3
$\begingroup$

As stated in Probabalistic Robotics, the proof for correctness of a Bayesian Filter relies on the fact that

$$p(x_{t-1}|z_{1:t-1},\ u_{1:t}) = p(x_{t-1}|z_{1:t-1},\ u_{1:t-1})$$

In order to justify this, they say

$u_t$ can be safely omitted ... for randomly chosen controls

Why is that required? Isn't that true because the control input at time step t cannot possibly effect the state at time t-1?

Improve this question
$\endgroup$
1
$\begingroup$

I think it is easy to see, when you take a look at the bayesian network: enter image description here

Now, we eliminate all the variables not given in your equation: enter image description here

Based on this baesian network, you can see that $u_t$ has no effect on the other variables and thus can be omitted.

This however would not work for $u_t$s that are not random. If for example $u_t$ would be dependent on $x_{t-1}$. Which means, based on the last position, the actual control is selected:

enter image description here

Here $u_t$ can not be omitted.

Improve this answer
$\endgroup$
7
  • $\begingroup$ Thanks for the detailed answer, but I don't agree with your last sentence. u_t does not influence u_{t-1}, and so it doesn't influence x_{t-1}. For what you're saying to be true, the arrow would have to point the other way $\endgroup$
    – Peter Mitrano
    Apr 10 '18 at 15:48
  • $\begingroup$ You are right. Check updated answer! $\endgroup$
    – Steffen
    Apr 11 '18 at 7:02
  • $\begingroup$ I accepted your answer hastily, but now I am again unsure. We only care about u_t effecting x_{t-1} right? Not the other way around $\endgroup$
    – Peter Mitrano
    Apr 11 '18 at 22:07
  • $\begingroup$ Is it, that the direction of the arrow confuses you? In the last picture: If you know $u_t$, you can reason about $x_{t-1}$. So, $u_t$ must not be omitted. You can search for "Markov blanket". This describes, when variables are independent. $\endgroup$
    – Steffen
    Apr 12 '18 at 7:19
  • 1
    $\begingroup$ It is not about one variable effecting another variable. It is about the knowledge of one variable giving you information about the other variable. $\endgroup$
    – Steffen
    Apr 12 '18 at 7:21

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.