Timeline for How to eliminate variable given conditional probabilities
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Jun 8, 2019 at 12:42 | vote | accept | user4046073 | ||
| May 27, 2019 at 16:56 | history | edited | Brent | CC BY-SA 4.0 |
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| May 27, 2019 at 16:53 | comment | added | Brent | You're right. Updated the answer to fix that and gave a brief explanation with a link for a deeper understanding | |
| May 27, 2019 at 16:47 | history | edited | Brent | CC BY-SA 4.0 |
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| May 26, 2019 at 10:51 | comment | added | user4046073 | I mean should it be +𝑃(𝑐=𝑡𝑟𝑢𝑒|𝑝=𝑡𝑟𝑢𝑒,𝑠=𝑓𝑎𝑙𝑠𝑒)∗𝑃(𝑠=𝑓𝑎𝑙𝑠𝑒,𝑝=true) ? | |
| May 26, 2019 at 8:32 | comment | added | user4046073 | should the last bit be P(s= false, p= true)? | |
| May 26, 2019 at 8:31 | comment | added | user4046073 | Thanks. could you explain a bit more why the first equation ? | |
| May 26, 2019 at 4:29 | comment | added | Brent | Sorry, read your question too quickly and assumed your table was the joint distribution. Have updated the answer | |
| May 26, 2019 at 4:29 | history | edited | Brent | CC BY-SA 4.0 |
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| May 26, 2019 at 4:18 | history | edited | Brent | CC BY-SA 4.0 |
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| May 26, 2019 at 3:58 | comment | added | user4046073 | I know I can sum out P(C, P,S) , can I sum out P(C | P, S) as well? How to calculate P(C=false | P=true) then? I am confused as if I use P(C,P,S) = P(P,S)*P(C|P,S) , and sum out P(C, P,S), the results are different from suming up P(C, P,S) directly. Which one is correct? Thanks! | |
| May 26, 2019 at 3:48 | history | answered | Brent | CC BY-SA 4.0 |