MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
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. – user13166 Feb 22 '11 at 12:14
@vinay For $x\ne0$ and $t>0$, let $p_t(x,\cdot)$ denote the Gaussian distribution with mean $x$ and variance $t$ and $p_t(0,\cdot)$ denote the Dirac measure at $0$. For every $x$, let $p_0(x,\cdot)$ denote the Dirac measure at $x$. Then, for every bounded measurable $\varphi$, initial distribution $\nu$ and times $0=t_0\le t_1\le \cdots\le t_n$, $E_\nu[\varphi(X(t_0),X(t_1),\ldots,X(t_n))]$ is the integral you know, involving $\varphi$, $\nu$ and the semi-group $(p_t)_{t\ge0}$. QED. In fact, a good way to understand this example is to try to prove that $(p_t)_{t\ge0}$ is indeed a semi-group. – Did Feb 22 '11 at 12:34
up vote 9 down vote accepted

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

  1. if x = 0 then Xt = 0 for all times.
  2. 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).

share|cite|improve this answer

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.

share|cite|improve this answer
A little complicated to follow, but quite neat. I see it works because the curve given by f intersects itself so, if you stop at an intersection point, you don't know which part of the curve the process is currently moving on. – George Lowther Oct 27 '10 at 17:29
Isn't important that $W_t$ is a Wiener process on $[0,\infty)$? Otherwise it seems that you can't define the familly of measures induced by $f(X_t)$. What is the probability measure on $W(0) = -\pi$ and $W(0) = \pi$? – Conrado Costa Nov 9 '15 at 9:17

A standard example is Exercise 6.17 in Michael Sharpe's book 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.

share|cite|improve this answer
Ah, this is quite a simple example. It fails the strong Markov property, but not as badly as Andrey's and my example. This one still satisfies a restricted version of the strong Markov property, where you only look at predictable stopping times. It is "moderate Markov" (…). – George Lowther Oct 27 '10 at 17:51
I can't resist giving this quote from Kai Lai Chung's "Lectures from Markov Processes to Brownian Motion (1982)" -- It may be difficult for the novice to appreciate the fact that twenty five years ago a formal proof of the strong Markov property was a major event. Who now is interested in an example in which it does not hold? – Byron Schmuland Oct 28 '10 at 5:02

Reflected Brownian motion on the slit domain $D\setminus [0,\infty)\times \{0\}$ where $D$ is the unit disc in the plane. It is Not strong Markov for hitting times on the slit, but it is Markovian.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.