Statistics & Probability Letters 45 (1999) 379–382www.elsevier.nl/locate/stapro

Shift invariance of the occupation timeof the Brownian bridge process(

Peter Howarda ; ∗, Kevin Zumbrunb

a Division of Applied Mathematics, Brown University, Providence, RI 02912, USAb Department of Mathematics, Indiana University, Bloomington, IN 47405, USA

Received August 1998; received in revised form July 1999

Abstract

In this note the distribution for the occupation time of a one-dimensional Brownian bridge process on any Lebesguemeasurable set between the initial and �nal states of the bridge is shown to be invariant under translation and re ection,so long as the translation or re ection also lies between the initial and �nal states of the bridge. The proof employsonly the strong Markov property and elementary symmetry properties of the Brownian bridge process. c© 1999 ElsevierScience B.V. All rights reserved

MSC: 60J65; 35K15

Keywords: Occupation time; Brownian bridge

Let Ws denote a standard, one-dimensional Wiener process. A Brownian bridge from a to b on the interval[0; t], denoted Ba→b

[0; t] (s), can be de�ned as (see Karatzas and Shreve, 1991)

Ba→b[0; t] (s) := a

(1− s

t

)+ b

st+(Ws − s

tWt

); 06s6t: (1)

Note that the Brownian bridge process is a standard, one-dimensional Wiener process constrained to particularstarting and ending points. An important aspect of this process, which we employ below, is the symmetrybetween going from a to b or from b to a.The Feynman–Kac formula for a solution to the Fokker–Planck equation leads naturally to a study of the

occupation time of such a process. In particular, for the equation

ut = 12uxx − IA(x)u; u(0; x) = �y(x);

( This work was partially supported by the ONR under Grant no. N00014-94-1-0456 and by the NSF under Grant no. DMS-9107990and DMS-9706842 and also by the Gaann fellowship.

∗ Corresponding author.

0167-7152/99/$ - see front matter c© 1999 Elsevier Science B.V. All rights reservedPII: S0167 -7152(99)00080 -2

380 P. Howard, K. Zumbrun / Statistics & Probability Letters 45 (1999) 379–382

Fig. 1. The Brownian bridge process.

where A is some Lebesgue measureable set, we have the solution (see Oksendal, 1995)

u(t; x) = E[e−∫ t

0IA(Xs) ds�y(Xt)];

where Xt satis�es the stochastic di�erential equation

dXt = dWt; X0 = x: (2)

The delta function constrains Xt to satisfy Xt = y, leading to

u(t; x) = E[e−∫ t

0IA(B

x→y[0; t] (s)) dsP(t; x; y)];

where P(t; x; y) is the Wiener transition function. The integral∫ t

0IA(B

x→y[0; t] (s)) ds

is the occupation time of the Brownian bridge B x→y[0; t] (s) in the set A.

Proposition 1. For any Lebesgue measurable set; A; such that A⊂ [a; b]; the distribution of the occupationtime;

∫ t0 IA(B

a→b[0; t] (s)) ds; of B

a→b[0; t] (s) in A is invariant under translations and re ections; so long as the

translation or re ection also lies in [a; b].

Proof. For simplicity, we �rst carry out the proof in the case of an interval. Let a6c¡d6b. We show thatthe distribution of time spent in the interval [c; d] is the same as the distribution of time spent in the interval[b− (d− c); b], that is, that this distribution depends only on the length of the interval [c; d].We observe �rst that this assertion holds trivially for the case d=b, and that by symmetry it must also hold

for c= a. Now, let Tc denote the random variable or hitting time, representing the �rst time Ba→b[0; t] (s) equals

c – an event that occurs with probability one, since a6c¡b (see Fig. 1). We remark that the distributionof Tc depends on a; b and t.We condition on Tc and (intuitively) consider the random variable

B� :=[ ∫ t

0I[c; d](Ba→ b

[0; t] (s)) ds∣∣∣∣ Tc = �

]

P. Howard, K. Zumbrun / Statistics & Probability Letters 45 (1999) 379–382 381

for all � ∈ [0; t]. For z ∈ R we can write the distribution function for the occupation time on [c; d] in termsof B� as

