Discrete Applied Mathematics

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Discrete Applied Mathematics

journal homepage:www.elsevier.com/locate/dam

Commute times of random walks on trees

Mokhtar Konsowa

^a,Ď

, Fahimah Al-Awadhi

^a

, András Telcs

^b,∗

aDepartment of Statistics and Operations Research, Faculty of science, Kuwait University, Safat 13060, P.O.Box 5969, Kuwait

bDepartment of Quantitative Methods, Faculty of Economics, Veszprém, Egyetem utca 10, 8200 Veszprém, Hungary

a r t i c l e i n f o

Article history:

Received 12 November 2011

Received in revised form 4 October 2012 Accepted 7 October 2012

Available online 8 November 2012 Keywords:

Random walk Commute time

Spherically symmetric tree Resistance

a b s t r a c t

In this paper we provide exact formula for the commute times of random walks on spherically symmetric random trees. Using this formula we sharpen some of the results presented in Al-Awadhi et al. to the form of equalities rather than inequalities.

©2012 Elsevier B.V. All rights reserved.

1. Introduction

The commute time is a particular measure of random walks on weighted graphs. It has several nice properties which has been revealed independently partly or fully by many authors, see for example [5,2,3,14,21,1]. It is still in the focus of the research of computer scientists, probabilists and physicists as well. As examples, consider the tasks of graph embedding [9,16,18,22], graph sparsification [20], social network analysis, [13], proximity search [19], collaborative filtering [8], clustering [23], semisupervised learning [24], dimensionality reduction [10] image processing [17], graph labeling [11], and theoretical computer science [4,6]. For an extensive list of literature we refer the reader to [15]. Random walks on random graphs have been subject of permanent interest in the last three decades. Interestingly enough, very little is published on commute times of random walks on random graphs. The present paper studies commute times on very simple random objects, on spherically symmetric random trees,SSRT. Explicit results are presented in the annealed case, averaged commute times over the probability field of trees.

2. Commute times

Consider a random walk on a weighted graphG

= (

^V

,

^E

)

where a weight (conductivity)c_xy

=

c_yxis assigned to edge xy

∈

E. The commute time between two verticesrandsis the mean number of steps it takes the random walk to go fromr tosand back torand will be denoted byE

(τ) =

_E

(τ

r,s

)

. We know, see [5], for a finite connected graph,

E

(τ) =

2

ρ

rs

µ

rs

,

^(2.1)

where

ρ = ρ

rsis the effective resistance betweenrandsand

µ = µ

rs

=

¹

2



e∈Ec_e. If the assigned weights are all equal 1, then

E

(τ) =

2

ρ

^m

,

∗Corresponding author. Tel.: +36 303753896; fax: +36 14633157.

E-mail addresses:fahimah@kuc01.kuniv.edu.kw(F. Al-Awadhi),telcs.szit.bme@gmail.com(A. Telcs).

ĎOur co-author and friend Mokhtar Konsowa passed away with tragic suddenness while we were revising this submission.

0166-218X/$ – see front matter©2012 Elsevier B.V. All rights reserved.

doi:10.1016/j.dam.2012.10.006

wherem

= |

E

|

is the number of undirected edges ofG. We confine our study to investigating the commute time of random walk on spherically symmetric random treesSSRTin which the degree of a vertex depends only on its distance from the rootr.

The second probability space is given on the spherically symmetric trees of infinite heights and the corresponding probability and expectation will be denoted byPandE.

This type of trees is completely determined by its degree sequence

{

d_n

;

n

≥

0

}

whered_nis the degree of every node at leveln. Let

ℑ

_n

= σ (

^d1

,

^d2

, . . . ,

^dn

)

. Then, for each realizationT

(ω)

of a random treeT,

E

(τ |ℑ

_n

) =

2

ρµ.

We are interested in the expected value with respect toP

(ω)

, probability distribution on the set of all possible treesT. In such a case,

E

(τ) =

2E

(ρµ).

