Consider the following equation to hold for the conditional pdfs of random variables $A,B,C$:
$p(A|B,C) = p(A|B)$ (1).
Certainly: $C$ being independent of both $A$ and $B$ implies (1) to hold.
But does also (1) imply that $C$ is independent of $A$ and $B$, so is it equivalence? Or under which circumstances are the two equivalent?
If I apply Bayes i get $\frac{p(B,C|A)p(A)}{p(B,C)} = \frac{p(B|A)p(A)}{p(B)}$ (1')
This looks like I would need also $B$ to be independent of $C$ so that $p(B)p(C) = p(B,C)$ and the denominator would be the same after expansion with $p(B)$. ...
To state my problem more clearly.
I. It seems to me that
$\frac{p(B,C|A)p(A)}{p(B,C)} \neq \frac{p(B|A)p(A)}{p(B)}$ $\implies$ $p(B,C|A) \neq p(B|A)p(C)$ or $p(B,C) \neq p(B)p(C)$.
Which again implies
$B$ and $C$ are not conditionally independent given $A$ or $B$ and $C$ are not conditionally independent. or $C$ not independent of $A$
(It is: $B$ and $C$ independent and $A$ and $C$ independent $\implies$ equality and by Bayes $p(A|B,C) = p(A|B)$)
II. Whereas
$p(A|B,C) \neq p(A|B)$ seems to imply definitely that $C$ and $A$ are not independent.
(It is: $C$ independent $A$ $\implies p(A|B,C) = p(A|B)$)
This seems somehow puzzling as those unequalities can be transormed into one anoter by bayes...
More precise, if
(1) $p(A|B,C) \neq p(A|B) \implies$ $C$ not independent of $A$ or $C$ not independent of $B$.
(2) $C$ independent of $A \implies p(A|B,C) = p(A|B)$
then there is something wrong. But both (1) and (2) seem to be true considering above thoughts.
Property (1) means that the process $C\to B\to A$ is a Markov chain. Let property (2) be that $C$ is independent of $(A,B)$. Then (2) implies (1) but (1) does not imply (2). Counterexample: $B=C$. On the other hand, if (1) holds and if $B$ and $C$ are independent then (2) holds.