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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?
I think it is easy to see, when you take a look at the bayesian network:
Now, we eliminate all the variables not given in your equation:
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:
Here $u_t$ can not be omitted.
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
