Reflected Brownian motion in generic triangles and wedges

arXiv:math/0410007v3 [math.PR] 18 May 2005

Re?ected Brownian motion in generic triangles and wedges
Wouter Kager? February 1, 2008

Abstract Consider a generic triangle in the upper half of the complex plane with one side on the real line. This paper presents a tailored construction of a discrete random walk whose continuum limit is a Brownian motion in the triangle, re?ected instantaneously on the left and right sides with constant re?ection angles. Starting from the top of the triangle, it is evident from the construction that the re?ected Brownian motion lands with the uniform distribution on the base. Combined with conformal invariance and the locality property, this uniform exit distribution allows us to compute distribution functions characterizing the hull generated by the re?ected Brownian motion.

1 1.1

Introduction and overview Motivation

Re?ected Brownian motions in a wedge with constant re?ection angles on the two sides were characterized by Varadhan and Williams in [15]. Recent work by Lawler, Schramm and Werner [9] on SLE establishes a connection between one of these re?ected Brownian motions (with re?ection angles of 60? with respect to the boundary), chordal SLE6 and the exploration process of critical percolation. They show that these three processes generate the same hull, using an argument that we shall outline in the following paragraph. The connection was carried even further by Julien Dub? edat [4, 5]. He compared SLE6 and the aforementioned re?ected Brownian motion in an equilateral triangle, started from a given corner and conditioned to cross the
Instituut voor Theoretische Fysica, Universiteit van Amsterdam, Valckenierstraat 65, 1018 XE Amsterdam, The Netherlands. E-mail: kager@science.uva.nl.
?

1

triangle to a given point X on the opposite side. For these two processes, he showed that the conditional probability that the side to the right of the starting point is the last side visited before reaching X , is the same. He also showed that, under the assumption that SLE6 is the scaling limit of the exploration process of critical percolation, this conditional probability can be used to prove Watts’ formula for critical percolation (see [5]). The argument used in [9] to establish the connection between the three models is as follows. Let (Zt ) be a stochastic process in an equilateral triangle T , started from a corner and stopped when it ?rst hits the opposite side S . De?ne the hull Kt as the compact set of points in T disconnected from S by the trace of Zt up to time t. Denote by X the hitting point of S , and by τ the hitting time. We shall refer to the probability distribution of the point X as the “exit distribution”. For all three processes mentioned above, the exit distribution is uniform on the side of the triangle. Together with conformal invariance and the locality property (see Section 4.4 below and [8]), shared by all three processes, this exit distribution determines the law of the hull Kτ . The three processes therefore generate the same hull. This argument can be generalized to stochastic processes in arbitrary triangles that are conformally invariant and have the locality property: if the exit distributions of two such processes are the same, in particular if they are both uniform, then the processes generate the same hull. The papers of Lawler, Schramm, Werner [9] and Dub? edat [4] show that in equilateral and isosceles triangles there exist re?ected Brownian motions that have uniform exit distributions. The purpose of this paper is to generalize this result to arbitrary triangles, to review properties of these re?ected Brownian motions and to discuss distribution functions associated with these processes and the hulls they generate.

1.2

Notations and overview

To present an overview of the present paper, we ?rst need to introduce some notation. Given two angles α, β ∈ (0, π ) such that α + β < π , we : α ? π < arg z < ?β }. We also de?ne the wedge Wα,β as the set {z ∈ de?ne Tα,β as the triangle in the upper half of the complex plane such that one side coincides with the interval (0, 1), and the interior angles at the corners 0 and 1 are equal to α and β , respectively. The third corner is at wα,β := (cos α sin β + i sin α sin β )/ sin(α + β ). When α + β ≥ π , the domain Tα,β is similarly de?ned as the (unbounded) polygon having one side equal to the interval (0, 1), and interior angles α and β at the corners 0 and 1. We then identify the point at ∞ with the “third corner” wα,β . 2

Suppose now that ?L and ?R are two angles of re?ection on the left and right sides of the wedge Wα,β , respectively, measured from the boundary with small angles denoting re?ection away from the origin (?L , ?R ∈ (0, π )). We shall use the abbreviation RBM?L ,?R to denote the corresponding re?ected Brownian motion in the wedge Wα,β . For a characterization and properties of these RBMs, see Varadhan and Williams [15] (note that here we use a di?erent convention for the re?ection angles, namely, the angles ?1 and ?2 of Varadhan and Williams correspond in our notation to the angles ?L ? π/2 and ?R ? π/2). The main goal of this paper is to show that in every wedge Wα,β there is a unique RBM?L ,?R with the following property: started from the origin, the ?rst hitting point of the RBM of any horizontal line segment intersecting the wedge is uniformly distributed. This special behaviour is obtained by taking the re?ection angles equal to the angles of the wedge, that is, ?L = α and ?R = β . Restricting the wedge to a triangle we can reformulate this result as follows: Theorem 1 Let α, β ∈ (0, π ), α + β < π , and let (Zt : t ≥ 0) be an RBMα,β in the triangle Tα,β started from wα,β and stopped when it hits [0, 1]. Set τ := inf {t > 0 : Zt ∈ [0, 1]} and X := Zτ . Then X is uniform in [0, 1]. To prove this theorem we will cover the wedge Wα,β with a well-chosen lattice, and then de?ne a random walk on this lattice. By construction, this random walk will have the desired property that it arrives on each horizontal row of vertices on the lattice with the uniform distribution. Taking the scaling limit then yields the desired result. The proof is split in two sections. In Section 2 we consider the easier case where the angles α and β are in the range (0, π/2] such that π/2 ≤ α + β < π . Section 3 treats the extension to arbitrary triangles, which is considerably more involved. Section 4 collects some properties of the two-parameter family of RBMs. In particular, as was noted by Dub? edat [4], we can show from the discrete approximations that the imaginary parts of the RBMs are essentially 3dimensional Bessel processes. This allows us to describe the time-reversals of the RBMs, and sheds some light on how the RBMα,β behaves in the domain Tα,β for angles α, β ∈ (0, π ) such that α + β ≥ π . Finally, in Section 5 we compute several distribution functions associated with the RBMs and the hulls they generate. This also reveals intriguing connections between RBMs started from di?erent corners of the same triangle.

3

sin ψ h ? u

sin ? ψ v b

c 0 a

a b c

Figure 1: Picture of the lattice, showing the dimensions on the left, and the transition probabilities for a step of the random walk (from the origin in this picture) on the right.

2

RBM in a restricted geometry

Throughout this section, we assume that the angles α and β are ?xed and restricted to the range (0, π/2] such that π/2 ≤ α + β < π . We shall construct a random walk in the wedge W = Wα,β whose scaling limit is an RBMα,β and whose horizontal coordinate is uniformly distributed all the time, and thereby prove Theorem 1 for these restricted values of α and β . The generalization to a generic triangle will be treated in Section 3.

2.1

The lattice and further notations

Throughout this paper we shall make use of a distorted triangular lattice, de?ned as follows. Let ? and ψ be two angles in the range (0, π/2] such that π/2 ≤ ? + ψ < π (as we shall see later on, the range for ? and ψ is chosen such that the transition probabilities for our random walk are positive). We de?ne Γ?,ψ as the set of vertices {j sin(? + ψ ) ? k exp(i?) sin ψ : j, k ∈ } building the triangular lattice depicted in Figure 1. Throughout the paper we shall make use of the variables u := cos ? sin ψ , v := sin ? cos ψ and h := sin ? sin ψ to denote the lattice dimensions. When we use these variables, the values of ? and ψ will always be clear from the context. To introduce some further notation, let us ?rst consider how one de?nes a random walk (Xn : n ≥ 0) on Γ?,ψ that converges to standard complex Brownian motion in the full plane, before we consider the random walk in the wedge W in the following subsection. We set X0 := 0 and for each 4

n > 0, Xn is chosen among the nearest-neighbours of Xn?1 according to the probabilities a, b and c as depicted in Figure 1. We may then write the position Xn of the random walk as a sum of steps S1 + S2 + · · · + Sn where each step Sn = Un + iVn is a complex-valued random variable taking on the possible values with probability a; Sn = ±(u + v ) with probability b; ? ? ±(v ? ih) with probability c.
? ? ? ±(u + ih)

(1)

To obtain a two-dimensional Brownian motion as the scaling limit of the random walk (Xn ), it is su?cient that the covariance matrix of the real and imaginary parts Un and Vn of each step is a multiple of the identity. This gives two equations for the probabilities a, b and c: (a + b) cot2 ? + 2b cot ? cot ψ + (b + c) cot2 ψ = a + c, a cot ? ? c cot ψ = 0, (2) (3)

where cot x = 1/ tan x. The probabilities a, b and c can be determined from these equations, yielding a = λ cot ψ (cot ? + cot ψ ), b = λ(1 ? cot ? cot ψ ), c = λ cot ?(cot ? + cot ψ ), (4) (5) (6)

?2 where λ = 1 ψ ]?1 is the normalization constant. 2 [cot ?(cot ? + cot ψ ) + sin One may verify that ? and ψ must satisfy π/2 ≤ ? + ψ < π to make all three probabilities nonnegative. We conclude this subsection with a short discussion of how one obtains the scaling limit of the random walk (Xn ). To do so, for every natural number N > 0 one may de?ne the continuous-time, complex-valued stochastic (N ) process (Zt : t ≥ 0) as the linear interpolation of the process

Yt

(N )

=

1 X 2 N σ ?N t?

(7)

making jumps at the times {k/N 2 : k = 1, 2, . . .}. Here, σ 2 is the variance of 2 ] = E[V 2 ]. the real and imaginary parts of the steps Sn , that is, σ 2 = E[Un n (N ) It is then standard that Zt converges weakly to a complex Brownian motion in the full complex plane when N → ∞ (topological aspects are as described in the Introduction of Varadhan and Williams [15]). 5

sin ψ h ? u

sin ? 1/2 ψ v b a+c a b c

0 1/2 c b a b a+c

Figure 2: De?nition of the re?ected random walk in a wedge.

