Speed of the biased random walk on a Galton--Watson tree

0.34 MB·English

by  Elie Aidekon

Speed of the biased random walk on a Galton–Watson
tree
Elie A¨ıd´ekon∗
1
1
0 December 19, 2011
2

c
e
D
Summary. We give an expression of the speed of the biased random

6 walk on a Galton–Watson tree. In the particular case of the simple ran-
1
dom walk, we recover the result of Lyons, Pemantle and Peres [7]. The

] proof uses a description of the invariant distribution of the environment
R
seen from the particle.
P
.
h Keywords. Random walk, Galton–Watson tree, speed, invariant mea-
t
a sure.
m
[ 2010 Mathematics Subject Classification. 60J80, 60G50, 60F15.

3
v
1 Introduction
3
1
3
4 Let T be a Galton–Watson tree with root e, and ν be its offspring distribution with values
.
1 in N. We suppose that m := E[ν] > 1, so that the tree is super-critical. In particular,
1
1 the event S that T is infinite has a positive probability, and we let q := 1−P(S) < 1 be
1
: the extinction probability. We call ν(x) the number of children of the vertex x in T. For
v
i x ∈ T\{e}, we denote by x the parent of x, that is the neighbour of x which lies on the
X ∗
r path from x to the root e, and by xi,1 ≤ i ≤ ν(x) the children of x. We call T the tree
a ∗
T on which we add an artificial parent e to the root e.
∗
For any λ > 0, and conditionally on T , we introduce the λ-biased random walk
∗
(X ) which is the Markov chain such that, for x (cid:54)= e ,
n n≥0 ∗
λ
(1.1) P(X = x |X = x) = ,
n+1 ∗ n
λ+ν(x)
1
(1.2) P(X = xi|X = x) = for any 1 ≤ i ≤ ν(x),
n+1 n
λ+ν(x)
∗LPMA, Universit´e Paris 6, 4, place Jussieu,75005 Paris, France. Email: [email protected]
1
and which is reflected at e . It is easily seen that this Markov chain is reversible. We de-
∗
note by P the quenched probability associated to the Markov chain (X ) starting from
x n n
x and by P the annealed probability obtained by averaging P over the Galton–Watson
x x
measure. They are respectively associated to the expectations E and E .
x x
When λ < m, we know from Lyons [6] that the walk is almost surely transient on the
event S. Moreover, if we denote by |x| the generation of x, Lyons, Pemantle and Peres
[8] showed that, conditionally on S, the limit (cid:96) := lim |Xn| exists almost surely, is
λ n→∞ n
determinist and is positive if and only if λ ∈ (λ ,m) with λ := E[νqν−1]. This is the
c c
regime we are interested in.
For any vertex x ∈ T , let
∗
(1.3) τ := min{n ≥ 1 : X = x}
x n
be the hitting time of the vertex x by the biased random walk, with the notation that
min∅ := ∞, and, for x (cid:54)= e ,
∗
β(x) := P (τ = ∞)
x x∗
bethequenchedprobabilityofneverreachingtheparentofxwhenstartingfromx. Notice
that we have β(x) > 0 if and only if the subtree rooted at x is infinite. Then, let (β ,i ≥ 0)
i
be, under P, generic i.i.d. random variables distributed as β(e), and independent of ν.
Theorem 1.1. Suppose that m ∈ (1,∞) and λ ∈ (λ ,m). Then,
c
(cid:44)
(cid:20) (cid:21) (cid:20) (cid:21)
(ν −λ)β (ν +λ)β
(cid:96) = E 0 E 0 .
