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Set 3 random variables A, B and C.
A and C are independent, but B and C are dependent.
Do we have $$P(A \mid B) = P(A \mid B, C)$$ (because A and C are dependent).
If yes, how to prove it?
Else, why not?

Because I have an example in my lesson like this: $$P(A).P(C \mid A).P(F \mid C) = P(A,C,F)$$ and we only know that A and F are independent.

moth
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  • What do you mean by $P(A|B, C)$? It's $P(A|B\cup C)$ or $P(A|B\cap C)$. Or you mean the conditional probability with respect to the sigma algebra generated by $1_{B}$ and $1_C$? – Petite Etincelle Nov 11 '14 at 12:37
  • I mean $P(A|B\cap C)$ – moth Nov 11 '14 at 12:39
  • @FarorTahal Sorry, I actually cannot prove/disprove the following which I think I need to answer your question: If A and B are independent then $A|C$ and $B|C$ are also independent for any set $C$. So, sorry I do not think I can help you... – Jimmy R. Nov 11 '14 at 14:41
  • C is separator on join tree of Shafer-Shenoy, so I think yes, A and B are independent conditionally to C. – moth Nov 11 '14 at 14:52
  • One must choose between "Set 3 random variables A, B and C" and "I mean P(A|B∩C)", these are not compatible. – Did Nov 11 '14 at 14:58

1 Answers1

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That the random variables A and C are independent does not imply that P(A|B)=P(A|B,C).

For a counterexample, consider some i.i.d. centered Bernoulli random variables A and B on {-1,+1}, and C=AB, again centered Bernoulli on {-1,+1}. Then the distribution of A conditionally on B is uniform Bernoulli on {-1,+1} hence A is independent of B. The distribution of A conditionally on {B,C} is Dirac at the point BC hence A is not independent of {B,C}, actually A is even measurable with respect to {B,C} since A=BC.

Nota: I wish math.SE askers of probability questions would stop using the symbols A, B, C, etc., which commonly denote events, to denote random variables. Every decent text I might have stumbled upon denotes random variables by X, Y, Z (next in line are U, V, W, possibly S and T).

Formulas such as P(A|B)=P(A|B,C) should refer to three events A, B and C, and to the conditional probabilities P(A|B)=P(A∩B)/P(B) and P(A|B,C)=P(A∩B∩C)/P(B∩C).

I cannot determine whether this strange interversion on several math.SE pages signals that the exercise was incorrectly copied here by some careless student, or if indeed some teachers are perverse enough to exchange the two notations on purpose.

Did
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  • This exercice deals with Shafer-Shenoy tree and C is separator on join tree of Shafer-Shenoy, so I think A and B are independent conditionally to C and in this case this formula is ok. – moth Nov 11 '14 at 14:54
  • @Faror OK. How is your question related to what you now explain in a comment? – Did Nov 11 '14 at 14:56
  • Because first, I think it was a general principle, but I just find out that separator in Shafer-Shenoy tree (rather I think, but someone could confirm that?)means conditionally independence and in the message deleted by Stefanos, he explained that we can have this formula in the conditionally independence case. – moth Nov 11 '14 at 14:59
  • Except that the user you mention deleted their answer because it was incorrect. Sorry but I find your thought process difficult to follow. – Did Nov 11 '14 at 15:01
  • It was incorrect for generally independence not for conditionally independence. He said: "@FarorTahal Sorry, I actually cannot prove/disprove the following which I think I need to answer your question: If A and B are independent then A|C and B|C are also independent for any set C. So, sorry I do not think I can help you... – ". I have A|C and B|C independant for C because it's a Shafer-Shenoy tree. – moth Nov 11 '14 at 15:02
  • Yeah, sure they wrote that. First problem: A|C and B|C do not exist (care to define them?). Second problem: that the events A and B are independent does not imply that P(A∩B|C)=P(A|C)P(B|C)... as the example in my answer shows. Did you try to think about said example? – Did Nov 11 '14 at 15:05