2.2

Re?ected random walk in a wedge

We now return to the wedge W = Wα,β with angles α and β ?xed in the range (0, π/2] such that π/2 ≤ α + β < π . Clearly, this wedge is covered nicely with vertices of the lattice Γ?,ψ when we set ? := α and ψ := β , see Figure 2. For the duration of this section we consider these values of ? and ψ to be ?xed. Later, when we generalize to arbitrary triangles, the relation between ?, ψ and α, β will not be so simple, which is why we already reserve the symbols ?, ψ to denote the angles of the lattice. We shall denote by G = Gα,β the set of vertices obtained by taking the intersection of Γ = Γ?,ψ with W . We shall call the vertices of G having six nearest neighbours along the lattice directions interior vertices. The origin will be called the apex of G, and the remaining vertices will be referred to as the boundary vertices. The set of boundary vertices may be further subdivided into left boundary vertices and right boundary vertices, with the obvious interpretation. x : n ≥ 0) on G Given a vertex x of G, a re?ected random walk (Xn x x is de?ned as follows. We set X0 := x and for each n > 0, if Xn ?1 is an x x interior vertex, then Xn is chosen among the six nearest-neighbours of Xn ?1 according to the probabilities a, b and c as before. This guarantees that the scaling limit of the random walk is Brownian motion in the interior of the wedge. It remains to specify the transition probabilities for the random walk from the boundary vertices and the apex. In this paper, we restrict ourselves to the case where the transition probabilities on all left boundary vertices are the same, and likewise for the right boundary vertices. We write Ez [S1 ] for the expected value of the ?rst step

6

of the random walk started from the vertex z , and assume that it is nonzero at all the boundary vertices and the apex. Then, as we shall prove in Section 2.3, the random walk converges to a re?ected Brownian motion. Moreover, the directions of re?ection on the sides of the wedge are given by the directions of Ez [S1 ] at the left and right boundary vertices. It follows that by playing with the transition probabilities from the boundary vertices we can obtain di?erent RBMs in the scaling limit. There is, however, only one choice of transition probabilities for which the random walk has the following special property: started from the origin, the random walk ?rst arrives on any row of the lattice with the uniform distribution on the vertices of that row. We will refer to this special case as the uniform (random) walk. To derive its transition probabilities one proceeds as follows. In a picture where we represent the steps of the walk by arrows, we have to make sure that every vertex in a given row has two incoming arrows with probability b from vertices in the same row, and two incoming arrows with probabilities a and c from vertices in the rows above and below. This completely determines the transition probabilities from the boundary vertices, see Figure 2 for a picture of the solution. In formula, if x is a left boundary vertex, then we have the following transition probabilities for the uniform walk: p[x, x] = p[x, x + (u + v )] = b, p[x, x + u + ih] = a, p[x, x ? u ? ih] = a + c, p[x, x + v ? ih] = c, (8) (9) (10) (11)

and the transition probabilities from the right boundary vertices are de?ned symmetrically, as shown in Figure 2. At the apex we simply choose the transition probabilities to each the two vertices directly below the apex equal to 1/2. x at each step It will now be convenient to decompose the position Xn of the uniform walk as Jn (u + v ) ? Kn (u + ih). Then Kn is a nonnegative integer denoting a row of vertices on the lattice, and Jn is a nonnegative integer denoting the position on the Kn th row. Observe that there are a total of N (k) = k + 1 vertices on the kth row, so that Jn ranges from 0 to N (Kn ) ? 1. Henceforth, we shall always adopt this convention of numbering rows on the lattice in top-down order, and vertices on each row from left to right. 0 ) started from the origin By construction, the uniform random walk (Xn has the property that at every time n ≥ 0, the position of the walker is 7

uniformly distributed on the rows of the lattice. More precisely, for the 0 ) the following lemma holds: uniform walk (Xn Lemma 2 For all n ∈ ?, if k0 , k1 , . . . , kn is a sequence of natural numbers such that k0 = 0 and |km ? km?1 | ≤ 1 for all m = 1, 2, . . . , n, then for each j = 0, 1, . . . , N (kn ) ? 1, P[Jn = j | K0 = k0 , . . . , Kn = kn ] = P[Jn = j | Kn = kn ] = 1 . N (kn )

In particular, if k ∈ ? and T := min{n ≥ 0 : Kn ∈ k} is the ?rst time at which the walk visits row k, then for each j = 0, 1, . . . , N (k) ? 1, P[JT = j ] = 1 . N (k)

Proof: The ?rst claim of the lemma is proved by induction. For n = 0, 1 the claim is trivial. For n > 1, P[Jn = j | Kn = kn ] =
kn?1

P[Jn = j, Kn = kn | Kn?1 = kn?1 ] P[Kn?1 = kn?1 ] P[Kn = kn | Kn?1 = kn?1 ] P[Kn?1 = kn?1 ] . (12)

kn?1

Using the induction hypothesis, it is easy to compute the conditional probabilities in the numerator and denominator of this expression from the transition probabilities given earlier. We ?rst assume kn > 1. Then for kn?1 = kn these conditional probabilities are 2b/N (kn ) and 2b, respectively. For kn?1 = kn ± 1 they are (a + c)/N (kn?1 ) and (a + c)N (kn )/N (kn?1 ), and for all other kn?1 they vanish. When kn ≤ 1 the contributions from the apex take a di?erent form, but the computation is equally straightforward. Now observe that the quotient in Equation (12) has the same value if the event {Kn?1 = kn?1 } is replaced by {K0 = k0 , . . . , Kn?1 = kn?1 }. The ?rst claim of the lemma follows. The second claim of the lemma is then a straightforward consequence. 2 To prove convergence of the uniform walk to a re?ected Brownian motion, we shall need an estimate on the local time spent by the walk on the boundary vertices of G. This estimate is provided by Lemma 3 below. The proof of Lemma 3 is postponed to Section 4.1, since the proof will give us some intermediate results that we will need only in Section 4. 8

0 ), killed when it reaches row M > 0, and Lemma 3 Consider the walk (Xn let x be a ?xed boundary vertex of G. Then the expected number of visits 0 ) to x before it is killed is smaller than (a + c)?1 . of (Xn

2.3

Identi?cation of the scaling limit

Given a point x in W , the scaling limit of the re?ected random walk started from a vertex near x is obtained in a similar way as described in Section 2.1. That is, if (xN ) is a sequence in G such that |xN ? N σx| ≤ 1/2 for all N ∈ ?, then for every natural number N > 0 we de?ne the continuous (N ) complex-valued stochastic process (Zt : t ≥ 0) as the linear interpolation of the process 1 (N ) X xN2 (13) = Yt N σ ?N t? making jumps at the times {k/N 2 : k = 1, 2, . . .}. Here, σ 2 is again the variance of the real and imaginary parts of the steps Sn = Xn ? Xn?1 of the random walk in the interior of the wedge. Then it is clear that the random walk converges to a complex Brownian motion in the interior, but the behaviour on the boundary is non-trivial. It is explained in Dub? edat [4] how one proves that the scaling limit of x the random walk (Xn ) is in fact re?ected Brownian motion with constant re?ection angles on the two sides, started from x. The core of the argument x) identi?es re?ected Brownian motion as the only possible weak limit (Zt (N ) of (Zt ). We repeat this part of the argument below, mainly to allow us in Section 3 to point out some technical issues that need to be taken care of when we generalize to arbitrary triangles. We use the submartingale characterization of re?ected Brownian motion 2 (W ) be the in W (see [15], Theorem 2.1), which states the following. Let Cb set of bounded continuous real-valued functions on W that are twice continuously di?erentiable with bounded derivatives. Assume that the two re?ection angles ?L and ?R on the sides of W are given. Then the RBM?L ,?R in W x) started from x ∈ W is the unique continuous strong Markov process (Zt 2 (W ) with nonnegative derivain W started from x such that for any f ∈ Cb tives on the boundary in the directions of re?ection, the process
x f (Zt )?

1 2

t 0

x ?f (Zs ) d s,

(14)

where ? is the Laplace operator, is a submartingale. To prove convergence of our random walk to an RBM, it is therefore su?cient to show that there are two angles ?L and ?R such that if f is a 9

function as described above, then for all 0 ≤ s < t
N) lim inf E( f (Zt ) ? f (Zs ) ? x N →∞

1 2

t s

?f (Zu ) d u ≥ 0.

