Simple probability question: Suppose P(B|A)=1. Does that mean that P(A|B)=1?

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SUMMARY

The discussion centers on the implications of conditional probabilities, specifically whether P(B|A)=1 implies P(A|B)=1. Participants clarify that while P(B|A)=1 indicates that event B occurs with certainty when A occurs, it does not necessarily mean A occurs when B is known to occur. The conversation references Bayes' Theorem and the definitions of causality, emphasizing that correlation does not imply causation. Key points include the distinction between necessary and sufficient causes and the role of probability measures in defining events.

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  • Understanding of conditional probability and notation (e.g., P(A|B))
  • Familiarity with Bayes' Theorem and its application
  • Knowledge of basic set theory and events in probability
  • Concept of causality in statistical contexts
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  • #31
entropy1 said:
But actually, if ##A \rightarrow B## AND ##C \rightarrow NOT(B)##, then I wonder if C=True results in A=NOT True (or, of course, A=True in C=NOT True).
We can do a short proof here:

To be proven: ##C \rightarrow \neg A##

##1: A \rightarrow B## assumption 1
##2: C \rightarrow \neg B## assumption 2
##3: C## assumption 3
##4: \neg B## modus ponens 2,3
##5: \neg A ## modus tollens 1,4
##6: C \rightarrow \neg A## hypothethical syllogism (1,2),3,(4),5

##-## and there you have it ##\dots##
 
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  • #32
entropy1 said:
No, ##A \rightarrow B## is equivalent to ##NOT(B) \rightarrow NOT(A)##.

But actually, if ##A \rightarrow B## AND ##C \rightarrow NOT(B)##, then I wonder if C=True results in A=NOT True (or, of course, A=True in C=NOT True).
I always wondered why probability and logic are in the same forum. Now I see why. :smile:
 
  • #33
sysprog said:
We can do a short proof here:

To be proven: ##C \rightarrow \neg A##

##1: A \rightarrow B## assumption 1
##2: C \rightarrow \neg B## assumption 2
##3: C## assumption 3
##4: \neg B## modus ponens 2,3
##5: \neg A ## modus tollens 1,4
##6: C \rightarrow \neg A## hypothethical syllogism (1,2),3,(4),5

##-## and there you have it ##\dots##
So does that mean that one of (1) or (2) gets "reversed"? Can we then speak of retrocausality? (reversed causality?)
 
  • #34
entropy1 said:
Can we then speak of retrocausality? (reversed causality?)
Please do not ask this question again without a clear and exact definition of retrocausality. Preferably one from the professional literature.

To all other participants: please do not respond to this question without such a definition.
 
  • #35
At this point we will go ahead and close this thread. I strongly recommend studying the existing literature in this topic, perhaps including the time symmetric formulation of quantum mechanics. It is best to use definitions from the literature as they are more likely to have addressed some of the basic issues mentioned so far.

For any future threads on this topic please start with a professional scientific reference that can serve as the basis of discussion
 

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