Randomwalksystems

Random Walk Systems on the Complete Graph

T. Kurtz1 E. Lebensztayn2 A. Leichsenring 2 F. Machado 2
1 Department of Mathematics
University of Wisconsin
2 Institutode Matemática e Estatística
Universidade de São Paulo

March 15, 2007

Kurtz, Lebensztayn, Leichsenring, Machado (Universidade de São on the and University of Wisconsin) 15, 2007
Random Walk Systems Paulo Complete Graph March 1 / 21

Introduction

Introduction

Version of the Frog Model on the complete graph
Let Kn be the complete graph with n vertices and L ≥ 1 a xed integer
Initially: a xed number of particles at each vertex of Kn
Some of them are active, the others inactive
Active particles:
realize continuous time random walk through the vertices of Kn
wait a rate 1 exponential time and jump uniformly to the vertices of Kn
have L lives each
- each particle loses 1 life when it jumps to a visited vertex
- dies when it reaches 0 (zero) lives
Inactive particles: become active when some active particle jumps over
the vertex where they are placed


Introduction

For a realization of the Frog Model on Kn , consider
V n (t ): Number of visited vertices at time t ,
An (t ), i = 1, . . . , L: Number of active particles with i lives at time t .
i
Initial conditions:
Vn (0) = ρn n, with ρn n integer and such that 0 ρn 1,
0 0 0 (1)
An (0)
i = ρn n, for i = 1, . . . , L, with ρn n integer and such
i i (2)
L
that ρn ≥ 0 and
i ρn 0.
i
i =1

γ n := inf {t : L An (t ) = 0}, the rst time there are no active particles.
i =1 i
We look for V n ) , the proportion of visited vertices at the end of the
(γ n n

process. We call it the cover of the process.


Introduction

Figure: Simulation for z 1000 (t ) starting with 1 active particle (with L = 2).


Introduction

Z
n (t ) = ( V
n (t ), n
A1 (t ) , n
A2 (t ) , ... , n
AL (t ) )

is a Markov chain in ZL+1 governed by jump intensities
qln (Z ), Z , l ∈ ZL+1 , representing the transition rates from state Z to
Z + l . We observe that qln (Z ) = 0 for l outside Λ, the set of the possible
transitions. Note that Λ = {l0 , l1 , . . . , lL }, where
l0 = (1, 0, 0, . . . , 0, 1), → a particle jumps to an unvisited vertex
l1 = (0, −1, 0, . . . , 0, 0), → death of a particle
lj = (0, . . . , 0, 1 , −1 , . . . , 0), j = 2, . . . , L.
An−1 Aj n
j
→ a particle with j lives misses one life


Introduction

If the process is in the state Z = (V , A1 , . . . , AL ) the rates qln (Z ) have the
following form:

L
n V
ql0 (Z ) = (1 − ) Ai
n
i =1
n V
ql (Z ) =
i
Ai , i = 1, . . . , L,
n

and qln (Z ) = 0 for l outside Λ.


Introduction

Then the distribution of Z n (t ) is determined by

P (Z n (t + h) = Z + l |Z n (t ) = Z ) = hqln (Z ) + o (h), l = 0
n n
P(Z (t + h ) = Z |Z (t ) = Z ) = 1 −
n
hql (Z ) + o (h).
l
Now, let Y0 , Y1 , . . . , YL be independet standard Poisson processes dened
for each of the transitions in Λ. Then Z n (t ) can be written as
t
Z
n (t ) = Z n (0) + lYl (
n
ql (Z
n (s ))ds ).
l ∈Λ 0

Given Z n (t ) = Z , the probability that there will be a jump in
t
Yl ( 0 qln (Z n (s ))ds ) during (t , t + h) is hqln (Z ) + o (h) since the integrand
is equal to qln (Z ) until the rst jump after t .


Introduction

Let
= Z n(t ) = (v n (t ), a1 (t ), a2 (t ), . . . , aln (t )), and
n
z n (t ) n n
an (t ) =
L an (t ).
i =1 i

v n (0)= ρn ∈ [0, 1],
0
Remeber that
ai (0) = ρn 0, i = 1, . . . , L.
n
i

Then z n (t ) is a Markov chain with state space { n z : z ∈ ZL+1 } and
1

satises

l
t
z
n (t ) = z
n (0) + Yl n fl (Z
n (s ))ds . (3)
n 0
l ∈Λ


Introduction

In other words:
t
v n (t ) = ρn + Y0 n (1 − v n (s ))an (s )ds
1
0 ,
n 0
t t
ain (t ) = ρn + Yi +1 n v n (s )ain+1 (s )ds − Yi n v n (s )ain (s )ds ,
1 1
i
n 0 n 0
i = 1, . . . , L − 1,
t t
aL (t ) = ρn + Y0 n
n
(1 − v n (s ))an (s )ds v n (s )ain (s )ds .
1 1
L − YL n
n 0 n 0

L
where an (s ) = n
i =1 ai (s ) and Yi is the Poisson process associated to the vector li .


Deterministic Limit

Deterministic Limit

Dene the drift function F by F (z ) = l ∈Λ lfl (z ), for z ∈ RL+1 . Let
t
z (t ) = z0 + F (z (s ))ds . (4)
0

Then, the following theorem guarantees the convergence of z n (t ) to the
deterministic model z (t ).
Theorem
Suppose limn→∞ z n (0) = z0 and that for each compact K ∈ RL+1 there is
a constant MK 0 such that |F (x ) − F (y )| ≤ MK |x − y | for all x , y ∈ K.
Then limn→∞ sups ≤t |z n (s ) − z (s )| =0 almost surely, where z (t ) is the
unique solution to (4).


