Suppose we have a sequence of independent binomial random variables $\xi \in B(1,p)$. We build a new sequence $\eta_n = 2 \xi_{n+1} + \xi_n$. Will it be a Markov chain?
My thoughts. Lets consider conditional probability:
$$\mathbb{P}\big(\eta_n = x_n \big\vert \eta_{n-1} = x_{n-1}, \eta_{n-2} = x_{n-2}, \dots \big) = $$$$ = \mathbb{P}\big(2 \xi_{n+1} + \xi_n = x_n \big\vert 2 \xi_n + \xi_{n-1} = x_{n-1}, 2 \xi_{n-1} + \xi_{n-2} = x_{n-2}, \dots \big)$$
And we see that according to independence of $\xi$, $\eta_n$ doesn't depend on the subsequence $\{\eta_{i}, i\ge n-2\}$. But $\eta_{n-1}$ does. Should we consider somehow such dependence in the conditional part of probability?