Equivalent probability with independence question $P(A,B,C) = P(A|B)P(B|C)P(C)$ So, the statement is that:
$$P(A,B,C) = P(A|B)P(B|C)P(C)$$
if ($A$ is independent of $B$ given $C$) and ($A$ is independent of $C$)
Here is how I tried to work it out:
$$P(A,B,C) = P(B)P(A|B)P(C|A,B) \text{ (using the chain rule)}$$
$$P(A,B,C) = P(A|B)P(B,C|A)  \text{ (using the product rule)}$$

But from here I can't seem to go anywhere. I've tried taking other routes as well, but I keep ending up at a dead end. I was trying to find a way to go from $P(B,C|A)$ to $P(B|C)P(C)$. However, this may not be the correct route to take in the first place.
I was wondering if anyone knew where I went wrong? Thank you.
 A: Suppose $A$ is independent of $C$ given $B$. Then
\begin{align}
\textbf{P}(ABC) &= \textbf{P}(AC|B)\textbf{P}(B) = \textbf{P}(A|B)\textbf{P}(C|B)\textbf{P}(B) = \textbf{P}(A|B)\textbf{P}(BC) \\
&= \textbf{P}(A|B)\textbf{P}(B|C)\textbf{P}(C)
\end{align}
But I am not sure whether those two conditions $A$ is independent of $B$ given $C$ and $A$ is independent of $C$ imply $A$ is independent of $C$ given $B$.
Based on given two conditions, what I can do best is
\begin{align}
\textbf{P}(ABC) &= \textbf{P}(AB|C)\textbf{P}(C) = \textbf{P}(A|C)\textbf{P}(B|C)\textbf{P}(C) = \textbf{P}(A)\textbf{P}(B|C)\textbf{P}(C)
\end{align}
Trace back to see under what condition the equality holds. We have to have
\begin{align}
\textbf{P}(AB|C) = \textbf{P}(A|B)\textbf{P}(B|C)
\end{align}
which is equivalent to say
\begin{align}
\frac{\textbf{P}(ABC)}{\textbf{P}(C)} = \frac{\textbf{P}(AB)}{\textbf{P}(B)}\frac{\textbf{P}(BC)}{\textbf{P}(C)}
\end{align}
and by multiplying $\textbf{P}(C)$, 
\begin{align}
\frac{\textbf{P}(AB)\textbf{P}(BC)}{\textbf{P}(B)} = \textbf{P}(ABC).
\end{align}
Hence any condition satisfies the above equality will be suffice.
