Bijections, Prufer Codes and Cayley’s Formula

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I’m currently at the training camp in Cambridge for this year’s UK IMO squad. This afternoon I gave a talk to some of the less experienced students about combinatorics. My aim was to cover as many useful tricks for calculating the sizes of combinatorial sets as I could in an hour and a half. We started by discussing binomial coefficients, which pleasingly turned out to be revision for the majority. But my next goal was to demonstrate that we are much more interested in the fact that we can calculate these if we want than in the actual expression for their values.

Put another way, my argument was that the interpretation of \binom{n}{m} as the number of ways to choose m objects from a collection of n, or the number of up-and-right paths from (0,0) to (m,n) is more useful than the fact that \binom{n}{m}=\frac{n!}{m!(n-m)!}. The opening gambit was to prove the fundamental result underlying the famous construction of Pascal’s triangle that

\binom{n+1}{m+1}=\binom{n}{m}+\binom{n}{m+1}.

This is not a hard result to prove by manipulating factorials, but it is a very easy result to prove in the path-counting setting, for example.

So it turned out that the goal of my session, as further supported by some unsubtly motivated problems from the collection, was to convince the students to use bijections as much as possible. That is, if you have to count something awkward, show that counting the awkward thing is equivalent to counting something more manageable, then count that instead. For many simpler questions, this equivalence is often drawn implicitly using words (“each of the n objects can be in any subset of the collection of bags so we multiply…” etc), but it is always worth having in mind the formal bijective approach. Apart from anything else, asking the question “is this bijection so obvious I don’t need to prove it” is often a good starting-point for assessing whether the argument is in fact correct!

Anyway, I really wanted to show my favouriite bijection argument, but there wasn’t time, and I didn’t want to spoil other lecturers’ thunder by defining a graph and a tree and so forth. The exploration process encoding of trees is a strong contender, but today I want to define quickly the Prufer coding for trees, and use it to prove a famous result I’ve been using a lot recently, Cayley’s formula for the number of spanning trees on the complete graph with n vertices, n^{n-2}.

We are going to count rooted trees instead. Since we can choose any vertex to be the root, there are n^{n-1} rooted trees on n vertices. The description of the Prufer code is relatively simple. Take a rooted tree with vertices labelled by [n]. A leaf is a vertex with degree 1, other than the root. Find the leaf with the largest label. Write down the label of the single vertex to which this leaf is connected, then delete the leaf. Now repeat the procedure, writing down the label of the vertex connected to the leaf now with the largest label, until there are only two vertices remaining, when you delete the non-root vertex, and write down the label of the root. We get a string of (n-1) labels. We want to show that this mapping is a bijection from the set of rooted trees with vertices labelled by [n] to [n]^{n-1}.

Let’s record informally how we would recover a tree from the Prufer code. First, observe that the label of any vertex which is not a leaf must appear in the code. Why? Well, the root label appears right at the end, if not earlier, and every vertex must be deleted. But a vertex cannot be deleted until it has degree one, so the neighbours further from the root (or ancestors) of the vertex must be removed first, and so by construction the label appears. So know what the root is, and what the leaves are straight away.

In fact we can say slightly more than this. The number of times the root label appears is the degree of the root, while the number of times any other label appears is the degree of the corresponding vertex minus one. Call this sequence the Prufer degrees.

So we construct the tree backwards from the leaves towards the root. We add edges one at a time, with the k-th edge joining the vertex with the k-th label to some other vertex. For k=1, this other vertex is the leaf with maximum label. In general, let G_k be the graph formed after the addition of k-1 edges, so G_1 is empty, and G_n is the full tree. Define T_k to be the set of vertices such that their degree in G_k is exactly one less than their Prufer degree. Note that T_1 is therefore the set of leaves suggested by the Prufer code. So we form G_{k+1} by adding an edge between the vertex with label appearing at position k+1 in the Prufer sequence and the vertex of T_k with maximum label.

Proving that this is indeed the inverse is a bit fiddly, more because of notation than any actual mathematics. You probably want to show injectivity by an extremal argument, taking the closest vertex to the root that is different in two trees with the same Prufer code. I hope it isn’t a complete cop out to swerve around presenting this in full technical detail, as I feel I’ve achieved by main goal of explaining why bijection arguments can reduce a counting problem that was genuinely challenging to an exercise in choosing sensible notation for proving a fairly natural bijection.

