Equation 13.22 from Probabilistic Robotics below: enter image description here

Here's how I get from first line to second line:

$$ p(x_{1:t}^{[k]} | z_{1:t},u_{1:t}, c_{1:t}) = \frac{p(x_{1:t}^{[k]}, z_{1:t},u_{1:t}c_{1:t}) }{ p(z_{1:t},u_{1:t}, c_{1:t})} \\ p(x_{1:t}^{[k]}, z_{1:t},u_{1:t}c_{1:t}) = p(z_t|x_{1:t}^{[k]} ,z_{1:t-1},u_{1:t}, c_{1:t})p(x_{1:t}^{[k]} ,z_{1:t-1},u_{1:t}, c_{1:t}) \\ p(x_{1:t}^{[k]} ,z_{1:t-1},u_{1:t}, c_{1:t}) = p(x_{1:t}^{[k]} | z_{1:t-1},u_{1:t}, c_{1:t}) p(z_{1:t-1},u_{1:t}, c_{1:t}) $$ Just setting up conditional probabilities above, then I'm subbing back to the first equation:

$$ \\ p(x_{1:t}^{[k]} | z_{1:t},u_{1:t}, c_{1:t}) = \frac{p(x_{1:t}^{[k]}, z_{1:t},u_{1:t}c_{1:t}) }{ p(z_{1:t},u_{1:t}, c_{1:t})} = \frac{p(z_t|x_{1:t}^{[k]} ,z_{1:t-1},u_{1:t}, c_{1:t})p(x_{1:t}^{[k]} ,z_{1:t-1},u_{1:t}, c_{1:t})}{p(z_{1:t},u_{1:t}, c_{1:t})} = \frac{p(z_t|x_{1:t}^{[k]} ,z_{1:t-1},u_{1:t}, c_{1:t})p(x_{1:t}^{[k]} | z_{1:t-1},u_{1:t}, c_{1:t}) p(z_{1:t-1},u_{1:t}, c_{1:t})}{p(z_{1:t},u_{1:t}, c_{1:t})} = \frac{p(z_{1:t-1},u_{1:t}, c_{1:t})}{p(z_{1:t},u_{1:t}, c_{1:t})} p(z_t|x_{1:t}^{[k]} ,z_{1:t-1},u_{1:t}, c_{1:t})p(x_{1:t}^{[k]} | z_{1:t-1},u_{1:t}, c_{1:t}) = \eta \ p(z_t|x_{1:t}^{[k]} ,z_{1:t-1},u_{1:t}, c_{1:t})p(x_{1:t}^{[k]} | z_{1:t-1},u_{1:t}, c_{1:t}) $$

how do I get to the third line from here?


1 Answer 1


The third line comes from what it is called Markov Assumption and it is Stochastic Processes stuff. Basically, it says that a distribution is not altered by the insertion and/or remotion of variables that the distribution does not really depend on. It goes like this:

Is assumed that $ z_t $ simply does not depend on the previous reading history, inputs $ u_{1:t} $ and $ c_{1:t-1} $, so one can write (Assuming Markov Process)

$$ p(z_t | x^{k}_{1:t}, z_{1:t-1}, u_{1:t}, c_{1:t}) = p(z_t | x^{k}_{t}, c_{t}) $$

which makes sense, since the probability of observing a reading depends on the robot pose itself and not on the commands used to get there.

The same goes for the other component. I just do not remember exactly what is $ c $. In the second term, $ c_{t} $ is considered to not alter the probability, therefore it is removed from the expression based on Markov Assumption.

  • 1
    $\begingroup$ Thanks! The equation is discussing the FastSLAM with correspondence, so that c is the correspondence variable, and my understanding is it's saying which observed feature corresponding to which known feature.So about dropping the $c_t$ in the second term, maybe it's because we don't have $z_t$ so the correspondence $c_t$ isn't affecting the probability? $\endgroup$
    – drerD
    Mar 8, 2019 at 8:07
  • $\begingroup$ This is a possible cause, also $ x^{k}_t $, from what I remember, only depends on past information and not current information. $\endgroup$
    – Akindart
    Mar 8, 2019 at 18:38
  • $\begingroup$ that makes sense too. $\endgroup$
    – drerD
    Mar 9, 2019 at 2:32

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