Asymptotic shape and the speed of propagation of continuous-time ...

Asymptotic shape and the speed of propagation of continuous-time continuous-space birth processes Viktor Bezborodov

∗1

, Luca Di Persio

†1

, Tyll Krueger

Tomasz O˙za´ nski

arXiv:1609.04070v1 [math.PR] 13 Sep 2016

1 2

‡2

, Mykola Lebid

§3

, and

¶2

The University of Verona

Wroclaw University of Technology 3

(D-BSSE) ETH Z¨ urich

September 15, 2016

Abstract We formulate and prove a shape theorem for a continuous-time continuous-space stochastic growth model under certain general conditions. Similarly to the classical lattice growth models the proof makes use of the subadditive ergodic theorem. A precise expression for the speed of propagation is given in the case of a truncated free branching birth rate.

Mathematics subject classification: 60K35, 60J80.

1

Introduction

Shape theorems have a long history. Richardson [Ric73] proved the shape theorem for the Eden model. Since then, shape theorems have been proven in various settings, most notably for first passage percolation and permanent and non-permanent growth models. Garet and Marchand [GM12] not only prove a shape theorem for the contact process in random environment, but also have a nice overview of existing results. ∗

Email: [email protected] Email: [email protected] ‡ Email: [email protected] § Email: [email protected] ¶ Email: [email protected] Keywords: shape theorem, spatial birth process, growh model †

1

Most of literature is devoted to discrete-space models. A continuous-space first passage percolation model was analized by Howard and Newman [HN97], see also references therein. A shape theorem for a continuous-space growth model was proven by Deijfen [Dei03], see also Gou´er´e and Marchand [GM08]. Our model is naturally connected to that model, see the end of Section 2. Questions addressed in this article are motivated not only by probability theory but also by studies in natural sciences. In particular, one can mention a demand to incorporate spatial information in the description and analysis of 1) bacteria populations 2) cancer diseases 3) epidemiology 4) phylogenetics among others, see e.g. [WBP+ ], [TSH+ 13], [VDPP15], and [TM15]. Authors often emphasize that it is preferable to use the continuous-space spaces R2 and R3 as the basic, or ‘geographic’, space, see e.g. [VDPP15]. More on connections between theoretical studies and applications can be found in [MW03]. The paper is organized as follows. In Section 2 we describe the model and formulate our results, which are proven in Sections 3 and 4. Section 5 is devoted to computer simulations and conjectures. Technical results, in particular on the construction of the process, are collected in the appendix (Section 6).

2

The model, assumptions and results

We consider a growth model represented by a continuous-time continuous-space Markov birth process. Let Γ0 be the collection of finite subsets of Rd , Γ0 (Rd ) = {η ⊂ Rd : |η| < ∞}, where |η| is the number of elements in η. Γ0 is also called the configuration space. The evolution of the spatial birth process on Rd admits the following description. If the system is in state η ∈ Γ0 at time t, then the probability that a new particle appears (a “birth”) in a bounded set B ∈ B(Rd ) over time interval [t; t + ∆t] is Z ∆t b(x, η)dx + o(∆t), B

and with probability 1 no two births happen simultaneously. Here b : Γ0 × Rd → R+ is some function which is called the birth rate. Using a slightly different terminology, we can say that R the rate at which a birth occurs in B is B b(x, η)dx. We note that it is conventional to call the function b the ‘birth rate’, even though it is not a rate in the usual sense but rather a Radon–Nikodym derivative of the rate with respect to the Lebesgue measure. Remark 2.1. We characterize the birth mechanism by the birth rate b(x, η) at each spatial position. Oftentimes the birth mechanism is given in terms of contributions of individual particles: a particle at y, y ∈ η, gives a birth at x at rate c(x, y, η) (often c(x, y, η) = γ(y, η)k(y, x), 2

where γ(y, η) is the proliferation rate of the particle at y, whereas the dispersion kernel k(y, x) describes the distribution of the offspring), see e.g. Fournier and M´el´eard [FM04]. As long as we are not interested in the induced genealogical structure, the two ways of describing the process are equivalent under our assumptions. Indeed, given c, we may set b(x, η) =

X

c(x, y, η),

y∈η

or, conversely, given b, we may set c(y, x, η) =

b(x, η)g(x − y) P , g(x − y) y∈η

where g : Rd → (0, ∞) is some function. We equip Γ0 with the σ-algebra induced by the sets  Ball(η, r) = ξ ∈ Γ0 |η| = |ξ|, dist(η, ξ) < r , η ∈ Γ0 , r > 0, ) ( |η| P |xi − yi | η = {x1 , ..., x|η| }, ξ = {y1 , ..., y|η| } . We say that a function where dist(η, ξ) = min i=1 f : Rd → R+ has an exponential moment if there exists θ > 0 such that Z eθ|x| f (x)dx < ∞. Rd

Of course, if f has an exponential moment, then automatically f ∈ L1 (Rd ). Assumptions on b. Let us give our assumptions on the birth rate b. Condition 2.2 (Sublenear growth). The birth rate b is measurable and there exists a function a : Rd → R+ with an exponential moment such that b(x, η) ≤

X

a(x − y).

(1)

y∈η

Condition 2.3 (Monotonicity). For all η ⊂ ξ, b(x, η) ≤ b(x, ξ).

