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:thompnicks...@gmail.com> thompnicks...@gmail.com

 <https://wordpress.clarku.edu/nthompson/> 
https://wordpress.clarku.edu/nthompson/

 

From: Friam <friam-boun...@redfish.com> On Behalf Of Frank Wimberly
Sent: Monday, December 6, 2021 3:07 PM
To: The Friday Morning Applied Complexity Coffee Group <friam@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 <thompnicks...@gmail.com 
<mailto:thompnicks...@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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