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Zbl 0849.60076
Eisenbaum, Nathalie; Kaspi, Haya
A counterexample for the Markov property of local time for diffusions on graphs. (English)
[A] Azéma, J. (ed.) et al., Séminaire de probabilités XXIX. Berlin: Springer-Verlag. Lect. Notes Math. 1613, 260-265 (1995). ISBN 3-540-60219-4/pbk

The authors try to answer the following question: ``whether the conditions for the Markov property of the local time, for processes on $E \subset R$, are sufficient for processes taking values in $E \subset R^d$, and admitting a local time at each point of $E$''. To this end they consider the following definition of the Markov property for processes indexed by $E \subset R^d$: A process $\{L^x : x \in E\}$ where $E \subset R^d$ has the Markov property provided for every open, relatively compact set $A$, contained in $E$, $\{L^x : x \in A^\sim\}$ and $[L^x : x \in A^c\}$ are conditionally independent given $\{L^x : x \in \partial A\}$.\par Based on this definition, the answer to the above question is that continuity and fixed birth and death points are not sufficient for $\{L^x : x \in E\}$ to be Markov. For this, they consider the Brownian motion $X$ on the unit circle $S^1$ born at (1,0) and killed when the local time at point $(-1,0)$ exceeds an exponential variable that is independent of $X$. This process has continuous sample paths the fixed birth and death points. For the points $(1,0)$, $(0,1)$, $(-1,0)$ and $(0,-1)$ they obtain the following result: Given $(L^{(0,1)}, L^{(-1,0)})$, $L^{(0,1)}$ and $L^{(0,-1)}$ are not conditionally independent.
[G.Orman (Braşov)]

MSC 2000:: *60J60 Diffusion processes
60C05 Combinatorial probability

Keywords: Markov property; local time; conditionally independent; Brownian motion

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