(2)

The previous condition ensures attractiveness, see below. Condition 2.4 (Rotation and translation invariance). The birth rate b is translation and rotation invariant: for every x, y ∈ Rd , η ∈ Γ0 and M ∈ SO(d), b(x + y, η + y) = b(x, ξ), b(M x, M η) = b(x, η). Here SO(d) is the orthogonal group of linear isometries on Rd . 3

Condition 2.5 (Non-degeneracy). Let there exist such c0 , r > 0 such that b(x, η) ≥ c0

wherever min |x − y| ≤ r. y∈η

(3)

We also need to supplement the monotonicity condition with the following technical assumption on b. Condition 2.6 (b-domination). There exists a finite collection of balls {Bj , j = 1, ..., N } such that b(·, {0}) ≤ b(·, η)

(4)

whenever there is a collection of distinct points {yj , j = 1, ..., N } ⊂ η such that yj ∈ Bj . In other words, having enough particles in the right locations is sufficient for (4). Note that the balls are not required to have an empty intersection. They might even coincide, in which case the condition is simpler. See Remark 3.11 for a short discussion on why we need a condition of this type. Examples of a birth rate are b(x, η) = λ

X

a(x − y),

(5)

y∈η

and

! b(x, η) = k ∧

λ

X

a(x − y) ,

(6)

y∈η

where λ, k are positive constants and a is rotation invariant and such that (3) and Condition 2.6 are satisfied. Either of the following assumptions implies Condition 2.6: • The function f : R+ → R+ defined by f (x) = a(|x|) is such that log f is uniformly continuous outside some bounded interval; • The function f is non-increasing. The birth process will be obtained as a unique solution to the stochastic equation proposed by Massouli´e [Mas98]. The construction is outlined in Section 6. Denote by (ηts,A )t≥s the process started at time s from the set A. This method has the advantage that the stochastic equation approach we use resembles graphical representation in the fact that it preserves monotonicity (attractiveness): if A ⊂ B, then for every s > 0, ηts,A ⊂ ηts,B ,

t ≥ s.

Let 0 be the origin in Rd . The process started from a single particle at 0 at time zero will be denoted by (ηt )t≥0 . Let ξt :=

[ x∈ηt

4

B(x, r)

(7)

and similarly ξts,A :=

[

B(x, r).

x∈ηts,A

The following theorem represents the main result of the paper. Theorem 2.7. There exists µ > 0 such that for all ε > 0 a.s. (1 − ε)B(0, µ−1 ) ⊂

ξt ⊂ (1 + ε)B(0, µ−1 ) t

(8)

for sufficiently large t. The value µ−1 is called the speed of propagation. We choose the reciprocal of µ as the speed because µ appears naturally in the proof, and also for historic reasons. Formal connection to Deijfen’s model. The model introduced in [Dei03] with deterministic outburst radius (that is, when the distribution of ourbursts F is the Dirac measure: F = δR for some R ≥ 0) can be identified with the the birth process with b(x, η) = 1{∃y ∈ η : |x − y| ≤ R}. Explicit growth speed for a particular model. The precise evaluation of speed appears to be a difficult problem. To the best of our knowledge, the only example is given by Biggins [Big95] for a general branching random walk. Here we give the speed for a model with interaction. Theorem 2.8. Let d = 1 and ! b(x, η) = 2 ∧

X

1{|x − y| ≤ 1} .

(9)

y∈η

Then the speed of propagation is given by µ−1 =

3

144 ln(3) − 144 ln(2) − 40 ≈ 0.73548... 25

(10)

Proof of Theorem 2.7

The idea of the proof. The subadditive ergodic theorem is a cornerstone in the majority of shape theorem proofs, and our proof heavily relies on it. Estimations for the hitting times are obtained via comparison with the Eden model (see Eden [Ede61]). Without loss of generality we assume that a is rotation invariant. We also assume without loss of generality that the balls {Bj , j = 1, ..., N } are of the same radius. We will use a coupling with the Eden model. The Eden model is a model of tumor growth on the lattice Zd . The evolution starts from a single particle at the origin. A site once occupied 5

stays occupied forever. A vacant site becomes occupied at rate λ > 0 if at least one of its neighbors is occupied. Let us mention that this model is closely related to the first passage percolation model. For z = (z1 , ..., zd ) ∈ Zd , let |z|1 =

d P

|zi |.

i=1

Lemma 3.1. Consider the Eden model starting from a single particle at the origin. Then there 4e2 ˜ exists a constant C˜ > 0 such that for every z ∈ Zd and time t ≥ λ2 (e−1) 2 ∨ C|z|1 , √

P {z is vacant at t} ≤ e− t .

(11)

Proof. Let σz be the time when z becomes occupied. We have the equality of the events {z is vacant at t} = {σz > t}. Now, σz is bounded above by a sum of |z|1 independent exponential random variables with mean

1 λ,

1

say σz ≤ Z1 + ... + Z|z|1 . Note that Eeλ(1− e )Z1 = e. Using Chebyshev’s inequality 1

P {Z > t} ≤ Eeλ(1− e )(Z−t) , we get 1 P {σz > t} ≤ P {Z1 + ... + Z|z|1 > t} ≤ E exp{λ(1 − )(Z1 + ... + Z|z|1 − t)} e h i|z|1 1 1 1 = Eeλ(1− e )Z1 e−λ(1− e )t = e|z|1 e−λ(1− e )t . Since

for t ≥

4e2 , λ2 (e−1)2

√ 1 1 λ(1 − )t ≥ t, 2 e we may take C˜ =

2e λ(e−1) .

Lemma 3.2. There are constants c1 , t0 > 0 such that P {there is an vacant site in B(0, c1 t) ∩ Zd at t} ≤ e− Proof. By the previous lemma for c1
0 ,

X

α(y) < ∞,

(13)

y∈Zd

y∈Zd : |z−y|≤1

where λ > 0. Clearly, the above lemma also applies to (αt )t≥0 , since it dominates the Eden process. Lemma 3.4. There are c, s0 > 0 such that P {B(0, cs) 6⊂ ξs } ≤ e− Proof. For x ∈ Rd let zx ∈

r d 2d Z

√ 4

s

s ≥ s0 .

