Clarification on Markov processes (without Measure Theory) I'm reading a book on Stochastic Processes(without Measure Theory) and having some difficulty in understanding the following statements 1) and 2). I have not learned Measure Theory, thus my choice in the book.
1) $P(X_{n+m}=j|X_n=k,X_0=i) = P(X_{n+m}=j|X_n=k)$ by Markov Property
Loosely speaking, does this mean that the probability of a future state given the current state and any past state is the probability of the future state given the current state?
Example, $P(X_5 = i | X_3 = j, X_1 = k) = P(X_5=i | X_3=j)$, where $X_5$ is the future, $X_3$ is the current, $X_1$ is the past.
2) $P(X_{n+m}=j|X_n=k)= P(X_m=j|X_0=k)$
Is this true because one-step transition probability from state $k$ to state $j$ is the same irregardless of the current state of the process?
 A: (1) Yes, you could think of it as knowing $X_0$ gives no new information about $X_{n+m}$ if $X_n$ is known. By the Markov property, one could say that, conditioned on the present, the future does not depend on the past.
(2) I think this is because you are assuming that the Markov process is stationary (i.e. it is time-homogeneous), which means the state-to-state transition probabilities do not change over time. It does also relate to the Markov property as well of course. In this case:
$$ P(X_{n+m}=j\,|\,X_n=k) = P(X_{n+m+s}=j\,|\,X_{n+s}=k) $$
In words, going from state $k$ to state $j$ in $m$ steps is always the same. Hence, set $s=-n$ to get:
$$ P(X_{n+m}=j\,|\,X_n=k) = P(X_{m}=j\,|\,X_{0}=k) $$
In essence, only the time difference matters. Hence, translation of both indices in time does not change the value.
With regards to your question, I would not say that it is irregardless of the current state; rather, it is because the process acts the same once it reaches a given state, no matter how it got there. Hence, going from $s_1$ to $s_2$ in $t$ steps is always the same; it doesn't matter if you arrived at $s_1$ at time $0$ or at time $n$ (due to stationarity), as the chance of reaching $s_2$ in $t$ steps remains the same.
