Let $N(t)$ be a Poisson Process. We know that $N(t)$ is a time-homogenous Markov chain where $N(t) \sim Pois(\lambda t)$.
If we set $X_t = N_t^2$, then would this process also be a time homogeneous Markov chain? If so - does this work for all transformations? If not, how come?
(A time homogeneous Markov chain satisfies the property that $P(X(t) = j | X(s) = i) = P(X(t-s) = j | X(0) = i) )$
Let $N=(N_t)$ be a poisson process with rate $\lambda$. Let $f$ be such function that $f|_{[0,\infty)}$ is invertible (that is if we restrict $f$ to only positive axis then it is invertible). Then $(f(N_t))$ is a markow chain.
$$ \mathbb P( f(N_t) = j | f(N_{s_1}) = i_1, ... f(N_{s_m}) = i_m) = \mathbb P(N_t = f^{-1}(j) | N_{s_1} = f^{-1}(i_1),..., N_{s_m}=f^{-1}(i_m)) = \mathbb P(N_t = f^{-1}(j) | N_{s_m}= f^{-1}(i_m)) = \mathbb P(f(N_t) = j | f(N_{s_m})=i_m) $$
As for homogenuity: $$ \mathbb P( f(N_t) = m | f(N_s) = n) = \mathbb P(N_t = f^{-1}(m) | N_s = f^{-1}(n)) = \mathbb P(N_{t-s} = f^{-1}(m) | N_0 = f^{-1}(n)) = \mathbb P(f(N_{t-s}) = m | f(N_0) = n) $$
The assumption of invertibility is needed, too. Indeed, consider $f(x) = \begin{cases} 0 & x \in [0,1] \\ x-1 & x \in [1,10) \\ 19-x & x \in [10,19) \\ 0 & x \ge 19 \end{cases}$. Then $$ 1=\mathbb P(f(N_3) = 0 | f(N_{2}) = 0 , f(N_{1}) = 2) \neq \mathbb P(f(N_3)=0 | f(N_2)=0) < 1 $$