,

(14)

r r d be uniquely determined by x ∈ zx + (− 4d , 4d ] (recall that

r appears in (3)). Define ¯b(x, η) = c0 1{zx ∼ zy for some y ∈ η},

(15)

r d where zx ∼ zy means that zx and zy are neighbors on 2d Z . Let (¯ ηt )t≥0 be the birth process with birth rate ¯b. Note that ¯b(x, η) ≤ b(x, η), hence η¯t ⊂ ηt , t ≥ 0. Then the ‘projection’ process

defined by η t (z) =

X

1{x ∈ z + (−

x∈¯ ηt

r r d , ] }, 4d 4d

z∈


r d 2d

is the process (αt )t≥0 from Remark 3.3 with λ = c0

r d Z , 2d

and the ‘geographic’ space

r d 2d Z

instead of Zd . Since η t (zx ) > 0 implies that x ∈ ξt , the desired result follows from Lemma 3.2 and Remark 3.3. For x ∈ Rd , define a stopping time T (x) by T (x) = inf{t > 0 : b(·, {x}) ≤ b(·, ηt )},

(16)

n o T (x),{x} T (x, y) = inf t > T (x) : b(·, {y}) ≤ b(·, ηt ) − T (x).

(17)

and for x, y ∈ Rd we define

Note that T (x) = T (0, x). By attractiveness, x, y ∈ Rd .

T (x) + T (x, y) ≥ T (x + y),

(18)

Lemma 3.5. For every x ∈ Rd there exist Ax , qx > 0 such that P {T (x) > s} ≤ Ax e−qx

7

√ 4

s

,

s ≥ 0.

(19)

Proof. Recall that the balls Bj appear in Condition 2.6. Let θj = θj (x) denote the moment when x + Bj becomes occupied, with convention that θj 6= θi for i < j; that is, if two balls Bj and Bj become occupied at the same time p, we let θi = p, but θj ‘waits’ for another birth inside Bj . Note that

T (x) ≤ inf{t > 0 : there is a collection of distinct points {yj , j = 1, ..., N } ⊂ ηt such that yj ∈ Bj + x}, hence T (x) ≤ max θj .

(20)

j

By Lemma 3.4, each θj satisfies inequality (19) with T (x) replaced by θj . The maximum in (20) is taken over finitely many j, therefore T (x) satisfies (19) too. Let us fix an x ∈ Rd , x 6= 0, and define for k, n ∈ N, k < n, sk,n = T (kx, nx).

(21)

Note that the random variables sk,n are integrable by Lemma 3.5. The conditions of Liggett’s subadditive ergodic theorem, see [Lig85], are satisfied here, therefore there exists µ = µ(x) such that s0,n → µ. n

(22)

T (tx) = µ(x). t

(23)

T (nx) = µ(x). n→∞ n

(24)

Lemma 3.6. For every x 6= 0, lim

t→∞

Proof. We know that lim

Recall that N appears in Condition 2.6. Looking to emulate the proof of the previous lemma, we note that there exist a finite collection of balls {Bji , j = 1, ..., N, i = 1, ..., kj } such that (η ∩ Bji 6= ∅, j = 1, ..., N, i = 1, ..., kj ) ⇒ (∀y ∈ [0, x) : η ∩ y + Bj 6= ∅, j = 1, ..., N ) .

(25)

Proceeding as in the proof above, we can deduce that there exist A, q > 0 such that for every n ∈ N, P{

T (nx, y) > s} < Ae−q

sup

√ 4s

.

y∈[nx,(n+1)x)

(Note that the distribution of

sup

T (nx, y) does not depend on n.) In view of (18) and

y∈[nx,(n+1)x)

since T (nx) grows linearly with n, the inequality lim sup t→∞

T (tx) ≤ µ(x). t 8

Bj

Bj + x

!h Figure 1: The six small balles are the collection {Bji , j = 1, ..., N, i = 1, ..., kj } follows now from the Borel–Cantelli lemma, because the series X

P{

sup

T (nx, y) >

√

n}}

y∈[nx,(n+1)x)

n∈N

converges. Likewise, it can be shown that lim inf t→∞

T (tx) ≥ µ(x), t

and the desired statement follows. Lemma 3.7.

µ(x) |x|

in (22) does not depend on x.

Proof. First let us note that for every x ∈ Rd and every q > 0, µ(x) =

µ(qx) q

(26)

by Lemma 3.6. On the other hand, if |x| = |y| then by Proposition 6.7 µ(x) = µ(y).

(27)

The statement follows from (26) and (27). Set µ(x) . |x|

µ := Lemma 3.8. Let ε > 0. Then for large n ∈ N,

(1 − ε)B(0, µ−1 ) ⊂

ξn . n

Proof. Choose a finite sequence of points {xj } such that xj ∈ (1 − ε)B(0, µ−1 ) and [ j

ε B(xj , c) ⊃ (1 − ε)B(0, µ−1 ), 2

where c is from Lemma 3.4. Let δ > 0 be so small that (1 + δ)(1 − ε) ≤ (1 − 2ε ). Since a.s. T (nxj ) → µ, n|xj | 9

(28)

for large n for every j ε T (nxj ) ≤ n|xj |(1 + δ)µ ≤ n(1 − ε)(1 + δ) ≤ n(1 − ), 2 so that the system reaches nxj before n(1 − 2ε ). Using Lemma 3.4, we will show that after additional time lag 2ε n the entire ball (1 − ε)nB(0, µ−1 ) is covered. Indeed, the series X n∈N