(15)

(N ) Here, Ex denotes the expectation operator for the N th approximate process started near x, and we have dropped the (N ) superscript on (Zt ) to simplify the notation. Note that the conditioning in the submartingale property is taken care of because the starting point x is arbitrary. In fact, it is su?cient to verify (15) up to the stopping time τ := inf {t ≥ 0 : Im Zt ≤ ?M } for some large number M . We will make use of this to have uniform bounds on the error terms in our discrete approximation. 2 (W ), and write u = k/N 2 We now show (15). Let f be a function in Cb k for the jump times of (Zt ). Then by Taylor’s theorem, there exist (random) times Tk between uk and uk+1 such that

f (Zuk+1 ) ? f (Zuk ) =

1 ?x f (Zuk )Uk+1 + ?y f (Zuk )Vk+1 + Nσ

1 2 2 2 ? 2 f (ZTk )Uk +1 + ?y f (ZTk )Vk +1 + 2?x ?y f (ZTk )Uk +1 Vk +1 , (16) 2N 2 σ 2 x where Uk and Vk are the real and imaginary parts of the kth step of the underlying random walk, as before. When we now apply the expectation operator, we distinguish between the behaviour on the boundary and in the interior (we may ignore what happens at the apex because of the negligible time spent there, see below). If B denotes the set of boundary vertices of G, then we may write
N) (N ) E( f (Zuk+1 ) ? f (Zuk ) = Ex x

1 ?f (ZTk ) + 2N 2

1 1 ?x f (Zuk )Uk+1 + ?y f (Zuk )Vk+1 + O(N ?2 ) Nσ Nσ

?{N σZuk ∈B} , (17)

where the O(N ?2 ) error term is uniform in {z ∈ W : Im z > ?M }. Summing over k from N 2 (s ∧ τ ) to N 2 (t ∧ τ ) yields 1 ?f (Zu )d u + o(1) + s∧τ 2 1 z 1 z (N ) [Lz ] Ex ?x f ( ) Ez [U1 ] + ?y f ( ) Ez [V1 ] + O(N ?2 ) .(18) N σ N σ N σ N σ z ∈B
N) (N ) E( x [f (Zt∧τ ) ? f (Zs∧τ )] = Ex t∧τ

In the last line, Lz is the local time at z (the number of jumps of the random walk to z ) between times s ∧ τ and t ∧ τ , and Ez is expectation with respect 10

0 Wα,β α α β β

Figure 3: The thick arrows in this ?gure represent the directions of re?ection of the re?ected Brownian motion described in the text. The re?ection angles are such that the Brownian motion will hit the dashed line with uniform distribution. to the underlying random walk started from z . Lemma 3 shows that when the starting point of the walk is 0, the expected local time spent on the boundary vertices up to time τ is of order O(N ). Then one can use the Markov property to see that this is enough for us to ignore the O(N ?2 ) term in the limit. Observe that the remaining term in braces is just the derivative of f at the boundary point z/N σ along the direction of Ez [S1 ]. Thus, if the function f has nonnegative derivatives along this direction on the boundary of W , then the term in braces in equation (18) is nonnegative, and the desired result (15) is obtained. This proves that the scaling limit of the re?ected random walk is a re?ected Brownian motion, and that the angles of re?ection with respect to the left and right sides are given by ?L + α = arg (? Ez [S1 ]) and ?R + β = ? arg (Ey [S1 ]) , (19)

where z is any left boundary vertex and y any right boundary vertex. It is clear from the computation that this result is valid generally for any choice of transition probabilities on the left and right sides. We shall however focus again on the special case of the uniform walk introduced in Section 2.2. The computation of the re?ection angles for the uniform walk is straightforward. By symmetry it is enough to compute only ?R . From the transition probabilities one may verify that cot(?R + β ) = ? (a ? b) cot ψ ? (a + b) cot ? Ez [Re S1 ] = . Ez [Im S1 ] 2a cot2 ψ ? 1 = cot(2ψ ) = cot(2β ). 2 cot ψ 11 (20)

Substituting equations (4)–(6) then yields cot(?R + β ) = (21)

Thus, the angle of re?ection of the Brownian motion with respect to the right side is simply ?R = β . Similarly, the angle of re?ection on the left side is given by ?L = α. We conclude that the scaling limit of the uniform random walk de?ned in Section 2.2 is a re?ected Brownian motion with ?xed re?ection angles on the two sides of the wedge. The angles of re?ection are α and β with respect to the left and right sides, respectively, as illustrated in Figure 3. It follows from Lemma 2 that the RBMα,β has the special property that in the wedge Wα,β , the RBM ?rst arrives on any horizontal cross-section of the wedge with the uniform distribution. By a simple translation it follows that in the triangle Tα,β , the RBMα,β started from the top wα,β will land on [0, 1] with the uniform distribution. This completes the proof of Theorem 1 for angles α, β ∈ (0, π/2] that satisfy π/2 ≤ α + β < π . The case of arbitrary angles will be considered in the following section.

3

RBM in a generic triangle

In the previous section we proved Theorem 1 for angles α, β ∈ (0, π/2] such that π/2 ≤ α + β < π . Here we generalize to generic triangles. That is, we now choose the angles α and β arbitrarily in the range (0, π ) such that α + β < π . These values of α and β are assumed ?xed for the remainder of this section. As in Section 2, we will de?ne a uniform walk in the wedge Wα,β and identify the scaling limit as a re?ected Brownian motion.

3.1

Choice of the lattice

The ?rst thing that we have to do is to ?nd a lattice covering the wedge W = Wα,β in a nice way. We will make use of the triangular lattices Γ?,ψ introduced in the previous section. Consider such a lattice Γ?,ψ , and choose two vertices xL and xR on the ?rst row of the lattice such that xR is to the right of xL . Then the two half-lines {txL : t ≥ 0} and {txR : t ≥ 0} de?ne the left and right sides of a wedge that is covered nicely by vertices of Γ?,ψ . We will show below that for any given wedge W there is a choice of the lattice angles ?, ψ and the two points xL and xR such that the wedge thus de?ned coincides with W . We start by introducing some notation. Given the lattice Γ?,ψ and the two vertices xL and xR , we may introduce two integers nL and nR that count the positions of xL and xR on the ?rst row of the lattice. More precisely, the integers nL and nR are de?ned such that xL = ?u ? nL (u + v ) ? ih and xR = v + nR (u + v ) ? ih (for an illustration, see Figure 4). Conversely, given 12

0 h nR = ? 3 or nL = 2 u v nR = 1 or nL = ?2

nR = ? 2 or nL = 1

nR = ? 1 or nL = 0

nR = 0 or nL = ?1

Figure 4: Di?erent wedges can be covered by the same triangular lattice, by changing the directions of the two sides as shown. These directions can be expressed in terms of integers nL and nR , as explained in the text. Each thick line in the ?gure can represent either the right side of a wedge (for which the corresponding value of nR is given in the ?gure), or the left side of a wedge (for which the corresponding value of nL is given). two integers nL and nR and the lattice Γ?,ψ , the vertices xL and xR are ?xed by these equations. Observe that nL and nR must satisfy nL + nR ≥ 0 to make sure that xR lies to the right of xL . Figure 4 shows the wedges one obtains for di?erent choices of nL and nR on a given lattice. The main claim of this subsection is that for any choice of the angles α and β , there is a choice of integers nL , nR and of the lattice angles ?, ψ such that the wedge one obtains as described above coincides with the wedge Wα,β . This results is a direct consequence of Lemma 4 below. The proof of the lemma, which will give explicit formulas for nL , nR and ?, ψ in terms of the angles α and β , is postponed to the end of this subsection. Lemma 4 Let α, β ∈ (0, π ) such that α + β < π . Then there is a choice of (possibly negative) integers nL and nR with nL + nR ≥ 0, and angles ?, ψ ∈ (0, π/2] with π/2 ≤ ? + ψ < π , such that cot α = nL (cot ? + cot ψ ) + cot ?; cot β = nR (cot ? + cot ψ ) + cot ψ. (22) (23)

We remind the reader that in Section 2 we considered a random walk on a wedge with angles α, β ∈ (0, π/2] such that π/2 < α + β < π . The situation 13

described there corresponds to the special choice of nL = nR = 0, ? = α and ψ = β in Lemma 4. Thus, the random walk construction of Section 2 is a special case of the more general construction we are considering in this section. We would also like to remark at this point that the choice of nL , nR and ?, ψ in Lemma 4 is not unique in general. For instance, we will see in the proof of Lemma 4 that it is always possible to choose the angles ?, ψ in the range [π/4, π/2]. Thus, there are angles α, β such that the choices made in the proof of the lemma and in the construction of Section 2 are di?erent. From now on, we will assume that the values of ? and ψ are ?xed as in Lemma 4, so that the lattice Γ = Γ?,ψ is ?xed. As we explained above, the lattice provides a nice covering of the wedge W = Wα,β . Following the conventions introduced in Section 2, we shall denote by G = Gα,β the set of vertices obtained by taking the intersection of Γ with W . We again subdivide the set G into the apex, interior vertices and left and right boundary vertices. We now conclude this subsection with the proof of Lemma 4. Proof of Lemma 4: Then we can take Assume ?rst that both α and β are smaller than nR = ?cot β ? ? 1,
π 2.

nL = ?cot α? ? 1,

(24)

and solve equations (22) and (23) for ? and ψ to obtain nR + 1 nL cot α ? cot β ; (25) nL + nR + 1 nL + nR + 1 nR nL + 1 cot β ? cot α. (26) cot ψ = nL + nR + 1 nL + nR + 1 Observe that since nL < cot α ≤ nL + 1 and nR < cot β ≤ nR + 1, the angles ? and ψ are in the range [π/4, π/2]. It remains to consider the case when either α or β is at least π/2, and by symmetry it su?ces to assume α ≥ π/2. Then, we can for instance set k = ?cot α + cot β ?, and let l be the smallest positive integer such that l/(k + l) > ? cot α/ cot β . We then set cot ? = nL := ?l, nR := k + l ? 1, (27)

and the angles ? and ψ are given by the equations (25) and (26) as before. From the inequalities cot α l?1 l >? ≥ , k+l cot β k+l?1 (28)

plus the fact that k ≥ cot α + cot β , it follows that ? and ψ are in the range [π/4, π/2]. This completes the proof. 2 14

