Linear Gaussian Signal and Observation Models

The Linear Gaussian State-Space Model

Filtering begins with a distinction between a hidden state and the data available to estimate it. The state evolves continuously and is not observed directly; instead, an observation process accumulates a signal-dependent drift together with measurement noise. In the linear Gaussian model, both equations have linear coefficients and the driving noises and initial state are Gaussian. These assumptions make the model mathematically tractable and are the setting for the Kalman–Bucy filter.

Fix a finite horizon $T>0$. Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ satisfying the usual conditions. Let $V_t$ be a $p$-dimensional standard Brownian motion and $W_t$ an $m$-dimensional standard Brownian motion, both Brownian motions with respect to this filtration. Assume that $V$ and $W$ are independent of one another and of the initial state $X_0$. The integrals below are Itô integrals: an integrand must be progressively measurable and have finite expected integrated squared norm on each finite time interval. In particular, deterministic continuous matrix integrands on $[0,T]$ satisfy this condition.

Definition: The $n$-dimensional signal $X_t$ and $m$-dimensional observation $Y_t$ are given by $$ \begin{aligned} dX_t&=\big(A(t)X_t+a(t)\big)\,dt+G(t)\,dV_t,\\ dY_t&=\big(C(t)X_t+c(t)\big)\,dt+D(t)\,dW_t,\qquad Y_0=0. \end{aligned} $$ Here $A(t)$ is $n\times n$, $a(t)$ is $n$-dimensional, $G(t)$ is $n\times p$, $C(t)$ is $m\times n$, $c(t)$ is $m$-dimensional, and $D(t)$ is $m\times m$. The observation-noise covariance rate is $R(t)=D(t)D(t)^{\mathsf T}$; the signal-noise covariance rate is $Q(t)=G(t)G(t)^{\mathsf T}$. The model is linear Gaussian when $X_0$ is Gaussian and the coefficients are deterministic and satisfy the conditions below.

The equation for $Y_t$ describes cumulative observations. In particular, $dY_t$ contains both the infinitesimal signal contribution $(C(t)X_t+c(t))dt$ and a Brownian measurement-noise contribution. It does not assert that $Y_t$ is a direct, noise-free reading of $X_t$.

Initial Law and Coefficient Assumptions

Assume $X_0$ is $\mathcal{F}_0$-measurable with distribution $N(m_0,P_0)$, where $m_0\in\mathbb{R}^n$ and $P_0$ is a symmetric positive semidefinite $n\times n$ matrix. Positive semidefinite includes the zero matrix, so a deterministic initial state is allowed. The independence assumptions mean that the entire initial state and the two Brownian motions are mutually independent; in particular, the initial uncertainty is not correlated with future driving noise.

For the standard finite-horizon model, take $A,a,G,C,c,D$ to be deterministic continuous functions on $[0,T]$. Continuity implies boundedness on this interval. The matrix $D(t)$ is assumed invertible for every $t\in[0,T]$, so $R(t)=D(t)D(t)^{\mathsf T}$ is positive definite. No full-rank assumption is imposed on $G(t)$: $Q(t)$ may be singular or zero. These hypotheses are sufficient for the equations stated here; they are not the only possible hypotheses under which filtering can be developed.

Coefficient check: Since $A$ is bounded on $[0,T]$, there is a finite constant $K$ such that $\lVert A(t)x+a(t)\rVert\le K(1+\lVert x\rVert)$ for every $x\in\mathbb{R}^n$ and $t\in[0,T]$. Also, $\lVert A(t)x-A(t)y\rVert\le K\lVert x-y\rVert$. These are, respectively, a linear growth bound and a global Lipschitz bound for the signal drift. Boundedness of $G$ gives the corresponding growth and Lipschitz bounds for the diffusion coefficient, which does not depend on the state.

In a scalar signal, $A(t)$ is a scalar drift coefficient and $G(t)$ controls the signal’s random fluctuations. In a multidimensional signal, $G(t)$ can mix several independent Brownian components into the state. The observation matrix $C(t)$ selects linear combinations of the state that enter the observation drift. The matrix $D(t)$ controls observation noise, and its full-rank condition ensures nonsingular instantaneous observation-noise covariance.

Explicit Signal Solution and Well-Posedness

The hypotheses give a unique strong signal solution: “strong” means that the solution is constructed on the given probability space using the given initial state and Brownian motion $V$, and is adapted to the filtration. In this linear setting, the solution can be written explicitly using a deterministic fundamental matrix.

Theorem: Let $\Phi(t)$ solve $\Phi'(t)=A(t)\Phi(t)$, $\Phi(0)=I_n$. Then $\Phi(t)$ is invertible, and the unique strong solution of the signal equation is $$ X_t=\Phi(t)\left(X_0+\int_0^t\Phi(s)^{-1}a(s)\,ds +\int_0^t\Phi(s)^{-1}G(s)\,dV_s\right),\qquad 0\le t\le T. $$ The stochastic integral is well-defined: its integrand is deterministic, continuous, and bounded on $[0,T]$, and hence has finite integrated squared norm.