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Diameters of Trees and Cycle Deletion

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In the past two posts, we introduced two models of random trees. The Uniform Spanning Tree chooses uniformly at random from the set of spanning trees for a given underlying graph. The Minimum Spanning Tree assigns IID weights to each edge in the underlying graph, then chooses the spanning tree with minimum induced total weight. We are interested to know whether these are in fact the same distribution, and if they are not, what properties can be used to distinguish them asymptotically.

While investigating my current research problem, I was interested in the diameter of large random trees under various models. Specifically, I am considering what happens if you take a standard Erdos-Renyi process on n vertices, where edges appear at constant rate between pairs of vertices chosen uniformly at random, and add an extra mechanism to prevent the components becoming too large. For this particular model, our mechanism consists of removing any cycles as they are formed. Thus all the components remain trees as time advances, so it is not unreasonable to think that there might be some sort of equilibrium distribution.

Now, by definition, any tree formed by the Erdos-Renyi process is a uniform tree. Why? Well, the probability of a configuration is determined entirely by the number of edges present, so once we condition that a particular set of vertices are the support of a tree, all possible tree structures are equally likely. Note that this relies on sampling at a single fixed time. If we know the full history of the process, then it is no longer uniform. For example, define a k-star to be a tree on k vertices where one ‘centre’ vertex has degree k-1. The probability that a uniform tree on k vertices is a k-star is \frac{k}{k^{k-2}}=k^{-(k-3)}. But a star can only be formed by successively adding single vertices to an existing star. That is, we cannot join a 3-tree and a 4-tree with a edge to get a 7-star. So it is certainly not immediately clear that once we’ve incorporated the cycle deletion mechanism, the resulting trees will be uniform once we condition on their size.

In fact, the process of component sizes is not itself Markovian. For a concrete example, observe first that there is, up to isomorphism, only one tree on any of {0,1,2,3} vertices, so the first possible counterexample will be splitting up a tree on four vertices. Note that cycle deletion always removes at least three edges (ie a triangle), so the two possibilities for breaking a 4-tree are:

(4) -> (2,1,1) and (4) -> (1,1,1,1)

I claim that the probabilities of each of these are different in the two cases: a) (4) is formed from (2,2) and b) (4) is formed from (3,1). This is precisely a counterexample to the Markov property.

In the case where (4) is formed from (2,2), the 4-tree is certainly a path of length 4. Therefore, with probability 1/3, the next edge added creates a 4-cycle, which is deleted to leave components (1,1,1,1). In the case where (4) is formed from (3,1), then with probability 2/3 it is a path of length 4 and with probability 1/3 it is a 4-star (a ‘T’ shape). In this second case, no edge can be added to make a 4-cycle, so after cycle deletion the only possibility is (2,1,1). Thus the probability of getting (1,1,1,1) is 2/9 in this case, confirming that the process is non-Markovian. However, we might remark that we are unlikely to have O(n) vertices involved in fragmentations until at least the formation of the giant component in the underlying E-R process, so it is possible that the cycle deletion process is ‘almost Markov’ for any property we might actually be interested in.

When we delete a cycle, how many vertices do we lose? Well, for a large tree on n vertices, the edge added which creates the cycle is chosen uniformly at random from the pairs of vertices which are not currently joined by an edge. Assuming that n is w(1), that is we are thinking about a limit of fairly large trees, then the number of edges present is much smaller than the number of possible edges. So we might as well assume we are choosing uniformly from the possible edges, rather than just the possible edges which aren’t already present.

If we choose to add an edge between vertices x and y in the tree, then a cycle is formed and immediately deleted. So the number of edges lost is precisely the length of the path between x and y in the original tree. We are interested to know the asymptotics for this length when x and y are chosen at random. The largest path in a graph is called the diameter, and in practice if we are just interested in orders of magnitude, we might as well assume diameter and expected path length are the same.

So we want to know the asymptotic diameter of a UST on n vertices for large n. This is generally taken to be n^{1/2}. Here’s a quick but very informal argument that did genuinely originate on the back of a napkin. I’m using the LERW definition. Let’s start at vertex x and perform LERW, and record how long the resultant path is as time t advances. This is a Markov chain: call the path length at time t X_t.

Then if X_t=k, with probability 1-\frac k n we get X_{t+1}=k+1, and for each j in {0,…,k-1}, with probability 1/n we have X_{t+1}=j, as this corresponds to hitting a vertex we have already visited. So

\mathbb{E}\Big[X_{t+1}|X_t=k\Big]=\frac{nk-k^2/2}{n}.