It was shown in [1] thatE

(τ) ≤

_2E

(ρ)

E

(µ)

. It can easily be seen that this inequality can not be strengthened to equality.

We first note that for positive nondegenerate random variableX, the functionf

(

^X

) =

¹

X is strictly convex and hence E

(

¹

/

^X

)

1

/

E

(

^X

).

Consider now a treeT of height 1 rooted atrwhich has random degreed₀. Thenm

=

d₀and

ρ

rs

=

1

/

^d0. Hence,E

(τ) =

2E

(ρ

^m

) =

2. On the other hand, 2E

(ρ)

E

(

^m

)

2

.

Now we seek for asymptotic equality forE

(τ)

^{. LetS}ibe the sphere of radiusiand centered atr; that is the set of vertices at distanceifromr. Let

ρ

i

= ρ (

^Si−1

,

^Si

)

^,

µ

i

= µ (

^Si−1

,

^Si

)

^{, and}E

(τ)

is the commute time between the root and the sphere of radiusnshorted in one vertex. Then

E

(τ) =

2E

(ρµ) =

2E



_n



i=1

µ

i

 

_n



j=1

ρ

j



=

2E



_n



i=1 n



j=1

µ

i

ρ

j



=

2E







n



i=1 n



j=1 j̸=i

µ

i

ρ

j





 +

2n

,

^(2.2)

where the last step uses the fact that

ρ

i

=

1

/µ

i. Belowd⁺_j will denote the outdegree of statej; that isd⁺_j

=

d_j

−

1. Now we concentrate on the double sum.

I_n

+

J_n

:=

_E



_n



i=2 i−1



j=1

µ

i

ρ

j

 +

_E



_n−1



i=1 n



j=i+1

µ

i

ρ

j

 .

Now,

I_n

=

_E



_n



i=2 i−1



j=1

µ

i

ρ

j



=

_E



_n−1



j=1

ρ

j



_n



i=j+1

µ

i



=

_E



_n−1



j=1

ρ

j

µ 

S_j

,

^Sn





=

n−1



j=1

E

 ρ

j

µ 

S_j

,

^Sn



=

n−1



j=1

∞



k=1

E

 ρ

j

µ 

S_j

,

^Sn

 | µ

j

=

k



P



µ

j

=

k



=

∞



k=1 n−1



j=1

E



1

k

(µ

j+1

+ µ

j+2

+ · · · + µ

n

) | µ

j

=

k



P



µ

j

=

k



=

∞



k=1 n−1



j=1

E



1

kk

(

^d+j

+

d⁺_j d⁺_j+1

+ · · · +

d⁺_jd⁺_j+1

· · ·

d⁺_n−1

)



P



µ

j

=

k



=

n−1



j=1 n−1



x=j

E



Πi^x=jd⁺_i



=

n−1



x=1 x



j=1

E



Πi^x=jd⁺_i



.

On the other hand, J_n

=

_E



_n−1



i=1 n



j=i+1

µ

i

ρ

j



=

∞



k=1

E



_n−1



i=1 n



j=i+1

µ

i

ρ

j

| µ

i

=

k



P

(µ

i

=

k

)

=

∞



k=1

E



_n−1



i=1

k

(ρ

i+1

+ ρ

i+2

+ · · · + ρ

n

) | µ

i

=

k



P

(µ

i

=

k

)

=

_E



_n−1



i=1

k1 k



1 d⁺_i

+

¹

d⁺_i d⁺_i+1

+ · · · +

¹ d⁺_i d⁺_i+1

· · ·

d⁺_n−1



=

n−1



i=1

E



1 d⁺_i

+

¹

d⁺_i d⁺_i+1

+ · · · +

¹ d⁺_i d⁺_i+1

· · ·

d⁺_n−1



=

n−1



i=1 n−1



x=i

E



Πj^x=i

1 d⁺_j



=

n−1



x=1 x



i=1

E



Πj^x=i

1 d⁺_j

 ,

where in the third step the condition

µ

i

=

kis used to calculate the resistance ofkparallel branches starting in leveli. Finally, E

(τ) =

2n

+

2

n−1



x=1 x



j=1



E



Πi^x=jd⁺_i

 +

_E



Πi^x=j

1 d⁺_i



.

