Filtering an Ornstein-Uhlenbeck Signal

Filtering a Mean-Reverting Signal

A mean-reverting signal tends to return toward a long-run level, while continuous noisy observations provide information about its current value. Filtering combines these effects: the signal dynamics propagate the estimate and its uncertainty, and each new observation adjusts the estimate according to the current uncertainty and observation noise. For an Ornstein-Uhlenbeck signal, the conditional mean satisfies a linear stochastic equation and the conditional variance satisfies a scalar Riccati differential equation that can be solved explicitly.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying independent standard Brownian motions $B_t$ and $V_t$. Let $X_0\sim N(m_0,P_0)$, with $P_0\ge0$, independent of both Brownian motions. Assume $a>0$, $\eta\ge0$, and $\rho>0$. The signal $X_t$ and observation $Y_t$ are

dX_t=-a(X_t-\theta)\,dt+\eta\,dB_t, \qquad dY_t=X_t\,dt+\rho\,dV_t, \qquad Y_0=0.

Here $a$ is the rate of mean reversion, $\theta$ is the signal's long-run level, $\eta$ is the signal diffusion coefficient, and $\rho$ is the observation-noise scale. The observation filtration is $\mathcal{F}_t^Y=\sigma(Y_s:0\le s\le t)$, with the usual completion. Define the conditional mean and variance by $\widehat X_t=\mathbb{E}[X_t\mid\mathcal{F}_t^Y]$ and $P_t=\mathbb{E}[(X_t-\widehat X_t)^2\mid\mathcal{F}_t^Y]$.

All stochastic integrals below are Itô integrals. On each finite time interval $[0,T]$, their integrands are progressively measurable and square-integrable: $\mathbb{E}\int_0^T H_s^2\,ds<\infty$. In particular, the constant coefficients in the signal and observation noise integrals meet this condition. The Kalman-Bucy filtering theorem for linear Gaussian models gives a deterministic conditional variance and the following equations. This is the scalar specialization of the Kalman-Bucy mean and Riccati equations.

Theorem: Under the model assumptions above, $X_t\mid\mathcal{F}_t^Y$ is Gaussian with mean $\widehat X_t$ and deterministic variance $P_t$. The mean, variance, and observation gain $K_t$ satisfy $$ K_t=\frac{P_t}{\rho^2},\qquad d\widehat X_t=-a(\widehat X_t-\theta)\,dt+K_t(dY_t-\widehat X_t\,dt), \qquad \frac{dP_t}{dt}=\eta^2-2aP_t-\frac{P_t^2}{\rho^2}, \quad \widehat X_0=m_0,\quad P(0)=P_0. $$

The term $dY_t-\widehat X_t\,dt$ is the observation innovation: it is the observed increment after subtracting the increment predicted from the current estimate. In the normalized form, define $I_t=\rho^{-1}(Y_t-\int_0^t\widehat X_s\,ds)$. The innovations theorem says that $I_t$ is Brownian motion in the observation filtration, and the mean equation becomes $d\widehat X_t=-a(\widehat X_t-\theta)\,dt+(P_t/\rho)\,dI_t$. The observation integral here is an ordinary time integral; the integral with respect to $I_t$ is an Itô integral. On finite intervals $P_t$ is bounded, so its coefficient is square-integrable.

Solving the Variance Equation

The variance equation has a useful balance of three effects. Signal noise adds uncertainty at rate $\eta^2$. Mean reversion reduces uncertainty through $-2aP_t$. Observations reduce it through $-P_t^2/\rho^2$: when the current uncertainty is large relative to observation noise, the data can reduce it more rapidly. The next result solves this balance for any nonnegative initial variance.

Theorem: Let $\lambda=\sqrt{a^2+\eta^2/\rho^2}$, $p_+=\rho^2(\lambda-a)$, and $p_-=-\rho^2(\lambda+a)$. For $P_0\ge0$, set $z_0=(P_0-p_+)/(P_0-p_-)$. The solution of the variance equation is $$ z(t)=z_0e^{-2\lambda t}, \qquad P_t=\frac{p_+-z(t)p_-}{1-z(t)}. $$ The formula includes the case $\eta=0,\ P_0=0$, for which $P_t=0$ at all times.

