Show that if event A is completely independent of Event B

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Discussion Overview

The discussion revolves around the independence of events in probability theory, specifically examining whether event A being completely independent of events B and C implies that A is also independent of the union of B and C (BUC). The scope includes mathematical reasoning and counterexamples related to probability spaces.

Discussion Character

  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Post 1 presents the problem of showing the independence of A from BUC given its independence from B and C.
  • Post 2 suggests using formulas related to independence to approach the problem.
  • Post 3 argues against the initial claim, providing a counterexample involving a unit square as a probability space, demonstrating that A is not independent of BUC despite being independent of B and C.
  • Post 4 offers a simpler counterexample with a probability space consisting of four elementary events, illustrating the same conclusion that A is not independent of BUC.

Areas of Agreement / Disagreement

Participants disagree on the validity of the claim that A is independent of BUC. Multiple counterexamples are provided to challenge the initial assertion, indicating that the discussion remains unresolved.

Contextual Notes

The counterexamples rely on specific constructions of probability spaces and may depend on the definitions of independence used. The implications of these examples are not universally accepted, and the discussion highlights the complexity of the independence concept in probability theory.

TomJerry
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Problem : Show that if event A is completely independent of Event B and C then A is independent of BUC?
 
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Show some work. There are a few formulas about independence you can use to get you started.
 
TomJerry said:
Problem : Show that if event A is completely independent of Event B and C then A is independent of BUC?
Not true!

Counter example: Probability space is unit square, with probability = area.
A: 0≤x≤1/2, 0≤y≤1
B: 0≤x≤1, 0≤y≤1/2
C: two pieces C1 + C2
C1: 0≤x≤1/2, 0≤y≤1/2
C2: 1/2<x≤1, 1/2≤y≤1
P(A)=P(B)=P(C)= 1/2
P(AandB)=1/4, P(AandC)=1/4, A independent of B and C
P(BUC)=3/4, P(AandBUC)=1/4, while P(A)P(BUC)=3/8, not independent!
 
Simpler counter example (actually a simplification of the above).
Consider a probability space with 4 elementary events k,l,m,n where each event has probability 1/4.
Let A={k,l}, B={k,m}, C={k,n}. Then:
P(A)=P(B)=P(C)= 1/2
A∩B={k}, A∩C={k}, A∩(BUC)={k}, while BUC={k,m,n}.
P(A∩B)=1/4, P(A∩C)=1/4, A independent of B and C
P(BUC)=3/4, P(A∩(BUC))=1/4 ≠ P(A)P(BUC)=3/8, not independent!
 

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