^(2.3)

3. Spherically symmetric trees

In this section we use

τ

rsto denote the commute time between the rootrof aSSRTΓand the levelnshorted in one node s, while

τ

rxndenotes the commute time betweenrand a leafx_nof leveln. We assign a unit resistance to every edge ofΓ^. We also used⁺_n to refer the outdegree of each node of levelnandZ_nto refer to the number of vertices in the leveln. Then Z_n

=

_Πkⁿ₌⁻₀¹_dk. We will assume thatd⁺_n’s are independent random variables. We need the following lemma from [12].

Lemma 1. Consider two nonnegative sequences a_nand b_nsuch that



nb_nis divergent. Iflim_n^a_bⁿ

n

=

L, thenlim_n

n k=1ak

_n

k=1bk

=

L.

Notation 1. We use a_n

=

_Θ

(

^bn

)

^forlimnan

bn

ϵ (

⁰

, ∞ )

^.

The following theorem strengthen Theorem 1 of [1] that gives only an upper bound for the commute time.

Theorem 1. Consider a spherically symmetric random treeΓ ^{such that} d⁺_n

=



1 with probab.1

−

q_n 2 with probab. q_n and



q_n

< ∞ .

^Then

τ

rs

=

_Θ



n²



P-a.s.

and

τ

rxn

=

_Θ



n²



P-a.s.

.

Proof. Applying the Borel Cantelli Lemma to the infinite outdegree sequence

{

d⁺_n

}

shows thatp

(

^dn

=

1 eventually

) =

1.

As such, there isNsuch thatp

(

^dn

=

1

,

ⁿ

≥

N

) =

1. Then

µ

rs

=

n



k=1

Zk

−

1

=

N



k=1

Zk

+

n



k=N+1

Zk

−

1

=

_Θ

(

ⁿ

)

^a.s.

whereZk

= |

S_k

|

. Similarly,

ρ

rs

=

_Θ

(

ⁿ

)

^a.s.

Therefore,

µ

rs

ρ

rs

=

_Θ



n²



a.s.

and the result follows from Eq.(2.1).

The following lemma is presented in [7, p. 63].

Lemma 2. For any c

>

^0, Πjⁿ=1



1

+

^c

j



∼

n^c

.

The following lemma is presented in [12, p. 66].

Lemma 3. For0

< α <

^1,

n



k=1

1 k^α

∼

ⁿ

1−α 1

− α .

Theorem 2.Consider a SSRTΓ ^{such that} d⁺_n

=



1 with probab.1

−

q_n 2 with probab. q_n where q_n

=

min

(

¹

,

^c

/

ⁿ

) ,

^c

>

^{0. Then}

E

(τ

rs

) =

_Θ



n²logn



if c

=

1

,

E

(τ

rs

) =

_Θ



n²



if c

<

¹ E

(τ

rs

) =

_Θ



n^c+1



if c

>

¹

.

Moreover, for any c

>

⁰

,

E

 τ

rxn

 =

_Θ



n^c+2



.

Proof. We first note that as long as theΘ-asymptotic behavior of the commute time is our concern and sinceI_nis greater thanJ_n, it is enough to calculateI_n. It follows forc

=

1 that

E



Π_i^x=jd⁺_i



=

_Πi^x_=j



1

+

¹

i



=

^x

+

1 j

.

Then,

I_n

=

n−1



x=1 x



j=1

x

+

1 j

=

n−1



x=1

(

^x

+

1

) α

xlogx

; α

j

−→

1

∼

n−1



x=₁

(

^x

+

1

)

^logx

=

_Θ



n²logn



,

where the last equality follows from the proof of Theorem 11 of [1]. Let us recall that inJ_nthe product of expected values is

x+¹₂

j while inI_n^x+_j¹. It follows that J_n

=

I_n

−

n−1



x=1 x



j=1

1 2j

which means that I_n

−

J_n

∼

n−1



x=1

1 2logx

.