Note that this drift is positive for k<< \sqrt n and negative for k>>\sqrt n, so we would expect n^{-1/2} to be the correct scaling if we wanted to find an equilibrium distribution. And the expected hitting time of vertex y is n, by a geometric distribution argument, so in fact we would expect this Markov chain to be well into the equilibrium window with the n^{-1/2} scaling by the time this occurs. As a result, we expect the length of the x to y path to have magnitude n^{1/2}, and assume that the diameter is similar.

So this will be helpful for calculations in the cycle deletion model, provided that the trees look like uniform trees. But does that even matter? Do all sensible models of random trees have diameter going like n^{1/2}? Well, a recent paper of Addario-Berry, Broutin and Reed shows that this is not the case for the minimum spanning tree. They demonstrate that the diameter in this case is n^{1/3}. I found this initially surprising, so tried a small example to see if that shed any light on the situation.

The underlying claim is that MSTs are more likely to be ‘star-like’ than USTs, a term I am not going to define. Let’s consider n=4. Specifically, consider the 4-star with centre labelled as 1. There are six possible edges in K_4 and we want to see how many of the 6! weight allocations lead to this star. If the three edges into vertex 1 have weights 1, 2 and 3 then we certainly get the star, but we can also get this star if the edges have weights 1, 2 and 4, and the edge with weight 3 lies between the edges with weights 1 and 2. So the total number of possibilities for this is 3! x 3! + 3! x 2! = 48. Whereas to get a 4-path, you can assign weights 1, 2 and 3 to the edges of the path, or weights 1, 2 and 4 provided the 4 is not in the middle, and then you have the 3 joining up the triangle formed by 1 and 2. So the number of possibilities for this is 3! x 3! + 4 x 2! = 44.

To summarise in a highly informal way, in a star-like tree, you can ‘hide’ some fairly low-scoring weights on edges that aren’t in the tree, so long as they join up very low-scoring edges that are in the tree. Obviously, this is a long way from getting any formal results on asymptotics, but it does at least show that we need to be careful about diameters if we don’t know exactly what mechanism is generating the tree!

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Mixing Times 6 – Aldous-Broder Algorithm and Cover Times

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In several previous posts, I’ve discussed the Uniform Spanning Tree. The definition is straightforward: we choose uniformly at random from the set of trees which span a fixed underlying graph. But for a dense underlying graph, there are a very large number spanning trees. Cayley’s formula says that the complete graph K_n has n^{n-2} spanning trees, so to select from this list is impractical.

We seek a better algorithm. In a post about a year ago, I presented the result that the path between two fixed points x and y in the UST is distributed as the path generated by Loop-Erased Random Walk, for which we start at x and delete cycles as they appear. An initial problem might be that this only gives us a single path, which might be enough in some contexts, but in general we will want to specify the whole tree. Wilson’s Algorithm is an unsurprising but useful extension to this equivalence which does just that. You start by constructing the LERW between two vertices, then you add the LERW which connects some other vertex to the path you already have. Then you take a further vertex not currently explored and start LERW there, continuing until you hit the tree that you already have. Iterate this process, which must terminate after at most n steps when there are no vertices which to start from. The tree thus obtained is the UST. The tricky part is proving that the method for selecting which unused vertices to start from has no effect on the distribution of paths between two fixed points.

I want to consider a different algorithm, discovered roughly simultaneously by Aldous and Broder. Start a random walk on the underlying graph at some particular vertex. Every time we traverse an edge which takes us to a vertex we haven’t yet explored, add this edge to the tree. For now I don’t want to give a proof that this algorithm works, but rather to talk about how fast it works, because it ties in nicely with something from the Mixing Times book we’ve been reading recently. It is clear that the algorithm terminates at the first time the random walk has visited every vertex. This is a stopping time, called the cover time of the Markov chain. If we are working with an underlying complete, then we notice that this is annoying, because it means that the cover time will increase like n.log n. That is, it will take an increasingly long time to gather the final few vertices into the tree. Perhaps some combination of Aldous-Broder initially then Wilson’s method for the final o(n) vertices might be preferable?

I want to discuss how to treat this cover time. Often we have information about the hitting times of states from other states \mathbb{E}_x T_y. A relationship between S, the hitting time, defined to be the maximum of the previous display over x and y, and the expected cover time would be useful, especially for a highly symmetric graph like the complete graph where the expected hitting times are all the same.