Proof. The ordinary linear differential equation for $\Phi$ has a unique solution because $A$ is continuous. To see it is invertible, solve $\Psi'(t)=-\Psi(t)A(t)$ with $\Psi(0)=I_n$. Differentiating the product gives $(\Psi(t)\Phi(t))'=0$, so $\Psi(t)\Phi(t)=I_n$ for every $t$. Thus $\Psi(t)=\Phi(t)^{-1}$. Define the expression on the right side of the theorem to be $X_t$. The term in parentheses is a continuous semimartingale. Applying the product rule with the deterministic differentiable matrix $\Phi(t)$ gives $$ dX_t=A(t)X_t\,dt+a(t)\,dt+G(t)\,dV_t. $$ At $t=0$ the expression equals $X_0$, so it is an adapted solution. For uniqueness, let $X$ and $\widetilde X$ be two solutions with the same initial state and driving Brownian motion. Their difference $U_t=X_t-\widetilde X_t$ satisfies $dU_t=A(t)U_t\,dt$, $U_0=0$. Multiplication by $\Phi(t)^{-1}$ and the product rule yield $d(\Phi(t)^{-1}U_t)=0$. Its initial value is zero, hence $U_t=0$ for all $t$. This proves existence and pathwise uniqueness, and the displayed construction proves the asserted strong-solution property. $\square$

Once $X$ is defined, the observation is $$ Y_t=\int_0^t\big(C(s)X_s+c(s)\big)\,ds+\int_0^tD(s)\,dW_s. $$ The first integral is an ordinary time integral and the second is an Itô integral with progressively measurable integrand $D(s)$. Because the coefficients are bounded and the explicit signal solution has finite second moments on finite intervals, both terms are well-defined.

Why the Model Is Jointly Gaussian

Proposition: For any finite collection of times in $[0,T]$, the corresponding signal and observation values form a jointly Gaussian vector. Gaussian distributions here may be degenerate.

Proof. The explicit signal solution is a deterministic linear function of $X_0$ and deterministic-integrand Itô integrals against $V$. For any finite set of times, these integrals are jointly Gaussian with $X_0$: for step-function integrands this follows because the integrals are linear combinations of Brownian increments, and for continuous deterministic integrands it follows by approximation in mean square by step-function integrands. The resulting Gaussian vectors converge in mean square, hence in distribution; their characteristic functions converge to that of a Gaussian vector with the limiting mean and covariance. Thus the finite collection of signal values is jointly Gaussian.

For observation values, approximate each time integral $\int_0^t(C(s)X_s+c(s))\,ds$ by Riemann sums. Each finite collection of these sums, together with finitely many signal values and observation-noise integrals, is jointly Gaussian: the sums are linear combinations of jointly Gaussian signal values, and the $W$-integrals are jointly Gaussian and independent of $X_0,V$. Continuity of the coefficients and mean-square continuity of the signal imply convergence of the sums in mean square. The Gaussian-vector limit argument just given shows that the limits, including the observation values, are jointly Gaussian. $\square$

This proposition concerns the joint distribution of the model variables, not a formula for conditioning on observations. Conditional means and covariance updates are a separate calculation. The key point here is that deterministic linear operations preserve Gaussianity, and the independent driving noises enter linearly.

Worked Example: A Scalar Signal with Noisy Observations

Consider $dX_t=-2X_t\,dt+\tfrac12\,dV_t$, $dY_t=3X_t\,dt+2\,dW_t$, with $X_0\sim N(1,\tfrac14)$. Let $V$, $W$, and $X_0$ be mutually independent. Verify the model assumptions and compute the mean and variance of $X_1$.

Here $n=m=p=1$, $A=-2$, $G=\tfrac12$, $C=3$, and $D=2$, all constant. The signal-noise covariance rate is $Q=G^2=\tfrac14$, while the observation-noise covariance rate is $R=D^2=4>0$. The initial mean is $1$ and initial covariance is $\tfrac14$, which is positive. All coefficient functions are continuous and bounded, and the observation noise coefficient is invertible.

The fundamental matrix is $\Phi(t)=e^{-2t}$. Therefore $$ X_1=e^{-2}X_0+\frac12\int_0^1e^{-2(1-s)}\,dV_s. $$ The Itô integral has mean zero and variance $\int_0^1e^{-4(1-s)}\,ds=(1-e^{-4})/4$. It is independent of $X_0$. Consequently, $$ \mathbb{E}[X_1]=e^{-2},\qquad \operatorname{Var}(X_1)=\frac{e^{-4}}4+\frac14\cdot\frac{1-e^{-4}}4 =\frac{1+3e^{-4}}{16}. $$ Check: the variance at time zero from the same expression is $(1+3)/16=\tfrac14$, the specified initial variance. The added variance from the signal noise over $[0,1]$ is $(1-e^{-4})/16>0$, as expected.