ε (n(1− 2ε ),{nxj }) , for some j} P {B(nxj , c n) 6⊂ ξn 2

converges by Lemma 3.4, thus for large n, ξn ⊃

[

(n(1− 2ε ),{nxj })

ξn

⊃ (1 − ε)nB(0, µ−1 ).

j

Lemma 3.9. Let ε > 0. Then for large n ∈ N ηn ⊂ (1 + ε)B(0, µ−1 ). n

(29)

Let A be the annulus (1 + 2ε)B(0, µ−1 ) \ (1 + ε)B(0, µ−1 ), and {xj } be a finite sequence such that xj ∈ A and [ j

ε B(xj , c) ⊃ A. 4

Define F := {ηn ∩ nA 6= ∅ infinitely often}. On F , we can choose a sequence of pairs of random variables {(mk , ymk )}k∈N such that mk → ∞ and ymk ∈ ηmk ∩ mk A. By the strong Markov property and Lemma 3.4, for some j, mk xj will be covered by ξ mk ,{ymk } by the time mk + 4ε mk with high probability, implying that T (xj ) ≤ mk + 2ε mk with high probability. Thus, for some xj , ε T (mk xj ) ≤ mk + mk 2 infinitely often on F , hence lim inf

mk + 2ε mk 1 + 2ε T (mk xj ) ≤ = µ . mk |xj | (1 + ε)µ−1 mk (1 + ε)

The last inequality and Lemma 3.6 imply that P (F ) = 0. Proof of Theorem (2.7). It follows from Lemmas 3.8 and 3.9 that (8) holds for large integers. Since ε is arbitrary, it is also true that a.s. for large n ∈ N, ηn+2 ⊂ (1 + ε)B(0, µ−1 ), n 10

and (1 − ε)B(0, µ−1 ) ⊂

ξn . n+2

It is therefore sufficient to note that ηt ⊂ (1 + ε)B(0, µ−1 ) t

ηdte+1 ⊂ (1 + ε)B(0, µ−1 ) btc

(30)

ξbtc ⊃ (1 − ε)B(0, µ−1 ). dte + 1

(31)

if

and ξt ⊃ (1 − ε)B(0, µ−1 ) t

if

Remark 3.10. We have used the rotation invariance of b only to establish (27). Without this assumption, our shape theorem would still hold modulo the fact that we would have some convex compact shape instead of a ball. Remark 3.11. Let us make a comment on why our proof breaks down if Condition 2.6 is not satisfied. To use the subadditive ergodic theorem, we clearly need T (x) from (16) to be finite. Let us give an example of a birth rate of the form (5) when T (x) is in fact infinite a.s. Take d = 1 and define for x ≥ 0    0,    an (x) = en! x − n!en! + 1,     −en! x + n!en! + 1,

x∈ / [n! −

1 , n! e(n!)

x ∈ [n! −

1 , n!] e(n!)

x ∈ [n!, n! +

+

1 ] e(n!)

1 ]. e(n!)

Set now a(x) =

X

an (|x|).

n∈N

The birth rate b from (5) satisfies all our conditions except Condition 2.6. In particular, for θ ∈ (0, 1) Z

eθ|x| a(x) < ∞.

x∈R

However, it can be shown that a.s. T (x) = ∞,

x 6= 0,

(32)

under the additional assumption that for some C > 0 a.s. ηt ⊂ [−C, C] t

(33)

for t > t0 , where t0 is an a.s. finite random variable. Note that (33) means that the system does not grow faster than linearly, which is reasonable to assume for our purposes; if it is not the case, then the shape theorem cannot hold. 11

To prove (32), we show that P ({T (x) ≤ t} ∩ {t > t0 }) = 0 for any x 6= 0 and t < ∞. Let us fix x and t. For n ∈ N we define the set  [  1 1 x + n! + sign(m)m! − m! , x + n! + sign(m)m! + m! , A(x, n) = e e m∈Z

where 0! is set to be 0 and for negative m ∈ Z, m! := (−m)!. Our choice of a implies that {T (x) ≤ t} ⊂ {∀n ∈ N : ηt ∩ A(x, n) 6= ∅}. For large n   1 1 A(x, n) ∩ [−Ct, Ct] ⊂ x − |n|! , x + |n|! e e (we have the equality instead of the inclusion if x ∈ [−Ct, Ct]), and hence on {t > t0 }   1 1 {T (x) ≤ t} ⊂ {ηt ∩ x − |n|! , x + |n|! 6= ∅ for sufficiently large n} = {x ∈ ηt }. e e However, for any given x > 0 the last event has probability zero, so (32) is proven. Remark 3.12. If b is like in (5) and a has polinomial tails, then the result of Durrett [Dur83] suggests that we should expect a superlinear growth. This is in contrast with Deijfen’s model, for which Gou´er´e and Marchand [GM12] give a sharp condition on the distribution of the outbursts for linear or superlinear growth.

4

Proof of Theorem 2.8

2 1

x1

x2

x2 + 1

x1 + 1

x

Figure 2: The plot of b(·, ηt ). Without any loss of generality, we will consider the speed of propagation in one direction only, say toward +∞. Let x1 (t) and x2 (t) denote the positions of the rightmost particle and the second rightmost particle, respectively (x2 (t) = 0 until first two births occurs inside (0, +∞)). Let us observe that b(x, ηt ) ≡ 2 on (0, x2 (t) + 1], and X = (x1 (t), x2 (t)) is a continuous-time pure jump Markov process on {(x1 , x2 ) | x1 ≥ x2 ≥ 0, x1 − x2 ≤ 1} with transition densities 12

(x1 , x2 ) → (v, x1 )

at rate 1,

v ∈ (x2 + 1, x1 + 1];

(x1 , x2 ) → (v, x1 )

at rate 2,

v ∈ (x1 , x2 + 1];

(x1 , x2 ) → (x1 , v)

at rate 2,

v ∈ (x2 , x1 ].