Matthews’ Method relates these two for an irreducible finite Markov chain on n states. It says:

t_{cov}\leq t_{hit}\left(1+\frac12+\ldots+\frac 1 n\right).

We first remark that this agrees with what we should get for the random walk on the complete graph. There, the hitting time of x from y is a geometric random variable with success probability 1/n, hence expectation is n. The cover time is the standard coupon collector problem, giving expectation n log n, and the sum of reciprocals factor is asymptotically a good approximation.

The intuition is that if we continue until we hit state 1, then reset and continue until we hit state 2, and so on, by the time we hit state n after (n-1) iterations, this is a very poor overestimate of the cover time, because we are actually likely to have hit most states many times. What we want to do really is say that after we’ve hit state 1, we continue until we hit state 2, unless we’ve already done so, in which case we choose a different state to aim for, one which we haven’t already visited. But this becomes complicated because we then need to know the precise conditional probabilities of visiting any site on the way between two other states, which will depend rather strongly on the exact structure of the chain.

Peres et al give a coupling proof in Chapter 11 of their book which I think can be made a bit shorter, at least informally. The key step is that we still consider hitting the sites in order, only now in a random order.

That is, we choose a permutation \sigma\in S_n uniformly at random, and we let T_k be the first time that states \sigma(1),\ldots,\sigma(k) have all been visited. This is a random time that is measurable in the product space, and for each \sigma it is a stopping time.

The key observation is that \mathbb{P}(T_{k+1}=T_k)=1-\frac{1}{k+1}. This holds conditional on any path of the Markov chain because the requirement for the event is that \sigma(k+1) is visited after \{\sigma(1),\ldots,\sigma(k)\}. The statement therefore holds as stated as well as just pathwise. Then, by the SMP, conditional on \{T_{k+1}>T_k\}, we have

T_{k+1}-T_k \leq_{st} t_{hit}.

Note that by the definition of t_{hit}, this bound on the hitting time T_{k+1} is unaffected by concerns about where the chain actually is at T_k (since it is not necessarily at \sigma(k)).

So, removing the conditioning, we have:

\mathbb{E}\Big[T_{k+1}-T_k\Big]\leq\frac{1}{k+1}t_{hit},

and so the telescoping sum gives us Matthews’ result.

One example is the cover time of random walk on the n x n torus, which turns out to be

O(n^2(\log n)^2).

If anyone remembers that Microsoft screensaver from many years ago which started with a black screen and a snake leaving a trail of white pixels as it negotiated the screen, this will be familiar. The last few black bits take a frustratingly long while to disappear. Obviously that isn’t quite a random walk, but it perhaps diminishes the surprise that it should take this long to find the cover time.

There are a couple of interesting things I wanted to say about electrical networks for Markov chains and analytic methods for mixing times, but the moment may have passed, so this is probably the last post about Mixing Times. Plans are in motion for a similar reading group next term, possible on Random Matrices.

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Minimum Spanning Trees

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In my last post, I discussed the Uniform Spanning Tree. To summarise very briefly, given a connected graph on n vertices, a tree is a subgraph, that is a subset of the edges, which is connected, but which contains no cycles. It turns out this requires the tree to have n-1 edges.

We are interested in natural mechanisms for generating randomly chosen spanning trees of a given graph. One way we can always do this is to choose uniformly at random from the set of possible trees. This UST is in some sense canonical, but it is worth knowing about some other measures on trees that might be of interest.

A family of natural problems in operations research concerns an arbitrary complex network, with some weight or cost associated to each connection. The question is how to perform some operation on the network so as to minimise the resulting cost. Perhaps the most famous such problem is that of the Travelling Salesman. The story is that a salesman needs to visit n locations and wants to do the trip as efficiently as possible. This might be thought of as some sort of financial or time cost, but proably the easiest way to set it up is to imagine he is trying to minimise the distance he has to travel. It is not hard to see why this problem might genuinely arise in plenty of real-world situations, where a organisation or agent is trying to be as efficient as possible.

It might be the case that it is not possible to travel between every pair of locations, but we needn’t assume that for now. So if he knows the distance between any pair of cities, he wants to know which of the possible routes gives the shortest overall distance. The problem is that there are n! routes, and this grows roughly like n^n, which is faster than exponential, so for as few as 20 cities it has turned into a comparison which is too large to compute.

