5.2 Special graphs
In this section we record results about some specific easy-to-analyze
graphs. As in Section 5.1.3,
we focus on the parameters
discussed in Chapter 4;
orders of magnitudes of these parameters (in the asymptotic setting
discussed with each example) are summarized in terms of , the number of
vertices, in the following table. A minor theme is that some of the graphs
are known or conjectured to be extremal for our parameters. In the context of
extremality we ignore the parameter since its exact definition is a
little arbitrary.
jjj David: (1) Shall I add complete bipartite to table? (2) Please fill in
missing entries for torus.
Orders of magnitude of parameters [] for
special graphs.
Example |
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5.7. cycle |
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5.8. path |
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5.9. complete graph |
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5.10. star |
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5.11. barbell |
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5.12. lollipop |
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5.13. necklace |
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5.14. balanced -tree |
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5.15. -cube () |
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5.16. dense regular graphs |
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5.17. -dimensional torus |
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jjj? |
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jjj? |
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jjj? |
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jjj? |
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5.19. rook’s walk |
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In simpler cases we also record the -step transition probabilities
in discrete and continuous time. In fact one could write out exact expressions
for and indeed for hitting time distributions in almost
all these examples, but complicated exact expressions
are seldom very illuminating.
qqq names of graphs vary—suggestions for “standard names” from
readers of drafts are welcome.
This is just the graph – – – – –
on vertices.
By rotational symmetry, it is enough to give formulas for random walk
started at .
If is random walk on (all) the integers, then
is random walk on the -cycle, for
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Thus results for the -cycle can be deduced from results for the
integers.
For instance,
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(5.24) |
by Lemma 5.2,
because this is the mean exit time from for random walk on
the integers.
We can now calculate
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(5.25) |
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(5.26) |
where for the final equality we used the formula
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As at (5.9) and (5.10) we can get an expression for the
distribution at time from the Binomial distribution (in discrete time)
or the Poisson distribution (in continuous time).
The former is
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A more useful expression is obtained from the spectral representation. The
eigenvalues of the transition matrix are , . That is, and (if is even) are simple eigenvalues, with
respective normalized eigenvectors and (). The multiplicity of is for ; the corresponding normalized eigenvectors
are and (). Thus
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a fact most easily derived using Fourier analysis.
jjj Cite Diaconis book [123]? Continue same paragraph:
So the relaxation time is
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As an aside, note that the eigentime identity (Chapter 3 yyy)
gives the curious identity
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whose limit is the well-known formula
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If is even, the discrete-time random walk is periodic. This parity
problem vanishes in continuous time, for which we have the formula
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(5.27) |
Turning to total variation convergence,
we remain in continuous time
and consider the distance functions and of
Chapter 4 yyy.
The reader familiar with the notion of weak convergence of random
walks to Brownian motion
(on the circle, in this setting)
will see immediately that
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where the limit is “ for Brownian motion on the circle”,
which can be written as
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where has the standard Normal distribution.
So
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for the constant such that ,
whose numerical value has no real significance.
jjj David: You got . Please check. Continue same paragraph:
Similarly
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where is the density of .
As for , the sup in its definition is attained by some
of the form , so
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As remarked in Chapter 4 yyy,
this provides a counter-example to reversing inequalities in Theorem yyy.
But if we consider ,
the max is attained with , say, where . By Lemma 5.2, for ,
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and so
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Thus
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consistent with Chapter 4 Open Problem yyy.
xxx level of detail for results, here and later.
Remark.
Parameters , , , and are all
in this example, and in Chapter 6 we’ll see that they are
over the class of regular graphs.
However, the exact maximum values
over all -vertex regular graphs
(or the constants in the asymptotics)
are not known.
See Chapter 6 for the natural conjectures.
This is just the graph – – – –
on vertices.
If is random walk on (all) the integers, then
is random walk on the -path, for
the “concertina” map
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Of course the stationary distribution is not quite uniform:
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Again, results for the -path can be deduced from results for the
integers.