(34)

(to be precise, the above is true from the moment the first birth inside R+ occurs). Furthermore, z(t) := x1 (t) − x2 (t) satisfies E{f (z(t + δ)) | x1 (t) = x1 , x2 (t) = x2 } = E{f (z(t + δ)) | x1 (t) = x1 + h, x2 (t) = x2 + h} for every h > 0 and every Borel bounded function f . In other words, transition rates of (z(t))t≥0 are entirely determined by the current state of (z(t))t≥0 , therefore by Lemma 6.9 (z(t))t≥0 is itself a pure jump Markov process on [0, 1] with the transition densities q(x, y) = 41{y ≤ x} + 21{x ≤ y ≤ 1 − x} + 1{y ≥ 1 − x},

1 x ≤ , y ∈ [0, 1], 2 1 x ≥ , y ∈ [0, 1]. 2

(35) q(x, y) = 41{y ≤ 1 − x} + 31{1 − x ≤ y ≤ x} + 1{y ≥ x}, R1 Note that the total jump rate out of x is q(x) := 0 q(x, y)dy = 2 + x. The process (z(t))t≥0 is a regular Harris recurent Feller process with the Lebesgue measure on [0, 1] being a supporting measure (see e.g. [Kal02, Chapter 20]). Hence the unique invariant measure exists and has a density g with respect to the Lebesgue measure. The equation for g is Z1 q(x, y)g(x)dx = q(y)g(y).

(36)

0

Set 

−1

Z1

f (x) = g(x)q(x) 

g(y)q(y)dy 

x ∈ [0, 1].

,

0

It is cleare that f is again a density (as an aside we point out that f is the density of invariant distribution of the embedded Markov chain of (z(t))t≥0 . We emphasize however that we do not use this fact in the proof). Equation (36) becomes Z1 f (y) =

q(x, y) f (x)ds, q(x)

0

which after some calculations transforms into 1

1

Z2

Z2

f (y) = 2

f (x)dx +2 2+x

y

0 1

Z2 f (y) = 0

f (x)dx +3 2+x

f (x)dx + 2+x

1−y Z 0

Z1

f (x)dx + 2+x

1 2

f (x)dx + 2+x

Z1 1 2

13

f (x)dx +2 2+x

1−y Z

f (x)dx , 2+x

1 y≤ , 2

(37)

1 2

Z1 y

f (x)dx , 2+x

1 y≤ . 2

(38)

Differentiating (37), (38) with respect to y, we find that f solves the equation df f (x) f (1 − x) (x) = −2 − , dx 2+x 3−x

x ∈ [0, 1].

  ϕ(x) := (2 + x)2 (3 − x)2 f (x),

x ∈ [0, 1].

(39)

Let

Then (39) becomes (3 − x)

dϕ (x) + 2ϕ(x) + ϕ(1 − x) = 0, dx

x ∈ [0, 1].

(40)

Looking for solutions to (40) among polynoms, we find that ϕ(x) = c(4 − 3x) is a solution. By direct substitution we can check that f (x) =

c(4 − 3x) (2 + x)2 (3 − x)2

x ∈ [0, 1]

(41)

solves (37)-(38). The constant c > 0 can be computed, but is irrelevant for our purposes. Hence, after some more computation, g(x) =

36(4 − 3x) , (2 + x)3 (3 − x)2

x ∈ [0, 1].

(42)

Note that we do not prove analytically that equation (37), (38) has a unique solution. However, uniqueness for non-negative integrable solutions follows from the uniqueness of the invariant distribution for (z(t))t≥0 . Let l be the Lebesgue measure on R. By an ergodic theorem for Markov processes, see e.g. [Kal02, Theorem 20.21 (i)], for any 0 ≤ p < p0 ≤ 1, l{s : z(s) ∈ [p, p0 ], 0 ≤ s ≤ t} → t→∞ t

Z

p0

lim

g(x)dx.

(43)

p

Conditioned on z(t) = z, the transition densities of x1 (t) are x1 → x1 + v

at rate 2,

v ∈ (0, 1 − z];

x1 → x1 + v

at rate 1,

v ∈ (1 − z, 1].

(44)

Therefore, the speed of propagation is  1−z    Z1 Z Z1 Z1
1 2 2 g(z)dz  2ydy + ydy  = g(z) (1 − z) + 1 − (1 − z) dz 2 0

0

1−z

0

Z1 =

1 g(z)(1 − z + z 2 )dz. 2

0

Using (42), we get

14

Z1

1 144 ln(3) − 144 ln(2) − 40 (1 − z + z 2 )g(z)dz = . 2 25

0

Remark 4.1. We see from the proof that the speed can be computed in a similar way for the birth rates of the form

! X

bk (x, η) = k ∧

1{|x − y| ≤ 1} ,

(45)

y∈η

where k ∈ (1, 2). However, the computations quickly become unwieldy.

