On explicit occupation time distributions for Brownian processes

dp

Applying Lemma 5.1 of the appendix and the identity 1 − E0 {e−β
1 0

(W0 (s)≥a) ds

1 }= √ i 2π

1+i∞ 1−i∞

√ 2λ dλ = i 2π , √ √ √ λ + β − λ eλ−2a 2λ dλ. √ √ √ 2λ λ + β + λ

Keywords: Brownian bridge, excursion, meander, occupation times, sojourn times AMS 1991 Subject Classification: 60J65

1

Introduction

In [8], Tak´ acs gave explicit expressions for the sojourn time distribution of Brownian bridge
∞ 0

2y 2 2 2 e−2a y x/(1−x) dy, 2 2 (y + 1)

from which it is straightforward to obtain Formula (6) of [8]. For the moments we obtain: 2r µr (a) = r x (1 − Fa (x)) dx = √ π 0 1 rΓ(r + 2 ) 1 r−1 = u ϕa (1 − u) du, (r + 1)! 0
+ W0 (t), 0 ≤ t ≤ 1, and excursion W0 (t), 0 ≤ t ≤ 1, by approximating these processes by suitable

conditioned simple random walk. Inspection of these results show a close connection between the sojourn time distribution for the bridge and that for the excursion. It is our goal to explain this connection and at the same time to give an independent derivation of the mentioned distributions. We also include an explicit expression for the sojourn time distribution of Brownian meander W + (t), 0 ≤ t ≤ 1, which was not contained in [8]. Kac’s formula (cf. [5], p. 54 or [7], p. 271) is used to derive transforms of the mentioned distributions for Brownian motion W (t), t ≥ 0. To step from motion to bridge we use that the bridge is pinned down Brownian motion in a weak convergence sense. In the Appendix we give
1+i∞ λ 1−i∞ e /

√

(5)

Theorem 2.1 For 0 < x < 1, the distribution function Fa of the occupation time above level a > 0 of Brownian bridge satisfies: 2x1/2 1 − Fa (x) = √ π with ϕa (t) given in (9). PROOF: From (5) and using (29.3.52) and (29.3.84) of [1], we obtain: √ √ √ 1 1+i∞ eλ−2a 2λ λ + β − λ 1 √ √ √ e−βx (1 − Fa (x)) dx = √ 0 iβ 2π 1−i∞ 2λ λ+β+ λ 2 /(1−u) −βu/2 − 2 a 1 e e I1 (βu/2) = du, βu 0 (1 − u)
Fra Baidu bibliotek

x ≥ a, x < a,

(3)

B=

−4β ( 2(λ + β ) + ip) (2λ + p2 )(2(λ + β ) + p2 )

 

e−a

√

2λ+ipa

2(λ + β ) +

√

. 2λ

(4)

Transforming back with respect to p, after putting x = 0, gives:
∗

Faculty ITS, Technical University Delft, Mekelweg 4, 2628 CD Delft, The Netherlands; e-mail:

G.Hooghiemstra@its.tudelft.nl

a new proof of this. We further use results of [4] that connect the bridge to the excursion and the meander, respectively. The paper is organized as follows. In Section 2 we derive the sojourn time distribution of the bridge. The form of the distribution function is slightly different from that in [8]. In Section 3, we show how this result can be used for the occupation time of the excursion; the resulting distribution function is somewhat simpler than the result of [8] because it does not contain a double sum. In Section 4 we treat the meander. All three distribution function can be expressed in the transition density of Brownian motion: p(t; 0, x) = (2πt)−1/2 e−x
∞ 0

e−λt E0 {(Wt ∈ dy )e−β

√

t 0

(Ws ≥a) ds

}dt =

e−y 2λ − = √ 2λ

√

√ √  √ λ + β − λ e−2a 2λ+y 2λ  dy, √ √ √ 2λ λ + β + λ

dy 2π

∞ −∞

e−ipy

2 +B 2λ + p2 for 0 < y < a.
   
1 u 2 1 u 2

(x) − βu(x) + f (x), (x) + f (x),

x ≥ a, x < a,

(2)

with unique bounded solution given by:  √  2eipx  Ae−x 2(λ+β ) + , 2(λ+β )+p2 u(x) =  √ ipx  Bex 2λ + 2e , 2λ+p2 satisfying u(a) = u(a−) and u (a) = u (a−). This yields:

2

The bridge
∞

We define for a > 0, u(x) = Ex
0

f (Wt )e−λt−β

t 0

(Ws ≥a) ds

dt,

(1)

where (A) denotes the indicator function of the set A and f (x) = eipx , p, x ∈ I R. Then according to Kac’s formula, u satisfies the second order differential equation: λu(x) = 
r −1 1 1

x
0

1 r− 2

1 x

(u − x) 2 ϕa (1 − u) du dx u2

1

(10)

r ≥ 1,

µ0 (a) = P( max W0 (t) ≥ a) = e−2a .
0≤t≤1

2

3

The excursion

Our approach to calculate the sojourn time distribution for the excursion uses Theorem 5.1 of [4], which theorem implies with m0 = min0≤t≤1 W0 (t):
0 1 1 −1

(1 − t2 )j −1/2 e−xt/2 dt (7)

{s(1 − s)}j −1/2 e−xs ds,

and hence
1 0

e−βx (1 − Fa (x)) dx =

1 0

2 ϕa (1 − u) √ π

1 0

s1/2 (1 − s)1/2 e−βus ds du,

(8)

where for t > 0, ϕa (t) = (πt)−1/2 e−2a /t .
On explicit occupation time distributions for Brownian processes
Gerard Hooghiemstra∗ November 30, 2000

Abstract Explicit expressions for the occupation time distribution of Brownian bridge, excursion and meander are derived using Kac’s formula and results of [4]. The first two distributions appeared in [8], and were derived using weak convergence of simple random walk.
2

(9)

In the right hand side of (8) we change variables us = x. Then (6) is obtained, after changing the order of integration, by uniqueness of Laplace transforms. Changing variables: u = (x + xy 2 )/(1 + xy 2 ), we obtain from (6): 1 − Fa (x) = 2(1 − x) −2a2 /(1−x) e π
1 x

ϕa (1 − u) (u − x)1/2 du, u2

(6)

dλ

where Iν (z ) denotes the modified Bessel function of order ν (cf. 9.6 of [1]). Using the integral representation (9.6.18) of [1] we obtain, after the change of variables t + 1 = 2s, √ Γ(j + 1/2) πe−x/2 Ij (x/2) = (x/4)j e−x/2 = xj
2 / 2t

, compare Formula (9).

The explicit forms of the sojourn times distributions are important in applied probability as limit distributions of functionals of queueing and or branching processes (cf. [8] and the references therein).