I need to proof the following statement:
If AA is independent of (B,C)(B,C) then A−B−CA-B-C is a Markov chain.
My current approach is to write the probability mass function:
p(a,b,c)=p(a)p(b|a)p(c|a,b)p(a,b,c) = p(a)p(b|a)p(c|a,b)
so that my goal is to proof that p(c|a,b)=p(c|b)p(c|a,b) = p(c|b). But apparently I need stronger assumptions to do so as for example that AA is independent of BB.
Please could you provide a proof of the statement above or a counterexample?