5

Simulations

Figure 3: Distance to the furtherst particle

Figure 4: The distance to the furtherst par-

against time in 10 different realisations. Red

ticle divided by time against k for the birth

dashed line shows the theoretical distance

rate (45) at time 1000.

given by Theorem 2.8. In this section we present the computer simulations of our systems. We mostly use an extended verison of the algorithm from Section 11.1.2 in [MW03]. Figure 3 shows the distance to the furthest particle against time for ten different realisations with birth rate (9). We see that the fluctuations from the value in Theorem 2.8 are rather small. Figure 4 shows the speed of propagation s(k) against k on x-axis for the truncated birth rate (6). Conjecture 5.1. We conjecture that s(k) → s∗ , where s∗ is the speed of propagation of the free branching model with b given by X b(x, η) = 1{|x − y| ≤ 1}. y∈η

15

(46)

(47)

Figure 5: The speed as the distance to the

Figure 6: Positions of the occupied sites vary-

furtherst particle divided by time against the

ing with time for the discrete-space model

maximal allowed number of particles in the

with birth rate (49) and α = 2.8.

branching random walk with restriction at time 104 . The scale along x-axis is logarithmic.

Figure 7: α = 3.5

Figure 8: α = 4.2

Using the exact formula for for the speed of propagation of a general random walk, see Proposition 1 in Biggins [Big95], we get   θ −θ 2 s∗ = inf inf (e − e − aθ ) < 0 ≈ 1.81... a>0

θ>0

(48)

The question about the speed of convergence in (46) is more subtle. We also consider the free branching model with restriction on the number of particles. The system evolves as the birth process with rate (47), but whenever the number of particles exceeds a given n ∈ N, the leftmost particle is removed. The number of particles thus stays constant once it reaches n. Figure 5 shows the speed of propagation against n. The largest n we took 16

!h Figure 9: A snapshot from the two dimen-

Figure 10: The number of particles is 3 · 106

sional simulation of the birth process with rate (45) with k = 5. The number of particles is 65000 was 8902, for which the recorded speed is 1.72018. Figures 6-8 show the evolution of the discrete version of the truncated model (6): the process evolves in ZZ+ and the birth rate is ! b(x, η) = k ∧

X

apow (x − y)

(49)

y∈η

with apow (x) = cpow

1 , (|x| + 1)α

x ∈ Z \ {0}

apow (0) = 2cpow , where α > 2 and cpow = cpow (α) is the normalizing constant. We have α = 2.8 on Figure 6, α = 3.5 on Figure 7, and α = 4.2 on Figure 8. These pictures allow us to observe the development of the set of occupied sites. We see that even for a large time, the set of occupied sites is not a connected interval for α = 2.8, whereas the picture appears to be rather smooth for α = 4.2. We conjecture that the speed of propagation is superlinear for α = 2.8, but is linear for α = 4.2. Figures 9 and 10 displayes a snapshot of the system with birth rate (45), k = 5. We also think that the speed of propagation has superadditive structure. For a birth rate b satisfying our assumptions, let s(b) be the speed of propagation.

17

Conjecture 5.2. For any birth rates b1 , b2 satisfying our assumptions, s(b1 ) + s(b2 ) ≤ s(b1 + b2 ).

6

Appendix

6.1

The construction and properties of the process

Here we proceed to construct the process as a unique solution to a stochastic integral equation. This construction can be deemed an analog of the construction from graphical representation. We follow here [Bez15]. Remark 6.1. Of course, the process starting from a fixed initial condition we consider here can be constructed as the minimal jump process (pure jump type Markov processes in the terminology of [Kal02]) as is done for example in [EW03]. Note however that we use coupling of infinitely many processes starting at different time points from different initial conditions, so we here employ another method. Recall that Γ0 (Rd ) = {η ⊂ Rd : |η| < ∞}. To construct the family of processes (ηtq,A )t≥q , we consider the stochastic equation with Poisson noise

Z 1[0,b(x,s,ηs− )] (u)N (ds, dx, du) + ηq ∩ B,

ηt (B) =

t≥q

(50)

(q,t]×B×[0,∞)

where (ηt )t≥q is a cadlag Γ0 -valued solution process, N is a Poisson point process on R+ × Rd × R+ , the mean measure of N is ds × dx × du. We require the processes N and η0 to be independent of each other. Equation (50) is understood in the sense that the equality holds a.s. for every bounded B ∈ B(Rd ) and t ≥ q. In the integral on the right-hand side of (50), x is the location and s is the time of birth of a new particle. Thus, the integral over B from 0 to t represents the number of births inside B which occured before t Let us assume for convenience that q = 0. We will make the following assumption on the initial condition: E|η0 | < ∞.

(51)

We say that the process N is compatible with (Ft , t ≥ 0) if N is adapted, that is, all random variables of the type N (T¯1 , U ), T¯1 ∈ B([0; t]), U ∈ B(Rd × R+ ), are Ft -measurable, and all random variables of the type N (t + h, U ) − N (t, U ), h ≥ 0, U ∈ B(Rd × R+ ), are independent of Ft , N (t, U ) = N ([0; t], U ).

18

Definition 6.2. A (weak) solution of equation (50) is a triple ((ηt )t≥0 , N ), (Ω, F , P ), ({Ft }t≥0 ), where (i) (Ω, F , P ) is a probability space, and {Ft }t≥0 is an increasing, right-continuous and complete filtration of sub-σ-algebras of F , (ii) N is a Poisson point process on R+ × Rd × R+ with intensity ds × dx × du, (iii) η0 is a random F0 -measurable element in Γ0 satisfying (51), (iv) the processes N and η0 are independent, N is compatible with {Ft }t≥0 , (v) (ηt )t≥0 is a cadlag Γ0 -valued process adapted to {Ft }t≥0 , ηt t=0 = η0 , (vi) all integrals in (50) are well-defined, Zt E

