Sum of two random variables in a Markov chain

Question

Assume that $X_i, Y_i$ are two random variables in the ith step of a Markov chain, is $Z_i=X_i+Y_i$ also a random variable of step $i$ in a Markov chain? ($X_i,Y_i$ are separate processes)

Answer 1

There is a counter-example.

Suppose the only permitted values for variables are $-1$ and $1$. And suppose that $X_i$ is a process that with $100\%$ probability gives $1$ when $X_{i-1}$ is $-1$, and $-1$ when $X_{i-1}$ is $1$. Obviously $\{ X_i \}$ is a Markov chain. And suppose $Y_i$ is a constant chain, $i.e.$, $Y_i = Y_{i-1} = \pm 1$. Then $\{ Y_i \}$ is also a Markov chain. Let $Z_i = X_i + Y_i$.

Now we have, if $Z_{i-1} = 0$, there are two possibilities: $X_{i-1} = 1, Y_{i-1} = -1$ or $X_{i-1} = -1, Y_{i-1} = 1$. In the former case, $X_i = -1, Y_i = -1, Z_i = -2$. In the latter case, $X_i = 1, Y_i = 1, Z_i = 2$. So the probability $0 < P(Z_i = 2 | Z_{i-1} = 0) < 1$, but $P(Z_i = 2 | Z_{i-1} = 0, Z_{i-2} = 2) = 1$ and $P(Z_i = 2 | Z_{i-1} = 0, Z_{i-2} = -2) = 0$. Therefore, $\{ Z_i \}$ is not a Markov chain.