I have given a Markov Chain $X_n$ with the state space $\{0,1,2\}$ and the transition Matrix $$P=
\begin{Bmatrix}
0.3 & 0.2 & 0.5 \\
0.5 & 0 & 0.5 \\
0.2 & 0.1 & 0.7
\end{Bmatrix}
$$
Given a function $f(x)=x^3$ for $x\in\{ 0,1,2 \}$. Is $f(X_n)$ a Markov Chain?
I don't know what to do here. I looked in my lectures but can't find anything on what a function does to a Markov Chain. I know that the state space changes to $\{0,1,8\}$ but what does the function do with the transition probabilities? How does the transition matrix change? Is $f(X_n)$ even a Markov chain? And if not how do you see this?
Best Answer
Since $f$ is one-to-one, the new sequence $f(X_n)$ is still a Markov chain. To prove this, argue that the transition matrix is the same as before, only the rows and columns are labelled with the new state space $f(0), f(1), f(2)$ instead of $0, 1, 2$.