Talk:Markov property/Archive 1

I suppose that some description of the formulas would be welcomed - like it is done for Poisson process.

I believe that the Markov property is only the left hand side of the 1st formula. The right hand side states that this is a first order Markov chain. Can somebody confirm this?

Brownian Motion

The most famous Markov processes are Markov chains, but many other processes, including Brownian motion, are Markovian.

What disqualifies Brownian motion from being a continuous-time Markov chain? Should the above say "are discrete-time Markov chains"? Josh Cherry 02:56, 15 Nov 2004 (UTC)

Upon further reading of continuous-time Markov chain, I suspect that I know the answer. But then isn't the first sentence of continuous-time Markov chain too loose a definition? Josh Cherry 03:03, 15 Nov 2004 (UTC)

I agree; the 'chain' part is poorly specified. In a Markov chain there is a sequence of states visited by the process (such as the count of a Poisson process), rather than a continuous path (as in Brownian motion). I'm planning some further edits to that page, so (if no one else does in the meantime) I'll incorporate this too. Ben Cairns 22:52, 27 Jan 2005 (UTC).

redirect

I've redirected this to Markov property because it was worthless. There is an article titled Markov chain that treats discrete-time Markov processes. Markov property does not assume discrete time. This article assumed (incorrectly) not only discrete time but also a state space that was not merely discrete but actually finite. In effect, this denies that the standard Wiener process (and many others) is a Markov process! Michael Hardy 14:51, 23 June 2006 (UTC)[reply]

continuous and discrete time

Hmm, this article gives the arkov property only for continuous time systems; it would be nice if it included the defn for discrete-time systems as well, or at leaast recapping sufficiently before telling the reader to go to read about markov chains. linas 23:59, 28 August 2006 (UTC)[reply]

Goblygook

Agree with the previous 100% with the previous. Without any discription of the formulas, this article looks like science classroom blackboard scribblings in a bad made-for-tv Sci-Fi movie. —The preceding unsigned comment was added by Bigjimslade (talk • contribs) 14:16, 17 April 2007 (UTC).[reply]

Totally agree... It's full of variables that aren't defined, and very difficult to understand for a novice in this field. --Mighty Jay 14:19, 15 June 2007 (UTC)[reply]

Second. Jm546 (talk) 16:39, 11 January 2008 (UTC)[reply]