MHB Conditional independence problem

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The discussion centers on the conditional independence of events A1 and A2 given event B, with the assertion that if A1 and A2 are conditionally independent, they should also be conditionally independent of the complement of B. A counterexample is provided using a probability space with three outcomes, illustrating that A1 and A2 can be conditionally independent given B but not necessarily independent of B's complement. Participants engage in clarifying the example and its implications for understanding conditional independence. The conversation highlights the complexity of conditional independence in probability theory.
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Let A1, A2 and B be events with P(B)>0. Events A1 and A2 are said to be conditionally independent given B if P(A1nA2|B)=P(A1|B)P(A2|B).

Prove or disprove the following statement:

Suppose 0<P(B)<1. If events A1 and A2 are conditionally independent then A1 and A2 are also condtionally independent of Bcomplement.
 
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Re: conditional independence problem

bl00d said:
where?

The figure with the squares. If you prefer, consider $\Omega=\{a,b,c\}$ and the probability $p$ on $\mathcal{P}(\Omega)$ defined by $p(a)=p(b)=p(c)=1/3$, and choose $A_1=\{a\}$, $A_2=\{a,b\}$ and $B=\{c\}$. Let's see what do you obtain.
 
Re: conditional independence problem

Fernando Revilla said:
The figure with the squares. If you prefer, consider $\Omega=\{a,b,c\}$ and the probability $p$ on $\mathcal{P}(\Omega)$ defined by $p(a)=p(b)=p(c)=1/3$, and choose $A_1=\{a\}$, $A_2=\{a,b\}$ and $B=\{c\}$. Let's see what do you obtain.

oh gosh.. it's right in front of me and i didnt see it
 
The standard _A " operator" maps a Null Hypothesis Ho into a decision set { Do not reject:=1 and reject :=0}. In this sense ( HA)_A , makes no sense. Since H0, HA aren't exhaustive, can we find an alternative operator, _A' , so that ( H_A)_A' makes sense? Isn't Pearson Neyman related to this? Hope I'm making sense. Edit: I was motivated by a superficial similarity of the idea with double transposition of matrices M, with ## (M^{T})^{T}=M##, and just wanted to see if it made sense to talk...

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