But



n−1

x=1 1

2logx

=

_o



n²logn



and since lim_nn2^Ilogⁿ n

∈ (

⁰

, ∞ )

we have that lim_nn^I2ⁿ⁺log^Jnn

=

_limn ^2In

n²logn

∈ (

⁰

, ∞ ) .

We consider now the casec

<

¹

.

It follows fromLemma 2that

Π_i^x=j



1

+

^c

i



=

^x

c

α

x

j^c

α

j

; α

x

−→ α ∈ (

⁰

, ∞ ) ,

we see

I_n

=

n−1



x=1 x



j=1

x^c

α

x

j^c

α

j

=

n−1



x=1

x^c

α

x x



j=1

1

j^c

α

j

.

^(3.1)

It follows fromLemmas 1and3that I_n

∼

n−1



x=1

α

xx^c

λ

x

x^−c+1

−

c

+

1

; λ

x

→ λ ∈ (

⁰

, ∞ )

=

_Θ



n²



.

While forc

>

¹

,

x



j=1

1 j^c

α

j

=

_Θ

(

¹

)

and then, from Eq.(3.1)andLemma 1, I_n

=

n−1



x=1

x^c

α

x x



j=1

1 j^c

α

j

∼

n−1



x=1

x^c

α

x

∼

n−1



x=1

x^c

.

Hence,

I_n

=

_Θ



n^c+1



.

The result forE

(τ

rxn

)

follows from the fact that

ρ

rxn

=

nand applyingLemma 2gives

n



k=1

E

(

^Zk

) =

n



k=1

Πi^k=1



1

+

^c

i



∼

n



k=₁

k^c

=

_Θ



n^c+1



.

The following lemma is analogous to Theorem 3, p. 64 of [12].

Lemma 4. Consider a positive decreasing function f and define a sequence a_k, k

=

1

,

²

, . . .

such that f

(

^t

) =

a_t. Let In

=



n

1 f

(

^t

)

^{dt and}Sn

= 

n

k=1a_k. If lim_nIn

= ∞

then limn

Sn

In

=

1

.

Proof. Sincefis decreasing, then forj

=

2

,

³

, . . .



j+1 j

f

(

^x

)

^dx

≤

a_j

≤



j j−1

f

(

^x

)

^dx

.

By summing overj, we obtain



n+1 2

f

(

^x

)

^dx

≤

_Sn

−

a₁

≤



n 1

f

(

^x

)

^dx

.

That is,

In+1

−

_I2

≤

_Sn

−

a₁

≤

_In

.

^(3.2)

It follows also that



n+1 n

f

(

^x

)

^dx

≤



2 1

f

(

^x

)

^dx

≤

C

.

As such,

lim

n



n+1 n f

(

^x

)

^dx

In

=

0

,

which implies that

limn

In+₁ In

=

lim

n



1

+



n+1 n f

(

^x

)

^dx

In



=

1

,

and the result follows from Eq.(3.2).

Theorem 3.Consider a SSRTΓ such that for0

< α <

¹ d⁺_n

=



 

 

1 with probab.1

−

¹ n^α 2 with probab. 1

n^α

.

Then for any

ϵ <

₁−¹α

,

there exists N such that for n

≥

N, the following inequalities hold

(

ⁱ

)

Θ



n^α+1exp



1 1

− α − ϵ



n^1−α



≤

_E

 τ

rxn

 ≤

_Θ



n^α+1exp



1 1

− α + ϵ



n^1−α



, (

ⁱⁱ

)

Θ



n^αexp



1 1

− α − ϵ



n^1−α



≤

_E

(τ

rs

) ≤

_Θ



n^αexp



1 1

− α + ϵ



n^1−α



.

Remark 1. The case

α >

1 is covered byTheorem 1and the case

α =

1 is covered byTheorem 2.

