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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 $m \leq n$.

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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 is 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 $w_k$.

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  • $\begingroup$ "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 $\endgroup$ – ABK Nov 27 '19 at 12:13
  • $\begingroup$ @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. $\endgroup$ – Kwin van der Veen Nov 27 '19 at 12:19
  • $\begingroup$ Dear @Kwin van der Veen, random means non-constant, i.e. having a certain distribution. This is a simple explanation. $\endgroup$ – ABK Nov 27 '19 at 12:31
  • $\begingroup$ @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. $\endgroup$ – Kwin van der Veen Nov 27 '19 at 12:35
  • $\begingroup$ 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. $\endgroup$ – ABK Nov 27 '19 at 13:20

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