There are various algorithms which reduce the number of routes that must be checked, and some approximation methods. But if you want the exact answer, it is not currently possible to calculate this in polynomial time.

Minimal Spanning Trees and Uniqueness

For the travelling salesman, we were looking for the minimal cost spanning path. In the case of the complete graph, this is the same as the minimal cost non-repeating path of length n-1. Such paths are a subset of the set of spanning trees on the underlying graph. So what if we look instead for the minimal cost spanning tree? This exists as after all, there are only finitely many spanning trees.

So far, this has been deterministic, but we were looking for a random spanning tree. We can achieve this by choosing the weights at random. Anything other than assigning the weights as an IID sequence seems likely to be complicated, but there isn’t a canonical choice of the distribution of the weights. Our first question will be whether the distribution of the weights affects the distribution of the induced MST. In fact it will turn out that so long as the distribution is continuous, it has no effect on the distribution of the MST. The continuous condition might seem odd, but it is present only to ensure that the weights almost certainly end up generating a unique MST.

It turns out that there is a straightforward greedy algorithm to find the MST once the weights are known. We will examine some consequences of this algorithm in the random setting. First we check uniqueness. The condition required for uniqueness is that the weights be distinct. Note that this is slightly weaker than the statement that all of sums of (n-1)-tuples be distinct, which immediately implies a unique MST.

We now prove this condition. Suppose we have distinct weights, and an associated MST. If the underlying graph is a tree, then the result is clear. Otherwise, add some extra edge e, with weight w(e). By the definition of a tree, this generates exctly one cycle. Consider the other edges, say e_1,\ldots,e_k in this cycle. If any of w(e_i)>w(e) then we can replace e_i with e to get a spanning tree with smaller weight, a contradiction of the claim that we started with an MST. So by distinctness of weights, we conclude that w(e)>w(e_i) for all i.

Conversely, suppose we remove some edge e which IS in the MST. We end up with exactly two connected components. Consider all the edges in the underlying graph between the two components, and suppose that one of these f satisfies w(f)<w(e). Then if we add in edge f, which is by construction not in the original MST, we end up with a smaller total weight than we started with, a further contradiction.

We can summarise this in a neat form. Given an edge e between x and y, consider the set of all edges in the underlying graph with weight LESS THAN w(e). Then if x and y are in different components, the edge e must be in the MST. Since we have an explicit description of which edges are present, it follows that the MST is unique. The problem is that working out the component structure of the graph with higher weights removed is computationally rather intensive. We want a slightly faster algorithm.

Kruskal’s Algorithm

Several rather similar algorithms were developed roughly simultaneously. Prim’s algorithm is a slight generalisation of what we will discuss. Anyway, for now we consider Kruskal’s algorithm which has the advantage that it can be described without really needing to draw a diagram.

We start by ordering the weights. Without loss of generality, we might as well relabel the edges so that

w(e_1)< w(e_2)<\ldots< w(e_{|E|}).

Now, by the condition derived in the argument for uniqueness, we must have e_1 and e_2 in any MST. Now consider e_3. Unless doing so would create a cycle, add e_3. Then, unless doing so would create a cycle, add e_4. Continue. It is clear that the result of this procedure is acyclic. To check it is actually a spanning tree, we show that it is also connected. Suppose not, and two of the components are A and B. Let e be the edge between A and B with minimal weight. According to the algorithm, we should have included e in our MST because at no point would adding it possibly have created a cycle. So we have proved that this greedy algorithm does indeed give the (unique) MST.

A useful consequence of this is that we know the two edges with overall minimum weight are definitely in the MST. In the search for a random measure on spanning trees, what is most important is that we didn’t use the actual values of the weights in this construction, only the order. In other words, we might as well have assumed the weights were a random permutation from S_{|E|}. This now answers our original question about how the random weight MST depends in distribution on the underlying edge weight distribution. So long as with probability one the weights are distinct (which holds if the distribution is continuous), then the distribution of the resulting spanning tree is constant.

It’s not too hard to show this isn’t the same as UST: n=4 suffices as a counterexample. But the difference in asymptotic behaviour of properties such as the diameter is of interest, and will be explored in the next post.