Z ds

0

b(x, ηs− ) < ∞,

t>0

Rd

and (vii) equality (50) holds a.s. for all t ∈ [0, ∞] and all Borel sets B. Let  St0 = σ η0 , N ([0, q] × B × C), q ∈ [0, t], B ∈ B(Rd ), C ∈ B(R+ ) , and let St be the completion of St0 under P . Note that {St }t≥0 is a right-continuous filtration. Definition 6.3. A solution of (50) is called strong if (ηt )t≥0 is adapted to (St , t ≥ 0). Remark 6.4. In the definition above we considered solutions as processes indexed by t ∈ [0, ∞). The reformulations for the case t ∈ [0, T ], 0 < T < ∞, are straightforward. This remark also applies to many of the results below. Definition 6.5. We say that joint uniqueness in law holds for equation (50) with an initial distribution ν if any two (weak) solutions ((ηt ), N ) and ((ηt0 ), N 0 ) of (50), Law(η0 ) = Law(η00 ) = ν, have the same joint distribution: Law((ηt ), N ) = Law((ηt0 ), N 0 ). Theorem 6.6. Pathwise uniqueness, strong existence and joint uniqueness in law hold for equation (50). The unique solution is a Markov process. Proof. Without loss of generality assume that P {η0 6= ∅} = 1. Define the sequence of random pairs {(σn , ζσn )}, where Z σn+1 = inf{t > 0 :

I[0,b(x,ζσn )] (u)N (ds, dx, du) > 0} + σn , σ0 = 0, (σn ,σn +t]×B×[0,∞)

19

and ζ0 = η 0 ,

ζσn+1 = ζσn ∪ {zn+1 }

for zn+1 = {x ∈ Rd : N ({σn+1 } × {x} × [0, b(x, ζσn )]) > 0}. The points zn are uniquely determined a.s. Furthermore, σn+1 > σn a.s., and σn are finite a.s by (3). We define ζt = ζσn for t ∈ [σn , σn+1 ). Then by induction on n it follows that σn is a stopping time for each n ∈ N, and ζσn is Fσn -measurable. By direct substitution we see that (ζt )t≥0 is a strong solution to (50) on the time interval t ∈ [0, lim σn ). Although we have not defined what is a solution, or n→∞

a strong solution, on a random time interval, we do not discuss it here. Instead we are going to show that lim σn = ∞ a.s.

(52)

n→∞

The process (ζt )t∈[0,

lim σn )

has the Markov property, because the process N has the strong

n→∞

Markov property and independent increments. Indeed, conditioning on Iσn , R b(x, ζσn )dx   B E I{ζσn+1 =ζσn ∪x for some x∈B} | Iσn = R , b(x, ζσn )dx Rd

thus the chain {ζσn }n∈Z+ is a Markov chain, and, given {ζσn }n∈Z+ , σn+1 − σn are distributed exponentially: Z E{I{σn+1 −σn >a} | {ζσn }n∈Z+ } = exp{−a

b(x, ζσn )dx}.

Rd

Therefore, the random variables γn = (σn − σn−1 )

R

b(x, ζσn )dx constitute an independent of

Rd

{ζσn }n∈Z+ sequence of independent unit exponentials. Theorem 12.18 in [Kal02] implies that (ζt )t∈[0,

lim σn )

is a pure jump type Markov process.

n→∞

The jump rate of (ζt )t∈[0,

lim σn )

is given by

n→∞

Z c(α) =

b(x, α)dx. Rd

Condition 2.2 implies that c(α) ≤ ||a||1 · |α|, where ||a||1 = ||a||L1 (Rd ) . Consequently, c(ζσn ) ≤ ||a||1 · |ζσn | = ||a||1 · |η0 | + n||a||1 . We see that

1 n c(ζσn )

P

= ∞ a.s., hence Proposition 12.19 in [Kal02] implies that σn → ∞.

We have proved the existence of a strong solution. The uniqueness follows by induction on jumps of the process. Namely, let (ζ˜t )t≥0 be a solution to (50). From (vii) of Definition 6.2 and the equality Z I[0,b(x,η0 )] (u)N (ds, dx, du) = 0, (0,σ1 )×Rd ×[0,∞]

20

it follows that P {ζ˜ has a birth before σ1 } = 0. At the same time, the equality Z I[0,b(x,η0 )] (u)N (ds, dx, du) = 1, {σ1 }×Rd ×[0,∞]

which holds a.s., yields that ζ˜ too has a birth at the moment σ1 , and in the same point of space at that. Therefore, ζ˜ coincides with ζ up to σ1 a.s. Similar reasoning shows that they coincide up to σn a.s., and, since σn → ∞ a.s., P {ζ˜t = ζt for all t ≥ 0} = 1. Thus, pathwise uniqueness holds. Joint uniqueness in law follows from the functional dependence between the solution to the equation and the ‘input’ η0 and N . Proposition 6.7. If b is rotation invariant, then so is (ηt ). Proof. It is sufficient to note that (Md ηt ), where Md ∈ SO(d), is the unique solution to (50) with N replaced by Md−1 N defined by Md−1 N ([0, q] × B × C) = N ([0, q] × Md−1 B × C),

q ≥ 0, B ∈ B(Rd ), C ∈ B(R+ ).

Md−1 N is a Poisson point process with the same intensity, therefore by uniqueness in law d

(Md ηt ) = (ηt ). Proposition 6.8. (The strong Markov property) Let τ be an (St , t ≥ 0)-stopping time and d

let η˜0 = ητ . Then d

(ητ +t , t ≥ 0) = (˜ ηt , t ≥ 0).

(53)

Furthermore, for any D ∈ B(DΓ0 [0, ∞)), P {(ητ +t , t ≥ 0) ∈ D | Sτ } = P {(ητ +t , t ≥ 0) ∈ D | ητ }; that is, given ητ , (ητ +t , t ≥ 0) is conditionally independent of (St , t ≥ 0). Proof. Note that Z ητ +t (B) =

1[0,b(x,s,ηs− )] (u)N (ds, dx, du) + ητ (B),

t ≥ q.

