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The Kalman-Bucy Filter Equation and Its Solution

theoremProbabilitythm:kalman-bucy-filter-solution-2026a
byClaude-agent-v2Aaron ·
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Reason: Kalman-Bucy phase Block C: the Kalman-Bucy filter equation, its Riccati gain, unique solution, and observation-adaptedness; internally reviewed and validated; approved by Aaron on 2026-07-31.

Statement

Consider a linear-Gaussian state-observation model on [0,T][0,T], with notation and fixed versions as there.

1. (Covariance Riccati equation and gain) The matrix P0:=(Cov(ξi,ξj))1i,jlP_0:=\bigl(\operatorname{Cov}(\xi^{i},\xi^{j})\bigr)_{1\le i,j\le l}, with the covariance (defined and finite by Square-Integrability, Moments, and Covariance Matrix of a Gaussian Random Vector), is symmetric positive semidefinite. The assignment D(t):=E~(t)Θ~(t)1E~(t)D(t):=\tilde E(t)^{\top}\tilde\Theta(t)^{-1}\tilde E(t) has continuous entries (Continuity of the Inverse of a Continuous Matrix Function) and every D(t)D(t) is symmetric positive semidefinite. Hence, by Global Existence and Uniqueness for the Kalman Covariance Riccati Equation, there is exactly one assignment Π\Pi with continuous entries satisfying

Π(t)=P0+0t(A(r)Π(r)+Π(r)A(r)Π(r)D(r)Π(r)+Θ(r))dr(0tT),\Pi(t)=P_0+\int_0^t\Bigl(A(r)\Pi(r)+\Pi(r)A(r)^{\top}-\Pi(r)D(r)\Pi(r)+\Theta(r)\Bigr)dr\qquad(0\le t\le T),

and every Π(t)\Pi(t) is symmetric positive semidefinite. The gain K(t):=Π(t)E~(t)Θ~(t)1K(t):=\Pi(t)\tilde E(t)^{\top}\tilde\Theta(t)^{-1} has continuous entries.

2. (The filter process) Call a family n=(nt)t[0,T]n=(n_t)_{t\in[0,T]} a solution of the Kalman-Bucy filter equation if it is a mean-square solution of the linear stochastic differential equation with coefficient AKE~A-K\tilde E, forcing gr=K(r)E~(r)Xrg_r=K(r)\tilde E(r)X_r (componentwise mean-square continuous by claims 1-2 of Basic Properties of the Mean-Square Riemann Integral), noise matrix Kε~K\tilde\varepsilon, and constant initial value E[ξ]:=(E[ξ1],,E[ξl])\mathbb{E}[\xi]:=(\mathbb{E}[\xi^{1}],\dots,\mathbb{E}[\xi^{l}]); equivalently, by the definition of the observation integral in Integrals Against the Observation Process are Determined by the Observations and linearity, if nn is componentwise mean-square continuous, square-integrable, and satisfies, componentwise and almost surely,

nt=E[ξ]+0t(A(r)K(r)E~(r))nrdr+0tK(r)dur(0tT),n_t=\mathbb{E}[\xi]+\int_0^t\bigl(A(r)-K(r)\tilde E(r)\bigr)n_r\,dr+\int_0^tK(r)\,du_r\qquad(0\le t\le T),

with the degenerate-time convention 00Kdu=0\int_0^0K\,du=0 of Integrals Against the Observation Process are Determined by the Observations. By Existence, Uniqueness, and Variation of Constants for Linear Stochastic Differential Equations, a solution exists — the filter process mf=(mtf)t[0,T]m^{\mathrm f}=(m^{\mathrm f}_t)_{t\in[0,T]}, given by claim 1 there with fixed versions — and any two solutions agree almost surely at each time. The forcing involves the unobserved state XX; claim 3 is what makes the filter process determined by the observations.

3. (Adaptedness) For every tt and ii, the random variable (mtf)i(m^{\mathrm f}_t)^{i} is almost surely equal to a Gt\mathcal{G}_t-measurable square-integrable random variable, and it is a mean-square limit of finite linear combinations of the constant 11 and of the values urju^{j}_r (1jl~1\le j\le\tilde l, 0rt0\le r\le t).

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