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Uniform Spanning Trees

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For applications to random graphs, the local binomial structure and independence means that the Galton-Watson branching process is a useful structure to consider embedding in the graph. In several previous posts, I have shown how we can set up the so-called exploration process which visits the sites in a component as if the component were actually a tree. The typical degree is O(1), and so in particular small components will be trees with high probability in the limit. In the giant component for a supercritical graph, this is not the case, but it doesn’t matter, as we ignore vertices we have already explored in our exploration process. We can consider the excess edges separately by ‘sprinkling’ them back in once we have the tree-like backbone of all the components. Again, independence is crucial here.

I am now thinking about a new model. We take an Erdos-Renyi process as before, with edges arriving at some fixed rate, but whenever a cycle appears, we immediately delete all the edges that make up the cycle. Thus at all times the system consists of a collection (or forest) of trees on the n vertices. So initially this process will look exactly like the normal E-R process, but as soon as the components start getting large, we start getting excess edges which destroy the cycles and make everything small again. The question to ask is: if we run the process for long enough, roughly how large are all the components? It seems unlikely that the splitting mechanism is so weak that we will get true giant components forming, ie O(n) sizes, so we might guess that, in common with some other split-merge models of this type, we end up with components of size n^{2/3}, as in the critical window for the E-R process.

In any case, the scaling limit process is likely to have components whose sizes grow with n, so we will have a class of trees larger than those we have considered previously, which have typically been O(1). So it’s worth thinking about some ways to generate random trees on a fixed number of vertices.

Conditioned Galton-Watson

Our favourite method of creating trees is inductive. We take a root and connect the root to a number of offspring given by a fixed distribution, and each of these some offspring given by an independent sample from the same distribution and so on. The natural formulation gives no control over the size of the tree. This is a random variable whose distribution depends on the offspring distribution, and which in some circumstances be computed explicitly, for example when the offspring distribution is geometric. In other cases, it is easier to make recourse to generating functions or to a random walk analogue as described in the exploration process discussion.

Of course, there is nothing to stop us conditioning on the total size of the population. This is equivalent to conditioning on the hitting time of -1 for the corresponding random walk, and Donsker’s theorem gives several consequences of a convergence relation towards a rescaled Brownian excursion. Note that there is no a priori labelling for the resulting tree. This will have to be supplied later, with breadth-first and depth-first the most natural choices, which might cause annoyance if you actually want to use it. In particular, it is not obvious, and probably not true unless you are careful, that the distribution is invariant under permuting the labels (having initially assumed 1 is the root etc) which is not ideal if you are embedding into the complete graph.

However, we would like to have some more direct constructions of random trees on n vertices. We now consider perhaps the two best known such methods. These are of particular interest as they are applicable to finding random spanning trees embedded in any graph, rather than just the complete graph.

Uniform Spanning Tree

Given a connected graph, consider the set of all subgraphs which are trees and span the vertex set of the original graph. An element of this set is called a spanning tree. A uniform spanning tree is chosen uniformly at random from the set of spanning trees on the complex graph on n vertices. A famous result of Arthur Cayley says that the number of such spanning trees is n^{n-2}. There are various neat proofs, many of which consider a mild generalisation which gives us a more natural framework for using induction. This might be a suitable subject for a subsequent post.

While there is no objective answer to the question of what is the right model for random trees on n vertices, this is what you get from the Erdos-Renyi process. Formally, conditional on the sizes of the (tree) components, the structures of the tree components are given by UST.

To see why this is the case, observe that when we condition that a component has m vertices and is a tree, we are demanding that it be connected and have m-1 edges. Since the probability of a particular configuration appearing in G(n,p) is a function only of the number of edges in the configuration, it follows that the probability of each spanning tree on the m vertices in question is equal.

Interesting things happen when you do this dynamically. That is, if we have two USTs of sizes m and n at some time t, and condition that the next edge to be added in the process joins them, then the resulting component is not a UST on m+n vertices. To see why, consider the probability of a ‘star’, that is a tree with a single distinguished vertex to which every other vertex is joined. Then the probability that the UST on m vertices is a star is \frac{m}{m^{m-2}}=m^{-(m-3)}. By contrast, it is not possible to obtain a star on m+n vertices by joining a tree on m vertices and a tree on n vertices with an additional edge.

However, I think the UST property is preserved by the cycle deletion mechanism mentioned at the very start of this post. My working has been very much of the back of the envelope variety, but I am fairly convinced that once you have taken a UST and conditioned on the sizes of the smaller trees which result from cycle deletion. My argument is that you might as well fix the cycle to be deleted, then condition on how many vertices are in each of the trees coming off this cycle. Now the choice of each of these trees is clearly uniform among spanning trees on the correct number of vertices.

