Does $(A \perp B)|X$ implies $(A|X)\perp B$ ? I managed to derive it, but it feels wrong.
Here is my derivation. Is it flawed?
If $(A \perp B)|X$ then $(A|X) \perp (B|X)$ and therefore $p[A|(X, (B|X) ) ]=p[A|X]$
in addition: $p[A|(X, (B|X) ) ] = p[A|(X, B) ]$, because both $X$ and $B$ are given.
We can conclude that $p[A|(X, B) ] = p[A|X]$ and therefore $(A|X)\perp B$
EDIT:
Here is a relevant use case: 1) Let's say we have a causal graph B->X->A, with the structured eq. model (SEM)
$B = Nb$, s.t. $Nb$ ~ $p(Nb)$
$X = f1(B, Nx)$, s.t. $Nx$ ~ $p(Nx)$
$A = f2(X, Na)$, s.t. $Na$ ~ $p(Na)$
Na, Nb, Nx may (or may not) be independent, yet we definitely know that the conditional independence $(A \perp B)|X$ holds, does it implies that $(A|X)\perp B$ ?
Note: Under the SCM, (A|X) is a random variable that depends on Na.
2) What happens if the graph (and SEM) is reversed A->X->B. Would it also imply $(A|X)\perp B$ ?