Help with Probabilistic Robotics Equation 13.22 detailed derivation - Robotics Stack Exchange most recent 30 from robotics.stackexchange.com 2019-12-07T11:08:47Z https://robotics.stackexchange.com/feeds/question/18351 https://creativecommons.org/licenses/by-sa/4.0/rdf https://robotics.stackexchange.com/q/18351 2 Help with Probabilistic Robotics Equation 13.22 detailed derivation drerD https://robotics.stackexchange.com/users/9972 2019-03-06T22:05:15Z 2019-03-07T00:50:56Z <p>Equation 13.22 from Probabilistic Robotics below: <a href="https://i.stack.imgur.com/X3xC1.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/X3xC1.png" alt="enter image description here"></a></p> <p>Here's how I get from first line to second line:</p> <p><span class="math-container">$$p(x_{1:t}^{[k]} | z_{1:t},u_{1:t}, c_{1:t}) = \frac{p(x_{1:t}^{[k]}, z_{1:t},u_{1:t}c_{1:t}) }{ p(z_{1:t},u_{1:t}, c_{1:t})} \\ p(x_{1:t}^{[k]}, z_{1:t},u_{1:t}c_{1:t}) = p(z_t|x_{1:t}^{[k]} ,z_{1:t-1},u_{1:t}, c_{1:t})p(x_{1:t}^{[k]} ,z_{1:t-1},u_{1:t}, c_{1:t}) \\ p(x_{1:t}^{[k]} ,z_{1:t-1},u_{1:t}, c_{1:t}) = p(x_{1:t}^{[k]} | z_{1:t-1},u_{1:t}, c_{1:t}) p(z_{1:t-1},u_{1:t}, c_{1:t})$$</span> Just setting up conditional probabilities above, then I'm subbing back to the first equation:</p> <p><span class="math-container">$$\\ p(x_{1:t}^{[k]} | z_{1:t},u_{1:t}, c_{1:t}) = \frac{p(x_{1:t}^{[k]}, z_{1:t},u_{1:t}c_{1:t}) }{ p(z_{1:t},u_{1:t}, c_{1:t})} = \frac{p(z_t|x_{1:t}^{[k]} ,z_{1:t-1},u_{1:t}, c_{1:t})p(x_{1:t}^{[k]} ,z_{1:t-1},u_{1:t}, c_{1:t})}{p(z_{1:t},u_{1:t}, c_{1:t})} = \frac{p(z_t|x_{1:t}^{[k]} ,z_{1:t-1},u_{1:t}, c_{1:t})p(x_{1:t}^{[k]} | z_{1:t-1},u_{1:t}, c_{1:t}) p(z_{1:t-1},u_{1:t}, c_{1:t})}{p(z_{1:t},u_{1:t}, c_{1:t})} = \frac{p(z_{1:t-1},u_{1:t}, c_{1:t})}{p(z_{1:t},u_{1:t}, c_{1:t})} p(z_t|x_{1:t}^{[k]} ,z_{1:t-1},u_{1:t}, c_{1:t})p(x_{1:t}^{[k]} | z_{1:t-1},u_{1:t}, c_{1:t}) = \eta \ p(z_t|x_{1:t}^{[k]} ,z_{1:t-1},u_{1:t}, c_{1:t})p(x_{1:t}^{[k]} | z_{1:t-1},u_{1:t}, c_{1:t})$$</span></p> <p>how do I get to the third line from here? </p> https://robotics.stackexchange.com/questions/18351/help-with-probabilistic-robotics-equation-13-22-detailed-derivation/18354#18354 2 Answer by Akindart for Help with Probabilistic Robotics Equation 13.22 detailed derivation Akindart https://robotics.stackexchange.com/users/20197 2019-03-07T00:50:56Z 2019-03-07T00:50:56Z <p>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:</p> <p>Is assumed that <span class="math-container">$z_t$</span> simply does not depend on the previous reading history, inputs <span class="math-container">$u_{1:t}$</span> and <span class="math-container">$c_{1:t-1}$</span>, so one can write (Assuming Markov Process)</p> <p><span class="math-container">$$p(z_t | x^{k}_{1:t}, z_{1:t-1}, u_{1:t}, c_{1:t}) = p(z_t | x^{k}_{t}, c_{t})$$</span></p> <p>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.</p> <p>The same goes for the other component. I just do not remember exactly what is <span class="math-container">$c$</span>. In the second term, <span class="math-container">$c_{t}$</span> is considered to not alter the probability, therefore it is removed from the expression based on Markov Assumption.</p>