However, it is my current belief that the combination of these two mechanisms does not give UST-like trees even after conditioning on the sizes at fixed time.

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Beyond Erdos-Renyi: more realistic models of networks

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The claim is often made that the study of random graphs such as the Erdos-Renyi model is worthwhile because it gives us information about complex systems which exist in the real world. The internet or social networks provide the example du jour at the moment, but it’s equally plausible to think about traffic flows, electrical systems or interacting biological processes too.

If this were entirely true, it would be great for two reasons. Firstly, in my opinion at least, it is a beautiful subject in its own right, and to have a concrete applicable reason to continue studying it would make it even better. (Not to mention the dreaded competition for funding…) Secondly, Erdos-Renyi is so simple. After all, it involves little more than adding some simple topology to a collection of IID Bernoulli random variables, and so it would surely be possible to draw some significant conclusions about how complicated real-world objects interact without too much mathematical effort.

Unfortunately, but unsurprising, this simplicity is a drawback as far as applications go. It is fairly clear that most real-world systems cannot offer any property even approaching the niceness of the independent, same probability edges condition. But rather than consign E-R to the ‘pretty but useless’ category of mathematical structures, we should think carefully about exactly why it fails to be a good model for real-world networks, and see whether there are any small adjustments that could be made to improve it.

This is something I’ve been meaning to read up about for ages and ages. What follows is based heavily on the Albert and Barabasi 2002 review paper. I suspect that many of the open problems and intuitive calculations have since been finished and formalised, but for an overview I hope that doesn’t matter hugely. I’ve also leafed through the relevant chapters of Remco Van der Hofstad’s notes, but am setting the details and the exercises aside for the holidays when I have a bit more time!

Problems with Erdos-Renyi

Recall that G(n,p) takes n vertices, and adds edges between any pair of vertices independently with probability p.

One property shared by most real-world networks is the scale-free phenomenon, which says that the degree distribution has a power law tail. The Albert-Barabasi papers gives a comprehensive survey of data verifying this claim. By contrast, G(n,p) has degree distribution which is approximately Poisson as n grows. This is concentrated near the average degree with a thin exponential tail, so does not satisfy this requirement. I was and still am a bit confused by the term ‘scale-free’. The idea is certainly that the local structure is independent of the size of the system, which seems to be true for the degree distributions in sparse ER, that is where p = O(1/n). But I think the correct heuristic is that it doesn’t matter how far zoomed in you are – the macroscopic structure looks similar for n vertices as for n^2 vertices. This certainly fails to be true for ER, where no vertex has O(n) neighbours, whereas with a power law tail, this does hold.

The main consequence of this is that there are a few vertices with very high degree. These are often called ‘hubs’ and parallels are drawn to the internet, where key websites and servers connect lots of traffic and pages from different areas. The idea is that the hubs are almost certainly well-connected to each other, and this offers a step towards a small-world phenomenon, where the shortest path between any two vertices is very small relative to the size of the system. This notion was introduced to mainstream culture by Stanley Milgram’s ‘Six degrees of separation’ experiment in the 60s, where it became clear that subjects were able to deliver a package to a complete stranger on the other side of the USA, using only personal contacts, in about six stages. The graph theoretic notion for this is the diameter, defined as the maximal graph distance between two points. Here, the graph distance means the length of the shortest path between the points. This definition, with the max-min formalism looks rather complicated, but isn’t really. The diameter of an Erdos-Renyi graph for fixed p, increases like log n, which is small relative to n, and so this property holds.

A quick glance at your list of Facebook friends will confirm that the independent edges condition in an Erdos-Renyi random graph is not a plausible model for social networks. How many friends do you have? Let’s say about 1000, more to make the calculation easier than because you’re necessarily very popular. How many does your friend Tom have? Let’s say 1000 again. As was in the news a few months ago, there are now over a billion people on Facebook. Let’s say exactly a billion (that is 10^9 for these purposes). So both you and Tom are friends with 1/10^6 of the total membership of the network. So how large would you expect the overlap of your friendships to be, if they were all chosen independently at random? Well, the probability that you are both friends with Alice is 10^-12, and so the expected number of your mutual friends is 10^-12 x 10^9 = 10^-3 which is substantially less than 1. Yet I imagine if you substituted names suitably, you and Tom might well have over 50 mutual friends if you were, say, in the same year at school or niversity and haven’t yet purged your list.

