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Occupation time distributions for Lévy bridges and excursions. (English) Zbl 0837.60071

Let \(X\) be a one-dimensional Lévy process and \(R\) be the length of an excursion of \(X\) from 0. Further let \(A_t= \int^t_0 1_{X_s> 0} ds\). Assume that the distribution \(\mu_t= P^0(X_t\in \cdot)\) can be written as \(\mu_t(dx)= p_t(x)dx\). Then the bridge law \(P^{x, y}_t(F)\) defined for \({\mathcal F}_s\)-measurable functions \(F\) by \(P^{x, y}_t(F)= P^x(Fp_{t-s}(y- X_s))/p_t(y- x)\) is well-defined. It is proved that the occupation time \(A_t\) under \(P^{0,0}_t\) is uniformly distributed on \([0, t]\). Under the additional assumption of the regularity of the state 0 for itself, the explicit expression of the Laplace transform of \((\tau(t), A_{\tau(t)})\) is obtained, where \(\tau(t)\) is the right-continuous inverse of the local time of \(L_t\) of \(X\) at 0. This further gives the Laplace transform of the joint distribution of the pair \((R, A_R)\). In particular, for the stable processes with index \(\alpha\in (1, 2)\), concrete expressions of the distributions of \(R\), \(A_R\) and \(A_R/R\) are given. If \(P^0 (A_\infty=\infty)= 1\), or equivalently \(\int^\infty_1 t^{-1} P^0 (X_t> 0) dt< \infty\), then they are related by \[ P^0(e^{- \lambda A_\infty})= \exp\Biggl(- \int^\infty_0 t^{- 1}(1- e^{-\lambda t}) P^0 (X_t> 0) dt\Biggr) \] which gives the Laplace transform of the distribution of the occupation time of \((0, \infty)\) during the final excursion.

MSC:

60J99 Markov processes
60J60 Diffusion processes
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