I am trying to use the Kalman filter (the scalar version) to estimate the steady state of a set measurements which is a random process. I have used a constant dynamic model as the state equation,

$$ x(t+1) = x(t) + w(t) $$

Where $ w(t) \sim N(0, Q) $.

The measurement equation

$$ y(t) = x(t) + v(t) $$

Where $ v(t) \sim N (0, R) $.

The measurement noise $ v(t) $ is taken as the uncertainty of the instrument and R is a known value with a assumed Gaussain distribution. The process error, $w(t) = x(t+1) - x(t)$ has high values as a constant variance (Calculated using the total error variance) or a time varying (Taking last 2, 5 readings as the set f sample). However, the state reaches the steady state when $ Q $ is very small or $ Q = 0 $.

I have seen several other theoretical examples uses $ Q = 0 $ and reaches the steady state. But I am working with actual data. Also I have seen some other posts explaining the similar behavior. Can some explain this behavior with reasons, please?

  • $\begingroup$ you are using a constant dynamic model but is the true signal actually an unknown constant ? $\endgroup$
    – Fat32
    Jul 24, 2019 at 22:05
  • $\begingroup$ Yes, trying to use this steady state estimation algirithm (Kalman) to estimate the unknown constant values of the steady state. $\endgroup$
    – user44324
    Jul 25, 2019 at 0:53

1 Answer 1


The process noise $w(t)$, which is typically assumed to be a zero mean, white Gaussian noise with (power) variance $\sigma^2$, is used to account for any mismatches between the assumed dynamic model of the states and the actual truth.

If your assumed constant state (DC) model perfectly matches with the underlying nature of the observed signal, then you should set the process noise to zero; i.e., your model is exact.

Then you only have measurement noise due to the sensor's intrinsic noise.

If you do not set your $\sigma^2$ to zero (adding non-zero process noise to a constant dynamic model) then your state estimation will never reach steady-state, but wander around it; as you are telling the Kalman filter that the true signal is not a constant...

  • $\begingroup$ Thanks a lot, this answers my question. $\endgroup$
    – user44324
    Jul 25, 2019 at 3:45
  • $\begingroup$ Actually what happens is the Kalman Filter tries to estimate the mean of this "Random Walk" in the case $ Q $ isn't vanished. $\endgroup$
    – Royi
    Jul 28, 2019 at 2:22

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