Assume sequence $(X_1,X_2, X_3, \ldots)$ is Markov sequence of real random variables where $X_i \in \mathcal{X}$ for some alphabet $\mathcal{X}$ of finite size $k$. Define random variable $Y_i = (X_{i-1},X_i)$ for $i \geq 2$. Is $Y_2, Y_3, \ldots$ a Markov sequnece?

In general, can we say that if $(A_1,A_2, A_3, \ldots)$ and $(B_1,B_2, B_3, \ldots)$ are Markov sequences where $A_i$ and $B_i$ coming from some finite alphabets, then $(C_1,C_2, C_3, \ldots)$ is Markov where $C_i = (A_i,B_i)$?


The requirement that the state space is finite (in which case there is no need to specify that your random variables are real) is completely irrelevant here (although technically it is easier to deal with countable state spaces).

The answer to the first question is "yes". One ways to see it consists in noticing that the path spaces for both chains are the same, and therefore the decompositions into the products of transition probabilities for the measures of cylinder sets will be the same for the original sequence $X_n$ and for the new sequence $Y_n=(X_{n-1},X_n)$.

The answer to the second question is "no". Of course, if the chains $(A_n)$ and $(B_n)$ are independent, then $(A_n,B_n)$ is also Markov (this is the so-called product Markov chain whose transition probabilities are the products of the transition probabilities of the original chains). However, in general $(A_n,B_n)$ is not Markov. For instance, take a stationary chain $(X_n)$. Its time reversal $(X_{-n})$ is also Markov (in the invariant form the Markov property means that the future and the past are conditionally independent with respect to the present). The sequence $Y_n=(X_n,X_{2-n})$ is then not Markov as $Y_2=(X_2,X_0)$ and $Y_0=(X_0,X_2)$ are not conditionally independent with respect to $Y_1$.


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