Propriétés de martingales, explosion et représentation de Lévy—Khintchine d'une classe de processus de branchement à valeurs mesures

Summary.A new approach is provided to the super-Brownian motion X with a single point-catalyst 8c as branching rate. We start from a superprocess U with l -stable subordinator. We constant branching rate and spatial motion given by the prove that the occupation density measure A c of X at the catalyst c is distributed as the total occupation time measure of U. Furthermore, we show that Xt is determined from A c by an explicit representation formula. Heuristically, a mass A~ of "particles" leaves the catalyst at time s and then evolves according to Itr's Brownian excursion measure. As a consequence of our representation formula, the density field x of X satisfies the heat equation outside of c, with a noisy boundary condition at c given by the singularly continuous random measure A c. In particular, x is W~ outside the catalyst. We also provide a new derivation of the singularity of the measure A c.

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“…We refer to Section 4 in E1 Karoui and Roelly [10] (see also Dawson and Perkins [7] and Le Gall [12] for related results).…”

Section: Canonical Measures and Campbell Measure Formulamentioning

confidence: 97%

A new approach to the single point catalytic super-Brownian motion

Fleischmann

Gall

1995

Probab. Theory Relat. Fields

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“…Notice that in this case the topology of E is a Ray topology. For other regularity conditions on and (V t ) t≥0 see also [12].…”

Section: Proposition 48 Each Of the Following Two Conditions Impliementioning

confidence: 99%

Potential-theoretical methods in the construction of measure-valued Markov branching processes

Beznea¹

2011

J. Eur. Math. Soc.

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“…From [8] and [4] we have that the (α, d, β)-superprocess X has the following decomposition: Let ϕ ∈ D(∆ α ), t ≥ 0, 9) where N X = N X −N X is a martingale measure and…”

Section: Introduction and Statement Of Resultsmentioning

confidence: 99%

“…Then, due to the uniqueness of the decomposition ( [8], Theorem 7) we conclude thatŸ K has the same distribution as Y K .…”

Section: Lemmamentioning

confidence: 98%