It may not be fun but it's essential in algorithms for inferring causation from correlation.
--- Frank C. Wimberly 140 Calle Ojo Feliz, Santa Fe, NM 87505 505 670-9918 Santa Fe, NM On Sat, Dec 4, 2021, 5:53 PM Frank Wimberly <[email protected]> wrote: > The point is if you know 2 has occurred you know the probability that 3 > will occur. 2 occurring makes 1 irrelevant. > > Very formally 1 is independent of 3 given 2. Please use my variable > names to avoid my making an error. > > Frank > --- > Frank C. Wimberly > 140 Calle Ojo Feliz, > Santa Fe, NM 87505 > > 505 670-9918 > Santa Fe, NM > > On Sat, Dec 4, 2021, 5:38 PM Nicholas Thompson <[email protected]> > wrote: > >> Frank, >> >> Still need help. Given events 1, 2, and 3, 3 has been screen off by 2 >> from 1, if the probability that 3 occurs given that 2 has occurred is >> equal to the probability that 3 occurs given that both 2 and one have >> occurred. As I understand mathematics this equality requires that the >> probability of 1 occurring is 1.00. Another way to say that is that the >> probability that 3 occurs if 2 has occurred is the same as the probability >> that 3 has occurred if 2 has occurred, and 1 has already occurred. What's >> the fun in that? In other words, given the possibility of other causes for >> 2, the fact that 2 occurs gives us relatively little evidence that 1 has >> occurred. Isn"t this true of all causal abduction? >> >> N >> >> .-- .- -. - / .- -.-. - .. --- -. ..--.. / -.-. --- -. .--- ..- --. .- - . >> FRIAM Applied Complexity Group listserv >> Zoom Fridays 9:30a-12p Mtn UTC-6 bit.ly/virtualfriam >> un/subscribe http://redfish.com/mailman/listinfo/friam_redfish.com >> FRIAM-COMIC http://friam-comic.blogspot.com/ >> archives: >> 5/2017 thru present https://redfish.com/pipermail/friam_redfish.com/ >> 1/2003 thru 6/2021 http://friam.383.s1.nabble.com/ >> >
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