# Definition of markov process that produces symbols (as an information source)

A finite markov chain (or markov process) is a stochastic process $X_1, X_2, X_3, \ldots$ where we have random variables $X_i : \Omega \to S$ with finite $S = \{0,\ldots,n-1\}$ such that for each $k$ $$P(X_k = x_k \mid X_{k-1} = x_{k-1}, X_{k-2} = x_{k-2},\ldots, X_1 = x_1) = P(X_k = x_k \mid X_{k-1} = x_{k-1}).$$ Now in a book about information theory I am reading, they talk about information sources with a finite number of states, transition between them labeleb by a probability and a symbol emitted when the transition is made. And these are also called finite markov process.

But now consider the following. (The transition probabilities do not matter) Then this would be a finite markov process with states $S_0, S_1, S_2$. But the outputs do not fulfill the markov property, as if an $b$ is emitted, what can follow depends on if we have before all the $b$'s emitted an $a$ or an $c$, and as the number of $b$ could be arbitrary large it dos not depend on the last symbol emitted. So in this case the probability of the next symbol might depend on an arbitrary "depth" past?

So in a strict sense this is not a markov process, or? But then why they call it that way? Or is there a different name for these kind of "markov process" in the mathematical world?

As I see it, if an emitted symbol corresponds one-to-one to its state, then the linked notion of markov process is the same as the above, but in general they do not seem to be equal to me.

• Are you sure that $a,b$ are emitted symbols, and not probabilities? Could you post a screenshot of the book page? – leonbloy Jun 24 '17 at 2:21
• Yes, I am sure. Its from the article "Three models for the description of langauge" by N. Chomsky, 4th page. See portfolio.colum.edu/downloads/pdfs/ionagroup_student_747.pdf – StefanH Jun 24 '17 at 11:39