Using Lemma 5.2,
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(5.28) |
¿From this, or from the more general results in Section 5.1.2, we obtain
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(5.29) |
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(5.30) |
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(5.31) |
We can also regard as being derived from random walk
on the -cycle via
.
So we can deduce the spectral representation from that in the
previous example:
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where, for ,
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and
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In particular, the relaxation time is
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Furthermore, and for all , so
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where the limits are those in the previous example. Thus , where is times the corresponding constant for the
-cycle.
xxx explain: BM on and circle
described in Chapter 16.
It is easy to see that
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In Section 5.3.2 we will see
that the -path attains the exact maximum values of
our parameters
over all -vertex trees.
For the complete graph on vertices,
the -step probabilities for the chain started at can be
calculated by considering the induced -state chain which
indicates whether or not the walk is at .
This gives, in discrete time,
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(5.32) |
and, in continuous time,
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(5.33) |
It is clear that the hitting time to has
Geometric() distribution
(in continuous time, Exponential() distribution),
and so in particular
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(5.34) |
Thus we can calculate the parameters
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(5.35) |
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(5.36) |
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(5.37) |
From (5.32) the discrete-time eigenvalues are
So the relaxation time is
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(5.38) |
The total variation functions are
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so
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(5.39) |
It is easy to check
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We have already proved (Chapter 3 yyy)
that the complete graph attains the exact minimum of
, and over all -vertex graphs.
The same holds for , by considering
(in an arbitrary graph)
a vertex of minimum degree.
This is the graph on
vertices with
edges – , – , – , …, – .
The stationary distribution is
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In discrete time the walk is periodic.
Starting from the leaf , the walk at even times is simply independent
and uniform on the leaves, so
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for . At odd times, the walk is at .
Writing these -step probabilities as
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we see that the discrete-time eigenvalues are
and hence the relaxation time is
The mean hitting times are
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where the latter comes from the fact that has
Geometric() distribution, using the “independent uniform on leaves
at even times” property.
Then
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We can calculate the parameters
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(5.40) |
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(5.41) |
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(5.42) |
In continuous time we find
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This leads to
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from which
Clearly (isolate a leaf)
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We shall see in Section 5.3.2
that the -star attains the exact minimum of
our parameters over all -vertex trees.
Here is a graph on
vertices ().
Start with two complete graphs on vertices.
Distinguish vertices in one graph (“the left bell”)
and vertices in the other graph (“the right bell”).
Then connect the graphs via a path
– – – – –
containing new vertices.
A point of the construction is that the mean time to go from
a typical point in the left bell to a typical point
in the right bell is roughly .
To argue this informally, it takes mean time about to hit ;
then there is chance to hit , so it takes mean time about
to hit ; and from there is chance about to
hit the right bell before returning into the left bell, so it takes
mean time about to enter the right bell.
The exact result, argued below, is
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(5.43) |
It is cleaner to consider asymptotics as
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with .
Then
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where is the asymptotic maximizer here and for the other
parameters below.
Similarly
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The stationary distribution puts mass on each
bell. Also, by (5.45)–(5.47) below, uniformly for vertices in the left bell and uniformly for vertices
in the right bell. Hence
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and so we have proved the “” part of
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(5.44) |
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Consider the relaxation time .
For the function defined to be on the left bell, on the
right bell and linear on the bar,
the Dirichlet form gives
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Since the variance of tends to ,
the extremal characterization of shows that
is an asymptotic lower bound for .
But in general ,
so having already proved (5.44) for we must have
the same asymptotics for .
