Assuming $X_1,...,X_4$ are discrete random variables, consider the following 2 Markov Chains:
- $C_1:X_1 \rightarrow X_2 \rightarrow X_3$
- $C_2:X_2 \rightarrow X_3 \rightarrow X_4$
I was trying to prove that $C_3: X_1 \rightarrow X_2 \rightarrow X_3 \rightarrow X_4$ is also a Markov Chain. (It is known that the inverse of this statement is true, if interested, see the relative post A subsequence of a Markov chain is a Markov chain)
After failing to prove this, I tried constructing a counterexample. I think I made it, but I would like some verification at least.
I will make the following assumptions
- $X_1 \perp X_2$ ($\perp$ denotes independence)
- $X_2 = X_3$
- $X_4 = X_3 + X_1$
With this setup, $C_1$ is a Markov chain since $X_3$ is a function of $X_2$. Hence, $p(x_3|x_2,x_1) = p(x_3|x_2)$.
Also since $X_2$ is a function of $X_3$, it holds: $$X_4 \rightarrow X_3 \rightarrow X_2 \iff X_2 \rightarrow X_3 \rightarrow X_4$$ Thus $C_2$ is also a Markov chain.
In order for $C_3$ to be a Markov chain, it suffices to show that $p(x_4|x_3,x_2,x_1) = p(x_4|x_3)$
- $p_1 = p(X_4 = x_4|X_3 = x_3) = p(X_1 + x_3 = x_4| X_3 = x_3) = p(X_1 = x_4 - x_3)$
- $p_2 = p(x_4|x_3,x_2,x_1) = p(\{x_1 + x_3 = x_4\}) \cdot \delta[x_3-x_2] = \delta[x_4-x_3-x_1] \cdot \delta[x_3-x_2]$
where $\delta[x] = \left\{ \begin{array}{ll} 1 ,& x = 0\\ 0 ,& x \ne0\\ \end{array} \right.$
Since in general $p_1 \neq p_2$(e.g. if $X_1 \sim \{1/2,1/2\}$), $C_3$ is not a Markov chain in general. Therefore, concatenation is probably prohibited, at least without further assumptions. $\square$
Thanks in advance, for the help!