Pr[∫ t

0I[c; d](Ba→b

[0; t] (s)) ds6z]=∫ t

0Pr

[ ∫ t

0I[c; d](Ba→b

[0; t] (s)) ds6z∣∣∣∣ Tc = �

]d�Tc(�); (3)

where �Tc(�) is the probability measure associated with the random variable Tc.Let Bc→b

[�; t] (s) represent the Brownian bridge on the interval [�; t] that goes from c to b. Since d6b we havec6b− (d− c), so that the amount of time spent by Bc→b

[�; t] (s) in either of the intervals [c; d] or [b− (d− c); b]is exactly the time spent by B� in that interval. But, by the symmetry discussed above, the distribution oftime spent by Bc→b

[�; t] (s) in [c; d] is the same as the distribution of time spent by Bc→b[�; t] (s) in [b− (d− c); b].

Hence, we can make the following computation, which begins with the right-hand side of (3).∫ t

0Pr

[ ∫ t

0I[c; d](Ba→b

[0; t] (s)) ds6z∣∣∣∣ Tc = �

]d�Tc(�)

=∫ t

0Pr

[∫ t

�I[c; d](Bc→b

[�; t] (s)) ds6z]d�Tc(�)

=∫ t

0Pr

[∫ t

�I[b−(d−c); b](Bc→b

[�; t] (s)) ds6z]d�Tc(�)

=∫ t

0Pr

[ ∫ t

0I[b−(d−c); b](Ba→b

[0; t] (s)) ds6z∣∣∣∣ Tc = �

]d�Tc(�)

=Pr[∫ t

0I[b−(d−c); b](Ba→b

[0; t] (s)) ds6z]:

We note that the �rst and third inequalities follow from the strong Markov property of the Brownian bridgeprocess (see Fitzsimmons et al., 1993). This completes the proof in the case of intervals, where re ectionsare equivalent to translations.The assertion regarding Lebesgue-measurable sets can be obtained similarly. Let Adc represent a Lebesgue

measurable set with endpoints c and d. Then∫ t

0IAdc (B

c→b[�; t] (s)) ds

d=∫ t

0IAbb−(d−c)

(Bc→b[�; t] (s)) ds;

by symmetry, where Abb−(d−c) is the re ection of A

dc with upper endpoint b and

d= denotes equality indistribution.Using this observation, we compute as before

Pr[∫ t

0IAdc (B

a→b[0; t] (s)) ds6z

]=∫ t

0Pr

[ ∫ t

0IAdc (B

a→b[0; t] (s)) ds6z

∣∣∣∣ Tc = �]d�Tc(�)

=∫ t

0Pr

[∫ t

�IAdc (B

c→b[�; t] (s)) ds6z

]d�Tc(�)

=∫ t

0Pr

[∫ t

�IAbb−(d−c)

(Bc→b[�; t] (s)) ds6z

]d�Tc(�)

=∫ t

0Pr

[ ∫ t

0IAbb−(d−c)

(Ba→b[0; t] (s)) ds6z

∣∣∣∣ Tc = �]d�Tc(�)

382 P. Howard, K. Zumbrun / Statistics & Probability Letters 45 (1999) 379–382

=∫ t

0Pr

[ ∫ t

0IAa+(d−c)

a(Ba→b[0; t] (s)) ds6z

∣∣∣∣ Tc = �]d�Tc(�)

= Pr[∫ t

0IAa+(d−c)

a(Ba→b[0; t] (s)) ds6z

];

where Aa+(d−c)a is the re ection of Abb−(d−c) (thus a translation of A

dc ) with lower endpoint a.

This yields the result for translations. The result regarding re ections follows similarly.

References

Karatzas, I., Shreve, S., 1991. Brownian Motion and Stochastic Calculus. Springer, Berlin.Oksendal, B., 1995. Stochastic Di�erential Equations. Springer, Berlin.Fitzsimmons, J., Pitman, J., Yor, M., 1993. Markovian bridges: Construction, palm interpretation, and splicing, in: E. �Cinlar, K.L. Chung,M.J. Sharpe (Eds.), Seminar on Stochastic Processes, Birkh�auser, Boston.