9
$\begingroup$

For a linear state-space model with independent Gaussian state and output noises and perfect guess for initial state, do Kalman estimates have the following properties: $$ E(\hat{x}_{k|k} - x_k) = 0 $$ $$ P_{k|k} = Var(\hat{x}_{k|k} - x_k),\text{ or }Var(\hat{x}_{k|k}),\text{ or }Var(x_k) ? $$ where

  • $x_k$ is the state at time $k$, which is random

  • $\hat{x}_{k|k}$ and $P_{k|k}$ are Kalman esitmates, i.e. outputs of Kalman filter.

Are there references mentioning these?

Thanks!

$\endgroup$
  • $\begingroup$ Is $P_{k|k}$ the a posteriori estimated covariance matrix at time $k$? There's not really a standard notation that is used, so it's not completely clear what you mean by "Kalman estimates." $\endgroup$ – Jason R Jun 6 '12 at 1:43
  • $\begingroup$ @Jason: yes, it is ... $\endgroup$ – Tim Jun 6 '12 at 2:43
3
$\begingroup$

The following two statements are equivalent of saying:

$$ E(\hat{x}_{k|k} - x_k) = 0 $$

(1) That the estimator is unbiased; and

$$ P_{k|k} = Var(\hat{x}_{k|k} - x_k) $$

(2) That the estimator is consistent.

Both of these conditions are necessary in order for the filter to be optimal - i.e. the best possible estimate of $\mathbf{x}_{k|k}$ with respect to some criteria.

If (1) is not true, then the mean-square error (MSE) would be the bias plus the variance (in the scalar case). Clear, this is larger than the variance only and hence suboptimal.

If (2) is not true (i.e. the filter-calculated covariance is different to the true covariance) then the filter will also be suboptimal. Since the Kalman Gain is based on the calculated state covariance, an error in the covariance will lead to an error in the gain. Error in the gain means a suboptimal weighting of the measurements.

(As it happens, both conditions are true for a properly modelled filter. Errors in modelling, such as the dynamic model or noise covariances will also render the filter suboptimal).

Source: Bar-Shalom, especially Section 5.4 on page 232-233.

$\endgroup$
2
$\begingroup$

It is important to note that $x_k$ is NOT a random variable. It is the system's state which is deterministic, which is in general, variable in $k$. $$ E(\hat{x}_{k|k}) = x_k $$ which is equivalent of saying $$ E(\hat{x}_{k|k} - x_k) = 0 $$

Also, $$Var(x_k) = 0$$

And,

$$P_{k|k} = Var(\hat{x}_{k|k})$$ which, given that $x_k$ is deterministic, happens to be also equal to $Var(\hat{x}_{k|k} - x_k)$

Background

$x_k$ is the system state which is deterministic. This is as opposed to the the system noise, which is represented in most literature as $w$ with variance $Q$. Even more, some literature models the system noise with a coefficents matrix; in which case the $Q$ matrix is replaced by $GQG^T$ in the propagation estimate, where $G$ is the noise coefficeint matrix. To elaborate, the system representation, in this case, is given by: $$ x_{k+1} = Ax_k + Bu_k +Gw $$

As a refernce: Kalman's paper himself:http://160.78.24.2/Public/Kalman/Kalman1960.pdf

$\endgroup$
  • $\begingroup$ As far as I know $ \left \{ x_k \right \}_{k=-\infty}^{\infty} $ is a random process. The variance of $ x_k $ is given by the process noise. For a given realization $ x_k $ is deterministic. $\endgroup$ – Royi Jun 10 '13 at 5:42
  • $\begingroup$ @Drazick The process noise is usually given the symbol w , with variance Q . x k is the system state, it wouldn't make any sense that the states are random; the estimate on the other, being a random variable, does make sense $\endgroup$ – aiao Jun 10 '13 at 13:39
  • $\begingroup$ I'm confused: How can $x_{k+1}$ be deterministic if $Gw$ (which is stochastic) is being added to form it? The only way $x_{k+1}$ can be deterministic is if the stochastic component is zero, yes? $\endgroup$ – Peter K. Jun 10 '13 at 19:00
  • $\begingroup$ @PeterK. because $w$ assumes a definite realization at every $k$ $\endgroup$ – aiao Jun 10 '13 at 19:32
  • 1
    $\begingroup$ Whilst Kalman himself never considered the state vector to be a stochastic variable (I think I can attribute this to Doucet, but I could be wrong), the Kalman Filter can be derived from Bayes Rule. In this instance, the state vector $\mathbf{x}_{k|k} \sim N\left( \hat{\mathbf{x}}_{k|k}, \mathbf{P}_{k|k}\right)$. See Wikipedia. $\endgroup$ – Damien Jun 11 '13 at 11:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.