CAPITULO 4: PROPUESTA ESTRATÉGICA DE NEGOCIO INCLUSIVO
4.10 Programa de capacitación en microempresa y producción sábila
11.3. Applications of the Matthews Method
11.3.1. Binary trees. Consider simple random walk on the rooted binary tree with depthkandn= 2k+1−1 vertices, which we first discussed in Section5.3.4.
The maximal hitting time will be realized by pairs of leavesa, bwhose most recent common ancestor is the root (see Exercise10.4). For such a pair, the hitting time will, by symmetry, be the same as the commute time between the root and one of the leaves. By Proposition10.6(the Commute Time Identity), we have
Eaτb= 2(n−1)k
(since the effective resistance between the root and the leaf is k, by Example 9.7, and the total conductancecG of the network is twice the number of edges). Hence
Theorem11.2gives tcov ≤2(n−1)k 1 +1 2 +· · ·+ 1 n = (2 +o(1))(log 2)nk2. (11.6)
What about a lower bound? We need an appropriate set A⊆X. Fix a level
hin the tree, and letA be a set of 2hleaves chosen so that each vertex at levelh
has a unique descendant inA. Notice that the largerhis, the more vertices there are in A—and the closer together those vertices can be. We will choose a value of
hbelow to optimize our bound.
Consider two distinct leaves a, b∈ A. If their closest common ancestor is at levelh′< h, then the hitting time from one to the other is the same as the commute
time from their common ancestor to one of them, saya. Again, by the Commute Time Identity (Proposition10.6) and Example9.7, this is exactly
Eaτb= 2(n−1)(k−h′),
which is clearly minimized whenh′ =h−1. By Proposition11.4,
tcov≥2(n−1)(k−h+ 1) 1 +1 2 +· · ·+ 1 2h−1 = (2 +o(1))(log 2)n(k−h)h. (11.7) Setting h=⌊k/2⌋in (11.7) gives tcov ≥1 4 ·(2 +o(1))(log 2)nk 2. (11.8)
There is still a factor of 4 gap between the upper bound of (11.6) and the lower bound of (11.8). In fact, the upper bound is sharp. See the Notes.
11.3.2. Tori. In Section 10.4 we derived fairly sharp (up to constants) esti- mates for the hitting times of simple random walks on finite tori of various dimen- sions. Let’s use these bounds and the Matthews method to determine equally sharp bounds on the expected cover times of tori. We discuss the case of dimension at least 3 first, since the details are a bit simpler.
When the dimensiond≥3, Proposition10.13tells us that there exist constants
cdandCd such that for any distinct verticesx, y ofZdn,
148 11. COVER TIMES
Remarkably, this bound does not depend on the distance between x and y! By Theorem11.2, the average cover time satisfies
tcov≤Cdnd 1 +1 2 +· · ·+ 1 nd (11.9) =Cddndlogn(1 +o(1)). (11.10)
To derive an almost-matching lower bound from Proposition 11.4, we must choose a set A large enough that the sum of reciprocals in the second factor is substantial. Let’s takeAto beZd
n itself (any set containing a fixed positive fraction
of the points of the torus would work as well). Then
tcov≥tAmin 1 + 1 2+· · ·+ 1 |A| −1 ≥cddndlogn(1 +o(1)),
which is only a constant factor away from our upper bound.
In dimension 2, Proposition10.13says that whenxandy are vertices ofZ2n at
distancek,
c2n2log(k)≤Ex(τy)≤C2n2log(k).
In this case the Matthews upper bound gives
E(τcov)≤2C2n2(logn)2(1 +o(1)), (11.11)
since the furthest apart two points can be isn.
To get a good lower bound, we must choose a setA which is as large as pos- sible, but for which the minimum distance between points is also large. Assume for simplicity that n is a perfect square, and let A be the set of all points in Z2
n
both of whose coordinates are multiples of√n. Then Proposition11.4and Propo- sition10.13imply E(τcov)≥c2n2log(√n) 1 + 1 2+· · ·+ 1 n−1 = c2 2n 2(logn)2(1 +o(1)).
Figure11.1shows the points of a 75×75 torus left uncovered by a single random walk trajectory at equally spaced fractions of its cover time.
Exercises11.4and11.5use improved estimates on the hitting times to get our upper and lower bounds for cover times on tori even closer together.
11.3.3. Waiting for all patterns in coin tossing. In Section 17.3.2, we will use elementary martingale methods to compute the expected time to the first occurrence of a specified pattern (such asHT HHT T H) in a sequence of indepen- dent fair coin tosses. Here we examine the time required for all 2k patterns of
lengthk to have appeared. In order to apply the Matthews method, we first give a simple universal bound on the expected hitting time of any pattern.