3.2

Discussion of the random walk construction

Now that we have chosen the lattice on the wedge W , the next task is to x ) on G whose scaling limit is re?ected Brownian de?ne a random walk (Xn motion. As in Section 2.2, we will focus on the special case of the uniform walk, i.e., the random walk on the lattice that stays uniform on the rows all the time. Earlier we explained how one de?nes this random walk in the case nL = nR = 0. In this subsection we will describe what one has to do to generalize to other wedges, i.e., to the case where nL or nR or both are nonzero. In the following subsection we will then spell out the transition probabilities of the uniform walk for this general case. It is clear that we should de?ne the transition probabilities from the interior vertices in the same way as before. This will guarantee that the random walk will converge to Brownian motion in the interior of the wedge. The nontrivial task is to de?ne the transition probabilities from the boundary vertices and at the top of the wedge. Once these are de?ned, we will use the strategy of Section 2.3 to identify the scaling limit as a re?ected Brownian motion. Let us therefore reconsider the arguments used there to identify the scaling limit, to see if they still apply to general wedges. Looking back at Section 2.3 we see that it is still su?cient to verify Equation (15) for appropriate functions f . Moreover, Equation (18) is still valid after making a Taylor expansion. However, we emphasize that in the previous section the sum over the boundary vertices picked up contributions from only one left boundary vertex and one right boundary vertex on every row of the lattice. This is no longer the case in the general situation, as we shall now discuss. Remember that the boundary vertices are de?ned as those vertices having less than six nearest-neighbours in G. Now consult Figure 4. Then we see that for a given value of nL , the number of left boundary vertices on each row of the lattice is ?xed and equals NL = |nL | + 1{nL ≥0} . Likewise, for given nR the number of right boundary vertices on every row is NR = |nR | + 1{nR ≥0} . (An observant reader will notice that if nL or nR is negative, then on the ?rst few rows of the lattice the number of boundary vertices is less than NL + NR . This is a complication that we will deal with later on.) We conclude that the sum in Equation (18) picks up NL contributions from the left boundary vertices in every row. Each of these contributions is proportional to the derivative of f near the boundary in the direction of Ez [S1 ]. We have to guarantee that these contributions add up to something positive if f has positive derivative along some (yet to be determined) ?xed 15

direction on the left side of the wedge. To do so, it is clearly su?cient to impose that Ez [S1 ] is the same at all the left boundary vertices, and this is what we shall do. The corresponding condition will be imposed at the right boundary vertices. It is clear from the discussion in Section 2.3 that the scaling limit of the random walk will be a re?ected Brownian motion. Moreover, the directions of re?ection on the two sides of W will be given by the values of Ez [S1 ] at the left and right boundary vertices. Let us remark that the condition that Ez [S1 ] is the same at all the boundary vertices may be more than necessary, but by imposing this condition we avoid having to estimate the di?erences in expected local time at di?erent boundary vertices. In summary, we will look for transition probabilities from the boundary vertices that satisfy the following conditions: 1. The summed transition probability to a given vertex from vertices in the row above is a + c, and likewise from vertices in the row below. 2. The summed transition probability to a given vertex from vertices in the same row is 2b. 3. The expected value of the ?rst step of the walk from every left boundary vertex is the same, and likewise for the right boundary vertices. Observe that condition 1 introduces an up-down symmetry which is not inherent in the geometry of the problem, but will be of importance later. Explicit expressions for all the transition probabilities of the uniform walk will be given in the following subsection. This will show that for any nL , nR it is possible to choose the transition probabilities such that they satisfy conditions 1–3. To conclude this subsection, we point out one further complication that we already commented on earlier. This complication arises when either nL or nR is negative, because then the ?rst few rows of the lattice will contain less than NL + NR vertices in total. Hence we will have to specify separately the transition probabilities on these ?rst rows of the lattice. As we shall discuss below, this complication is quite easy to handle. Consider once again Figure 4 and recall the de?nition of the integers nL and nR in Section 3.1, illustrated in the ?gure. Then one notes that each row of the lattice contains exactly nL + nR + 1 vertices more than the row above. In other words, the total number of vertices on row k of the lattice is N (k) = (nL + nR + 1)k + 1. From this one can compute that if either nL or nR is negative, then the ?rst row of the lattice that contains at least all 16

of the NL + NR boundary vertices is row k0 , where k0 := |nL ? nR | . nL + nR + 1 (29)

For example, for nL = 2 and nR = ?1 the numbers of left and right boundary vertices are NL = 3 and NR = 1, respectively. However, the ?rst two rows of G contain only N (0) = 1 and N (1) = 3 vertices in total, so that the ?rst row of the lattice that contains at least NL + NR vertices is row 2. The remaining problem is then to de?ne the transition probabilities from the vertices in the ?rst k0 rows of the lattice. It is not very di?cult to choose the transition probabilities on these ?rst few rows such that the walk will trivially stay uniform on the rows of the lattice. In fact, we can choose the transition probabilities such that conditions 1 and 2 above are also satis?ed at rows 1 up to k0 ? 1, as we shall see below. This guarantees that the walk does indeed stay uniform on rows. We can not make condition 3 hold on the ?rst k0 rows. This is no problem, since this condition was only introduced to prove convergence to re?ected Brownian motion, as we discussed above, and the local time spent at the ?rst k0 rows is negligible.

3.3

Transition probabilities and scaling limit

In the previous subsection we described in words how one should de?ne the transition probabilities of the uniform walk in the generic wedge W = Wα,β . Here we shall complete the uniform walk construction by providing explicit expressions for all of the transition probabilities. We then complete the proof of Theorem 1 by computing the directions of re?ection for the RBM obtained in the scaling limit. x = x. In the interior of To get us started, given a vertex x of G we set X0 x the wedge we de?ne the transition probabilities as before. That is, when Xn x is an interior vertex, Xn+1 is to be chosen from the six neighbouring vertices with the probabilities a, b and c as in Figure 1. At the apex, that is, if x = 0, the walk moves to either of the vertices on row 1 of the lattice with Xn probability 1/N (1). If nL and nR are both nonnegative, then it only remains to specify the transition probabilities from the boundary vertices. However, when either nL or nR is negative a little more work needs to be done near the top of the wedge, as explained in the previous subsection. In that case, we recall the de?nition of row k0 from the previous subsection. For every k = 1, . . . , k0 ? 1, we set the transition probability from each vertex in row k to each vertex in rows k ± 1 equal to (a + c)/N (k). Furthermore, for every vertex in row k 17

that have both a left and a right neighbour, the transition probabilities to these two neighbours are equal to b. To the two vertices on row k on the sides of W (that have only one neighbour) we assign transition probabilities to the single neighbour and to the vertices themselves equal to b. This takes care of all the nonzero transition probabilities near the top of the wedge. It remains to de?ne the transition probabilities from the boundary vertices. As we explained in the previous subsection, we are looking for transition probabilities that satisfy conditions 1–3 on page 16. Below we shall give explicit expressions for these transition probabilities from the left boundary vertices for arbitrary nL > 0. This is in fact su?cient to allow us to derive the transition probabilities for all possible wedges (i.e. for all combinations of nL and nR ), as we shall explain ?rst. Observe that by left-right symmetry we can derive the transition probabilities from the right boundary vertices for any given nR , if we know the corresponding transition probabilities from the left boundary vertices for nL = nR . Because the case nL = 0 was already covered in Section 2.2, it only remains to show that one can obtain the transition probabilities from the left boundary vertices for negative nL from those for positive nL . To do so, observe from Figure 4 that the left side of a wedge W with given nL < 0 coincides with the right side of a (di?erent) wedge W ′ with nR = ?nL ? 1. We propose that at the j th vertex on row k of W one can take the probability of a step S equal to the probability of the step ?S at the j th vertex on row k of W ′ , counted from the right side. Here we exploit the up-down symmetry inherent in condition 1 on page 16. It follows that it is indeed suf?cient to provide the transition probabilities from the left boundary vertices for positive nL only. So let nL > 0 be ?xed. To specify the transition probabilities we will need some notation. We write p0 [j, l] for the transition probability from the j th vertex on a row k to the lth vertex on the same row. By p± [j, l] we denote the transition probability from the j th vertex on a row k to the lth vertex on the row k ± 1 (vertices on a row k are numbered 0, 1, . . . , N (k) ? 1 from left to right). If nR and nL are both nonnegative, then these transition probabilities are to be used at all rows k > 0 of the lattice, otherwise they are valid for the rows k ≥ k0 . Finally, we write qj [S ] for the probability of a step S from the j th vertex on a row k. Remember from the previous subsection that there are NL = nL + 1 left boundary vertices on the rows of G, so that we have to give the transition probabilities for j = 0, 1, . . . , nL . First we specify the transition probabilities from a given row to the row

18

above. For all j = 0, 1, . . . , nL , p? [j, 0] = qj [u + (u + v )(nL ? j ) + ih] = 1 a. nL + 1 (30)

Second, for all j = 0, 1, . . . , nL the transition probabilities to the same row are given by p0 [j, j + 1] = qj [u + v ] = j+1 b, nL + 1 j p0 [j, j ? 1] = qj [?(u + v )] = b, nL + 1 2(nL ? j ) + 1 b. p0 [j, j ] = qj [0] = nL + 1 (31) (32) (33)

