Timeline for Does $P(R\vert Q,P) = P(R\vert Q)$ implies that $R$ is independent of $P$?
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Jan 24, 2021 at 18:00 | history | tweeted | twitter.com/StackStats/status/1353402262240989184 | ||
| Dec 7, 2020 at 17:08 | vote | accept | Three Diag | ||
| Dec 7, 2020 at 15:46 | comment | added | fblundun | Oops, that should have read "$P$ and $Q$ are not independent and are disjoint". | |
| Dec 7, 2020 at 14:29 | comment | added | fblundun | That isn't the only problematic case though - for example, suppose $P$ is "we roll a 1", $Q$ is "we roll a 2", and $R$ is "we roll a 1 or a 2". Then $P$ and $Q$ are independent and disjoint. The flaw in your derivation is that if two fractions are equal, it need not follow that their numerators are equal and their denominators are equal. | |
| Dec 7, 2020 at 14:19 | history | edited | Three Diag | CC BY-SA 4.0 |
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| Dec 7, 2020 at 14:18 | comment | added | Three Diag | I see what you mean, I will edit accordingly to rule this case out. Thanks! | |
| Dec 7, 2020 at 14:08 | comment | added | fblundun | I still don't believe the conclusion $P \perp Q$. For example, suppose we are rolling a die and $P$ is "we roll a 3 or lower", $Q$ is "we roll a 2 or lower", and $R$ is "we roll a 1". Or in general any case where $P$ is always true when $Q$ is true. | |
| Dec 7, 2020 at 14:03 | comment | added | Three Diag | Ok, got rid of this edge case. | |
| Dec 7, 2020 at 14:02 | history | edited | Three Diag | CC BY-SA 4.0 |
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| Dec 7, 2020 at 13:46 | comment | added | fblundun | I don't agree with your penultimate sentence - consider the case where $P$, $Q$, and $R$ are all the same event. | |
| Dec 7, 2020 at 13:44 | answer | added | gunes | timeline score: 3 | |
| Dec 7, 2020 at 13:36 | history | asked | Three Diag | CC BY-SA 4.0 |