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The definition of the Markov property is typically that the next state depends only on the present state and no past states. However, the mathematical definition I usually see (e.g. https://stats.stackexchange.com/a/2463/225553) says that P(X_t = x | X_{t-1}, X_{t-2}, X_{t-3}...X_t0) = P(X_t | X_{t-1}). That seems to say that the present state depends on only the immediate prior state. Are these equivalent definitions or am I misunderstanding the notation?

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You correctly wrote... $$P(X_t = x | X_{t-1}, X_{t-2}, X_{t-3},...X_{t0}) = P(X_t | X_{t-1})$$ However this is equivalent to the following by virtue of shifting the time index by 1. $$P(X_{t+1} = x | X_{t}, X_{t-1}, X_{t-2},...X_{t0}) = P(X_{t+1} | X_{t})$$

In words: Given the present, the future is independent of the past.

Hope this helps. If not, feel free to comment and ask for clarification.

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  • $\begingroup$ Thanks for answering. The shifted time index makes sense. I'm still a bit confused. Does this mean that $$P(X_t = x | X_{t-1}, X_{t-2}, X_{t-3},...X_{t0}) = P(X_t | X_{t-1})$$ $$X_{t-1}$$ is considered the "present"? I am still unsure if the statements "The future state depends only on the present state" is equivalent to "The present state depends only on the immediate past state". $\endgroup$ – coderunner Nov 2 '18 at 5:29
  • $\begingroup$ @coderunner, if you were at $t-1$ in time then at that moment it would be the present, right? Similarly, if you were at $t-5$ in time, $t-6$ would be the immediate past for the process and $t-4$ would be the future. I'm not sure if that helps... $\endgroup$ – SecretAgentMan Nov 2 '18 at 5:33
  • $\begingroup$ e.g. We have states A and B that directly lead to state C. Is it still correct to say the present state C depends only on the immediate prior state(s) A and B? If no, I think I am confusing myself from translating the math notation to english words. $\endgroup$ – coderunner Nov 2 '18 at 5:35
  • $\begingroup$ sorry about that. Had to add another comment because the first was running too long... $\endgroup$ – coderunner Nov 2 '18 at 5:35
  • $\begingroup$ The Markov property is talking about the state of the process in time, not just the states themselves. $\endgroup$ – SecretAgentMan Nov 2 '18 at 5:37

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