Worked Example: A Two-Dimensional Signal and One Observation

Let $X_t=(X_t^{(1)},X_t^{(2)})^{\mathsf T}$, with $$ A=\begin{pmatrix}-1&0\\0&-2\end{pmatrix},\quad G=\begin{pmatrix}1&0\\0&2\end{pmatrix},\quad C=\begin{pmatrix}1&-1\end{pmatrix},\quad D=(3). $$ Take $X_0\sim N((0,0)^{\mathsf T},\operatorname{diag}(1,4))$, independent of the two-dimensional $V$ and one-dimensional $W$. Find the state covariance at time $1$, and verify the observation-noise condition.

The dimensions are $n=p=2$ and $m=1$. The signal-noise covariance rate is $$ Q=GG^{\mathsf T}=\begin{pmatrix}1&0\\0&4\end{pmatrix}, $$ and the observation-noise covariance rate is $R=DD^{\mathsf T}=(9)$, which is positive definite in dimension one. The initial covariance is positive definite. All coefficients are constant, so the continuity and boundedness requirements hold.

The two state coordinates evolve independently. For the first coordinate, the initial contribution to its variance at time $1$ is $e^{-2}$, and the noise contribution is $\int_0^1e^{-2(1-s)}ds=(1-e^{-2})/2$. Their sum is $(1+e^{-2})/2$. For the second coordinate, the initial contribution is $4e^{-4}$, and the noise contribution is $4\int_0^1e^{-4(1-s)}ds=1-e^{-4}$. Their sum is $1+3e^{-4}$. Independence gives zero cross-covariance, so $$ \operatorname{Cov}(X_1)= \begin{pmatrix}(1+e^{-2})/2&0\\0&1+3e^{-4}\end{pmatrix}. $$ Check: if both signal-noise coefficients were set to zero, the same initial covariance would evolve to $\operatorname{diag}(e^{-2},4e^{-4})$. The two positive terms $(1-e^{-2})/2$ and $1-e^{-4}$ in the computed diagonal entries are precisely the contributions from signal noise.

Worked Example: Random Constant Signal

Let $X_0\sim N(1,1)$, set $A=0$ and $G=0$, and use the scalar observation equation $dY_t=X_t\,dt+2\,dW_t$. Assume $X_0$ and $W$ are independent. Compute $\mathbb{E}[Y_1]$, $\operatorname{Var}(Y_1)$, and $\operatorname{Cov}(X_1,Y_1)$.

The signal is $X_t=X_0$: it has a random initial value but no drift or signal noise. The observation coefficient $D=2$ is invertible and has covariance rate $4$, while the signal covariance rate $Q=0$ is allowed. Integrating the observation equation gives $Y_1=X_0+2W_1$, and $X_1=X_0$. Hence $\mathbb{E}[Y_1]=\mathbb{E}[X_0]+2\mathbb{E}[W_1]=1$. Independence and $\operatorname{Var}(W_1)=1$ give $\operatorname{Var}(Y_1)=1+4=5$, and $$ \operatorname{Cov}(X_1,Y_1) =\operatorname{Cov}(X_0,X_0+2W_1)=1+0=1. $$ Check: the observation’s variance decomposes into initial-state variance $1$ and measurement-noise variance $2^2\operatorname{Var}(W_1)=4$, giving $5$. The zero signal-noise covariance does not make the model non-Gaussian or invalidate its assumptions.

Common Pitfall: The Observation Noise Is Not Its Coefficient

A frequent error is to treat the observation coefficient $D$ as the observation-noise variance rate. For the scalar equation $dY_t=3X_t\,dt+2\,dW_t$, the noise increment is $2\,dW_t$, so its variance over a time interval of length $h$ is $2^2h=4h$, not $2h$. Equivalently, its quadratic variation over $[0,t]$ is $4t$. The coefficient $2$ is the diffusion coefficient; the covariance rate is its square.

Another modelling error is to omit the time differential in the signal contribution and write $dY_t=3X_t+2\,dW_t$. The terms then have inconsistent meanings: the observation drift must multiply $dt$, so the correctly specified equation is $dY_t=3X_t\,dt+2\,dW_t$. In multiple dimensions, check matrix dimensions before interpreting the equation: $C(t)X_t$ must be $m$-dimensional, and $D(t)dW_t$ must also be $m$-dimensional.

The model assumptions have distinct roles. Gaussian initial data and deterministic linear coefficients produce joint Gaussian signal and observation values. Independence separates signal noise from measurement noise. The full-rank observation coefficient ensures positive definite observation-noise covariance, while the signal-noise covariance may be degenerate. Together, these conditions define the standard setting in which the Kalman–Bucy filtering equations are developed.

Check Your Understanding

Use the stated model assumptions and distinguish the signal covariance rate from the observation covariance rate.

For a two-dimensional signal and three-dimensional observation, what are the dimensions of $C(t)$, $D(t)$, and $Y_t$?
Why is $P_0=0$ permitted, and what does it mean for the initial state?
If $D= \operatorname{diag}(2,1)$, compute the observation-noise covariance rate $DD^{\mathsf T}$ and state whether it is positive definite.
In the explicit signal solution, which independence assumption allows the initial-state and signal-noise contributions to the covariance to be added?
Can $G(t)$ be singular under the stated assumptions? Can $D(t)$ be singular? Explain the different consequences.