TITLE: When the sum of independent Markov chains is a Markov chain?
QUESTION [7 upvotes]: I try to find as much as possible cases, when the chain $Z(t) = |X_1(t)-X_2(t)|$ is Markov, where $X_1(t)$ and $X_2(t)$ are independent, discrete-time and space, preferably non-homogeneous Markov chains.
I started to search for the sum of independent Markov chains and I found this statement in Stoyanov J. - Counterexamples in Probability (2ed., Wiley, 1997)(p.229, one can google it and find in google books):

... the sum of two Markov processes need not be a Markov process. Note, however, that that the sum of two independent Markov processes preserves this property.

That seems very strange to me and I wish to find the proof or at least the statement elsewhere.

EDIT: The way of thinking how it may look like for a sum of two independent Markov chains (If I want to prove that the sum of two independent Markov chains is again a Markov chain):
Let $Y(n) = X_1(n)+X_2(n)$. Then
$P(Y(n+1) = i_{n+1} \ | \ Y(n) = i_n, ..., Y(0)=i_0) =$
$= P(X_1(n+1)+X_2(n+1)=i_{n+1}\ | \ X_1(n)+X_2(n)=i_n,...,X_1(0)+X_2(0)=i_0)=$
$=/ (?) / = \sum_{j+k=i_{n+1}}P(X_1(n+1)=j,X_2(n+1)=k \ | \ \cdot)=$
$=/\text{X's are independent}/ = \sum_{j+k=i_{n+1}}P(X_1(n+1)=j \ | \ \cdot)\cdot P(X_2(n+1)=k \ | \ \cdot) = $
$= /\text{Markov property + (??) } / =  \sum_{j+k=i_{n+1}}P(X_1(n+1)=j \ | \ X_1(n)+X_2(n)=i_n)\cdot P(X_2(n+1)=k \ | \ X_1(n)+X_2(n)=i_n)=$
$=P(X_1(n+1)+X_2(n+1)=i_{n+1} \ | \ X_1(n)+X_2(n)=i_n) = P(Y(n+1)=i_{n+1}\ | \ Y(n)=i_n)$.
Here I am uncertain about (?) and (??) steps.

REPLY [3 votes]: I know this is an old post, but it's the top result on google when one searches for "sum of Markov chains" -- at least it is for me at the time of writing.
I just wanted to suggest a much easier and more intuitive solution.
Let $U$ and $V$ be independent Bernoulli$(\tfrac12)$.
Let $X_n = U$ for all $n$.
Let $Y_n = -(-1)^n V$ for all $n$
Let
$$Z_n = U - (-1)^n V.$$
In particular, $U-V = Z_0 = Z_2 = Z_4 = \cdots$ and also $U+V = Z_1 = Z_3 = Z_5 = \cdots$.
Suppose that the 'present' is time $n = 2$, and we wish to consider the distribution of $Z_3$. Note that $Z_2 = 0$ implies $U = V$, and so $Z_1 = Z_3 = 2U$. Knowing $Z_1$ now determines precisely $Z_3$. For example,
$$ P(Z_3 = 2 \mid Z_2 = 0, Z_1 = 2 ) = 1 \neq P(Z_3 = 2 \mid Z_2 = 0, Z_1 = 0). $$
(In particular, if one sees consecutive $0$s, then one knows that $U = V = 0$.)