Finally, without going into details, it is not hard to show that
for fixed ,
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from which the “” assertion of (5.44) follows.
jjj Proof? (It’s not so terrifically easy, either! How much do we want to
include?) I’ve (prior to writing this) carefully written out an argument
similar to the present one, also involving approximate exponentiality of a
hitting time distribution, for the balanced -tree below. Here is a rough
sketch for the argument for here; note that it uses results about
the next example (the lollipop). (The argument for is similar.) The pair
of initial states achieves for every (although
the following can be made to work without knowing this “obvious fact” a
priori). Couple chains starting in these states by having them move
symmetrically in the obvious fashion. Certainly these copies will couple by
the time the copy started at has reached the center vertex of the bar. We claim that the distribution of is approximately
exponential, and its expected value is by the first displayed result for the
lollipop example, with changed to there. (In keeping with
this observation, I’ll refer to the “half-stick” lollipop in the next
paragraph.)
jjj (cont.) To get approximate exponentiality for the distribution of ,
first argue easily that it’s approximately the same as that of
for the half-stick lollipop started in stationarity. But that distribution
is, in turn, approximately exponential by Chapter 3 Proposition yyy,
since
for the half-stick lollipop.
Proof of (5.43).
The mean time in question is the sum of the following mean times:
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(5.45) |
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(5.46) |
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(5.47) |
Here (5.45) is just the result (5.34)
for the complete graph.
And (5.46) is obtained by summing over the edges of the “bar”
the expression
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(5.48) |
obtained from the general formula for mean hitting time across an
essential edge of a graph
(Lemma 5.1), where and .
To argue (5.47), we start with the -step recurrence
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where denotes a vertex of the right bell distinct from
and .
Now
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using the formula (5.48) for the mean passage time from
to .
Starting from , the time until a hit on either or
has Geometric() distribution, and the two vertices
are equally likely to be hit first. So
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The last three expressions give an equation for
whose solution is (5.47).
And it is straightforward to check that does achieve
the maximum, using (5.45)–(5.47) to bound the general
.
It is straightforward to check
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This is just the barbell without the right bell.
That is, we start with a complete graph on vertices and add
new vertices in a path.
So there are vertices, and is now a leaf.
In this example, by (5.45) and (5.46), with
in place of , we have
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In the asymptotic setting with
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where ,
we have
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(5.49) |
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where gives the asymptotic maximum.
Let us discuss the other parameters only briefly, in the asymptotic
setting.
Clearly
and it is not hard to check
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(5.50) |
whence
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Because the stationary distribution puts mass on the
“bar”, (5.50) is also enough to show that .
So by the general inequalities between our parameters, to show
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(5.51) |
it is enough to show
that .
But for the function defined to be on the “bell”, at
the end of the “bar,” and linear along the bar,
a brief calculation gives
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so that ,
as required.
Finally, in the asymptotic setting it is straightforward to check that
is achieved by , giving
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Remark.
The barbell and lollipop are the natural candidates for the -vertex
graphs which maximize our parameters.
The precise conjectures and known results will be discussed in Chapter 6.
jjj We need to put somewhere—Chapter 4 on ? Chapter 6 on max parameters over -vertex graphs? in the barbell example?—the fact
that is attained, when is even, by the barbell
with , the max value being .
Similarly, when is odd, the maximizing graph is formed by joining
complete graphs on and vertices
respectively by a single edge, and the max value is easy to write down
(I’ve kept a record) but not so pretty; however, this value too is , which is probably all we want to say anyway. Here is the first draft
of a proof:
For random walk on an unweighted graph, is the maximum over nonempty
proper subsets of the ratio
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(5.52) |
where is defined to be the sum of the degrees of vertices in
and is the number of directed cut edges from to .
jjj Perhaps it would be better for exposition to stick with undirected edges and introduce factor ?
Maximizing now over choice of graphs, the max in question is no
larger than the maximum , over all choices of , , ,
, and satisfying and for and , of the ratio
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(5.53) |
(We don’t claim equality because we don’t check that each -graph is
connected. But we’ll show that is in fact achieved by the connected graph
claimed above.)
Simple calculus shows that the ratio (5.53) is (as one would
expect) increasing in and and decreasing in . Thus, for
given , (5.53) is maximized by considering complete graphs of
size and joined by a single edge. Call the maximum
value . If is even, it is then easy to see that is
maximized by , giving , as desired.
For the record, here are the slightly tricky details if is odd. Write
and and put . A short
calculation gives , where with , , and . Easy calculus shows that is -shaped over and then that
. Thus is maximized when .