λ λ−1+(cid:80)ν β λ−1+(cid:80)ν β
i=0 i i=0 i
The speed in the case λ = 1 was already obtained by Lyons, Pemantle and Peres [7],
who found that (cid:96) = E[ν−1] whereas Ben Arous, Hu, Olla and Zeitouni [2] computed the
1 ν+1
derivative of (cid:96) at the point λ = m. In this paper, the authors give another represen-
λ
tation of the speed (cid:96) , at least when λ is close enough to m. In the zero speed regime
λ
λ ≤ λ , Ben Arous, Fribergh, Gantert and Hammond [1] showed tightness of the properly
c
rescaled random walk, though a limit law fails. A central limit theorem was obtained by
Peres and Zeitouni [10], by means, in the case λ = m, of a construction of the invariant
distribution on the space of trees. The invariant distribution in the case λ > m was
given in [2]. We mention that, so far, the only case in the transient regime λ < m for
which such an invariant distribution was known was the simple random walk case λ = 1
studied in [7]. Theorem 4.1 in Section 4 gives a description of the invariant measure for
2
all λ ∈ (λ ,m). In the setting of random walks on Galton–Watson trees with random con-
c
ductances, Gantert, Mu¨ller, Popov and Vachkovskaia [5] obtained a similar formula for
the speed via the construction of the invariant measure in terms of effective conductances.
The paper is organized as follows. Section 2 introduces some notation and the concept
of backward tree seen from a vertex. Section 3 investigates the law of the tree seen from
a vertex that we visit for the first time. Using a time reversal argument, we are able to
describe the distribution of this tree in Proposition 3.2. Then, we obtain in Section 4 the
invariant measure of the tree seen from the particle. Theorem 1.1 follows in Section 5.
2 Preliminaries
2.1 The space of words U
We let U := {e} ∪ (cid:83) (N∗)n be the set of words, and |u| be the length of the word u,
n≥1
where we set |e| := 0. We equip U with the lexicographical order. For any word u ∈ U
with label u = i ...i , we denote by u ∈ U the word with letters in reversed order
1 n
u := i ...i (and e := e). If u (cid:54)= e, we denote by u the parent of u, that is the word
n 1 ∗
i ...i , and by u the word i ...i , which stands for the ancestor of u at generation
1 n−1 ∗ 1 n−k
k
|u|−k. We have u := e if k = |u| and u := u if k = 0. Finally, for u,v ∈ U, we denote
∗ ∗
k k
by uv the concatenation of u and v. We add to the set of words the element e , which
∗
stands for the parent of the root and we write U := U ∪{e }. We set |e | = −1, hence
∗ ∗ ∗
u = e for k = |u|+1 for any u ∈ U. We denote by S := {x , 1 ≤ k ≤ |x|+1} the set
∗ ∗ x ∗
k k
of strict ancestors of x.
2.2 The space of trees T
Following Neveu [9], a tree T is defined as a subset of U such that
• e ∈ T,
• if x ∈ T\{e}, then x ∈ T,
∗
• if x = i ...i ∈ T\{e}, then any word i ...i j with j ≤ i belongs to T.
1 n 1 n−1 n
We call T the space of all trees T. For any tree T, we define T as the tree on which we
∗
add the parent e to the root e. Then, let T := {T ,T ∈ T }. For a tree T ∈ T , and a
∗ ∗ ∗
vertex u ∈ T , we denote by ν (u) = ν (u) the number of children of u in T , and we
∗ T T∗ ∗
3
notice that ν (e ) = ν (e ) = 1. We will write only ν(u) when there is no doubt about
T ∗ T∗ ∗
which tree we are dealing with.
We introduce double trees. For any u ∈ U, let u− := (u,−1) and u+ := (u,1). Given
two trees T,T+ ∈ T , we define the double tree T−•T+ as the tree obtained by drawing an
edge between the roots of T and T+. Formally, T−•T+ is the set {u−, u ∈ T}∪{u+, u ∈
T+}. We root the double tree at e+. Given r an element of T, we say that X is the
r-parent of Y in T−•T+ if either
• Y = y+ and X = y+,
∗
• Y = e+ and X = e−,
• Y = y− with y ∈/ S ∪{u ∈ U : u ≥ r} and X = y−,
r ∗
• Y = r− and X = r− for some k ∈ [1,|r|].
