# Why does a Bayesian Filter require random controls?

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.

• 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 7:02
• 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 7:19
• 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 7:21