Consider the Markov chain whose state space is the collection Ω = {0,1}k
of binary k-tuples and whose transitions are as follows: at each step, delete the leftmost bit and append on the right a new fair random bit independent of all earlier bits. We can also view this chain as sliding a window of widthkfrom left to right along a stream of independent fair bits. (In fact, the winning streak chain of
11.3. APPLICATIONS OF THE MATTHEWS METHOD 149
Figure 11.1. Black squares show the states unvisited by a single trajectory of simple random walk on a 75×75 torus. This tra- jectory took 145,404 steps to cover. The diagrams show the walk after 10%, 20%,. . ., 100% of its cover time.
Example 4.15is a lumping of this chain—see Lemma2.5.) We call this the shift chain on binary k-tuples.
In the coin tossing picture, it is natural to consider the waiting time wx for
a pattern x ∈ {0,1}k, which is defined to be the number of steps required for x
to appear using all “new” bits—that is, without any overlap with the initial state. Note that
wx≥k and wx≥τx for allx∈ {0,1}k. (11.12)
Also,wxdoes not depend on the initial state of the chain. Hence
Ewx≥Exτx+= 2k (11.13)
(the last equality follows immediately from (1.26), since our chain has a uniform stationary distribution).
Lemma 11.6. Fix k≥1. For the shift chain on binaryk-tuples,
Hk := max
x∈{0,1}kEwx= 2
k+1
−2.
Proof. Whenk= 1,wx is geometric with parameter 2. HenceH1= 2.
Now fix a pattern xof length k+ 1 and let x− be the pattern consisting of
the firstkbits of x. To arrive atx, we must first build upx−. Flipping one more coin has probability 1/2 of completing patternx. If it does not, we resume waiting for x. The additional time required is certainly bounded by the time required to constructxfrom entirely new bits. Hence
Ewx≤Ewx−+ 1 +1
2Ewx. (11.14)
To boundHk+1 in terms of Hk, choose anxthat achievesHk+1 =Ewx. On the
right-hand-side of (11.14), the first term is bounded byHk, while the third is equal
to (1/2)Hk+1. We conclude that
Hk+1 ≤Hk+ 1 +
1 2Hk+1, which can be rewritten as
150 11. COVER TIMES
This recursion, together with the initial conditionH1= 2, impliesHk≤2k+1−2.
Whenxis a constant pattern (all 0’s or all 1’s) of lengthkandyis any pattern ending in the opposite bit, we have Eyτx = Hk = 2k+1−2. Indeed, since one
inappropriate bit requires a copy ofxto be built from new bits, equality of hitting time and waiting time holds throughout the induction above.
We can now combine Lemma11.6with (11.12) and the Matthews upper bound of Theorem11.2, obtaining Ex(τcov)≤Hk 1 +1 2 +· · ·+ 1 2k = (log 2)k2k+1(1 +o(1)).
Looking more closely at the relationship between hitting times and waiting times will allow us to improve this upper bound by a factor of 2 and to prove a matching lower bound.
Lemma 11.7. Let θ=θa,b=Pa(τb+ < k). Then for anya, b∈ {0,1}k we have
Ewb≤
Eaτb++kθ
1−θ .
Proof. The following inequality is true:
wb≤τb++1{τ+
b<k}(k+w ∗
b), (11.15)
wherew∗
b is the amount of time required to buildbwith all new bits, starting after
thek-th bit has been added. (Note thatw∗
b has the same distribution aswb.) Why?
Whenτb+ ≥k, we havewb =τb+. Whenτb+ < k, we can ensure a copy ofb made
from new bits by waiting for kbits, then restarting our counter and waiting for a copy ofbconsisting of bits added after the firstk.
Since w∗
b in independent of the event {τb+ < k}, taking expectations on both
sides of (11.15) yields
Ewb≤Eaτb++θ(k+Ewb)
(since Eawb does not depend on the initial state a, we drop the subscript), and
rearranging terms completes the proof.
Proposition11.8. The cover time satisfies
tcov≥(log 2)k2k(1 +o(1)).
Proof. Fixj =⌈log2k⌉and let A⊆ {0,1}k consist of those bitstrings that
end with j zeroes followed by a 1. Fixa, b∈A, wherea6=b. By Lemma 11.7, we have
Eaτb+>(1−θ)Ewb−kθ,
where our choice ofA ensures
θ=Pa(τb+ < k)<2− (j+1)+ · · ·+ 2−(k−1)<2−j. By (11.13) we may conclude Eaτb+>2 k(1 +o(1)).
Now apply Proposition11.4. Since|A|= 2k−j−1, we get
tcov≥(k−j−1)(log 2)2k(1 +o(1)) = (log 2)k2k(1 +o(1)).
