Steady-State Kalman-Bucy Gain

From a Time-Dependent Covariance to a Steady State

For a scalar linear Gaussian filter, the error covariance changes over time according to a deterministic Riccati differential equation. With constant model coefficients, a steady state is a constant solution of that equation. Solving for this limit gives a time-independent Kalman-Bucy gain, useful when the filter has run long enough that its initial covariance has little effect. The limiting gain is not, in general, the exact gain at every finite time.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying independent standard Brownian motions $B_t$ and $V_t$. Consider the signal and observation model

dX_t=-\alpha X_t\,dt+\sqrt{q}\,dB_t,\qquad dY_t=cX_t\,dt+\sqrt{R}\,dV_t,

where $\alpha>0$ is the signal's mean-reversion rate, $q\ge0$ is its process-noise variance rate, $R>0$ is the observation-noise variance rate, and $c\in\mathbb{R}$ is the observation coefficient. All stochastic integrands in these equations and below are understood to be adapted, progressively measurable, and in $\mathcal{L}^2$ on every finite time interval: $\mathbb{E}\int_0^T H_s^2\,ds<\infty$. In integral form, the observation records the signal plus independent observation noise.

Let $\widehat X_t=\mathbb{E}[X_t\mid\mathcal{F}^Y_t]$ denote the conditional mean given the observation history, and let $P_t=\mathbb{E}[(X_t-\widehat X_t)^2]$ be its error variance. Here $\mathcal{F}^Y_t$ is the filtration generated by the observations (with any initial information included). The earlier material on the Kalman-Bucy Riccati equation gives, for this model,

\frac{dP_t}{dt}=q-2\alpha P_t-\frac{c^2}{R}P_t^2,\qquad K_t=\frac{cP_t}{R}.

The first equation balances process noise, mean reversion, and the information supplied by observations. The second defines the gain: it multiplies the innovation increment $dY_t-c\widehat X_t\,dt$ in the conditional-mean equation. Thus a constant limiting covariance gives a constant limiting gain.

Solving the Algebraic Riccati Equation

Theorem: Suppose $\alpha>0$, $q\ge0$, $R>0$, and $c\in\mathbb{R}$. The unique nonnegative constant solution $P_\infty$ of the scalar algebraic Riccati equation is $$ P_\infty= \begin{cases} \dfrac{\sqrt{\alpha^2+(c^2/R)q}-\alpha}{c^2/R},&c\ne0,\\[6pt] \dfrac{q}{2\alpha},&c=0. \end{cases} $$ For $c\ne0$, the corresponding steady-state gain is $K_\infty=cP_\infty/R$. For $c=0$, $K_\infty=0$.

Proof. A constant covariance must satisfy $0=q-2\alpha P-(c^2/R)P^2$. If $c\ne0$, put $\beta=c^2/R>0$. The quadratic equation $\beta P^2+2\alpha P-q=0$ has roots $(-\alpha\pm\sqrt{\alpha^2+\beta q})/\beta$. Since $\sqrt{\alpha^2+\beta q}\ge\alpha$, the root with the plus sign is nonnegative. The other root is negative when $q>0$; when $q=0$, it equals $-2\alpha/\beta<0$, while the nonnegative root is zero. Thus exactly one root is nonnegative. If $c=0$, the equation reduces to $q-2\alpha P=0$, whose unique solution is $q/(2\alpha)\ge0$. Finally the gain formula follows by substituting this covariance into $K=cP/R$; for $c=0$, that formula itself gives zero. $\square$

When $c\ne0$, the same expression for the gain can be written as

K_\infty=\frac{\sqrt{\alpha^2+c^2q/R}-\alpha}{c}.

This signed gain has the sign of $c$ when $q>0$. If $q=0$, it is zero. The covariance remains nonnegative in all the stated cases.

Worked Example: A Steady-State Covariance and Gain

Take $\alpha=1$, $q=8$, $c=1$, and $R=1$. Then $\beta=c^2/R=1$, and the algebraic Riccati equation is $0=8-2P-P^2$. Rearranging gives $P^2+2P-8=0$, whose roots are $2$ and $-4$. The nonnegative covariance is therefore $P_\infty=2$, and $K_\infty=cP_\infty/R=2$.