This graph, pictured below, is -regular with vertices,
consisting of subgraphs linked in a line, the two end subgraphs
being different from the intervening ones.
This is an artificial graph designed to mimic the -path while being
regular, and this construction (or some similar one) is the
natural candidate for the -vertex
graph which asymptotically maximizes our parameters
over regular graphs.
This example affords a nice illustration of use of the commute interpretation
of resistance. Applying voltage at vertex and voltage at , a
brief calculation gives the potentials at intervening vertices as
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and gives the effective resistance . Since the effective
resistance between and equals , we see the maximal effective
resistance is
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So
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One could do elementary exact calculations of other parameters,
but it is simpler to get asymptotics from the Brownian motion limit,
which implies that the asymptotic ratio of each parameter (excluding )
in this example and the -path is the same for each parameter.
So
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jjj I haven’t checked this carefully, and I also have abstained from writing
anything further about .
Finally, it is clear that , achieved by breaking a “link”
between “beads” in the middle of the necklace.
Take and .
The balanced -tree is the rooted tree where all leaves
are at distance from the root, where the root has degree ,
and where the other internal vertices have degree .
Call the height of the tree. For we have the
-star, and for we have the balanced binary tree.
The number of vertices is
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The chain induced (in the sense of Chapter 4 Section yyy)
by the function
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is random walk on , biased towards , with reflecting
barriers, as in Example 5.5 with
In fact, the transition
probabilities for can be expressed in terms of as
follows. Given vertices and with and , the paths
and
intersect in the path
, say, with . Then
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As a special case, suppose that is on the path from the root
to ; in this case .
Using the essential edge lemma
(or Theorem 5.20 below)
we can calculate
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(5.54) |
Using this special case we can deduce the general formula for mean hitting
times.
Indeed, , which leads to
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(5.55) |
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The maximum value is attained when
and are leaves and is the root.
So
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(5.56) |
(The part is simpler via (5.88) below.)
Another special case is that, for a leaf ,
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(5.57) |
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(5.58) |
where asymptotics are as for fixed .
Since is decreasing in ,
it follows that
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On the other hand, we claim , so that
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To sketch the proof, given a vertex , let be a leaf such that lies
on the path from to the root. Then
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and by (5.54). But the stationary
distribution puts nearly all its mass on vertices with of
constant order, and .
We claim next that
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Since , it is enough to show
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(5.59) |
and
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(5.60) |
Proof of (5.59).
Put
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for brevity.
We begin the proof by recalling the results (5.22) and (5.19)
for the induced walk :
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(5.61) |
By Proposition yyy
of Chapter 3,
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(5.62) |
For started at , let be a stopping time at which the
chain has exactly the stationary distribution. Then, for ,
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Since by (5.23), we can
arrange to have . Fixing and choosing
and (say) , (5.62)
and (5.61) in conjunction with Markov’s inequality yield
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Returning to the continuous-time walk on the tree, for sufficiently large
we have
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for every vertex . Now a simple coupling argument (jjj spell out
details?: Couple the induced walks and the tree-walks will agree when the
induced walk starting farther from the origin has reached the origin) shows
that
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for all large . Hence for all
large , and (5.59) follows.
jjj [This requires further exposition in both Chapters 3 and 4-1. In
Chapter 3, it needs to be made clear that one of the inequalities having to do
with CM hitting time distributions says precisely that . In Chapter 4-1 (2/96 version), it
needs to be noted that Lemma 2(a) (concerning for the joining of two
copies of a graph) extends to the joining of any finite number of copies.]
Let denote a balanced -tree of height . Let denote a
balanced -tree of height with root and construct a tree from
by adding an edge from to an additional vertex . We can
construct by joining copies of at the vertex , which becomes
the root of . Let and denote the respective stationary
distributions for the random walks on and , and use the notation
and , respectively, for hitting times on these graphs.
By Chapter 4 jjj,
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(5.63) |
where is the quasistationary distribution on associated
with the hitting time .
