[FRIAM] Screening off
thompnickson2 at gmail.com
thompnickson2 at gmail.com
Mon Dec 6 18:22:38 EST 2021
Frank, Thanks for sticking with me on this.
My comment was not conceptual, merely algebraic: Can
Pr(R at t3 | I at t2) = Pr(R at t3 | I at t2 & S at t1).
…be true if the value of S at t1.is anything other than one?
n
Nick Thompson
<mailto:ThompNickSon2 at gmail.com> ThompNickSon2 at gmail.com
<https://wordpress.clarku.edu/nthompson/> https://wordpress.clarku.edu/nthompson/
From: Friam <friam-bounces at redfish.com> On Behalf Of Frank Wimberly
Sent: Monday, December 6, 2021 3:07 PM
To: The Friday Morning Applied Complexity Coffee Group <friam at redfish.com>
Subject: Re: [FRIAM] Screening off
The question isn't whether 1 has occurred but what is the value of 1 (it's a variable) and is the distribution of 3 is independent of that observed value.
---
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 <thompnickson2 at gmail.com <mailto:thompnickson2 at gmail.com> > 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
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