Distribution of the supremum of increments of Brownian local time
Zapiski Nauchnykh Seminarov POMI, Problems of the theory of probability distributions. Part IX, Tome 142 (1985), pp. 6-24
A. N. Borodin. Distribution of the supremum of increments of Brownian local time. Zapiski Nauchnykh Seminarov POMI, Problems of the theory of probability distributions. Part IX, Tome 142 (1985), pp. 6-24. http://geodesic.mathdoc.fr/item/ZNSL_1985_142_a0/
@article{ZNSL_1985_142_a0,
     author = {A. N. Borodin},
     title = {Distribution of the supremum of increments of {Brownian} local time},
     journal = {Zapiski Nauchnykh Seminarov POMI},
     pages = {6--24},
     year = {1985},
     volume = {142},
     language = {ru},
     url = {http://geodesic.mathdoc.fr/item/ZNSL_1985_142_a0/}
}
TY  - JOUR
AU  - A. N. Borodin
TI  - Distribution of the supremum of increments of Brownian local time
JO  - Zapiski Nauchnykh Seminarov POMI
PY  - 1985
SP  - 6
EP  - 24
VL  - 142
UR  - http://geodesic.mathdoc.fr/item/ZNSL_1985_142_a0/
LA  - ru
ID  - ZNSL_1985_142_a0
ER  - 
%0 Journal Article
%A A. N. Borodin
%T Distribution of the supremum of increments of Brownian local time
%J Zapiski Nauchnykh Seminarov POMI
%D 1985
%P 6-24
%V 142
%U http://geodesic.mathdoc.fr/item/ZNSL_1985_142_a0/
%G ru
%F ZNSL_1985_142_a0

Voir la notice du chapitre de livre provenant de la source Math-Net.Ru

The joint distribution of the variables $\hat t(t,r)$, $\hat t(t,r)$ and $\sup_{0\le s\le t}(\hat t(s,q)-\hat t(s,r))$, where $\hat t(t,x)$ is Brownian local time, is determined uniquely by the Laplace transform $\int_0^\infty e^{-\lambda t}E\{e^{-\mu\hat t(t,r)-\eta\hat t(t,q)},\sup_{0\le s\le t}(\hat t(s,q)-\hat t(s,r))>h|w(0)=x\}\,dt.$ The computation of this transform constitutes the basic content of this paper. The obtained expression is used for the derivation of the exact modulus of continuity of the process $\hat t(t,x)$ with respect to the variable $x$: $$ P\Big\{\limsup_{\substack{|y-x|=\Delta\downarrow0\\y,x\in R^1}}\frac{\sup_{0\le s\le t}|\hat t(s,y)-\hat t(s,x)|}{((\hat t(t,x)+\hat t(t,y))\Delta\ln 1/\Delta)^{1/2}}=2\Bigl\}=1. $$