Proof. LetSn

=

logE

(

Zn

) = 

n−1 k=0log



1

+

¹

kα



.

We first show that limn

Sn

n^1−α

=

¹

1

− α .

^(3.3)

Since In

=



n 1

log



1

+

¹

x^α



dx

=

nlog



1

+

¹

n^α



−

log 2

+ α 

n 1

1

1

+

x^αdx

,

^(3.4)

and lim

n

nlog



1

+

¹

nα



n^1−α

=

1

,

and also

limn



n 1

1 1+xαdx n^1−α

=

lim

n 1 1+nα

(

¹

− α)

^n−α

=

¹ 1

− α ,

then, from(3.4),

limn

In

n^1−α

=

1

+ α

1

− α =

¹ 1

− α .

It follows fromLemma 4that lim

n

Sn

n^1−α

=

lim

n

Sn

In

·

^In n^1−α

=

lim

n

In

n^1−α

=

¹ 1

− α .

As such,

E

(

Zn

) =

exp

 γ

nn^1−α



; γ

n

→

¹

1

− α .

^(3.5)

That is, for arbitrary small

ϵ >

0, there is a sufficiently largeNsuch that forn

≥

N

,

exp



1 1

− α − ϵ



n^1−α



≤

_E

(

Zn

) ≤

exp



1 1

− α + ϵ



n^1−α



,

^(3.6)

and

n



k=N

exp



1 1

− α − ϵ



k^1−α



≤

n



k=N

E

(

Zk

) ≤

n



k=N

exp



1 1

− α + ϵ



k^1−α

 .

Since,

lim

n



n

Nexp



₁

1−α

+ ϵ 

x^1−α



dx n^αexp



₁

1−α

+ ϵ 

n^1−α

 =

¹ 1

+ ϵ (

¹

− α) ,

and



n

Nexp



₁

1−α

− ϵ 

x^1−α



dx n^αexp



₁

1−α

− ϵ 

n^1−α

 =

¹ 1

− ϵ (

¹

− α) ,

then

Θ



n^αexp



1 1

− α − ϵ



n^1−α



≤

n



k=N

E

(

Zk

) ≤

_Θ



n^αexp



1 1

− α + ϵ



n^1−α



and the result forE

 τ

rxn



follows since

ρ

rxn

=

n

.

ForE

(τ

rs

)

, we follow the same argument of computingI_nas in the proof ofTheorem 2. From Eq.(3.5), we see that E

(

Πi^x=jd⁺_i

) =

_Πi^x_=j



1

+

¹

i^α



=

exp



γ

xx^1−α

− γ

jj^1−α



; γ

x

→

¹ 1

− α .

As such,

x



j=1

E

(

Πi^x=jd⁺_i

) =

x



j=1

exp



γ

x

(

^x1−α

) − γ

j

(

^j1−α

) 

= 

exp

γ

x



x^1−α



x



j=1

exp



− γ

j

(

^j1−α

)  .

Using the fact that the two series



a_nand



2^νa₂ν have the same convergence behavior, we can see that

x



j=1

exp



− γ

j

(

^j1−α

) 

=

_Θ

(

¹

)

and hence,

x



j=1

E

(

Π_i^x=jd⁺_i

) ∼

exp

(γ

x

(

^x1−α

)).

It follows then that for arbitrary small

ϵ >

0, and sufficiently largeN, exp



1 1

− α − ϵ



x^1−α



≤

x



j=1

E

(

Πi^x=jd⁺_i

) ≤

exp



1 1

− α + ϵ



x^1−α



;

x

≥

N

n



x=N

exp



1 1

− α − ϵ



x^1−α



≤

n−1



x=N x



j=1

E

(

Πi^x=jd⁺_i

) ≤

n



x=N

exp



1 1

− α + ϵ



x^1−α



.

The same argument of part (i) shows that Θ



n^αexp



1 1

− α − ϵ



n^1−α



≤

I_n

≤

_Θ



n^αexp



1 1

− α + ϵ



n^1−α



,

and this proves part (ii).

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