Third, specifying the transition probabilities to the row below is a bit more complicated. For each of the boundary vertices j = 0, 1, . . . , nL we have p+ [j, 2nL + 1] = qj [?u + (u + v )(nL + 1 ? j ) ? ih] = However, for j = 0 we have p+ [0, nL ] = qj [?u ? ih] = (a + c)nL , nL + 1 p+ [0, 0] = qj [?u ? (u + v )nL ? ih] = a + c, (35) (36) c . nL + 1 (34)

whereas for j = 1, 2, . . . , nL ? 1, (a + c)nL , nL + 1 (a + c)nL , p+ [j, nL + j ] = qj [?u ? ih] = nL + 1 a+c , p+ [j, 2j ] = qj [?u ? (u + v )(nL ? j ) ? ih] = nL + 1 p+ [j, j ] = qj [?u ? (u + v )nL ? ih] = and ?nally, for j = nL we have (40) a+c p+ [nL , 2(nL ? n) ? 1] = qj [?u ? (u + v )(2n + 1) ? ih] = . (41) nL + 1 In the last equation, n is an integer taking values in {0, 1, . . . , nL ? 1}. The above list speci?es all the nonzero transition probabilities from the left 19 p+ [nL , 2nL ] = qj [?u ? ih] = a + c, (37) (38) (39)

sin ψ h ? u

sin ? 1/3 ψ v 3b/2 a+c
a+c 2

0 1/3 1/3 a/2 b/2 a/2 b/2 b c/2 a+c c b a b a+c

Figure 5: Transition probabilities for the re?ected random walk in a wedge with nL = 1 and nR = 0. The inset shows the lattice dimensions. boundary vertices. Figure 5 shows an example of the transition probabilities in the case nL = 1. We deliberately gave both the transition probabilities and the corresponding step probabilities at the boundary vertices, to make it easy to verify that conditions 1–3 on page 16 are indeed satis?ed. To verify conditions 1 and 2, one simply has to add up the relevant transition probabilities. To check condition 3, one may compute the real and imaginary parts of the ?rst step of the random walk from each of the nL + 1 boundary vertices from the step probabilities (the result is given in the following paragraph). As we explained in the previous subsection, the random walk de?ned above will converge in the scaling limit to a re?ected Brownian motion. Moreover, the walk is uniform by construction and satis?es Lemma 2. Thus, the RBM will hit any horizontal cross-section of W with the uniform distribution. The proof of Theorem 1 will therefore be complete if we can show that the angles of re?ection of the RBM are ?L = α and ?R = β . As in Section 2, the direction of re?ection with respect to the left side of W is given by the expected value of the ?rst step from a left boundary vertex. From the step probabilities we compute Ez [Re S1 ] = 1 c(v ? u) + b(u + v ) nL + 1 ? (a + c)(2nL u + n2 L (u + v )) , Ez [Im S1 ] = 1 ? h(2c + 2(a + c)nL ) , nL + 1 20 (42) (43)

where z is any left boundary vertex. Observing that u/h = cot ? and v/h = cot ψ , this gives us the following expression for the re?ection angle ?L with respect to the left side of the wedge:
2 cot ?[(a + c)(n2 L + 2nL ) ? b + c] + cot ψ [(a + c)nL ? b ? c] . 2c + 2(a + c)nL (44) Substituting Equations (4)–(6) yields

cot(?L + α) =

cot(?L + α) =

[(cot ? + cot ψ )nL + cot ?]2 ? 1 . 2(cot ? + cot ψ )nL + 2 cot ?

(45)

According to Equation (22), the result simpli?es to cot(?L + α) = cot2 α ? 1 = cot(2α). 2 cot α (46)

Thus, the re?ection angle of the Brownian motion with respect to the left side is simply ?L = α. By symmetry, the angle of re?ection with respect to the right side is β . This is consistent with the results obtained in Section 2 and completes the proof of Theorem 1.

4

Properties of the RBMs

This section reviews several properties shared by the re?ected Brownian motions with constant re?ection angles, and sheds some light on the RBMα,β when α + β ≥ π . The properties discussed below will be used in the following section to derive distribution functions for the RBMs.

4.1

Intertwining relations for the uniform walk

From the de?nitions of the uniform walks in Sections 2 and 3 the following interesting picture arises. Started with the uniform distribution from the vertices in a given row of the lattice, the walk can be seen as a walk from row to row on the lattice that remains uniform on each row all the time. This can be stated more precisely in the form of an intertwining relation, as was noted for the case of symmetric wedges by Dub? edat [4]. Here we shall describe the generalization of his result to generic wedges. First let us explain what is meant by an intertwining relation. Let (Pt : t ≥ 0) and (Pt′ : t ≥ 0) be two Markovian semigroups with discrete or continuous time parameter and corresponding state spaces (S, S ) and (S ′ , S ′ ). Suppose that Λ is a Markov transition kernel from S ′ to S . 21

That is, Λ is a function Λ : S ′ × S → [0, 1] such that (1) for each ?xed x′ ∈ S ′ , Λ(x′ , · ) is a probability measure on S , and (2) for each ?xed A ∈ S , Λ( · , A) is S ′ -measurable. The two semigroups (Pt ) and (Pt′ ) are said to be intertwined by Λ if for all t ≥ 0 and every pair (a′ , A) ∈ S ′ × S , Λ(a′ , d x)Pt (x, A) = Pt′ (a′ , d x′ )Λ(x′ , A). (47)

In a more compact notation, the semigroups are intertwined if for all t ≥ 0 the identity ΛPt = Pt′ Λ between Markov transition kernels from S ′ to S holds. Examples of such intertwining relations have been studied before, see for instance [2, 13]. In our case, we are interested in the uniform random walk (Xn ) on the graph G = Gα,β covering the wedge W = Wα,β , as de?ned in Section 3. ′ ) be the semigroup of the random Let (Pn ) be its semigroup, and let (Pn walk on the set ? of natural numbers with transition probabilities p′ (k, k) = 2b (k ±1) p′ (k, k ± 1) = (a + c) NN (k ) (48)

for k > 0 and p′ (0, 1) = 1. Observe that p′ (k, l) is just the conditional probability that Xn+1 will be on row l of G, given that Xn is uniformly distributed on the vertices of row k. Now consider the Markov transition kernel Λ from ? to G such that for each k ∈ ?, Λ(k, · ) is the uniform measure on the vertices of row k of G. Recall that for the walk (Xn ), if X0 is uniformly distributed on row k, the walk will stay uniform on the rows of G afterwards (Lemma 2). It follows that for each k ∈ ?, A ? G and n ≥ 0, Λ(k, x)Pn (x, A) =
x∈G l∈

?

′ Pn (k, l)Λ(l, A).

(49)

′ Λ. The reader should Hence, we have the intertwining relation ΛPn = Pn compare this with the statement and proof of Lemma 2. This discrete intertwining relation may be used to compute the Green 0 ), killed when it reaches the row M > 0. function for the uniform walk (Xn The computation of this Green function gives us the expected local time spent by the walk on a given boundary vertex of G, and thereby proves Lemma 3. ′ ?1 = Proof of Lemma 3: Consider the Green function G′ M = (I ? PM ) ∞ ′ ′ n n=0 (PM ) for the random walk on ? with transition matrix PM as introduced in Equation (48), except that now the walk is killed as soon as it

22

reaches level M . It is not di?cult to verify that the ?rst row of this Green function is given by G′ M [0, k ] =
? ? ?
1 (N (1)?1)(a+c) N (k ) a+c N (1) ′ [0, 1] gM

1?

N (k ) N (M )

for k > 0; (50) for k = 0.

0 ) on the Now consider the Green function GM for the uniform walk (Xn graph G in W , killed when it reaches {z : Im z = ?M h}. Let us denote the j th vertex on row k of G by (j, k). Then, by the intertwining relation (49) (or by Lemma 2), the Green function GM is related to G′ M by

? ? 1+

GM [(0, 0), (j, k)] =

1 G′ [0, k] N (k) M

for j = 0, 1, . . . , N (k) ? 1,

(51)

0 ) before since the expected local time spent at vertex (j, k) by the walk (Xn it is killed at row M is the same for all j = 0, 1, . . . , N (k) ? 1. It follows that the expected local time spent at any vertex (j, k) by the walk before it is killed at row M is smaller than (a + c)?1 . 2

4.2

Intertwining relations and time reversal of the RBMs

In the previous subsection we considered the intertwining relation between the uniform walk on G and a random walk on the integers. Here we will turn our attention to the scaling limit. This time, let (Zt ) be the RBMα,β in W = Wα,β , and let (Pt ) be its semigroup. Consider the Markov transition kernel Λ from the positive reals ?+ to W which for each ?xed y ∈ ?+ assigns the uniform measure to the horizontal interval [?y cot α ? iy, y cot β ? iy ]. It is clear from the intertwining relation (49) between the random walks, that (Pt ) and the semigroup of the scaling limit of the random walk on ? will be intertwined by Λ. It remains to identify this scaling limit. We claim that this is a 3-dimensional Bessel process, or in other words, it is a Brownian motion on ?+ conditioned not to hit the origin. This generalizes Proposition 1 in Dub? edat [4] to the following statement: Theorem 5 Let (Zt ), (Pt ) and Λ be as above and let (Pt′ ) be the semigroup of the 3-dimensional Bessel process taking values in ?+ . Then (Pt ) and (Pt′ ) Λ(y, · ) are intertwined by Λ. In particular, for all y ≥ 0, the process ? Im Zt is a 3-dimensional Bessel process started from y .

