Does the function of a random variable have the same transition matrix as the variable itself?

Question

If I have a variable X, that follows a Markov Chain with a transition density $\rho(X)$ does a function of that variable f(X) have the same density or is there a one to one mapping to the density of f(X) from $\rho(X)$?

Answer 1

Let $X$ be a Markov process. First of all, $f(X)$ may not be a Markov process: think of $X$ being a skewed random walk on $\Bbb Z$ which moves to the right with probability $\frac23$. Then if $f(x) = x^2$ and you know that $X^2_n = 1$ you can't really say what's the probability that $X^2_{n+1} = 2$: for that you need to know the sign of $X_n$ which you can't deduce from $X^2_n$ alone.

You can always show that $f(X)$ is Markov whenever $f$ is injection, if $f$ is not an injection, then $f(X)$ will be Markov just because the transition probabilities of $X$ are "same" on the level sets of $f(X)$. There is a whole area of lumping/bisimulation where they study how to find such non-injective $f$ for a given Markov process. Anyways, when $f$ is injective you shall be able to derive the transition matrix for $f(X)$ from that of $X$.

Edit: Due to the issue with a biased random walk mentioned by @Did, consider another example. Let's $X_{n+1} = 2X_n$ and $X_0$ is uniformly distributed on $\{-1,1\}$. Let $f(x) = x1_{|x|\neq 2} + |x|1_{|x| = 2}$. Then we obtain that $\mathsf P(f_2 = 4|f_0,f_1) = 1_{f_0 = 1}$ whereas $\mathsf P(f_2 = 4|f_1) = \frac12$.

Answer 2

If $f$ is one-to-one, $P(f(X_{n+1}) = f(t) | f(X_n) = f(s)) = P(X_{n+1} = t | X_n = s)$. So the transition matrix for $f(X_n)$ is obtained from that for $X_n$ by relabeling the rows and columns.