# Kalman filter: asymptotic of state estimate

Assume we have a linear state-space model: $$z_{k} = Hx_{k} + v_{k}\\ x_{k} = F x_{k-1} + Bu + w_{k},$$ where $$u$$ is some control variable (constant intercept is the simplest case).

Kalman filter, for example, provides us the estimate of $$E[x_{n}|z_{0}, \dots, z_{n}]$$.

Assume that the state process is weakly stationary.

The question: what can one say about provided by Kalman filter estimate $$\hat{x}_{k|k}$$ of $$E[x_{n}|z_{0}, \dots, z_{t}]$$, if $$k\to\infty$$?

The first wrong idea was that it should converge (in probability) to $$E[x]$$. This is wrong and can be check on simulations given in

Kalman filter for AR(1) plus noise

• Add self-study tag? – Jarle Tufto Nov 27 '19 at 16:16
• Before voting to close as off topic, why not clarify whether this actually is self-study or not? I also do not get another close vote regarding subject-matter off-topicness; this looks entirely on topic to me. – Richard Hardy Nov 28 '19 at 8:53

As $$k\to\infty$$, we have an asymptotic gain, $$K$$ and we end up with linear time-invariant filter: $$\hat{x}_{k|k} = F \hat{x}_{k-1|k-1} + K[z_{k} - HF\hat{x}_{k-1|k-1}],$$ see, for example, in wiki: https://en.wikipedia.org/wiki/Kalman_filter#Underlying_dynamical_system_model. Conclusion: $$\hat{x}_{k|k}$$ converges in distribution to some normal random variable.