $m_2^{-}=278/21$ but adds fresh process noise variance, giving $P_2^{-}=89/84$. If $y_2=12$, thenThus new uncertainty is introduced at each transition and reduced by each measurement. It does not shrink to zero merely because many measurements have arrived: the state continues to change.

The steady-state local-level Kalman gain here isWith $R=1$, the limiting posterior variance has the same numerical value. By contrast, the static-state model $Q=0$ with $P_0=4$ gives $P_t=1/(t+1/4)$ and $K_t=1/(t+1/4)$. That decreasing gain is appropriate for a state that truly remains constant. Applied to the changing random walk, it builds confidence in the wrong model. The choice of process noise is therefore a statement about state evolution, rather than just an instruction to make a plot look smoother.

A constant posterior variance must be a fixed point of the recursion in part (b). Multiplying $P=V(P+W)/(P+W+V)$ by its positive denominator givesConversely, if a nonnegative $P$ solves this equation and the filter is initialized with $P_0=P$, the recursion preserves $P_t=P$ at every step. The fixed-point equation alone does not force an arbitrary initial covariance to be constant.

For $W,V>0$, the admissible root and its steady-state local-level Kalman gain areThe state estimate then obeys exponential smoothing:Iteration makes the weights explicit:Thus older observations receive geometrically decreasing weights. Consistently with part (a), the innovations variance is $q=P+W+V$ and the differenced MA coefficient is $\vartheta =-(1-K)$. The degeneracies $W=0,P=0$ and $V=0,P=0$ give gain zero and gain one respectively when the denominator is positive.

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.