Check: substitution gives $8-2(2)-1(2^2)=8-4-4=0$. The other root is inadmissible as an error variance because $-4<0$.

Why the Covariance Converges

The algebraic equation identifies candidate steady states; convergence from an initial covariance is a separate question. For this scalar model it can be proved directly from the sign of the Riccati differential equation.

Proposition: Under the theorem's assumptions, if $P_0\ge0$, the solution of $\dot P_t=q-2\alpha P_t-(c^2/R)P_t^2$ remains nonnegative and converges to $P_\infty$ as $t\to\infty$.

Proof. First suppose $c\ne0$, set $\beta=c^2/R>0$, and let $p_+=P_\infty$ and $p_-=(-\alpha-\sqrt{\alpha^2+\beta q})/\beta<0$. Factoring the right-hand side gives $\dot P_t=-\beta(P_t-p_+)(P_t-p_-)$. If $q>0$, then $p_+>0$. For $0\le P_t<p_+$, the derivative is positive; for $P_t>p_+$, it is negative. A solution cannot cross the equilibrium $p_+$: the right-hand side is a polynomial and hence locally Lipschitz, so uniqueness for this ordinary differential equation implies that a solution reaching $p_+$ agrees thereafter with the constant solution. Consequently, if $P_0<p_+$, the solution is increasing and bounded above by $p_+$; if $P_0>p_+$, it is decreasing and bounded below by $p_+$; and if $P_0=p_+$, it is constant. In either nonconstant case, monotonicity and boundedness give a limit $L$. If the right-hand side at $L$ were nonzero, continuity would make the derivative bounded away from zero with a fixed sign for all sufficiently large times, contradicting convergence to $L$. Hence $L$ is an equilibrium. The only nonnegative equilibrium is $p_+$, so $L=p_+$. If $q=0$, then $p_+=0$, and for $P_0\ge0$ the equation is $\dot P=-2\alpha P-\beta P^2\le0$. The solution stays nonnegative and decreases to a limit; the same derivative argument forces that limit to be an equilibrium, and the only nonnegative one is zero. Finally, if $c=0$, the equation is $\dot P=q-2\alpha P$, with solution $P_t=q/(2\alpha)+(P_0-q/(2\alpha))e^{-2\alpha t}$, which is nonnegative for $P_0\ge0$ and converges to $P_\infty=q/(2\alpha)$. $\square$

Worked Example: Exact Convergence from Zero Initial Error

Use $\alpha=1$, $q=8$, $c=1$, $R=1$, and $P_0=0$. The equation factors as $\dot P=8-2P-P^2=(2-P)(P+4)$. For $0\le P<2$, separate variables. Since $\dfrac{d}{dt}\log\!\left(\dfrac{P+4}{2-P}\right)=\dfrac{\dot P}{P+4}+\dfrac{\dot P}{2-P}=(2-P)+(P+4)=6$, integration and the initial value give $(P_t+4)/(2-P_t)=2e^{6t}$. Solving for $P_t$ yields

P_t=\frac{4(e^{6t}-1)}{1+2e^{6t}}.

At $t=0$, this is $4(1-1)/(1+2)=0$. As $t\to\infty$, division of numerator and denominator by $e^{6t}$ gives $P_t\to4/2=2$. Differentiating the displayed expression at $t=0$ gives $\dot P_0=8$, agreeing with the original equation evaluated at $P_0=0$. Thus the covariance approaches the steady-state value, while at finite times it is generally different.

How Model Parameters Affect the Limit

Assume for the comparisons in this section that $q>0$ and $c\ne0$, and write $\beta=c^2/R$. The positive solution satisfies $q=2\alpha P_\infty+\beta P_\infty^2$. Implicit differentiation, holding the other parameters fixed, gives

\frac{\partial P_\infty}{\partial q} =\frac{1}{2\alpha+2\beta P_\infty}>0,\qquad \frac{\partial P_\infty}{\partial \alpha} =-\frac{2P_\infty}{2\alpha+2\beta P_\infty}<0,\qquad \frac{\partial P_\infty}{\partial \beta} =-\frac{P_\infty^2}{2\alpha+2\beta P_\infty}<0.