∗ ∗
k k−1
In words, the r-parent of a vertex x is the vertex which would be the parent of x if we
were ’hanging’ the tree at r. Notice that we defined the r-parent only for the vertices
which do not belong to {u− : u ∈ U, u ≥ r}.
e−
e+
T T+ T−•T+
Figure 1: A double tree
4
2.3 The backward tree B (T )
x ∗
Let δ be some cemetery tree. For a tree T ∈ T and a word x ∈ U, we define the tree
∗ ∗
T ≤x ∈ T ∪{δ} cut at x by
∗ ∗
(cid:40)
δ if x ∈/ T ,
T ≤x := ∗
∗
T \{u ∈ U : x < u} if x ∈ T .
∗ ∗
e e
∗ ∗
e e
x = 1121
x = 1211
T B (T )
∗ x ∗
Figure 2: The backward tree at x
In other words, if x ∈ T , then T ≤x is the tree T in which you remove the strict descen-
∗ ∗ ∗
dants of x. We call U≤x the set of words U \{u ∈ U : x < u}. We now introduce the
∗ ∗
’backward’ tree at x. For any word x ∈ U, let Ψ : U≤x → U≤x such that:
x ∗ ∗
• for any k ∈ [0,|x|+1], Ψ (x ) = x ,
x ∗ ∗
k |x|−k+1
• for any k ∈ [1,|x|] and v ∈ U such that x v is not a descendant of x , Ψ (x v) =
∗ ∗ x ∗
k k+1 k
Ψ (x )v.
x ∗
k
The application Ψ is a bijection, with inverse map Ψ . For any tree T ∈ T , we call
x x ∗ ∗
backward tree at x the tree
(2.4) B (T ) := Ψ (T ≤x),
x ∗ x ∗
image of T ≤x by Ψ , with the notation that Ψ (δ) := δ. This is the tree obtained by
∗ x x
cutting the descendants of x and then ’hanging’ the tree T at x. We observe that,
∗
5
• ν (e ) = 1,
Bx(T∗) ∗
• ν (x) = 0,
Bx(T∗)
• for any other u ∈ B (T ), we have ν (u) = ν (Ψ (u)).
x ∗ Bx(T∗) T∗ x
Recall that T is a Galton–Watson tree with offspring distribution ν.
Lemma 2.1. Let x ∈ U. The distributions of the trees B (T ) and T≤x are the same.
x ∗ ∗
Proof. For any sequence (k ,u ∈ U) ∈ NU, denote by M(k ,u ∈ U) ∈ T the unique tree
u u ∗
such that for any u ∈ M(k ,u ∈ U) the number of children of u is 1 if u = e and k
u ∗ u
otherwise. Take (κ(u),u ∈ U) i.i.d. random variables distributed as ν. Then notice that
the tree M(κ(u),u ∈ U) is distributed as T . Therefore, we set in this proof
∗
T := M(κ(u),u ∈ U).
∗
We check that we can extend the map Ψ to a bijection on U by letting Ψ (xv) := xv
x ∗ x
for any strict descendant xv of x. Suppose that x ∈ T . We know that if u ∈ B (T ),
∗ x ∗
then the number of children of u is 1 if u = e , 0 if u = x and κ(Ψ (u)) otherwise. By
∗ x
definition, this yields that
B (T ) = M(κ(Ψ (u))1 ,u ∈ U).
x ∗ x {u(cid:54)=x}
Let T(cid:101) := M(κ(Ψ (u)),u ∈ U). We notice that M(κ(Ψ (u))1 ,u ∈ U) = T(cid:101)≤x.
∗ x x {u(cid:54)=x} ∗
Therefore, if x ∈ T , then
∗
B (T ) = T(cid:101)≤x.
x ∗ ∗
We check that the equality holds also when x ∈/ T . Observe that T(cid:101) is distributed as T
∗ ∗ ∗
to complete the proof. (cid:50)
3 The environment seen from the particle at first-
visit times
For any tree T ∈ T , we denote by PT∗ a probability measure under which (X ) is a
∗ ∗ n n≥0
Markov chain on T with transition probabilities given by (1.1) and (1.2). For any vertex
∗
x ∈ T , we denote by PT∗ the probability PT∗(·|X = x). We will just write P if the tree
∗ x 0 x
T is clear from the context.
