# Help with Probabilistic Robotics Equation 13.22 detailed derivation

Equation 13.22 from Probabilistic Robotics below:

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?

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.

• 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? – drerD Mar 8 '19 at 8:07
• This is a possible cause, also $x^{k}_t$, from what I remember, only depends on past information and not current information. – Akindart Mar 8 '19 at 18:38
• that makes sense too. – drerD Mar 9 '19 at 2:32