By Chapter 3 jjj, the expectation (5.63) is no
smaller than , which by the collapsing principle equals
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But it is easy to see that this last quantity equals , which is asymptotically equivalent to by (5.61).
¿From the discussion at the beginning of Section 5.3.1, it follows
that is achieved at any of the subtrees of the root. This gives
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An extension of the balanced -tree example is treated in
Section 5.2.1 below.
This is a graph with vertex-set and hence with vertices. Write for a vertex, and write
.
Then is an edge if and only if ,
and in general
is the graph-distance between i and j. Thus discrete-time
random walk proceeds at each step by choosing a coordinate at random and
changing its parity.
It is easier to use the continuized walk
,
because the component processes are independent as
varies, and each is in fact just the continuous-time random walk on the
-path with transition rate . This follows from an elementary fact
about the superposition of (marked) Poisson processes.
Thus, in continuous time,
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(5.64) |
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By expanding the right side,
we see that the continuous-time eigenvalues are
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(5.65) |
(Of course, this is just the general fact that the eigenvalues of
a -fold product of continuous-time chains are
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(5.66) |
where are the eigenvalues of the
marginal chain.)
In particular,
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(5.67) |
By the eigentime identity (Chapter 3 yyy)
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(5.68) |
the asymptotics being easy analysis.
From (5.64) it is also straightforward to derive the discrete-time
-step transition probabilities:
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Starting the walk at
, let
.
Then is the birth-and-death chain on states
with transition rates (transition probabilities, in discrete time)
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This is the Ehrenfest urn model mentioned in many textbooks.
In our terminology we may regard as random walk on the weighted
linear graph (Section 5.1.2) with weights
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In particular, writing for hitting times
for , symmetry and (5.13) give
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On the -cube, it is “obvious” that
is maximized by ,
and this can be verified by observing in (5.64)
that is minimized by ,
and hence is minimized by ,
so we can apply Chapter 2 yyy.
Thus
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(5.69) |
The asymptotics are the same as in (5.68).
In fact it is easy to use (5.64) to show
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and then by Chapter 2 yyy
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Since
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it follows that
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xxx refrain from write out exact —refs
To discuss total variation convergence, we have by symmetry
(and writing to distinguish from dimension )
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Following Diaconis et al [114]
we shall sketch an argument leading to
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(5.70) |
where has the standard Normal distribution.
This implies
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(5.71) |
For the discrete-time walk made aperiodic by incorporating chance
of holding, (5.70) and (5.71) remain true,
though rigorous proof seems complicated: see [114].
Fix , and consider such that
.
Using
as in (5.64),
we can calculate for with fixed
that
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Note the limit is when .
Now
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where the second sum is over with . But from (5.64) we can write this sum as
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where denotes a Binomial random variable.
By the Normal approximation to Binomial, this converges to
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as stated.
As an aside, symmetry and Chapter 4 yyy give
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and so the difference
is ,
which is much smaller than what the series
expansions (5.68) and (5.69) imply.
The fact that the “half-cube” ,
yielding
achieves the sup in the definition of can be proved using a
slightly tricky induction argument. However, the result follows immediately
from (5.67) together with the general inequality .
Consider an -regular -vertex graph with .
Of course here we are considering a class of graphs rather than
a specific example.
The calculations below show that these graphs necessarily mimic the
complete graph (as far as smallness of the random walk parameters is
concerned) in the asymptotic setting .
The basic fact is that, for any pair of vertices, there must
be at least other vertices such that is a path.
To prove this, let (resp., ) be the number of vertices
such that exactly (resp., ) of the edges
exist. Then by counting vertices, and by counting edges, and these inequalities imply .
Thus, by Thompson’s principle (Chapter 3, yyy) the effective resistance
and so the commute interpretation of resistance
implies
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(5.72) |
A simple “greedy coupling” argument (Chapter 14, Example yyy) shows
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(5.73) |
This is also a bound on and on ,
because always, and special case 2 below
shows that this bound on cannot be
improved asymptotically (nor hence can the bound on or ).