Proof. The right-hand side of the Riccati equation factors as $$ \eta^2-2aP-\frac{P^2}{\rho^2} =-\frac{1}{\rho^2}(P-p_+)(P-p_-), $$ because $p_++p_-=-2a\rho^2$ and $p_+p_-=-\eta^2\rho^2$. Also $p_+-p_-=2\rho^2\lambda>0$. Since $p_-\!<0$ and $P_0\ge0$, $P_0-p_-\!>0$, so the ratio $z=(P-p_+)/(P-p_-)$ is defined at the initial time. Differentiating this ratio along a solution gives $$ \frac{dz}{dt} =\frac{(p_+-p_-)P'}{(P-p_-)^2} =-\frac{p_+-p_-}{\rho^2}\frac{P-p_+}{P-p_-} =-2\lambda z. $$ Solving this linear equation yields $z(t)=z_0e^{-2\lambda t}$. Solving the defining relation $z=(P-p_+)/(P-p_-)$ for $P$ gives $P=(p_+-zp_-)/(1-z)$, as claimed. To check the denominator, $z_0<1$ because $P_0-p_+<P_0-p_-$; if $z_0\ge0$, then $z(t)\le z_0<1$, and if $z_0<0$, then $z(t)<0$. Thus $1-z(t)>0$ for every $t\ge0$. The formula starts at $P_0$ and was obtained by differentiating the variance equation, so it gives its solution; uniqueness also follows from the local Lipschitz property of the polynomial right-hand side. If $\eta=0$ and $P_0=0$, then $p_+=0$, $z_0=0$, and the formula gives $P_t=0$. $\square$

The nonnegative equilibrium of the variance equation is $p_+$. The explicit solution also shows how the initial uncertainty affects the transient: the factor $e^{-2\lambda t}$ governs the decay of the transformed initial condition. The equilibrium is mentioned here to help interpret the solution; the focus is the time-dependent variance, not a separate steady-state gain analysis.

Worked Example: A Signal Starting with Known Value

Take $a=1$, $\eta^2=3$, $\rho^2=1$, and $P_0=0$. Find the variance and gain at $t=(\ln 3)/4$.

First, $\lambda=\sqrt{1+3}=2$, $p_+=2-1=1$, and $p_-=-(2+1)=-3$. Thus $z_0=(0-1)/(0-(-3))=-1/3$, and $z(t)=-e^{-4t}/3$. Substitution into the solution gives $$ P_t=\frac{1-e^{-4t}}{1+e^{-4t}/3}. $$ At $t=(\ln 3)/4$, $e^{-4t}=1/3$, so $$ P_t=\frac{1-1/3}{1+(1/3)/3} =\frac{2/3}{10/9} =\frac35. $$ Since $\rho^2=1$, the gain is $K_t=P_t/\rho^2=3/5$.

Check: Differentiate the displayed variance expression, writing $u=e^{-4t}$. Then $du/dt=-4u$ and $dP/du=-(4/3)/(1+u/3)^2$, so $dP/dt=(16u/3)/(1+u/3)^2$. At $u=1/3$, this is $36/25$. The Riccati right-hand side at $P=3/5$ is $3-2(3/5)-(3/5)^2=75/25-30/25-9/25=36/25$, as required.

Worked Example: Uncertainty from the Initial State Only

Suppose $a=1$, $\eta=0$, $\rho=1$, and $P_0=1$. Find $P_t$ and evaluate it at $t=(\ln 3)/2$.

Here $\lambda=1$, $p_+=0$, and $p_-=-2$. Therefore $z_0=(1-0)/(1-(-2))=1/3$, and $$ P_t=\frac{2z(t)}{1-z(t)} =\frac{2}{3e^{2t}-1}. $$ At $t=(\ln 3)/2$, $e^{2t}=3$, so $P_t=2/(9-1)=1/4$. The signal has no new process noise, but its uncertain initial value is not known from the start; observations gradually reduce that uncertainty.