More process noise increases the steady-state error variance. Faster mean reversion, which makes the signal less persistent, reduces it. Increasing $\beta$, by increasing the observation strength $c^2/R$, also reduces it. In particular, more observation noise $R$ means smaller $\beta$ and a larger covariance. The gain has a different response to $R$, because it also contains the factor $1/R$. For $c>0$, its formula shows that $K_\infty$ decreases as $R$ increases and increases as $q$ increases; it decreases as $\alpha$ increases. These statements concern fixed values of the other parameters.

Worked Example: Increasing the Observation Noise

Keep $\alpha=1$, $q=8$, and $c=1$, but increase the observation-noise variance rate to $R=4$. Now $\beta=1/4$, so $0=8-2P-P^2/4$. Multiplying by $4$ gives $P^2+8P-32=0$, with roots $-4\pm4\sqrt{3}$. The nonnegative root is $P_\infty=4\sqrt{3}-4$, and the gain is $K_\infty=P_\infty/4=\sqrt{3}-1$.

Check: substituting $P=4\sqrt{3}-4$ into the Riccati right-hand side gives $8-2(4\sqrt{3}-4)-(4\sqrt{3}-4)^2/4$. The terms are $8-8\sqrt{3}+8$ and $(4\sqrt{3}-4)^2/4=16-8\sqrt{3}$, so the result is $16-8\sqrt{3}-(16-8\sqrt{3})=0$. The gain formula gives $(\sqrt{1+8/4}-1)/1=\sqrt{3}-1$, as above. Since $4\sqrt{3}-4>2$ and $\sqrt{3}-1<2$, the covariance is larger but the gain is smaller than in the preceding example.

Worked Example: Making the Signal Less Persistent

Return to $q=8$, $c=1$, and $R=1$, but increase the mean-reversion rate from $\alpha=1$ to $\alpha=2$. The equation is $0=8-4P-P^2$, or $P^2+4P-8=0$. Its roots are $-2\pm2\sqrt{3}$, so $P_\infty=2\sqrt{3}-2$ and $K_\infty=2\sqrt{3}-2$.

Check: substituting gives $8-4(2\sqrt{3}-2)-(2\sqrt{3}-2)^2$. The linear term is $-8\sqrt{3}+8$, and the square is $16-8\sqrt{3}$, so the total is $16-8\sqrt{3}-(16-8\sqrt{3})=0$. Because $\sqrt{3}<2$, $2\sqrt{3}-2<2$, the steady-state covariance and gain are both smaller than in the $\alpha=1$ example. Faster mean reversion continually pulls the signal toward its mean, reducing the uncertainty that observations must correct.

Common Mistake: Confusing Covariance and Gain

The covariance and gain are related but are not the same quantity. A common error is to omit $R$ from the Riccati term or to calculate the gain as $cP_\infty$ when the observation-noise variance rate is not one. For example, with $\alpha=1$, $q=8$, $c=1$, and $R=4$, incorrectly using $\beta=c^2=1$ gives the wrong covariance $P=2$ and the wrong gain $K=2/4=1/2$. The correct Riccati coefficient is $\beta=c^2/R=1/4$, giving $P_\infty=4\sqrt{3}-4$ and $K_\infty=\sqrt{3}-1$. Substitution into $8-2P-P^2/4$ gives zero for the corrected covariance, whereas the mistaken value gives $8-4-4/4=3\ne0$.

The distinction also explains the observation-noise comparison: noisier observations raise the error covariance but, in this model, lower the gain multiplying each innovation increment. The steady-state calculation describes the long-run limit under constant coefficients and a nonnegative initial covariance; it does not replace the time-dependent Riccati equation when transient filtering behavior matters.

Check Your Understanding

Use the scalar Riccati equation and gain formula to answer the following.

For $\alpha=3$, $q=10$, $c=1$, and $R=1$, solve for the nonnegative steady-state covariance and gain.
With $c=0$, what are the steady-state covariance and gain, and why does the observation provide no correction?
For $q>0$ and $c\ne0$, explain from the Riccati equation why increasing $q$ raises $P_\infty$.
For fixed $\alpha,q,c$ with $q>0$ and $c\ne0$, explain why increasing $R$ raises the covariance but lowers the gain when $c>0$.
In the exact-convergence example, verify that $P_t$ stays between $0$ and $2$ for $t\ge0$.

What You'll Learn