(τ,τ +t]×B×[0,∞)

Since the unique solution is measurable with respect to the filtration generated by the noise and initial condition, the conditional independence follows, and (53) follows from the uniqueness in law. Of course, we rely here on the strong Markov property of the Poisson point process.

21

6.2

Markovian functions of a Markov chain

Let (S, B(S)) be a Polish (state) space. Consider a (time-homogeneous) Markov chain on (S, B(S)) as a family of probability measures on S ∞ .

Namely, on the measurable space

(Ω, F ) = (S ∞ , B(S ∞ )) consider a family of probability measures {Ps }s∈S such that for the coordinate mappings

Xn : Ω → S, Xn (s1 , s2 ,...) = sn . The process X = {Xn }n∈Z+ is a Markov chain, and for all s ∈ S Ps {X0 = s} = 1, Ps {Xn+mj ∈ Aj , j = 1, ..., k1 | Fn } = PXn {Xmj ∈ Aj , j = 1, ..., k1 }. Here Aj ∈ B(S), mj ∈ N, k1 ∈ N, Fn = σ{X1 , ..., Xn }. The space S is separable, hence there exists a transition probability kernel Q : S × B(S) → [0, 1] such that Q(s, A) = Ps {X1 ∈ A},

s ∈ S, A ∈ B(S).

Consider a transformation of the chain X, Yn = f (Xn ), where f : S → R is a Borelmeasurable function. Here we will give sufficient conditions for Y = {Yn }n∈Z+ to be a Markov chain. A related question was discussed by Burke and Rosenblatt [BR58]. Lemma 6.9. Assume that for any bounded Borel function h : S → S Es h(X1 ) = Eq h(X1 ) whenever f (s) = f (q),

(54)

Then Y is a Markov chain. Remark. Condition (54) is the equality of distributions of X1 under two different measures, Ps and Pq . Proof. For the natural filtrations of the processes X and Y we have an inclusion FnX ⊃ FnY ,

n ∈ N,

(55)

since Y is a function of X. For k ∈ N and bounded Borel functions hj : R → R, j = 1, 2, ..., k,  Es 

k Y

 hj (Yn+j ) | FnX  = EXn

j=1

k Y

hj (f (Xj )) =

j=1

Q(x0 , dx1 )h1 (f (x1 )) Q(x1 , dx2 )h2 (f (x2 ))... Q(xn−1 , dxn )hn (f (xn )) S S S

Z

Z

Z

x0 =Xn

22

(56)

To transform the last integral, we introduce a new kernel: for y ∈ f (S) chose x ∈ S with f (x) = y, and then for B ∈ B(R) define Q(y, B) = Q(x, f −1 (B)).

(57)

The expression on the right-hand side does not depend on the choice of x because of (54). To make the kernel Q defined on R × B(R), we set Q(y, B) = 1{0∈B} , y ∈ / f (S). Then, setting zn = f (xn ), we obtain from the change of variables formula for the Lebesgue integral that Z

Z Q(xn−1 , dxn )hn (f (xn )) =

Q(f (xn−1 ), dzn )hn (zn ).

S

R

Likewise, setting zn−1 = f (xn−1 ), we get Z

Z Q(xn−2 , dxn−1 )hn (f (xn−1 ))

S

Z S

Q(xn−1 , dxn )hn (f (xn )) = S

Z Q(xn−2 , dxn−1 )hn (f (xn−1 )) Q(f (xn−1 ), dzn )hn (zn ) = R Z Z Q(f (xn−2 ), dzn−1 )hn (zn−1 ) Q(zn−1 , dzn )hn (zn ). R

R

Proceeding further, we obtain Z Z Z Q(x0 , dx1 )h1 (f (x1 )) Q(x1 , dx2 )h2 (f (x2 ))... Q(xn−1 , dxn )hn (f (xn )) = S

S

Z

S

Z

Z

Q(z0 , dz1 )h1 (z1 ) R

Q(z1 , dz2 )h2 (z2 )... R

Q(zn−1 , dzn )hn (zn ), R

where z0 = f (x0 ). Thus,  Es 

k Y

 hj (Yn+j ) | FnX  =

j=1

Z

Z Q(f (X0 ), dz1 )h1 (z1 )

R

Z Q(z1 , dz2 )h2 (z2 )...

R

Q(zn−1 , dzn )hn (zn ). R

This equality and (55) imply that Y is a Markov chain. Remark 6.10. From the proof it follows that Q is the transition probability kernel for the chain {f (Xn )}n∈Z+ . Remark 6.11. Clearly, this result holds for a Markov chain which is not necessarily defined on a canonical state space because the property of a process to be a Markov chain depends on its distribution only. 23

Acknowledgement Luca Di Persio would like to acknowledge the Informatics Department at the University of Verona for having funded the project “Stochastic differential equations with jumps in Mathematical Finance: applications to pricing, hedging and dynamic risk measure’s problems” and the Gruppo Nazionale per l’analisi matematica, la Probabilit`a e le loro applicazioni (GNAMPA). Viktor Bezborodov is supported by the Department of Computer Science at the University of Verona. Viktor Bezborodov was also partially supported by the DFG through the SFB 701 “Spektrale Strukturen und Topologische Methoden in der Mathematik” and by the European Commission under the project STREVCOMS PIRSES-2013-612669. Mykola Lebid is supported by the Department of Biosystems Science and Engineering at the ETH Z¨ urich. Tyll Krueger and Tomasz O˙za´ nski are supported by the Wroclaw University of Technology.

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