Baye's Theorem w/ Multiple Events

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Discussion Overview

The discussion revolves around the application of Bayes' Theorem to multiple events, specifically how to correctly formulate the theorem when considering hypotheses that consist of two events, S and O. Participants explore different formulations and interpretations of the theorem, including joint and conditional probabilities.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants propose two potential formulations for P(S,O|D) based on Bayes' Theorem, questioning whether to include terms for all combinations of S and O in the denominator.
  • Others argue that the second formulation, which includes all combinations of S and O, appears to be the correct one.
  • Several participants discuss the relationship between joint probabilities and conditional probabilities, noting that P(D) can be expressed in different ways depending on the context.
  • One participant suggests that using probability notation P(1-H) is clearer than using logical notation P(~H) in the context of conditional probability.
  • There is a mention of the equivalence of two expressions involving P(D) and how they relate to Bayes' Theorem, with some participants questioning the assumptions behind these expressions.
  • Another participant notes that while D can be expressed in terms of joint probabilities, it cannot be directly written in terms of Bayes' Theorem without involving conditional probabilities.

Areas of Agreement / Disagreement

Participants express differing views on the correct formulation of Bayes' Theorem for multiple events, with no consensus reached on which formulation is definitively correct. The discussion remains unresolved regarding the best approach to apply the theorem in this context.

Contextual Notes

Participants highlight the importance of distinguishing between joint and conditional probabilities, and the implications of using different notations. There are unresolved assumptions regarding the definitions and interpretations of the terms used in the formulations.

dnbwise
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So, Bayes' Theorem states

P(H|D) = P(D|H) X P(H) / [P(D|H) X P(H) + P(D|~H) x P(~H)], where ~ = negation

If my hypothesis consists of two events - S and O - would this be

P(S,O|D) = P(D|S,O) X P(S,O) / [P(D|S,O) X P(S,O) + P(D|~S,~O) x P(~S,~O)]

or this

P(S,O|D) = P(D|S,O) X P(S,O) / [P(D|S,O) X P(S,O) + P(D|~S,O) x P(~S,O) + P(D|S,~O) x P(S,~O) + P(D|~S,~O) x P(~S,~O)] ?
 
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dnbwise said:
So, Bayes' Theorem states

P(H|D) = P(D|H) X P(H) / [P(D|H) X P(H) + P(D|~H) x P(~H)], where ~ = negation

If my hypothesis consists of two events - S and O - would this be

P(S,O|D) = P(D|S,O) X P(S,O) / [P(D|S,O) X P(S,O) + P(D|~S,~O) x P(~S,~O)]

or this

P(S,O|D) = P(D|S,O) X P(S,O) / [P(D|S,O) X P(S,O) + P(D|~S,O) x P(~S,O) + P(D|S,~O) x P(S,~O) + P(D|~S,~O) x P(~S,~O)] ?

the second formulation looks right.
 
dnbwise said:
So, Bayes' Theorem states

P(H|D) = P(D|H) X P(H) / [P(D|H) X P(H) + P(D|~H) x P(~H)], where ~ = negation

Show how P(D)=P(D|H)P(H)+P(D|1-H)P(1-H)
 
Last edited:
SW VandeCarr said:
Show how P(D)=P(D|H)P(H)+P(D|1-H)(P|1-H)

HnH' = 1

D = HnD+ H'nD

P(D)= P (HnD) + (H'nD)= P(D | H) P(H) + P(D | H') P(H')
 
ych22 said:
HnH' = 1

D = HnD+ H'nD

P(D)= P (HnD) + (H'nD)= P(D | H) P(H) + P(D | H') P(H')

Can you derive this from Bayes' Theorem?:

P(D)=P(D|H)P(H)/P(H|D) is equivalent to P(D)=P(D|H)P(H)+P(D|1-H)P(1-H)?

EDIT: I think it's more clear to use probability notation P(1-H) rather than logical notation P(~H) in what is essentially a conditional probability problem.
 
Last edited:
I think you meant

HuH' = 1

Given that D = HnD + H'nD can we assume that P(D) = P(HnD) + P(H'nD) ?

HuH' = 1
or H' = 1 - H
HnD = P(D|H)P(H)
H'nD = P(D|H - 1)(1 - P(H))

P(D) = P(D|H)P(H) + P(D|H - 1)(1 - P(H))

but bayes theorem says P(D)=P(D|H)P(H)/P(H|D)

so i think it's Ok
 
Note:

D=P(D|H)P(H)+P(D|1-H)P(1-H) is in terms of joint probabilities (terms cancel);

D=P(D|H)P(H)/P(H|D) is in terms of a conditional probability.

You can express D either way and you can write a conditional probability in terms of joint probabilities as the OP has done. However, you cannot write D in terms of Bayes Theorem without a conditional probability.
 
Sorry, that's P(D) where D appears in the previous post.
 

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