EXERCISES 151
To improve the upper bound, we apply a variant on the Matthews method which, at first glance, may seem unlikely to help. For anyB⊆Ω, the argument for the Matthews bound immediately gives
ExτcovB ≤ max b,b′∈BEbτ ′ b 1 + 1 2+· · ·+ 1 |B| . (11.16)
Certainly the total cover time τcov is bounded by the time taken to visit first all
the states inB and then all the states inBc. Hence
Exτcov≤ExτcovB + max
y∈ΩEyτ
Bc
cov. (11.17)
If the states that take a long time to hit form a small fraction of Ω, then separating those states from the rest can yield a better bound ontcovthan direct application of
Theorem11.2. For the current example of waiting for all possible patterns in coin tossing, we improve the bound by a factor of 2—obtaining an asymptotic match with the lower bound of Proposition11.8.
Proposition11.9. The cover time satisfies
tcov≤(log 2)k2k(1 +o(1)).
Proof. We partition the state space{0,1}k into two sets. Fix j =
⌈log2k⌉
and let B be the set of all strings b ∈ {0,1}k with the following property: any
bitstring that is both a suffix and a prefix ofbmust have length less thank−j. (In other words, elements ofB must be shifted bymore than j bits in order to agree with themselves. For any stringb∈B, we must haveτb+> j.)
Since form < kthere are only 2mstrings of lengthkthat agree with themselves
after shifting bymbits, we have|Bc|= 2 + 4 +· · ·+ 2j≤2j+1≤4k.
Fora, b∈B, we can use Lemma11.7to bound the maximum expected hitting time. We have
Eaτb≤Ewb≤
Ebτb++kθ
1−θ .
(SinceEwb does not depend on the initial state, we have taken the initial state to
beb as we apply Lemma11.7.)
Since our chain has a uniform stationary distribution, (1.26) implies thatEbτb+=
2k. By our choice ofB, we haveθ=P
b(τb+< k)<1/k. Thus
Eaτb≤
2k+k(1/k)
1−1/k = 2
k(1 +o(1)). (11.18)
For a, b ∈ Bc, we again use Lemma 11.6 to bound E
aτb. Finally we ap-
ply (11.17), obtaining
tcov ≤(log|B|+o(1)) 2k(1 +o(1)+ (log|Bc|+o(1)) 2k+1+o(1))
= (log 2)k2k(1 +o(1)).
Exercises
Exercise11.1. LetY be a random variable on some probability space, and let
B =SjBj be a partition of an eventB into (finitely or countably many) disjoint
subeventsBj.
152 11. COVER TIMES
(b) Give an example to show that the conclusion of part (a) can fail when the eventsBj are not disjoint.
Exercise11.2. What upper and lower bounds does the Matthews method give for cycleZn? Compare to the actual value, computed in Example11.1, and explain
why the Matthews method gives a poor result for this family of chains.
Exercise 11.3. Show that the cover time of them-dimensional hypercube is asymptotic tom2mlog(2) asm
→ ∞.
Exercise 11.4. In this exercise, we demonstrate that for tori of dimension
d≥3, just a little more information on the hitting times suffices to prove a matching lower bound.
(a) Show that when a sequence of pairs of points xn, yn ∈ Zdn has the property
that the distance between them tends to infinity withn, then the upper-bound constantCd of (10.18) can be chosen so that Exn(τyn)/n
d
→Cd.
(b) Give a lower bound on tcov that has the same initial constant as the upper
bound of (11.9).
Exercise 11.5. Following the example of Exercise11.4, derive a lower bound forE(τcov) on the two-dimensional torus that is within a factor of 4 of the upper
bound (11.11).
Notes
The Matthews method first appeared in Matthews (1988a). Matthews (1989) looked at the cover time of the hypercube, which appears in Exercise11.3.
The argument we give for a lower bound on the cover time of the binary tree is due to Zuckerman (1992). Aldous (1991a) shows that the upper bound is asymp- totically sharp;Peres (2002) presents a simpler version of the argument.
In theAmerican Mathematical Monthly, Herb Wilf (1989) described his surprise at the time required for a simulated random walker to visit every pixel of his com- puter screen. This time is, of course, the cover time for the two-dimensional finite torus. The exact asymptotics of the expected cover time on Z2
n have been deter-
mined. Zuckerman (1992) estimated the expected cover time to within a constant, whileDembo, Peres, Rosen, and Zeitouni (2004) showed that
E(τcov)∼ 4
πn
2(logn)2.
M´ori (1987) found the cover time for all patterns of lengthk using ideas from
Aldous (1983a). The collectionGodbole and Papastavridis (1994) has many fur- ther papers on this topic. A single issue of the Journal of Theoretical Prob- ability contained several papers on cover times: these include Aldous (1989a),
Aldous (1989b), Broder and Karlin (1989), Kahn, Linial, Nisan, and Saks (1989), andZuckerman (1989).
Aldous (1991b) gives a condition guaranteeing that the cover time is well- approximated by its expected value. See Theorem19.8for a statement.
CHAPTER 12