Because for regular graphs (Chapter 3 yyy),
we get
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This implies
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which also follows from (5.72) and .
We can also argue, in the notation of Chapter 4 yyy, that
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Special case 1.
The orders of magnitude may change for .
Take two complete -graphs, break one edge in each
(say edges and ) and add edges
and . This gives an -vertex -regular graph for which all our parameters are .
jjj I haven’t checked this.
Special case 2.
Can
the bound be asymptotically improved? Eric Ordentlicht
has
provided the following natural counter-example.
Again start with two -complete graphs on vertices
and .
Now add the edges for which
.
This gives an -vertex -regular graph.
By considering the
set consisting of the vertices ,
a brief calculation gives
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Example 5.17
The -dimensional torus .
The torus is the set
of -dimensional integers
modulo ,
considered in the natural way as a -regular graph on vertices.
It is much simpler to work with
the random walk in continuous time,
,
because its component processes are independent as
varies; and each is just continuous-time random walk on the -cycle,
slowed down by a factor .
Thus we can immediately write the time- transition probabilities
for in terms of the
corresponding probabilities
for continuous-time random walk on the -cycle
(see Example 5.7 above) as
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Since the eigenvalues on the -cycle are
,
by (5.66)
the eigenvalues of are
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In particular, we see that the relaxation time satisfies
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where here and below asymptotics are as for fixed .
This relaxation time could more simply be derived from the -cycle
result via the general “product chain” result of Chapter 4 yyy.
But writing out all the eigenvalues enables us to use the eigentime
identity.
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(the sum excluding ),
and hence
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(5.74) |
where
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(5.75) |
provided the integral converges. The reader who is a calculus whiz will see that in fact
for only, but this is seen more easily in the alternative
approach of Chapter 15, Section yyy.
xxx more stuff: connection to transience, recurrent potential, etc
xxx new copy from lectures
xxx
jjj David: I will let you develop the rest of this example. Note that
is considered very briefly in Chapter 15, eq. (17) in 3/6/96
version. Here are a few
comments for . First suppose that is even and .
Presumably, is achieved by the following half-torus:
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In the notation of (5.52) observe
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whence
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[By Example 5.15 (the -cube) this last result is also true for , and (for even ) it is by Example 5.7 (the
-cycle) also true for .] If we have correctly conjectured the
maximizing , then
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and presumably(??)
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in any case.
Here is a classic homework problem for an undergraduate Markov chains
course.
Start a knight at a corner square of an otherwise-empty chessboard.
Move the knight at random, by choosing uniformly from the legal knight-moves
at each step.
What is the mean number of moves until the knight returns to the starting
square?
It’s a good question, because if you don’t know Markov chain theory it
looks too messy to do by hand, whereas using Markov chain theory it becomes
very simple.
The knight is performing random walk on a graph
(the 64 squares are the vertices, and the possible knight-moves are the edges).
It is not hard to check that the graph is connected, so by the elementary
Chapter 3 yyy for a corner square the mean return time is
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and by drawing a sketch in the margin the reader can count the
number of edges to be .
Other chess pieces—queen, king, rook—define different graphs
(the bishop’s is of course not connected, and the pawn’s not undirected).
One might expect that the conventional ordering of the “strength” of the
pieces as (queen, rook, knight, king) is reflected in parameters and
(jjj how about the other taus?) being increasing in this
ordering. The reader is invited to perform the computations. (jjj: an
undergraduate project?) We have done so only for the rook’s move, treated in
the next example.
The computations for the queen, knight, and king are simplified if the walks
are made on a toroidal chessboard. (There is no difference for the rook.)
jjj Chess on a bagel, anyone? Continue same paragraph:
Then Fourier analysis (see Diaconis [123]) on the abelian
group (with ) can be brought to bear, and the eigenvalues are
easy to compute. We omit the details, but the results for
(queen, rook, knight, king) are asymptotically
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as , in conformance with our expectations,
and numerically
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for . The only surprise is the inverted ordering for
(rook, knight).