∗
6
Lemma 3.1. Suppose that λ > 0. Let T be a tree in T , x be a vertex in T and
∗ ∗ ∗
(e = u ,u ,...,u = x) be a nearest-neighbour trajectory in T such that u ∈/ {e ,x} for
∗ 0 1 n ∗ j ∗
any j ∈ (0,n). Then,
PT∗(X = u , ∀j ≤ n) = PBx(T∗)(X = Ψ (u ), ∀j ≤ n).
e∗ j j e∗ j x n−j
Proof. We decompose the trajectory (u ,j ≤ n) along the ancestral path S . Let j := 0.
j x 0
Supposing that we know j , we define j as the smallest integer j > j such that u
i i+1 i+1 i ji+1
is an ancestor of x different from u . Let m be the integer such that u = x. We
ji jm+1
see that necessarily j = 1, (u ,u ) = (e ,e) and (u ,u ) = (x ,x). For i ∈ [1,m],
1 j0 j1 ∗ jm jm+1 ∗
let c be the cycle (u ,u ,...,u ). Notice that in this cycle, the vertex u is the
i ji ji+1 ji+1−1 ji
unique element of S visited, at least twice at times j and j −1. We set for any cycle
x i i+1
c = (z ,z ,...,z ),
0 1 k
k−1
(cid:89)
PT∗(c) := PT∗(X = z )
z 1 (cid:96)+1
(cid:96)
(cid:96)=0
(cid:81)
with the notation that := 1. Using the Markov property, we see that
∅
m m
(cid:89) (cid:89)
(3.5) PT∗(X = u , ∀j ≤ n) = PT∗(c ) PT∗ (X = u ).
e∗ j j i uji 1 ji+1
i=1 i=1
For any vertex z, let a(z) := (λ+ν (z))−1. Notice that the term corresponding to i = m
T∗
in the second product is
PT∗(X = x) = a(x ).
x∗ 1 ∗
For any z (cid:54)= e , let N (z) be the number of times the oriented edge (z,z ) is crossed by
∗ u ∗
the trajectory (u ,j ≤ n). Notice that the oriented edge (z ,z) is crossed 1+N (z) times
j ∗ u
when z ∈ S . Using the transition probabilities (1.1) and (1.2), we deduce that
x
m−1 |x|−1
(cid:89) PT∗ (X = u ) = (cid:89) (cid:0)λa(x )a(x )(cid:1)Nu(x∗k)a(x ).
uji 1 ji+1 ∗k ∗k+1 ∗k+1
i=1 k=1
Therefore, we can rewrite (3.5) as
(3.6) PT∗(X = u , ∀j ≤ n) = Π Π
e∗ j j 1 2
where
m
(3.7) Π := (cid:89)PT∗(c ),
1 i
i=1
|x|−1
(3.8) Π := a(x ) (cid:89) (cid:0)λa(x )a(x )(cid:1)Nu(x∗k)a(x ).
2 ∗ ∗ ∗ ∗
k k+1 k+1
k=1
7
We look now at the probability PBx(T∗)(X = v , ∀j ≤ n), where v := Ψ (u ). We
e∗ j j j x n−j
decompose the trajectory (v ,j ≤ n) along S . Observe that (v ,j ≤ n) is the time-
j x j
reversed trajectory of (u ,j ≤ n) looked in the backward tree. Therefore, the cycles of
j
(v , j ≤ n) are the image by Ψ of the time-reversed cycles of (u , j ≤ n). We need some
j x j
(cid:16) (cid:17)
← ←
notation. Let c be the path c time-reversed, and Ψ c be its image by Ψ , that is
i i x i x
(cid:16) (cid:17)
←
Ψ c = (Ψ (u ),Ψ (u ),...,Ψ (u )).
x i x ji+1−1 x ji+1−2 x ji
Let
PBx(T∗)(cid:16)Ψ (cid:16)←c (cid:17)(cid:17) := ji(cid:89)+1−2PBx(T∗)(X = Ψ (u )|X = Ψ (u )).
