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I am trying to wrap my head around Partially Observed Markov Decision Process (POMDP). However, I am unable to understand the application of the Bayes Rule in following equation (Step Nr. 2):

enter image description here

Can anyone help me understand how Bayes rule can be applied for multiple conditions like this ? or at least point me towards the right literature. Thank you.

The picture is taken from here: Acting Optimally in Partially Observable Stochastic Domains

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1 Answer 1

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Basically, going from line 1 to line 2 requires applying repeatedly the following rule $$ P(A, B) = P(B | A) P(A) $$ Indeed: $$ P(s^\prime | a, o, b) = \frac{P(s^\prime, a, o, b)}{P(a, o, b)} \\ = \frac{P(o | s^\prime, a , b) P(s^\prime, a, b)}{P(a, o, b)} \\ = \frac{P(o | s^\prime, a , b) P(s^\prime| a, b) P(a, b)}{P(o| a, b) P(a, b)} \\ $$ which is the expression on the second line once you erase $P(a,b)$ at the numerator and denominator.

Going from the second to the third line is easy as well, the observations are independent on the belief $b$, conditioning to the state and the action, so that $$ P(o | s^\prime , a, b) = P(o | s^\prime , a) := O(a, s^\prime, o) $$ while one can write $$ P(s^\prime | a, b) = \sum_{s \in S} P(s^\prime | a, b, s) P(s) $$ and again observing that the transition from $s$ to $s^\prime$ does not depend on $b$ given the action, one has $$ \sum_{s \in S} P(s^\prime | a, b, s) P(s) = \sum_{s \in S} T(s, a, s^\prime) b(s) $$ where $b(s) = P(s)$ in the authors' notation

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