Let $\lbrace X_n \rbrace$ be a Markov Chain and $A$ be a subset of its state space. Let $T_1 = \inf \lbrace k \geq 1 : X_k \in A \rbrace$ and $T_n =\inf \lbrace k > T_{n-1} : X_k \in A \rbrace$. Assume that $T_n< \infty$. I want to prove that $Y_n = X_{T_n}$ is also a markov chain.
This is was i've done so far:
I want to prove that $$ \mathbb{P}(Y_{n+1} = i \ | \ Y_1, Y_2, \cdots, Y_{n}) = \mathbb{P}(Y_n = i \ | \ Y_n) $$ We have that $$ \mathbb{P}(Y_{n+1} = i \ | \ Y_1, Y_2, \cdots, Y_{n})= \mathbb{P}(X_{T_{n+1}} = i \ | \ X_{T_1}, X_{T_2}, \cdots, X_{T_n}) $$ Since $X_{T_n} \in A$ for all $n \in \mathbb{N}$ by definition, $X_{T_n}$ is independent of $X_i$ for all $i$ such that $i \neq T_n$, and then we can write $$ \mathbb{P}(x_{T_{n+1}} = i \ | \ X_{T_1}, X_{T_2}, \cdots, X_{T_n}) = \mathbb{P}(X_{T_{n+1}} = i \ | \ X_1, X_2, \cdots, X_{T_{n+1}-1}) $$ Now, since $T_n$ is a stopping time, using the Strong Markov Property we have that $$ \mathbb{P}(X_{T_{n+1}} = i \ | \ X_1, X_2, \cdots, X_{T_{n+1}-1}) = \mathbb{P}(X_{T_{n+1}} = i \ | \ X_{T_n}, X_{T_n +1 }, \cdots, X_{T_{n+1} - 1}) $$ Using again the fact used above about independence of $X_{T_n}$ and $X_i$ we have finally that $$ \mathbb{P}(X_{T_{n+1}} = i \ | \ X_{T_n}, X_{T_n +1 }, \cdots, X_{T_{n+1} - 1}) = \mathbb{P}(X_{T_{n+1}} = i \ | \ X_{T_n}) = \mathbb{P}(Y_{n+1}=i \ | \ Y_n) $$ And it´s done.
Am I right? Is there some flaw in the proof? Thanks!