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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?

Peter Mitrano
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1 Answers1

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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.

Steffen
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  • 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 – Peter Mitrano Apr 10 '18 at 15:48
  • You are right. Check updated answer! – Steffen Apr 11 '18 at 07:02
  • 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 – Peter Mitrano Apr 11 '18 at 22:07
  • 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. – Steffen Apr 12 '18 at 07:19
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    It is not about one variable effecting another variable. It is about the knowledge of one variable giving you information about the other variable. – Steffen Apr 12 '18 at 07:21
  • Similarly, $x_t$ and $x_{t-1}$ are not independent and knowledge of one affects the probability distribution of the other. Also, knowledge of $z_t$ informs about distribution of $x_t$. Arrows seem to indicate cause and effect (in time or not); from the effect one can infer about the cause and vica-versa. – Don Slowik Oct 28 '19 at 15:53
  • Isn’t the assumption that $u_t$ independent of $x_{t-1}$ bad? Most control would depend on state of the world... – Don Slowik Oct 28 '19 at 16:00