Deterministic Limit

Supposing that (1) and (2) hold with ρn → ρi , i = 0, . . . , L, it results that
i
the corresponing deterministic model for z n (t ) is governed by

t
v (t ) = ρ0 + (1 − v (s ))a(s )ds ,
0
t
ai (t ) = ρi + v (s )(ai +1 (s ) − ai (s ))ds , i = 1, . . . , L − 1, (5)
0
t
aL (t ) = ρL + (1 − v (s ))a(s ) − v (s )aL (s )ds ,
0

where a(s ) = L a (s ).
i =1 i


Deterministic Limit

For the one-particle-per-site initial conguration, for example, the
corresponding initial conditions on the deterministic model are
ρ0 = ρ 1 = . . . = ρL = 0 . (6)
Since the solution of (5)-(6) is the limit of the proportions as n → ∞ in
any bounded time interval [0, T ], we have that for any t ≥ 0,

limn→∞ v
n (t ) = 0, limn→∞ ai (t )
n = 0, ∀i = 1, . . . , L, a.s.

Don't say much about the behaviour of the process for large values of
n;
γ n is going to innity witn n and the limit equations are valid only for
bounded intervals;
We are interested in the state where the process is absorbed.
⇒ We can speed up the process without aecting where it hits the
boundary.

Time Change

Time Change

Make the time change dened by
τ
τ n (t ) = inf {τ : a
n (s )ds = t },
0
and let us observe the process at time τ n (t ), dening z n (t ) = z n (τn (t )).
˜

Proposition
This change of scale speeds up the process in such a way that the times
between two jumps of the process are always rate n exponentials.


Time Change

The speeded system z n (t ) = z n (τn (t )) is described by the following
˜
equations:

1 t
v n (t ) = v n (τ (t ))
˜ = ρn + Y0 n (1 − v n (s ))ds
˜ ,
0
n 0
t ˜in+1 (s )
a 1 t ˜in (s )
a
˜in (t ) = ain (τ (t ))
a = ρ n + Yi + 1
i v n (s )
˜ ds − Yi n v n (s )
˜ ds ,
0 ñ (s )
a n 0 ñ (s )
a
i = 1, . . . , L − 1,
1 t 1 t ˜L (s )
an
˜L (t ) = aL (τ (t ))
an n = ρn +
L Y0 n (1 − v n (s ))ds
˜ − YL n v n (s )
˜ ds ,
n 0 n 0 ñ (s )
a
1 t 1 t ñ (s )
a
ñ (t ) = an (τ (t ))
a = ρn + Y0
i n (1 − v n (s ))ds
˜ − Y1 n v n (s ) 1 ds
˜ .
i
n 0 n 0 ñ (s )
a


Time Change

The time-changed system has nontrivial limit,

t
v (t ) = ρ0 + (1 − v (s ))ds ,
0
t ai +1 (s )
t ai (s )
ai (t ) = ρi + v (s ) ds − v (s ) ds ,
0 a(s ) 0 a(s )
= 1, . . . , L − 1,
i (7)
t t aL (s )
aL (t ) = ρL + (1 − v (s ))ds − v (s ) ds ,
0 0 a(s )
t t a1 (s )
a(t ) = ρi + (1 − v (s ))ds − v (s ) ds ,
0 0 a(s )
i
dened until γ = inf {t : a(t ) = 0}.
˜


Convergence of v n (γ n )


Let γ n = inf {t : ˜n (t ) = 0}. We are looking for v = limn→ v n (γ n ), which
˜ a
is the same as the limit of v n (˜ n ).
˜ γ

Theorem
The limit cover v of the process is given by

v = 1 − (1 − ρ0 ) exp {−LambertW (−cρ0 ,L exp{−cρ0 ,L − ρ}) − cρ0 ,L − ρ} ,

= (1 − ρ0 )(L + 1) L iρ .
where cρ0 ,L and ρ= i =1 i

Obs. The LambertW function is the inverse of the function x → xe x .



Proof. We prove the convergence of v n (γ n ) through proving the
convergence of γ n .
˜

Theorem
γn
˜ converges almost surely to the solution of

γ − ρ = (L + 1)(1 − e −γ )(1 − ρ0 ). (8)



As ilustration we return to the case ρ0 = ρ = 0 and plot γ and v for some
˜
values of L.

Figure: γ (L)
˜ Figure: v (L)
An interesting fact is that for these values of ρ0 and ρ, when L = 1 the limit
cover is approximately 0, 798612, and for all nit L it is less than 1.

Number of Jumps of the Model

Number of Jumps of the Model

From the convergence of γ n results the following corolary:
˜

Corolary
Let Nn be the total number of jumps the original model makes until the
end of the process. Then,

E (Nn )
lim = γ.
˜
n→∞ n


Central Limit Theorem


We can also derive a central limit theorem for vn (γ n ) for the case
ρ0 = ρ = 0:

Theorem
√
n(vn (γ
n ) − v ) n→∞ N (0, σ 2 ),
⇒

with

γeγ
˜ ˜
σ2 = .
(L + 1)(˜ − L)
γ



For L = 2, σ2
is approximately 0.06815. Bellow we present the histogram of
10000 simulations of the chain for this case with n = 1000. The curve in red is
the normal distribution with mean 0 and variance 0.06815.

√

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