x i 1 x (cid:96) 0 x (cid:96)+1
(cid:96)=ji
We introduce for any vertex z ∈ B (T ),
x ∗
(cid:0) (cid:1)−1
a (z) := λ+ν (z)
B Bx(T∗)
and, for z (cid:54)= e , N (z) the number of times the trajectory (v ,j ≤ n) crosses the directed
∗ v j
edge (z,z ). Equation (3.6) reads for the trajectory (v , j ≤ n),
∗ j
(3.9) PBx(T∗)(X = v , ∀j ≤ n) = Π Π
e∗ j j B,1 B,2
where
m
Π := (cid:89)PBx(T∗)(cid:16)Ψ (cid:16)←c (cid:17)(cid:17),
B,1 x i
i=1
|x|−1
Π := a (x ) (cid:89) (cid:0)λa (x )a (x )(cid:1)Nv(x∗k)a (x ).
B,2 B ∗ B ∗ B ∗ B ∗
k k+1 k+1
k=1
Going from T to B (T ), we did not change the configuration of the subtrees located
∗ x ∗
(cid:16) (cid:16) (cid:17)(cid:17) (cid:16) (cid:17)
← ←
outside the ancestral path S of x. This yields that PBx(T∗) Ψ c = PT∗ c which
x i i
Π
is PT∗(c ) since the Markov chain (X ) is reversible. By definition of in (3.7), we
i n n≥0 1
get
Π Π
= .
B,1 1
We observe that a (z) = a(Ψ (z)) whenever z ∈/ {e ,x}, and Ψ (x ) = x by
B x ∗ x ∗ ∗
k |x|−k+1
definition. Moreover, for any k ∈ [1,|x|−1], we have N (x ) = N (x ). This gives
v ∗ u ∗
k |x|−k
8
that
|x|−1
Π = a(e) (cid:89) (cid:16)λa(x )a(x )(cid:17)Nu(x∗|x|−k)a(x )
B,2 ∗ ∗ ∗
|x|−k+1 |x|−k |x|−k
k=1
|x|−1
= a(x ) (cid:89) (cid:16)λa(x )a(x )(cid:17)Nu(x∗|x|−k)a(x ),
∗ ∗ ∗ ∗
|x|−k+1 |x|−k |x|−k+1
k=1
Π Π
hence, recalling (3.8), = . Equations (3.6) and (3.9) lead to
B,2 2
PT∗(X = u , ∀j ≤ n) = PBx(T∗)(X = v , ∀j ≤ n)
e∗ j j e∗ j j
which completes the proof. (cid:50)
Weintroduceξ , thek-thdistinctvertexvisitedbythewalk, andθ := τ theso-called
k k ξ
k
first-visit time. They can be defined by θ = 1, ξ = X and for any k ≥ 2 by
1 1 1
(3.10) θ := min{i > θ such that X ∈/ {X ,1 ≤ j < i}},
k k−1 i j
(3.11) ξ := X .
k θ
k
Wegivethedistributionofthetreeseenatafirst-visittimeθ , conditionallyon{θ < τ }.
k k e∗
Proposition 3.2. Suppose that λ > 0. Let k ≥ 1. Under P (·|θ < τ ), we have
e∗ k e∗
(B (T ),(Ψ (X )) ) (=d) (cid:0)T≤ξk,(X ) (cid:1).
ξk ∗ ξk θk−j j≤θk ∗ j j≤θk
Proof. For any relevant bounded measurable map F and any word x ∈ U, we have
E (cid:2)F (B (T ),(Ψ (X )) )1 (cid:3)
e∗ ξk ∗ ξk θk−j j≤θk {ξk=x,θk<τe∗}
= E (cid:2)F (B (T ),(Ψ (X )) )1 (cid:3)
e∗ x ∗ x θk−j j≤θk {ξk=x,θk<τe∗}
(cid:104) (cid:16) (cid:17) (cid:105)
= E F B (T ),(X(cid:101) ) 1
e∗ x ∗ j j≤θ(cid:101)k {ξ(cid:101)k=x,θ(cid:101)k<τ(cid:101)e∗}
by Lemma 3.1, where (X(cid:101) ) is the λ-biased random walk on the tree B (T ), and the
n n≥0 x ∗
variables θ(cid:101) , ξ(cid:101) and τ are the analogues of θ , ξ and τ for the Markov chain (X(cid:101) ) .
