What can we infer if P(A|B,C) = P(A|B)? I want to understand what we can infer when the following equation holds:
$$P(A|B,C) = P(A|B)$$
My understanding is that $C$ does not give us any extra information about $A$ that $B$ does not give us already or in other words that $A$ is conditionally independent from $C$ given $B$. Another explanation could be that A and C are independent.
Which of the above is true (if any)? Is there anything else we can say about the relationship of $A$, $B$ and $C$?
 A: It turns out that this is the definition of Conditional independence. Thus the equality means that $A$ and $C$ are conditionally independent given $B$. 
Source: http://en.wikipedia.org/wiki/Conditional_independence
A: Using the definition of conditional probability, we have that 
$$\begin{align} 
P(A\mid B \cap C) &= P(A\mid B)\tag{1}\\
&\Downarrow\\
\frac{P(A\cap B\cap C)}{P(B\cap C)} &=   \frac{P(A\cap B)}{P(B)}\\
&\Downarrow\\
\frac{P(A\cap B\cap C)}{P(A\cap B)} &= \frac{P(B\cap C)}{P(B)}\\
&\Downarrow\\
P(C\mid A\cap B) &= P(C\mid B)\tag{2}
\end{align}$$
which is eerily reminiscent of $(1)$ since $(2)$ is just $(1)$ with $A$ and $C$
interchanged. In fact, 

$A$ and $C$ are conditionally independent given $B$.

We can proceed from $(1)$ above and write
$$\begin{align} 
P(A\mid B \cap C) &= P(A\mid B)\tag{1}\\
&\Downarrow\\
\frac{P(A\cap B\cap C)}{P(B\cap C)} &= P(A\mid B)\\
&\Downarrow\\
\frac{P(A\cap C\mid B)P(B)}{P(C\mid B)P(B)} &= P(A\mid B)\\
&\Downarrow\\
\frac{P(A\cap C\mid B)}{P(C\mid B)} &= P(A\mid B)\\
P(A\cap C\mid B) &= P(A\mid B)P(C\mid B)\tag{3}
\end{align}$$
where $(3)$ can be recognized as the litmus test for declaring
that $A$ and $C$ are conditionally independent given $B$.
However, $A$ and $C$ need not be (unconditionally) independent as 
you think they might be. As a
counterexample, suppose that
$$P(ABC)=P(ABC^c) = P(A^cBC) = P(A^cBC^c) = 0.1; ~~P(A^cB^cC^c) = 0.6$$
Then, we readily get that $P(AB) = P(BC) = 0.2; ~ P(B) = 0.4$ so that
the right sides of $(1)$ and $(2)$ have value $\frac{0.2}{0.4}=\frac 12$
while the left sides have value $\frac{0.1}{0.2} = \frac 12$. But,
$$P(AC) = 0.1 \neq P(A)P(C) = 0.2\times 0.2$$ and so $A$ and $C$
are not independent events.
