Some Local Asymptotic Laws for the Cauchy Process on the Line A .

This paper investigates the lim inf behavior of the sojourn time process and the escape rate process associated with the Cauchy process on the line. The monotone functions associated with the lower asymptotic growth rate of the sojourn time are characterized and the asymptotic size of the large values of the escape rate process is developed.


Introduction
Let X(t) = {X(t, ω),t ≥ 0} be a Levy process on a probability space (Ω, f ,P) with values in R n .We are interested in the sample path properties of the function X(t) = X(t,ω) for a fixed ω ∈ Ω.Let X(t), t ≥ 0, denote a Levy process and define where Let B(0,r) denote the ball in R n of radius r centered at zero, then, T(r) is the sojourn time of X(t) in B(0,r) up to time τ.It is well known (see [1,2]) that the sojourn time forms a useful tool in studying the local geometric properties of fractal sets determined by the sample paths of Levy processes in R n .
For example, the application of the density theorem of Taylor and Tricot [3] which remains one of the main tools for establishing packing measure and packing dimension results relies on the lower growth rate of T 1 (r) + T 2 (r).Here T 1 and T 2 are independent copies of T.
Results in [4] show that the lower growth rate of T 1 (r) + T 2 (r) is of higher order of magnitude than that of T(r) for the symmetric stable processes of index α>1 in R n .Pruitt and Taylor [5] further observed that they may differ by a factor | logr|(log|logr|) 1/2 .
The aim of this note is to investigate the lower asymptotic behaviour of T(r) for the Cauchy process on the line.This may serve as a useful tool to characterize the geometric structure of the random set determined by the symmetric Cauchy processes and also we consider asymptotic result, which may be related to the sojourn time process.

Preliminaries
A symmetric Cauchy process on the line, which we will denote by X(t), is a Levy process which is uniquely determined by its Fourier transform where g(t,x) = (1/π)(t/(t 2 + x 2 )).t > 0, x ∈ R 1 and satisfies the scaling property that c −1 X(ct) is a version of the same process X(t) for every c > 0. X(t) is recurrent, that is, {t : X(t) ∈ G} is unbounded for an open interval G containing the origin.In this case, T(r) is almost surely infinite as r → ∞.Thus we instead consider the process so that we have the relationship The first passage time is, as usual, defined by whose distribution is obtained from that of by the means of an obvious relationship ω : P(a) < r = ω : M(r) > a , a > 0, r > 0. (2.6) We will need the estimates for the distribution of the following events which we state as lemmas.
Lemma 2.2 [7]. ) , then for the symmetric Cauchy process on the line where P x is the conditional probability under the condition X(0) = x.Assume X(0) = 0 with probability one, and use the abbreviation P 0 = P.The c 1 ,c 2 ,... will denote positive constants whose values are not important.
Lemma 2.3 [8].Let {E n } be a sequence of events and suppose that 2 .Assume also that a version of the process is dealt with, which is strong Markov.

The lower asymptotic behaviour of the sojourn time for the symmetric Cauchy process on the line
In [9], Ray obtained a function ψ for which for the symmetric Cauchy process on the line.Here we consider the liminf behaviour of T(r) and state the following.
For a symmetric Cauchy process on the line . Set N k = number of returns of the process, started at x with |x| = ρ, a k+1 /2 ≤ ρ ≤ a k+1 , makes from S(a k+1 ) to S(a k+1 /2) before {S(1)} c .We set up sequences of stopping times: This continues until the process enters {S(1)} c at σ 2Nk+2 .There exists contribution to T(a k+1 ) from σ 2i to σ 2i+1 which is greater than the first passage time for X out of the sphere of radius a k+1 /2.Each time the process returns to S(a k+1 /2) from S(a k+1 ), since the process is recurrent, the event occurs for the restarted process a.s.Thus for j returns, M k/2 happens j times a.s.so that since there exists c 3 such that (see [6, page 353]), where (3.9) A. Chukwuemeka Okoroafor 5 Thus so P(G k ,i • 0) = 0 by Borel Cantelli lemma.Hence a.s.G k happens for at most a finite number of k for each λ, so that we can find r ∈ [a k+1 ,a k ] for which In the opposite direction, set a k = ρ −k , ρ > 1 and for ∈=∈ (ω) > 0. Suppose (h(a k )/ k) = ∞, then for any fixed λ, choose ρ large enough so that P(B k ) is close to 1 whenever (3.17) , similar arguments as in [5, page 140] suffice to show that where T k j = {the process X(s) started at X(λψ(a k )) enters {S(a j )} c at a time t before entering S(( by Blumenthal zero-one law.Hence for each λ, E k and therefore G k happen infinitely often a.s., which in turn implies that lim r→0 inf(T(r)/ψ(r)

The asymptotic size of the large values of f (s) as s → 0
The asymptotic size of the small values of f (s) as s → 0 was obtained by Takeuchi and Watanabe [10], where f (s) is as in (2.2).
In this section we obtain the asymptotic size of the large values of f (s) as s → 0.
Our basic arguments will follow those in [11,Lemmas 3.3 and 3.4], although some modifications are necessary.Theorem 4.1.For the symmetric Cauchy process X(t) on the line, according as ) is finite or infinite, where ϕ(s) = sg(s) and g(s) is a monotone decreasing function and For any fixed λ, and some ∈> 0, define so that when B k = {X(s) does not enter S(λ(1− ∈)ϕ(a k )) after a k before τ}, E k ⊂ A k ∩ B k .Thus where (where In the opposite direction, set a k = ρ −k , ρ > 1 and suppose For any fixed λ, and some ∈> 0, define where by (Lemma 2.1). (4.14)

( 4 .
23) by Lemmas 2.1 and 2.2. and P(E k ) < ∞.Thus P(E k ,i • o) = 0 by Borel Cantelli lemma.Therefore there exists k 0 such that for k > k 0 , f (a k ) ≤ λϕ(a k ) so that By the Blumenthal zero-one law, we have lim k→0 sup( f (a k )/ϕ(a k )) = 0 a.s. with .15)If ρ is chosen large enough so that P(A k ) is close to one, we have P T k j P B j P C j , (4.17)where T k j = {X(s) started at X(a k ) enters {S(∈ ϕ(a j ))} c at time t before (1 − 3 ∈)ϕ(a k )} and for k ≥ j + 1, ∞.Thus by (Lemma 2.3),P D k ,i • o ≥ C −1 > 0. (4.20) Therefore P(D k ,i • o) = 1 by Blumenthal zero-one law.Therefore we can find s ∈ (a k+1 ,a k ] for which k ) = ∞.Similar arguments in [11, Lemma 3.5] suffice to show that P D k ∩ D j ≤ P B k