Given markov chain $X \rightarrow Y \rightarrow Z$, $X^2 \rightarrow Y^2 \rightarrow Z^2$ is not a Markov chain

Question

Consider the markov chain $X \rightarrow Y \rightarrow Z$. $f(X) \rightarrow f(Y) \rightarrow f(Z)$ is not a Markov chain in general.

I want to find an example for this. I managed to figure out that if $f$ injective, then $f(X) \rightarrow f(Y) \rightarrow f(Z)$ is a markov chain.

I found $f(a) = a^2$, a non-injective function and this example:

$P(X = 1) = 1/4, P(X = -1) = 3/4$, the same for $Y$ and $Z$.

Then I managed to show that $X \rightarrow Y \rightarrow Z$ is a Markov chain because $P(X|Y) \cdot P(Z | Y) = P(X,Z | Y)$. Now I want to show that: $P(f(X)|f(Y)) \cdot P(f(Z)|f(Y)) \neq P(f(X), f(Z) | f(Y))$ but I don't know how to go on. No matter what I try, equality holds.

Answer 1

This can be done even in the setting of deterministic dynamics. Consider the deterministic dynamics $-2 \to -1 \to 1 \to 2 \to -2$. Consider $f(x)=x^2$. Notice that

$$P(X_{n+1}^2=4 \mid X_n=-1)=0 \\ P(X_{n+1}^2=4 \mid X_n=1)=1.$$

Yet $X_n^2$ is the same in both cases. If we can separate these cases by conditioning on older values of $X_n^2$, then $X_n^2$ is not Markov. And indeed:

$$P(X_{n+1}^2=4 \mid X_n^2=1,X_{n-1}^2=4)=0 \\ P(X_{n+1}^2=4 \mid X_n^2=1,X_{n-1}^2=1)=1$$

so $X_n^2$ is not Markov.