Note that if the error variances associated with different observations
are the same, then a simple average will be fine. However, in the case where
the observed values have different levels of accuracy, we would like to assign
more weight to the observations with greater accuracy. In such cases, instead
of taking a simple average that weights each point uniformly, a weighted av-
erage solution would be more appropriate. We will illustrate how to address
the weighted average situation using the Kalman filter concepts and thereby
hope to provide some insight to the calculation of the Kalman gain.
In the estimation of the constant value, the system state is the one-
dimensional constant value itself. Let us say that our current estimate of the
constant value is with error in the estimate being ei, that is,
(4.3)
xis the true value of the constant. Although we do not know what the exact
value of eiis, we do know that it has a zero mean and known variance given
by. Let us now do the prediction step. Since the value stays a constant,
our prediction for the next state is the current value itself. Therefore, we have
(4.4)Next we make a measurement. Let us call the measurement of the constant
valueyiand the error associated with it hiwith zero mean and known vari-
ance given as. Therefore at the end of the measurement step, we have the
observation
(4.5)We now have a prediction and an observation, and we wish to reconcile the
two values. We do this in the reconciliation/correction step where we correct
the predicted value based on the observation. Writing out the correction
equation, we have
(4.6)Note that this can be rewritten as
(4.7)That is, the corrected state estimate is a weighted average of the two state
estimates, with k, the Kalman gain, being the weight. The choice for the
xkxkyˆˆii++ 11 ||=−() (^1) ii+ 1 + i
xxkyxˆˆii++ 11 ||=+−ii+ 1 ()iˆii+ 1 |
yxiiii=+ησ, var()y= η^2 ,
ση^2 ,i
xxxˆˆii++ 11 ||,= i,var()ˆii=σ^2 εi
σε^2 ,i
xxˆii| =+εi
xˆi
58 BACKGROUND MATERIAL