## A Markov process which is not a strong markov process?

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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

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## 4 Answers

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).

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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.

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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 2010 at 17:29

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.

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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" (books.google.co.uk/…). – George Lowther Oct 27 2010 at 17:51

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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 2010 at 5:02

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.

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@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. – Didier Piau Feb 22 2011 at 12:34
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