Simplifying the Conditional probability

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SUMMARY

The discussion focuses on simplifying the conditional probability equation P(S1 ∩ S2 ∩ S3 | r) under the condition that S2 and S3 are independent of r. Participants emphasize the application of probability laws, specifically P((X|Y)|Z) = P(X | (Y ∩ Z)), and the importance of context, whether it is measure theory or simpler approaches. The conversation highlights the necessity of applying conditional probability laws consistently, such as P(A ∩ B | Z) = P((A|B)|Z) P(B | Z), to derive meaningful results.

PREREQUISITES
  • Understanding of conditional probability
  • Familiarity with independence in probability theory
  • Knowledge of measure theory concepts
  • Proficiency in applying probability laws
NEXT STEPS
  • Study the principles of conditional independence in probability
  • Learn about measure theory and its applications in probability
  • Explore advanced probability laws, including Bayes' theorem
  • Investigate practical examples of simplifying conditional probabilities
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Students of statistics, mathematicians, and professionals in data science who are looking to deepen their understanding of conditional probability and its simplifications.

bhathi123
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P(S1 [itex]\cap[/itex] S2 [itex]\cap[/itex] S3 | r)
How do I simplfy the above equation if S2 and S3 are independent of r ?
 
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I've don't recall seeing a book that gives the probability law
[itex]P((X|Y)|Z) = P(X | (Y \cap Z) )[/itex]
but I think its true. How to argue it depends on whether you are approaching probability as measure theory or something simpler.

You can also apply the usual probability laws with a condition lilke "[itex]| Z[/itex]" tagged onto every term. For example,
[itex]P(A \cap B) = P(A|B) P(B)[/itex]
so
[itex]P(A \cap B | Z) = P( (A|B)|Z) P(B | Z)[/itex]

See if you can make progress by applying those ideas.
 

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