k k (cid:101)e∗ k k e∗ n n≥0
By Lemma 2.1, this yields that
E (cid:2)F (B (T ),(Ψ (X )) )1 (cid:3) = E (cid:2)F (cid:0)T≤x,(X ) (cid:1)1 (cid:3)
e∗ ξk ∗ ξk θk−j j≤θk {ξk=x,θk<τe∗} e∗ ∗ j j≤θk {ξk=x,θk<τe∗}
= E (cid:2)F (cid:0)T≤ξk,(X ) (cid:1)1 (cid:3).
e∗ ∗ j j≤θk {ξk=x,θk<τe∗}
We complete the proof by summing over x ∈ U. (cid:50)
9
The last lemma gives the asymptotic probability that n is a first-visit time. To state it,
we introduce the regeneration epochs (Γ ,k ≥ 1) defined by Γ := inf{(cid:96) ∈ {0}∪{θ ,k ≥
k 1 k
1} : X (cid:54)= (X ) ∀j ≥ (cid:96), X (cid:54)= e } and for any k ≥ 2,
j (cid:96) ∗ (cid:96) ∗
(3.12) Γ := inf{(cid:96) > Γ : (cid:96) ∈ {θ ,k ≥ 1},X (cid:54)= (X ) ∀j ≥ (cid:96)},
k k−1 k j (cid:96) ∗
where (X ) stands for the parent of the vertex X . For any k ≥ 1, it is well-known
(cid:96) ∗ (cid:96)
that, under P, the random walk after time Γ is independent of its past. Moreover,
k
the walk (X , (cid:96) ≥ Γ ) seen in the subtree rooted at X is distributed as (X ,(cid:96) ≥ 0)
(cid:96) k Γ (cid:96)
k
under P (·|τ = ∞). We refer to Section 3 of [8] for the proof of such facts. We
e e∗
have that Γ < ∞ for any k ≥ 1 almost surely on the event S when λ < m, and
k
E [Γ |τ = ∞] < ∞ if and only if λ ∈ (λ ,m).
e 2 e∗ c
Lemma 3.3. Suppose that m > 1 and λ ∈ (0,m). We have
1
lim P (n ∈ {θ , k ≥ 1},τ > n) = .
n→∞ e∗ k e∗ Ee[Γ2|τe∗ = ∞]
Proof. By the Markov property at time n and the branching property at vertex X , we
n
observe that
P (n ∈ {θ , k ≥ 1},τ > n)P (τ = ∞) = P (n ∈ {Γ ,k ≥ 1}, τ = ∞)
e∗ k e∗ e e∗ e∗ k e∗
hence
P (n ∈ {θ , k ≥ 1},τ > n) = P (n ∈ {Γ ,k ≥ 1}|τ = ∞).
e∗ k e∗ e∗ k e∗
We notice that P (n ∈ {Γ ,k ≥ 1}|τ = ∞) = P (n − 1 ∈ {Γ ,k ≥ 1}|τ = ∞).
e∗ k e∗ e k e∗
We mention that Γ = 0 on the event that τ = ∞, when starting from e. Since
1 e∗
(Γ − Γ ,k ≥ 1) are a sequence of i.i.d random variables under P (·|τ = ∞) with
k+1 k e e∗
mean E [Γ |τ = ∞], the lemma follows from the renewal theorem pp. 360, XI.1 [4]. (cid:50)
e 2 e∗
4 Asymptotic distribution of the environment seen
from the particle
We equip T and T with the topology generated by finite subtrees. For any tree T ∈ T
∗
and any x ∈ T , let
∗
T := {u ∈ T : u ≥ x}
x
be the subtree composed of x and its descendants. We recall that B (T ) defined in (2.4)
x ∗
is the tree backward. This section is devoted to the asymptotic distribution of the tree
10