23

Proof: As we remarked above, it is su?cient to identify the scaling limit ′ ) on ?. Since the rows of the lattice have spacing h, of the random walk (Xn the proper scaling limit is obtained by considering the linear interpolations ′ 2 2 of the processes hX? n2 t? /nσ where σ = 2(a + c)h is as before, and then taking n → ∞. We may derive the in?nitesimal generator for the limit by computing, for a su?ciently di?erentiable function f on ?+ , h h a+c f x+ N (nσx/h + 1) ? f x ? N (nσx/h ? 1) N (nσx/h) nσ nσ 1 1 1 ′ f (x) + f ′′ (x) + o(n?2 ). (52) + 2bf (x) ? f (x) = 2 n x 2 Here, 1/n2 is the time scaling, and we recognize the generator of the 3dimensional Bessel process (see Revuz and Yor [12] Chapter VI, §3 and Chapter III, Exercise (1.15) for background on the 3-dimensional Bessel process and its semigroup). 2 We now turn our attention to the time-reversal of the RBMs. Precisely, let (Zt ) be an RBMα,β in the triangle T = Tα,β started from the top, and ?t ) of stopped when it hits [0, 1]. We are interested in the time-reversal (Z this process. From the time-reversal properties of the 3-dimensional Bessel ?t ) ?rst hits the process [12, Proposition VII(4.8)] we know that until (Z boundary of T , it is a complex Brownian motion started with the uniform distribution from [0, 1] and conditioned not to return to [0, 1]. In fact, we can ?t ) in terms of a conditioned re?ected Brownian describe the full process (Z motion. This is again a generalization of an earlier result of Dub? edat [4, Proposition 2]: Theorem 6 The time-reversal of the RBMα,β in the triangle T = Tα,β , started from the top and stopped when it hits [0, 1], is the RBMπ?α,π?β in T started with the uniform distribution from [0, 1], conditioned not to return to [0, 1], and killed when it hits the top of the triangle.
0 ) of EquaProof: Recall the Green function of the uniform walk (Xn tion (51). By Nagasawa’s formula (Rogers and Williams [14, III.42]) the time-reversal of this random walk is a Markov process with transition probabilities

qM [(j, k), (l, m)] =

GM [(0, 0), (l, m)] p[(l, m), (j, k )] . GM [(0, 0), (j, k)]

(53)

24

0) Here, the p[(l, m), (j, k)] are the transition probabilities for the walk (Xn killed at row M , as speci?ed in Section 3.3. We are interested in the transition probabilities for the reversed process in the limit M → ∞. From the expression for the Green function it is clear that in the limit one gets for k, m > 0 q [(j, k), (l, m)] = p[(l, m), (j, k)]. (54)

Observe in particular that in the interior of the wedge we recover the transition probabilities of the original walk. Hence, the reversed walk converges to a Brownian motion in the interior. Moreover, by condition 1 on page 16 it is clear that at every vertex on any row k > 0 of the lattice, the probability that the reversed walk will step to the row k + 1 is equal to the probability to step to the row k ? 1. Note especially that this is not only true at the interior vertices but also at the boundary vertices. Therefore, the expected step of the random walk from any vertex on the rows k > 0 is real. In particular, on the sides of W the walk must receive an average re?ection in the real direction toward the interior of W . It follows that the time-reversal of the re?ected Brownian motion in the wedge is a re?ected Brownian motion with re?ection angles π ? α and π ? β with respect to the sides of W . 2

4.3

Re?ected Brownian motions for α + β ≥ π

The fact that the RBMα,β is intertwined with a three-dimensional Bessel process sheds some light on the behaviour of the re?ected Brownian motions with re?ection angles satisfying α+β ≥ π . It is the purpose of this subsection to look at these RBMs more closely. For the duration of this subsection we will ?x α, β ∈ (0, π ) such that α + β ≥ π . Then an RBMα,β in the domain T = Tα,β may be described by considering an RBMπ?α,π?β in the wedge Wπ?α,π?β and putting it upside-down, as we shall explain below. We set x := cos α sin β/ sin(α + β ) and z := ? sin α sin β/ sin(α + β ). Let (Zt ) denote RBMπ?α,π?β in the wedge Wπ?α,π?β , and consider in particular Λ(y +z, · ) the process Zt with y ∈ ?+ . Here, Λ is the Markov transition kernel from ?+ to Wπ?α,π?β introduced in the previous subsection. Let f be the transformation f : w → w ? + x ? iz which puts the wedge upside-down Λ(y +z, · ) as illustrated in Figure 6, and set Yty := f Zt . Then the process y (Yt ), stopped when it hits the interval [0, 1], is an RBMα,β in Tα,β started with the uniform distribution from the horizontal line segment at altitude y . From the intertwining relation of Theorem 5 we conclude that the imaginary part of (Yty ) is a Brownian motion conditioned not to hit ?z . It 25

y α 0 x z β 1

Figure 6: By putting a wedge upside-down we can shed some light on the RBMα,β when α + β ≥ π . follows (see [12, Corollary VI(3.4)]) that (Yty ) has positive probability to never reach [0, 1]. In fact, the probability that (Yty ) does reach [0, 1] is z/(z + y ). It is furthermore clear from Theorem 5 that given the event that (Yty ) does reach [0, 1], it will arrive there with the uniform distribution. Now let Py denote probability with respect to the process (Yty ). Then by what we said above, (y/z + 1) Py is a probability measure on re?ected Brownian paths in Tα,β started from the horizontal line segment at altitude y that end on [0, 1]. Taking the limit y → ∞ we obtain a conformally invariant probability measure on paths of RBMα,β in Tα,β that start “with the uniform distribution from in?nity” and arrive on [0, 1] with the uniform distribution. Henceforth, when we speak about RBMα,β with α + β ≥ π , we shall always assume that we restrict ourselves to this collection of Brownian paths and the corresponding probability measure introduced above.

4.4

Conformal invariance and locality

Two elementary properties shared by the RBMs are conformal invariance and the locality property . To explain what we mean by these properties, let us ?rst consider an RBMα,β in the wedge W = Wα,β started from the x ), and let v := exp((2α ? π )i) and point x ∈ W . Call this process (Zt 1 v2 := exp(?2β i) denote the re?ection ?elds on the left and right sides of W . Then it is well-known (compare with Equation (2.4) in [15]) that we can uniquely write x x Zt = Bt + v1 Yt1 + v2 Yt2 , (55)
x ) is a complex Brownian motion started from x, and (Y 1 ) and (Y 2 ) where (Bt t t x ) such that are real-valued continuous increasing processes adapted to (Bt

26

w

g:T →D g (0)

w

T

D?T

g (1)

0

1

0

1

Figure 7: The locality property says that the RBMα,β in T started from w and stopped when it exits from the subset D behaves just like an RBMα,β in D started from w.
x is on the left side of W Y01 = Y02 = 0. Moreover, Yt1 increases only when Zt 2 x and Yt increases only when Zt is on the right side of W (Yt1 and Yt2 are x ) on the two sides of W ). essentially the local times of (Zt Now let g be a conformal transformation from W onto a domain D with smooth boundary. Consider the sum N ?1 x x g(Zt ) ? g(Z0 )= j =0 x x g(Z( j +1)t/N ) ? g (Zjt/N ) .

(56)

To compute the sum we use Taylor’s theorem to expand each term. The computation is very similar to the one in Section 2.3. In particular, on the boundary we may keep only the ?rst-order terms. Then letting N → ∞ and using the fact that the real and imaginary parts of g are harmonic, just as in the proof of It? o’s formula (see e.g. Sections 4.2 and 4.3 in Gardiner [7] for a nice discussion) one obtains
x x g(Zt ) ? g(Z0 )= t 0 x x g′ (Zs ) d Bs + v1 t 0 x g′ (Zs ) d Ys1 + v2 t 0 x g′ (Zs ) d Ys2 . (57)

The ?rst integral in (57) is the usual expression for the conformal image t ′ x )|2 d s of Brownian motion. Making the usual time-change u(t) := 0 |g (Zs (see Revuz and Yor [12, Theorem V(2.5)]) and denoting its inverse by t(u), ?u := g(Z x ) is a rewe conclude from Equation (57) that the process Z t(u) ?ected Brownian motion in D with re?ection vector ?elds v1 g′ (g?1 ( · )) and v2 g′ (g?1 ( · )) on the two “sides” of D (i.e. the images of the two sides of W ). Note in particular that because g is an angle-preserving transformation, the ?u ) is also re?ected at the angles α and β with respect to the process (Z boundary of D . This shows that the RBMα,β is conformally invariant. 27

We may use the same reasoning to explain what we mean by the locality property of the RBMα,β . We change the setup to one that will be more useful later. Indeed, we now let (Zt ) be an RBMα,β in the triangle T = Tα,β started from the top w = wα,β , and set τ := inf {t ≥ 0 : Zt ∈ [0, 1]}. Furthermore, we take g to be a conformal map of T onto an open subset D of T that ?xes w, and such that the left and right sides of T are mapped onto subsets of themselves. Finally, we set σ := inf {t ≥ 0 : Zt ∈ T \ D }, the exit time of the RBMα,β from the subset D . See Figure 7 for an illustration. From the calculation above we conclude that up to the stopping time τ , the process g(Zt ) is just a time-changed RBMα,β in the subset D of the triangle T . In particular, modulo a time-change the laws of (g(Zt ) : t ≤ τ ) and (Zt : t ≤ σ ) are the same. This is what is called the locality property. For more background and for consequences of the locality property we refer to the SLE literature [8, 16].

5

Distribution functions

In this section we compute several distribution functions associated with the family RBMα,β of re?ected Brownian motions. We ?x α and β in (0, π ) for the duration of the section, with no further restrictions on α, β (recall that when α + β ≥ π , we assume that we work with the probability measure of Section 4.3). Furthermore, we ?x two angles λ, ? ∈ (0, π ) such that λ + ? < π . These two angles de?ne the domain T = Tλ,? in this section. To be somewhat more general, we will derive distribution functions for the RBMα,β in the triangle T = Tλ,? rather than in the domain Tα,β .

