Combining Markov chains If the following Markov chain relations hold:
$$X \rightarrow  Y  \rightarrow Z,$$
$$Z \rightarrow  W  \rightarrow Y,$$
can we combine them to have 
$$X \rightarrow  Y  \rightarrow Z \rightarrow W \rightarrow  Y?$$
If so,  what does $$Y  \rightarrow Z \rightarrow W \rightarrow  Y$$ say about $Y$?
I'd appreciate any references with similar problems. 
 A: First of all $Y-Z-W-Y$ would imply that $Y$ is independent of itself given $Z$ or $W$, i.e., both $Z$ and $W$ are sufficient statistics for $Y$: $I(Y;Y|Z)=H(Y|Z)=0$.
But I do not think that this claim holds: The statements $X-Y-Z$ and $Y-W-Z$ neither imply $Y-Z-W$ (leading to your claim) nor $X-Y-W$ (leading to $X-Y-W-Z$).
A: In general, from $X\rightarrow Y \rightarrow Z$ and $Z\rightarrow W \rightarrow U$ we cannot conclude $X\rightarrow Y\rightarrow Z\rightarrow W\rightarrow U$. To have this last Markovity, we need to have $Y\rightarrow Z \rightarrow W$, too. That is, $p(w,y|z)=p(w|z)p(y|z)$ is also required too. 
Regarding $ Y \rightarrow Z \rightarrow W \rightarrow Y$, when we write the definition of Markovity for $ Y \rightarrow Z \rightarrow W$  and $ Z \rightarrow W \rightarrow Y$, we conclude that $ Y$ must be constant. 
A: Nope, you cannot combine them like that, because there would actually be a loop in the dependency graph (the two Y's are the same node), and the resulting graph does not supply the necessary Markov relations X->Y->Z and Y->W->Z.
The proper conclusion to draw from the two Markov relations can only be:
X->Y->W->Z
because p(x,y,z) = \sum_w p(x,y,w,z) = p(x) p(y|x) \sum_w p(w|y) p(z|w), where the last term is a function of (y,z) only.
