Conditional probability with joint condition

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Discussion Overview

The discussion revolves around the application of conditional probability, specifically focusing on the expression Pr(Z|X&Y) and its relationship to Bayes' Rule. Participants explore the implications of independence between events X and Y in the context of conditional probabilities.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant proposes that Pr(Z|X&Y) can be expressed using Bayes' Rule as Pr(Z|X&Y)=[Pr(X&Y|Z)Pr(Z)]/Pr(X&Y).
  • Another participant challenges the assumption that independence of X and Y implies Pr(X&Y|Z)=Pr(X|Z)Pr(Y|Z), providing a counterexample involving coin tosses where Z creates dependence between A and B.
  • A later reply acknowledges a mistake in dividing by the conditional and questions whether Bayes' Rule still applies in the context of joint events X and Y.
  • One participant clarifies that Bayes' Rule is valid as long as P(A) is not zero, noting potential issues with division by zero in certain cases.

Areas of Agreement / Disagreement

Participants express differing views on the implications of independence given a condition Z, with some asserting that independence does not hold under conditioning, while others maintain that Bayes' Rule is applicable in the presented scenario. The discussion remains unresolved regarding the specific application of independence in this context.

Contextual Notes

There are limitations regarding the assumptions made about independence and the conditions under which Bayes' Rule is applied. The discussion does not resolve the mathematical steps involved in the application of these concepts.

Cognac
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So say I have Pr(Z|X&Y)

I'm guessing that it follows the standard Pr(A|B)=[Pr(B|A)Pr(A)]/Pr(B)

So Pr(Z|X&Y)=[Pr(X&Y|Z)Pr(Z)]/Pr(X&Y)?Also, if X&Y are independent, then would I get Pr(X&Y|Z)=Pr(X|Z)Pr(Y|Z)?
 
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Cognac said:
Pr(A|B)=Pr(B|A)Pr(A)
What? If A=B, this would say that Pr(A)=1.
Also, if X&Y are independent, then would I get Pr(X&Y|Z)=Pr(X|Z)Pr(Y|Z)?
No. A and B may be independent in general but not independent given Z. Example: toss two coins. A=coin 1 is heads. B=coin 2 is heads. Z=coins 1 and 2 match. Obviously Z forces a dependence between A and B.
 
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FactChecker said:
What? If A=B, this would say that Pr(A)=1.

No. A and B may be independent in general but not independent given Z. Example: toss two coins. A=coin 1 is heads. B=coin 2 is heads. Z=coins 1 and 2 match. Obviously Z forces a dependence between A and B.
Thanks, that example about independence makes things make more sense now.

Oh sorry about that. I forgot to divide by the conditional. But does Bayes Rule still work for the case I presented? Given two joint events, X&Y?
 
Cognac said:
I forgot to divide by the conditional. But does Bayes Rule still work for the case I presented? Given two joint events, X&Y?
By "divide by the conditional", I assume you mean "divide the right side by P(A)". Yes, Bayes' Rule always works. The only exception to the form of Bayes' Rule that you are using is if P(A) = 0. Then both P(B|A) and the division of the right side by 0 are problems. (Some statements of Bayes' Rule talk about a partitioning of the space. If you use ever that form, make sure that condition is met.)
 

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