5.1

Notations

First we introduce some notations. We shall call the set of points disconnected from {0, 1} in the triangle T by the path of the RBM up to the time when it hits [0, 1] the hull K of the process. This hull has exactly one point in common with the real line, which we denote by X . We shall denote by |w|Y the distance of the lowest point of the hull on the left side to the top w = wλ,? , and by |w ? 1|Z the distance of the lowest point of the hull on the right side to the top w. Thus, all three random variables X , Y and Z take on values in the range [0, 1]. See Figure 8 for an illustration of the de?nitions. Below we shall compute the marginal and joint distributions of the variables X , Y and Z . We shall see that these can be expressed in terms of the conformal transformations of the upper half of the complex plane onto the 28

w |w|Y K λ 0 X ? 1 |w ? 1|Z

Figure 8: De?nition of the hull K of the re?ected Brownian motion with parameters α and β in the triangle T , and of the random variables X , Y and Z . triangles Tγ,δ . Thus it is useful to review some properties of these transformations ?rst. We simplify the notation somewhat by writing γ ′ as an abbreviation for γ/π whenever γ denotes an angle. By the Schwarz-Christo?el formula of complex analysis (see [1, Section 6.2.4] or [6, Section XI.3]), the unique conformal transformation of the upper half-plane ? onto Tγ,δ that ?xes 0 and 1 and maps ∞ to wγ,δ is given by ′ z γ ′ ?1 Bz (γ ′ , δ′ ) (1 ? t)δ ?1 d t 0 t Fγ,δ (z ) = 1 ′ = , (58) γ ?1 (1 ? t)δ′ ?1 d t B (γ ′ , δ′ ) 0 t where B (γ ′ , δ′ ) = Γ(γ ′ )Γ(δ′ )/Γ(γ ′ + δ′ ) is the beta-function, and Bz (γ ′ , δ′ ) the incomplete beta-function (see e.g. [3] for background on these special functions). Symmetry considerations show that the transformations satisfy
?1 ?1 (1 ? x). (x) = 1 ? Fδ,γ Fγ,δ (u) = 1 ? Fδ,γ (1 ? u) and Fγ,δ

(59)

See Figure 9 for an illustration. A di?erent kind of distribution that we can compute is the conditional probability that the last side of the triangle visited by the RBM, given that it lands at X = x, is the right side. As we shall see, this distribution can also be expressed in term of triangle mappings. In fact, it turns out that there is a remarkable resemblance between this conditional probability and the marginal distribution functions of the variables X , Y and Z .

5.2

Characteristics of the hull

Here we derive the (joint) distribution functions of X , Y and Z , which are characteristics of the hull generated by the RBM. Remember that we are considering an RBMα,β in the triangle T = Tλ,? started from w = wλ,? . For 29

wγ,δ

Fγ,δ

Fδ,γ

wδ,γ

γ 0 x

δ 1 0 u 1?u 1 0

δ

γ 1?x 1

Figure 9: Transformations of the upper half-plane onto triangles. convenience, let us also introduce the angle ν := π ? λ ? ?. Then our main conclusion may be formulated as follows: Proposition 7 Let a = a(y ) and b = b(z ) be the points on the left and right sides of T at distances |w|y and |w ? 1|z from w, respectively. Then the joint distribution of X , Y and Z is given by P[X ≤ x, Y ≤ y, Z ≤ z ] Fα,β = ? Fα,β
?1 (a) ?Fλ,? ?1 ?1 (a) (b) ? Fλ,? Fλ,?

(60)

?1 ?1 (a) (x) ? Fλ,? Fλ,? ?1 ?1 (a) (b) ? Fλ,? Fλ,?

?1 can be expressed in terms where the images of a and b under the map Fλ,? of y and z as 1 ?1 (a) = 1 ? ?1 ; Fλ,? (61) Fν,λ (y ) ?1 (b) = Fλ,?

1 1?
?1 F?,ν (1

? z)

=

. ?1 Fν,? (z )

1

(62)

Note that in the last equation we used the symmetry property (59). Proof: The idea of the computation of P[X ≤ x, Y ≤ y, Z ≤ z ] is illustrated in Figure 10. Consider an RBMα,β in the triangle T started from the top w, and stopped as soon as it hits the counter-clockwise arc from a to b on the boundary (the thick line in the ?gure). Then the probability P[X ≤ x, Y ≤ y, Z ≤ z ] is just the probability that this process is stopped in the interval (0, x). We now use conformal invariance and locality. Let g = ga(y),b(z ) be the conformal map of T onto Tα,β that sends a to 0, b to 1 and w to wα,β , as illustrated in Figure 10. Then the probability that we are trying to compute 30

w |w|y a λ 0 x T |w ? 1|z b ? 1 g a→0 b→1 w → wα,β 0

wα,β

Tα,β α g (0) g (x) β 1

Figure 10: This ?gure illustrates how the joint distribution function of the random variables X , Y and Z can be computed. As explained in the text, the joint probability P[X ≤ x, Y ≤ y, Z ≤ z ] is just g(x) ? g(0). is exactly the probability that an RBMα,β in Tα,β started from wα,β and stopped when it hits [0, 1], is stopped in the interval (g(0), g(x)). But since the exit distribution of the RBM is uniform in Tα,β , this probability is simply g(x) ? g(0). Thus, P[X ≤ x, Y ≤ y, Z ≤ z ] = g(x) ? g(0). (63)

Our next task is to ?nd an explicit expression for this joint probability by deriving the explicit form of the map g = ga(y),b(z ) . The explicit form of g is obtained by suitably combining conformal self-maps of the upper half-plane with triangle mappings. How this is done exactly is described in Figure 11. The expression for the joint distribution follows. 2 By sending one or or two of the three variables x, y and z to 1, and using the symmetry property (59), one may derive the following corollaries of Proposition 7: Corollary 8 We have the following joint distributions:
?1 ?1 (x)Fν,? (z ) ; P[X ≤ x, Z ≤ z ] = Fα,β Fλ,? ?1 ?1 ?1 (y ) ; (1 ? x)Fν,λ (y ) ? Fβ,α F?,λ P[X ≤ x, Y ≤ y ] = Fβ,α Fν,λ

(64) (65) (66)

P[Y ≤ y, Z ≤ z ] = Fα,β + Fβ,α

?1 (z ) Fν,? ?1 ?1 ?1 ?1 (y ) (y ) ? Fν,? (z )Fν,λ Fν,? (z ) + Fν,λ ?1 (y ) Fν,λ

?1 ?1 ?1 ?1 (y ) (y ) ? Fν,? (z )Fν,λ Fν,? (z ) + Fν,λ

? 1.

31

Fλ,? a λ 0 g Fα,β ? x b a ? 1 0 w→ x ? 1

w→ ? b

w ?1 w

Fν,λ

? λ

0 w→
1 1?w

1 F?,ν

0

y

1

w ?a ? ? b?a ?

λ α β 0 g (0) g (x) 1 0 1 0 1 0 ? 1?z 1

Figure 11: This illustration shows schematically how one obtains an explicit form for the map g in terms of the variables y and z . The notations a ?, ? b ?1 ?1 ?1 and x ? in the ?gure are short for Fλ,? (a), Fλ,? (b) and Fλ,? (x). Corollary 9 The marginal distributions of X , Y and Z are given by
?1 (x) ; P[X ≤ x] = Fα,β Fλ,? ?1 (y ) ; P[Y ≤ y ] = Fβ,α Fν,λ ?1 P[Z ≤ z ] = Fα,β Fν,? (z ) .

(67) (68) (69)

Observe that the marginal distributions of the variables X , Y and Z take on particularly simple forms. These marginal distribution functions have a nice geometric interpretation. For instance, P[Y ≤ y ] is just the image of y under the transformation that maps the triangle Tν,λ onto Tβ,α , ?xes 0 and 1 and takes wν,λ onto wβ,α . Similar observations hold for P[X ≤ x] and P[Z ≤ z ]. These observations lead to some intriguing conclusions. First of all, we conclude that for an RBMπ/3,π/3 in an equilateral triangle (so that α = β = λ = ? = ν = π/3) all three variables X , Y and Z are uniform. This is not so surprising when we realize that the hull in this case is the same as that of the exploration process of critical percolation, as we noted in the introduction. Indeed, in the case of percolation X and Y can be interpreted as the endpoints of the highest crossing of a given colour between the sides (0, w) and (0, 1) of the triangle. Thus by symmetry, if X is uniform, then so are Y and Z . 32

w |w|Y K0 α λ 0 X β ? 1 0 λ |w ? 1|Z |w|X ′

w |w ? 1|Y ′
′ K0

β α Z′ ? 1

′ of the shaded sets on the left and right Figure 12: The unions K0 and K0 are generated by di?erent RBMs, but have the same law.

In other triangles, similar but more intricate connections exist. For example, let (Zt ) be an RBMα,β in T = Tλ,? started from w = wλ,? and stopped when it hits [0, 1], as before. Compare this process with an ′ ) in T started from 1, stopped on [0, w] and re?ected on (0, 1) at RBM (Zt an angle α and on (w, 1) at an angle β with respect to the boundary. Here it is assumed that small angles denote re?ection away from 1. To this second RBM we can associate normalized random variables X ′ , Y ′ and Z ′ , measuring the distances of the exit point on [0, w] to w and of the “lowest points” of the hull on [w, 1] and [0, 1] to w and 0, respectively. See Figure 12. It follows from Corollary 9 that X and Z ′ have the same distribution (and so do Y and X ′ ). This result has an interesting interpretation in terms of the hulls generated by the two processes, as we shall now describe. We write ? for the collection of closed, connected subsets C of T such that the right side of T is in C and T \ C is connected. We further de?ne Q as the collection of compact A ? T such that A = A ∩ T , and T \ A is simply connected and contains the right side of T . We then endow ? with the σ -?eld generated by the events {K ∈ ? : K ∩ A = ?} for all A ∈ Q. This setup is similar to the one in Lawler, Schramm and Werner [9, Sections 2 and 3]. In particular, a probability measure P on ? is characterized by the values of P[K ∩ A = ?] for A ∈ Q, see [9, Lemma 3.2].
′ ) stopped on [0, 1] and Theorem 10 Consider the processes (Zt ) and (Zt ′ [0, w] as described above. Let K0 and K0 denote the sets of points in T that ′ ), respectively. Then the are disconnected from 0 by the paths of (Zt ) and (Zt ′ laws of K0 and K0 on the space ? are the same.

