# Kalman filter: state estimate

Assume we have a linear state-space model: $$z_{k} = Hx_{k} + v_{k}\\ x_{k} = F x_{k-1} + w_{k}$$

If I understand correctly, having observations $$z_{0}, \dots, z_{k}$$, in filtering problem the goal is to get the estimate $$\hat{x}_{k|k}$$ of $$E[x_{k}|z_{k}, \dots, z_{0}]$$.

The question: why do we call $$\hat{x}_{k|k}$$ the state estimate if state is itself random?

In wiki, it is even more confusing:

In what follows, the notation $$\hat{\mathbf{x}}_{n|m}$$ represents the estimate of $$x$$ at time $$n$$ given observations up to and including at time $$m \leq n$$.

Each measurement is associated with a certain time index and this then also fixes the associated estimate of the state with respect to an index. For example $$\hat{x}_{k|k} = E[x_{k}|z_{k}, \dots, z_{0}]$$ is a different problem compared to $$\hat{x}_{k+1|k} = E[x_{k+1}|z_{k}, \dots, z_{0}]$$. Namely, in that case the measurement $$z_{k+1}$$ is not used, so during that time step one can only predict. For example if $$w_k\sim\mathcal{N}(0,W)$$ then the expected value of the prediction would yield $$\hat{x}_{k+1|k} = F\,\hat{x}_{k|k}$$.

Also note that the state itself is a stochastic process, which does not have to be completely "random" since it can be modeled by $$x_{k} = F\,x_{k-1} + w_{k}$$. Only if $$F=0$$ would the state inherit all the stochastic properties from $$w_k$$.

• "the state itself is a stochastic process, which is does not have to be completely "random"" - I am totally confused... In a bayesian framework one can estimate the random parameters... but in the description of Kalman procedure one does not write it explicitely
– ABK
Nov 27, 2019 at 12:13
• @ABK How do you define "random"? Namely, each $v_k$ and $w_k$ are usually a sample from a certain distribution, independent of time. And frequently is assumed that this distribution is zero mean Gaussian white noise. Nov 27, 2019 at 12:19
• Dear @Kwin van der Veen, random means non-constant, i.e. having a certain distribution. This is a simple explanation.
– ABK
Nov 27, 2019 at 12:31
• @ABK as a couter example would you consider $x_k=k$ random? Also, it would be a possibility that $w_k=0$ for all $k$, in which case $x_k$ would be deterministic if you would know the initial conditions. Nov 27, 2019 at 12:35
• of course you can think of it as a degenerate case of random variable! According to definition of estimator of parameter, the parameter is not a random variable.
– ABK
Nov 27, 2019 at 13:20

The notation $$\hat{ x}_{n|m}$$ means $$E(x_n|z_m,\cdots,z_0)$$. When $$m=n$$, estimating $$x_t$$ is the filtering problem, when $$m>n$$, it is the prediction problem, and it is the smoothing problem for the cases where $$n.