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It is a well-known result that, if the pair $(A,Q^{1/2})$ is stabilizable and the pair $(A, C)$ is detectable, the solution to the discrete-time Riccati recursion

$P(t+1) = A P(t) A^T - A P(t) C^T\ (CP(t)C^T + R)^{-1}\ CP(t) A^T + Q$

converges to the unique stabilizing solution of the Algebraic Riccati Equation

$P = A P A^T - A P C^T(CPC^T + R)^{-1}CP A^T + Q$

from any initial positive semidefinite matrix $P(0)=P_0$.

Do you know any result on the behavior of $P(t)$ during the transient, or more generally for all $t \geq 0$? More precisely, I need some result that ensures that $P(t)$ is stabilizing for all $t \geq 0$, under fair conditions on $A$, $C$, $Q$, $R$, and for arbitrary $P_0$.

Thank you very much in advance,


EDIT. Some background: We consider a discrete-time, linear, time invariant system of the form

$x(t+1) = Ax(t) + v(t)$

$y(t) = Cx(t) + w(t)$

where $x(t) \in R^n$, $A\in R^{n\times n}$, $y(t)\in R^p$, $C\in R^{p\times n}$, and $v$ and $w$ are zero-mean, uncorrelated white noises (say, Gaussian) with appropriate dimensions, with variances $Q$ and $R$ respectively. The predictor $\hat{x}(t+1|t)$, that is, the best linear estimator of $x(t+1)$ given $y(0), \cdots, y(t)$, is given by the Kalman filter. It can be expressed recursively substituting the equations of the Kalman filter into each other. The variance $P(t+1)$ of the corresponding prediction error $\tilde{x}(t+1|t) = \hat{x}(t+1|t)-x(t+1)$ is then given (recursively) by the Riccati equation.

The "dual" of the linear estimation problem above is the "linear quadratic regulator" problem of optimal control, and the Riccati equation is fundamental also in this context.

Given matrices $A\in R^{n\times n}$ and $B \in R^{n\times m}$ with $m\leq n$, to say that the pair $(A, B)$ is reachable means that the matrix $[B\ AB\ \cdots\ A^{n-1}B]$ has full rank ($=n$). The pair $(A, B)$ is stabilizable if with a suitable "change of base" $(A,B)\mapsto(T^{-1} A T, T^{-1}B)$ it can be put in the form

$\left(\left[\begin{matrix} A_{11} & A_{12} \\\\ 0 & A_{22} \\\\ \end{matrix}\right], \left[\begin{matrix} B_1 \\\\ 0 \\\\ \end{matrix}\right]\right)$

where $(A_{11}, B_1)$ is reachable and $A_{22}$ is Hurwitz (all eigenvalues in the interior of the unit disk).

Dually, given matrices $A\in R^{n\times n}$ and $C \in R^{p\times n}$ with $p\leq n$, to say that the pair $(A, C)$ is observable means that the matrix

$\left[\begin{matrix} C \\\\ CA \\\\ \vdots \\\\ CA^{n-1} \end{matrix}\right]$

has full rank. The pair $(A, C)$ is detectable if with a suitable "change of base" $(A,C)\mapsto(T^{-1} A T, CT)$ it can be put in the form

$\left(\left[\begin{matrix} A_{11} & 0 \\\\ A_{21} & A_{22} \\\\ \end{matrix}\right], \left[\begin{matrix} C_1 & 0 \end{matrix}\right]\right)$

where $(A_{11}, C_1)$ is observable and $A_{22}$ is Hurwitz.

Finally, the matrix $P=P^T\geq 0$ is stabilizing if the "closed loop" matrix $A - KC$ is Hurwitz, where $K = A P C^T(CPC^T + R)^{-1}$.

For more details, see for example the Wikipedia pages on Kalman filter, controllability, observability, and Kalman decomposition. For a full reference, see e.g. A. H. Jazwinski, Stochastic Processes and Filtering Theory.

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Federico -- you'll need to add some more background, I think. You don't seem to have said what sort of objects A, Q and C are, although perhaps to an expert it's obvious from "stabilizable" and "detectable". – Scott Morrison Dec 11 '09 at 21:20

The following paper:

provides a geometric positive answer to your question. (The paper discusses the private case where the measurement noise is of unit variance, but this isn't really a shortcoming, because the covariance matrix of the measurement noise can be diagonalized and then all the measurement equation scaled). The answer is based on the fact that geometrically the time evolution of the Riccati equation is a flow on a Grassmann manifold. In the paper, it is proved that this flow preserves the positivity and of P, and that the time evolution matrix of the estimated state vector X is decaying.

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