A Markov process which is not a strong markov process? - MathOverflow

A Markov process which is not a strong markov process?

2010-10-27T17:00:37Z

Can anyone give an example of a Markov process which is not a strong Markov process? The Markov property and strong Markov property are typically introduced as distinct concepts (for example in Oksendal's book on stochastic analysis), but I've never seen a process which satisfies one but not the other.

Many thanks -Simon

Answer by Byron Schmuland for A Markov process which is not a strong markov process?

2010-10-27T17:17:16Z

A standard example is Exercise 6.17 in Sharpe's book The general theory of Markov processes. The process stays at zero for an exponential amount of time, then moves to the right at a uniform speed.

Answer by Andrey Rekalo for A Markov process which is not a strong markov process?

2010-10-27T17:20:50Z

Let $X(t) = f(W(t) + \pi)$, where $W(t)$ is a standard Wiener process and $$f(x) = \begin{cases} (x,0), & x\leq 0 \\ \\ (\sin x,1-\cos x), & 0 < x < 2\pi \\ \\ (x-2\pi,0), & x\geq 2\pi \end{cases} $$ is a map from $\mathbb R$ to $\mathbb R^2$. $X(t)$ is an $\mathbb R^2$-valued Markov process on $\mathbb R_+$ which is not strongly Markovian. See "A Modern Approach to Probability Theory" by Fristedt and Gray (1997, pp. 626–627).

If the time set is discrete, the ordinary Markov property implies the strong Markov property.

Answer by George Lowther for A Markov process which is not a strong markov process?

2010-10-27T17:24:15Z

Consider the following continuous Markov process X, starting from position x

if x = 0 then X_t = 0 for all times.
if x ≠ 0 then X is a standard Brownian motion starting from x.

This is not strong Markov (look at times at which it hits zero).

Answer by vinay for A Markov process which is not a strong markov process?

2011-02-22T12:14:11Z

I did not quite get the first answer(the one using Brownian Motion). If the process starts at x(not equal to 0), the distribution of X(0) is delta(x) and transition kernels are that of brownian motion and if x = 0 then distribution of x(0) is delta(0) and transition kernels according as a constant stochastic process. How do we mix the 2 processes? Sorry if I am missing something silly.