Example 5.19
Rook’s random walk on an -by- chessboard.
jjj Do we want to do this also on a -dimensional grid? We need to mention
how this is a serious example, used with Metropolis for sampling from log
concave distributions; reference is [33]? [32]?
Number the rows and columns of the chessboard each through in
arbitrary fashion, and denote the square of the chessboard at row and
column by . In continuous time, the rook’s random
walk is the product of two continuous-time random walks on the
complete graph on vertices, each run at rate . Thus (cf. Example 5.9)
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(5.76) |
which can be expanded to get the discrete-time multistep transition
probabilities, if desired. We recall that the eigenvalues for
discrete-time random walk on are with multiplicity and
with multiplicity . It follows [recall (5.66)]
that the eigenvalues for the continuous-time rook’s walk are
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In particular,
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(5.77) |
which equals for and converges to as grows.
Applying the eigentime identity, a brief calculation gives
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(5.78) |
which equals for and for large.
Starting the walk at , let denote the
Hamming distance of from ,
i.e., the number of coordinates (, , or ) in which
differs from . Then is a birth-and-death chain with transition
rates
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This is useful for computing mean hitting times.
Of course
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Since
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it follows that
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Finally, it is clear that , so that
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whence
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These calculations show
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which equals for , and they provide another proof
of (5.78).
From (5.76) it is easy to derive
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and thence
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which rounds to for and converges to as becomes large.
Any set of the form with either or
and a nonempty proper subset of
achieves the value
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A direct proof is messy, but this follows immediately from the general
inequality , (5.77), and a brief calculation
that the indicated indeed gives the indicated value.
xxx other examples left to reader?
complete bipartite;
ladders
jjj Note: I’ve worked these out and have handwritten notes. How much do we
want to include, if at all? (I could at least put the results in the table.)
5.2.1 Biased walk on a balanced tree
Consider again the balanced -tree setup of Example 5.14. Fix a
parameter . We now consider biased random
walk on the tree, where from each non-leaf vertex other than the root
the transition goes to the parent with probability
and to each child with probability . As in
Example 5.14 (the case ), the chain induced by
the function
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is (biased) reflecting random walk on with respective
probabilities and of moving to
the right and left from any ; the ratio of these two
transition probabilities is
The stationary distribution for is a modified
geometric:
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where
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Since the stationary distribution for is assigns the same
probability to each of the vertices with a given value of
, a brief calculation shows that for any edge in the
tree. In the same notation, it follows that is random walk on the
balanced -tree with edge weights and total
weight .
The distribution concentrates near the root-level if
and near the leaves-level if ; it is nearly uniform on the
levels if . On the other hand, the weight assigned by the
distribution to an individual vertex is a decreasing function of
(thus favoring vertices near the leaves) if (i.e.,
) and is an increasing function (thus favoring vertices near the
root) if ; it is uniform on the vertices in the unbiased case
.
The mean hitting time calculations of Example 5.14 can all be extended
to the biased case. For example, for the general
formula (5.55) becomes [using the same notation as
at (5.55)]
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(5.79) |
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if and
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if .
The maximum value is attained when and are leaves and is
the root. So if ,
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(5.80) |
The orders of magnitude for all of the -parameters (with and
, and hence , fixed as , and hence , becomes large) are
summarized on a case-by-case basis in the next table. Following are some of
the highlights in deriving these results; the details, and derivation of exact
formulas and more detailed asymptotic results, are left to the reader.
Orders of magnitude of parameters []
for -biased walk on a balanced -tree of height ().
Value of |
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( Example 5.14) |
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For we have . If , this bound is tight:
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for a more careful calculation is required.
If , then the same arguments as for the unbiased case () show
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In this case it is not hard to show that
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as well. If , then it is not hard to show that
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with achieved at a branch of the root (excluding the root), and so
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as well. If , then since has positive drift equal to
, it follows that
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The value is achieved by isolating a leaf, giving
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and so, by the inequalities of
Chapter 4, Section yyy,
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as well.
jjj Limiting value of when is that of for biased
infinite tree? Namely?