Timeline for Proof that any $\epsilon$-greedy policy is an improvement over any $\epsilon$-soft policy
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 12, 2023 at 17:11 | comment | added | Michael Hardy | I changed $\varepsilon-$greedy and $\varepsilon-$soft to $\varepsilon$-greedy and $\varepsilon$-soft. The correct usage here is a hyphen, not a minus sign. | |
| Jul 12, 2023 at 17:09 | history | edited | Michael Hardy | CC BY-SA 4.0 |
This calls for a hyphen, not a minus sign.
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| Oct 16, 2020 at 16:56 | comment | added | FantasticAI | why the first equality holds? | |
| S May 24, 2020 at 19:46 | history | suggested | Jayanth | CC BY-SA 4.0 |
It is action a given state s and not the other way round.
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| May 23, 2020 at 18:15 | review | Suggested edits | |||
| S May 24, 2020 at 19:46 | |||||
| Jul 21, 2019 at 9:36 | vote | accept | robertspierre | ||
| Jul 21, 2019 at 2:12 | answer | added | M.Q | timeline score: 3 | |
| Jul 16, 2019 at 11:48 | comment | added | CloseToC | The weighted average does not sum to 1, the weights in front of $q_\pi(s,a)$ do, here's why: $\pi(a|s)$ presumably sums to 1 over all $a$. Then the sum of the weights can be written as $\frac{1}{1-\epsilon} - |A| \cdot\frac{\epsilon}{ (1-\epsilon) \cdot |A|}$ which is 1 | |
| Jul 16, 2019 at 9:23 | comment | added | robertspierre | Well the first part, why it is a weighted average that sums to 1. Thank you | |
| Jul 16, 2019 at 8:09 | comment | added | CloseToC | The note tells you why it follows from the previous line, so the question is which parts of the note you need help understanding with | |
| Jul 16, 2019 at 8:07 | comment | added | robertspierre | I am asking why the last inequality is true | |
| Jul 16, 2019 at 8:02 | comment | added | CloseToC | Are you asking why the claim in note is true, or how to verify that these weights sum up to 1? | |
| Jul 16, 2019 at 7:31 | history | asked | robertspierre | CC BY-SA 4.0 |