33

Proof: Let P be the law of K0 , and let A ∈ Q. Denote by a and b the points of A ∩ ?T closest to w and 1, respectively. Let g : T → T \ A be the conformal transformation that ?xes 1 and w, and maps 0 onto a if Im a > 0, and onto 0 otherwise. Then, by conformal invariance of the RBMα,β in T , P[K0 ∩ A = ?] = P[X ∈ (g?1 (b), 1)] (compare this with our discussion of ′ satis?es the locality property in Section 4.4). Likewise, the law P′ of K0 ′ ′ ′ ? 1 ′ ′ P [K0 ∩ A = ?] = P [Z ∈ (g (b), 1)]. Since X and Z have the same distribution by Corollary 9, the theorem follows. 2

5.3

Last-visit distribution

As we discussed in Section 4.2, the imaginary part of the RBMα,β in Tα,β stopped when it hits [0, 1] is a three-dimensional Bessel process, and so is the imaginary part of its time-reversal. In particular, the time-reversed process, considered up to the ?rst time when it hits the left or right side of the triangle, is a complex Brownian motion started with uniform distribution from [0, 1] and conditioned not to return to the real line. In other words, up to its ?rst contact with the left or right side, this process is a Brownian excursion of the upper half-plane, started with uniform distribution from [0, 1] (for background on Brownian excursions, see [8] and [10]). This fact allows us to derive the following result: Proposition 11 Let (Zt ) be an RBMα,β in the triangle T = Tλ,? started from w = wλ,? . Let τ := inf {t ≥ 0 : Zt ∈ [0, 1]}, and let σ be the last time before τ when (Zt ) visited the boundary of T . Let E denote the event that Zσ is on the right side of the triangle. Then
?1 (x) . P[E | Zτ = x] = Fπ?α,π?β Fλ,?

It is shown in Dub? edat [4] how one derives this result in the special case where α = β = λ = ? = π/3. Here, for the sake of completeness, we repeat the computation for the general case. Proof: We want to use the fact that the time-reversal of the RBMα,β in the triangle Tα,β starts out as a Brownian excursion. So we ?rst map Tλ,? ?1 . This maps the point x onto onto Tα,β by the transformation Fα,β ? Fλ,?
?1 (x) . y := Fα,β Fλ,? Next, let (Yt : t ≥ 0) be a Brownian excursion of the upper half-plane, and let S be the ?rst time when (Yt ) visits either the left or right side of Tα,β .

34

wα,β M z

Bt α ε y

β 1

?

0

Figure 13: A complex Brownian motion Bt started from y + iε in the triangle Tα,β . This process conditioned to exit the strip {z : 0 < Im z < M } through the top boundary and stopped when it hits the boundary of the triangle Tα,β corresponds in the limit M → ∞, ε ↓ 0 to the time-reversal of the RBMα,β in the triangle, as explained in the text. Then P[E | Zτ = x] = lim
ε↓0 1

wα,β

Py+iε [YS ∈ d z ],

(70)

where the integrals runs over the right side of the triangle Tα,β , and Pz denotes probability with respect to the Brownian excursion started from z . Now let (Bt : t ≥ 0) be a complex Brownian motion, let U be the ?rst time when (Bt ) visits either the left or right side of Tα,β , and let UM be the ?rst time when (Bt ) exits the strip {z : 0 < Im z < M }. Suppose that P′ z denotes probability with respect to the Brownian motion started from z . Then, using the strong Markov property of Brownian motion (see Figure 13 for a sketch), we have Py+iε [YS ∈ d z ] =
M →∞

lim P′ y +iε [BU ∈ d z | Im BUM = M ]

= P′ y +iε [BU ∈ d z ] lim

P′ z [Im BUM = M ] ′ M →∞ Py +iε [Im BUM = M ] Im z = P′ , (71) y +iε [BU ∈ d z ] ε

where in the last step we have used [12, Proposition II(3.8)]. Combining Equations (70) and (71), we see that we have to compute the limit of P′ y +iε [BU ∈ d z ] Im z/ε as ε → 0. This computation can be done by using the conformal invariance of Brownian motion. Remember that the probability that a complex Brownian motion started from a + ib leaves the upper half-plane through (?∞, x), is given by the

35

wα,β z α 0 y β 1

Fα,β

Fπ?α,π?β



α 0 a 1 t 0 1

β

Figure 14: Conformal transformations between the upper half-plane and the domains Tα,β and Tπ?α,π?β . harmonic measure ω (x) of (?∞, x) at the point a + ib. It is straightforward to verify that
x

ω (x) =
?∞

dt 1 1 x?a b = + arctan . π b2 + (t ? a)2 2 π b

(72)

Thus, mapping the triangle Tα,β conformally to the upper half-plane by the ?1 as in Figure 14, we can write transformation Fα,β lim P′ y +iε [BU ∈ d z ]
ε↓0 ?1 ′ ) (z )d z (Fα,β Im z Im z = ? 1 ′ ε πFα,β (a) (Fα,β (z ) ? a)2

(73)

?1 ?1 (x). Therefore, using (58), (y ) = Fλ,? where a = Fα,β

P[E | Zτ = x] 1 = ′ (a) πFα,β =
′ a1?α (1

(74)
∞ 1

dt Im Fα,β (t) (t ? a)2


? a)1?β π

∞ 1

dt Im (t ? a)2


t 1

uα ?1 (1 ? u)β ?1 d u






′ sin β 1?α′ a (1 ? a)1?β = π sin β 1?α′ ′ = a (1 ? a)1?β π sin β 1?α′ ′ (1 ? a)1?β a = π

du u
1 ∞ 1 1 0


α′ ?1

(u ? 1)β ?1
′ ?1

∞ u

dt (t ? a)2

uα ?1 (u ? 1)β
′ ′

(u ? a)?1 d u

t1?α ?β (1 ? t)β ?1 (1 ? at)?1 d t,



where in the last step we have made the substitution t = 1/u. Using equations 15.3.1 and 15.2.5 for the hypergeometric function in [11] and the formulas 6.1.15 and 6.1.17 for the gamma function from [3], we

36

?nally derive P[E | Zτ = x] (75) sin β Γ(2?α′ ?β ′ )Γ(β ′ ) 1?α′ ′ (1 ? a)1?β 2 F1 (1, 2?α′ ?β ′ ; 2?α′ ; a) a = π Γ(2?α′ ) a ′ ′ Γ(2 ? α′ ? β ′ ) t?α (1 ? t)?β d t = ′ ′ Γ(1 ? α )Γ(1 ? β ) 0
?1 (x) . = Fπ?α,π?β (a) = Fπ?α,π?β Fλ,?

This is the desired result.

2

In words, we have considered the conditional probability that the last side visited by an RBMα,β in Tλ,? started from wλ,? is the right side, given that the exit point X equals x ∈ (0, 1). This conditional probability is exactly given by the image of x under the transformation that maps Tλ,? onto Tπ?α,π?β , ?xing 0 and 1 and sending wλ,? onto wπ?α,π?β . Acknowledgements. The author wishes to thank Julien Dub? edat for a useful discussion on the subject of this paper. Thanks are also due to Remco van der Hofstad for his comments and aid in preparing the manuscript. This research was supported ?nancially by the Dutch research foundation FOM (Fundamenteel Onderzoek der Materie).

References
[1] Ahlfors, L. V. Complex analysis: an introduction to the theory of analytic functions of one complex variable . New York: McGraw-Hill, second edition (1966). [2] Carmona, P., Petit, F. and Yor, M. Beta-gamma random variables and intertwining relations between certain Markov processes. Rev. Mat. Iberoamericana 14 (1998), no. 2, pp. 311–367. [3] Davis, P. J. Gamma function and related functions . In Abramowitz, M. and Stegun, I. (editors), Handbook of mathematical functions with formulas, graphs, and mathematical tables , chapter 6, pp. 253–293. New York: John Wiley & Sons, 10th edition (1972). [4] Dub? edat, J. Re?ected planar Brownian motions, intertwining relations and crossing probabilities (2003). Ann. Inst. H. Poincar? e Probab. Statist. 40 (2004), no. 5, pp. 539–552, arXiv:math.PR/0302250. 37

[5] Dub? edat, J. Excursion decompositions for SLE and Watts’ crossing formula (2004), arXiv:math.PR/0405074. [6] Gamelin, T. W. Complex analysis . New York: Springer-Verlag (2000). [7] Gardiner, C. W. Handbook of stochastic methods for physics, chemistry and the natural sciences . Berlin: Springer-Verlag (1983). [8] Lawler, G. F., Schramm, O. and Werner, W. Values of Brownian intersection exponents I: Half-plane exponents . Acta Math. 187 (2001), no. 2, pp. 237–273, arXiv:math.PR/9911084. [9] Lawler, G. F., Schramm, O. and Werner, W. Conformal restriction: the chordal case . J. Amer. Math. Soc. 16 (2003), no. 4, pp. 917–955, arXiv:math.PR/0209343. [10] Lawler, G. F. and Werner, W. Intersection exponents for planar Brownian motion . Ann. Prob. 27 (1999), no. 4, pp. 1601–1642. [11] Oberhettinger, F. Hypergeometric functions . In Abramowitz, M. and Stegun, I. (editors), Handbook of mathematical functions with formulas, graphs, and mathematical tables , chapter 15, pp. 555–566. New York: John Wiley & Sons, 10th edition (1972). [12] Revuz, D. and Yor, M. Continuous martingales and Brownian motion . Berlin: Springer-Verlag (1991). [13] Rogers, L. C. G. and Pitman, J. W. . Ann. Prob. 9 (1981), no. 4, pp. 573–582. [14] Rogers, L. C. G. and Williams, D. Di?usions, Markov processes, and martingales. Volume 1: foundations . New York: John Wiley & Sons, 2nd edition (1993). [15] Varadhan, S. R. S. and Williams, R. J. Brownian motion in a wedge with oblique re?ection . Comm. Pure App. Math. 38 (1985), pp. 405– 443. [16] W. Werner. Random planar curves and Schramm-Loewner Evolutions. Lecture notes from the 2002 Saint-Flour summer school, Springer, 2003 (to appear), arXiv:math.PR/0303354.

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