I am trying to verify if $A \perp C | H$ and $B \perp C | H$ implies $A\cap B \perp C | H$.

So far, I did the below -- but feel I'm skipping something in the step indicated (*) below. Any help would be greatly appreciated (and this is not any homework).

$P(A,B |C,H) = \frac{P(A,B,C,H)}{P(C,H)}$

$= \frac{P(A|B,C,H)P(B|C,H)P(C,H)}{P(C,H)}$

$= P(A|B,C,H)P(B|C,H)$

$= P(A|B,C,H)P(B|H) \hspace{2em} \because B \perp C | H$

$= P(A|B,H)P(B|H) \hspace{1em} ... step (*)$

$= P(A,B|H)$


1 Answer 1


When $A, B, C, H$ are events, this property seems wrong. For example, when $H = \Omega$, then it amounts to say "if $A$ and $C$ are independent, $B$ and $C$ are independent, then $A \cap B$ and $C$ are independent." But this is clearly not the case.

As a counterexample, set $\Omega = \{(u_1, u_2, u_3): u_i \in \{0, 1\}\}$ such that $P(\{u_1, u_2, u_3\}) = \frac{1}{8}$ for any $(u_1, u_2, u_3) \in \Omega$ (that is, $\Omega$ is the sample space of outcomes of tossing a coin for three times). Let

\begin{align*} & A = \{(u_1, u_2, u_3): u_1 = u_2\}, \\ & B = \{(u_1, u_2, u_3): u_1 = u_3\}, \\ & C = \{(u_1, u_2, u_3): u_2 = u_3\}. \end{align*} Then it is easy to verify that $P(A \cap C) = P(B \cap C) = P(A)P(C) = P(B)P(C) = \frac{1}{4}$, i.e., $A$ and $C$ are independent, $B$ and $C$ are independent. However, $$P(A \cap B \cap C) = \frac{1}{4} \neq P(A \cap B)P(C) = \frac{1}{4} \times \frac{1}{2} = \frac{1}{8}.$$


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