# independence property

Is this statement true: if $P(A|B \cup C) = P(A)$, i.e. $A$ is independent of $B \cup C$, then $P(A|B \cap C)=P(A)$, i.e. $A$ is also independent of $B \cap C$

Intuitively it makes much sense to me because if $A$ is independent of a bigger set, then $A$ must be independent of a set which belongs to the bigger set. But I could not able to prove it from definition.

Can some one help me to get the proof?

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This isn't true. For example: The probability of drawing a red card from a standard deck of cards is $1/2$. The probability that it is red given that it is a king is $1/2$. The probability that it is red given that it is a king and a spade is $0$. ($A$ is drawing a red card, $B$ is drawing a king, $C$ is drawing the king of spades.) – David Mitra Jan 31 '13 at 20:54
Perhaps, more general example: $B,C$ are any such that $B\cup C = \Omega$, the whole sample space. No reasons for $B\cap C$ to be independent from some set $A$. – Ilya Jan 31 '13 at 20:58

Denote $B \cup C = V, B \cap C= W$, we know $W \subset V$. Assume $P(A|W)=P(A)$. Then $P(A|W)=P(A|V)=\frac{P(A|V)P(V)}{P(V)} \Leftrightarrow P(A|W)P(V)=P(A|V)P(V)=P(A \cap V)$ which is possible iff $W=V.$

EDIT: also note $P(V) \geq P(W)$, hence $P(A|W)P(V) \geq P(A)|W)P(W)=P(A \cap W)$, hence you get $P(A \cap V) \geq P(A \cap W)$ and the equality holds iff $V=W$.

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how you showed P(A|W)=P(A|V)? – user42574 Jan 31 '13 at 21:16
Assuming the statement holds, $P(A|W)=P(A)=P(A|V)$, by independence, the first condition in your question. – Alex Jan 31 '13 at 21:18

Letting $\Omega$ denote the sample space we have $P(A \mid A \cup \Omega) = P(A)$ but $P(A \mid A \cap \Omega) = 1$.

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Thanks for the prompt replies. The motivation to ask this question raised from studying conditional independence property in Bayesian network. recently I cam across a resource which first says $I(X,Z,Y \cup W) -> P(X|Z,Y,W)=P(X|Z)$, where I(A,B,C) means A is independent of C when conditioned on B. My question is in the above definition how the union of Y and W changed to intersection in the conditional definition? – user42574 Jan 31 '13 at 21:36
@user42574 Oh, I see. This isn't really my area, so you might want to edit your question to include this clarification so that other people will see it. I just saw an easy counterexample to the question as you asked it and thought I might as well post it as an answer. – Trevor Wilson Jan 31 '13 at 22:10