Check: Differentiating $2/(3e^{2t}-1)$ gives $-12e^{2t}/(3e^{2t}-1)^2$, which at this time is $-36/64=-9/16$. The Riccati right-hand side is $-2(1/4)-(1/4)^2=-1/2-1/16=-9/16$. Also, the formula gives $P_0=2/(3-1)=1$.

What the Gain Does

The gain multiplies the innovation, not the entire observation. A large innovation moves the estimate more, but the multiplier depends on both the current uncertainty and the observation-noise scale. In raw observation units the coefficient is $K_t=P_t/\rho^2$. If the normalized Brownian innovation is used instead, its coefficient is $P_t/\rho$, because normalization divides the raw innovation by $\rho$.

Proposition: At fixed variance $P\ge0$, the raw observation gain $K=P/\rho^2$ is nondecreasing in $P$ and nonincreasing in $\rho>0$. The gain is zero when $P=0$.

Proof. For fixed $\rho>0$, $K=P/\rho^2$ has derivative $1/\rho^2>0$ with respect to $P$, so increasing uncertainty cannot decrease the gain. For fixed $P\ge0$, its derivative with respect to $\rho$ is $-2P/\rho^3\le0$, so increasing observation noise cannot increase the gain. When $P=0$, the formula gives $K=0$. These are properties of the coefficient in the Kalman-Bucy mean equation; they do not say that a particular realized estimate must move in a particular direction, since the innovation itself can be positive or negative. $\square$

Worked Example: Raw and Normalized Innovation Gains

At an instant when $P_t=2$, let the observation-noise scale be $\rho=2$. Compute both forms of the gain and check that they give the same correction to the mean.

The raw gain is $K_t=P_t/\rho^2=2/4=1/2$. The normalized innovation is $dI_t=(dY_t-\widehat X_t\,dt)/2$, and its coefficient in the mean equation is $P_t/\rho=2/2=1$. Therefore the observation correction can be written either as $$ \frac12(dY_t-\widehat X_t\,dt) \quad\text{or}\quad 1\cdot dI_t =\frac12(dY_t-\widehat X_t\,dt). $$ The two expressions agree exactly. If instead $\rho=1$ at the same variance, the raw gain is $2/1=2$; less observation noise gives a larger coefficient on the raw innovation.

A Common Mistake

A frequent error is to omit the mean-reversion contribution $-2aP_t$ from the variance equation. That term comes from the contraction of signal errors under mean-reverting dynamics; it is not part of the observation update and cannot be dropped when the signal is an Ornstein-Uhlenbeck process.

For a concrete check, take $a=1$, $\eta^2=3$, $\rho^2=1$, and $P=1$. The correct variance derivative is $3-2(1)-1^2=0$, so $P=1$ is an equilibrium. Omitting mean reversion gives the incorrect derivative $3-1^2=2$, falsely predicting that uncertainty increases at this point. Substituting $P=1$ into the factored equation confirms the correct value: its roots are $p_+=1$ and $p_-=-3$, and $- (1-1)(1-(-3))=0$.

The filtering equations separate two roles. Mean reversion pulls the estimate toward $\theta$ even without new data, while the innovation term uses observations to correct the estimate. In the variance equation, signal noise creates uncertainty, mean reversion contracts it, and observation quality determines how effectively it is reduced. Keeping these effects distinct is essential when interpreting or implementing the filter.

Check Your Understanding

Use the model assumptions and equations above. In each question, take $\rho>0$ and $P_0\ge0$.

Write the variance differential equation for $a=2$, $\eta^2=5$, and $\rho^2=4$.
For $a=1$, $\eta=0$, $\rho=1$, and $P_0=0$, explain from the variance equation why the conditional variance remains zero.
For fixed $P>0$, what happens to the raw gain $P/\rho^2$ when the observation-noise scale is doubled?
Explain why the mean equation uses $dY_t-\widehat X_t\,dt$ rather than $dY_t$ alone.
For $a=1$, $\eta^2=3$, and $\rho^2=1$, verify directly that $P=1$ is an equilibrium of the variance equation.

Filtering an Ornstein-Uhlenbeck Signal

What You'll Learn