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when toggle format what by license comment
S Jan 18 at 0:57 history suggested bilibraker CC BY-SA 4.0
deleting redundant line, changing 2 "\mid" to "|" for consistency
Jan 18 at 0:46 review Suggested edits
S Jan 18 at 0:57
Jan 18 at 0:43 comment added bilibraker superb answer, cleared up a lot of things!
Jul 13, 2023 at 20:39 review Suggested edits
Jul 13, 2023 at 20:45
Jul 13, 2023 at 20:28 history edited Finncent Price CC BY-SA 4.0
the edits that were made 14 hours ago are not mathematically correct, the short proof has been changed to be correct
S Jul 13, 2023 at 6:39 history edited utobi CC BY-SA 4.0
Short helper proof for a part of answer
S Jul 13, 2023 at 6:39 history suggested Kanishk CC BY-SA 4.0
Short helper proof for a part of answer
Jul 13, 2023 at 6:00 review Suggested edits
S Jul 13, 2023 at 6:39
Dec 17, 2020 at 22:42 comment added Fabian Werner but only on the variables $S_{t-1}, A_{t-1}, R_{t-1}$... ). So if you say that then you have a different definition of 'markovian' than everybody else I guess...
Dec 17, 2020 at 22:41 comment added Fabian Werner Two comments on this interesting answer: 1. Even if the reward is either 0 or 1, the values that $G_t$ may take can nevertheless be bigger so we need an integral instead of a sum over $\gamma$. 2. What exactly is $p(g|...)$ supposed to be? Let's start simple: what is $p(g)$ supposed to be? It is the density of $G_{t+1}$, right? Why does this random variable even have a density? Even if it had a density, why exactly does it not depend on $s$ in the sense that you wrote down? This is not the Markov property (Markov says that the single variables $S_t, A_t, R_t$ do not depend on the whole past...
Dec 8, 2020 at 20:57 vote accept Amelio Vazquez-Reina
Jun 27, 2019 at 15:04 history edited Finncent Price CC BY-SA 4.0
Clarified a point about the Markovian property
Jun 20, 2019 at 15:51 history answered Finncent Price CC BY-SA 4.0