We want a statistic that records this idea quantitatively. There are various candidates for such a clustering coefficient. The underlying notion that we might expect there to be greater connectivity between neighbours of some fixed point v than in the graph as a whole gives an intuition for a possible definition. Compare the proportion of triangles in the graph to the cube of the proportion of edges. When this ratio is large, then there is a lot of clustering. In the E-R case, we would expect these to be equal, as the probability of forming a triangle is equal to the cube of the probability of the presence of each of the three independent edges that make up the triangle.

So we have three properties of real networks that we would like to incorporate into a model: small diameter, power-law degree distribution, and high clustering. To avoid this turning into a book, I’m going to write a paragraph about each of the possibilities discussed by Albert and Barabasi.

Generalised Random Graph

The degree distribution will typically emerge as a consequence of the construction of a given model. The general idea here is to condition on the degree distribution having the form we want, and see what this does to the structure. Of course, the choice of how to do this conditioning is absolutely key. It certainly isn’t obvious what it means to ‘condition G(n,p) to have power-law distribution’, since the very idea of a power-law vs exponential tail requires the number of vertices to be large.

The first idea for achieving this gives the vertices ‘stubs’, which join up in pairs to form edges. We decide on the distribution of stubs according to this power law, then pair them up uniformly at random. Obviously, there is a possibility of getting some loops, but this is not going to happen so often as to be a genuine problem in the limit. This construction is similarly open to the branching process exploration ideas well covered for the E-R random graph, though we have to be careful to size-bias the degree distributions when necessary. There is still an underlying independence in the location of edges though, so it is reasonably clear that the amount of clustering may be closer to E-R than to the real examples cited.

The other possibility suggested is to retain the independent edge property, but give the vertices weights, and let the probability of an edge between two vertices be some sensible function of the weights. In the end it turns out to make little difference whether the weights are chosen deterministically or randomly, but by taking the weights i.i.d. with infinite mean, we can generate a so-called generalised random graph where the degree distribution has a power law.

Watts-Strogatz

In the WS model, the idea is to interpolate between a graph with maximal clustering and a random graph. A d-regular graph, say on a ring, where every vertex is connected to its d nearest neighbours has high clustering, but large diameter, as for example it takes roughly n/2d steps to get to the other side of the ring. Whereas in the standard E-R model we add edges with some fixed probability p, here we replace edges with some fixed probability p. That is, we take an edge in the regular graph and with some small probability we remove it and instead add an edge between two vertices chosen uniformly at random. The theoretical motivation is that removing a few edges doesn’t destroy the high clustering evident in the regular graph, but even a sparse random graph has small diameter, so adding a few ‘long-range’ edges should be enough to decrease the diameter significantly.

It obviously needs to be checked that a substantial drop in diameter occurs before a substantial decrease in clustering, and there is a calculation and diagram to support this intuitive idea in the paper. The one drawback of this model is that it fails to provide the power-law degree distributions we want. After all, an E-R graph has a concentrated degree distribution, and a d-regular graph has all degrees the same, so we would expect some interpolation between the two to have a concentrated distribution as well. Nonetheless, this model accords well with an idea of how complex networks might form, particularly if there is some underlying geometry. It is reasonable to assume that an initial setup for a network would be that people are connected to those closest to them, and then slowly acquire distant contacts as time progresses.

Preferential Attachment – Barabasi-Albert model

Most of our intuition for networks can be extended to an intuition for the formation of networks. The idea of prescribing a degree distribution is neat, but it doesn’t give any account to the mechanism of formation. Complexity emerges over time, and a good model should be able to describe why this happens. The Barabasi-Albert model takes this as its starting point, with the aim of producing a highly clustered system dynamically. Recall that we can describe G(n,p) as a process by coupling, then increasing p from 0 to 1, and seeing edges emerge. The independence assumption can be lifted through the coupling, and so which edge appears next is independent of the current state of the system.

This is what we need to relax. Recall the motivating idea of ‘hubs’, where a small collection of vertices have very high connectivity across the whole system, as observed in several real situations. A consequence of this is that new edge is more likely to be attached to a hub, than to a pair of poorly connected vertex elsewhere. But it turns out that this idea of preferential attachment isn’t enough by itself. Because as a network forms, it is not just the connectivity that increases, but also the size of the system itself. So in fact it makes sense to add vertices rather than edges, and join the new vertices to existing vertices in proportion to the degrees of the existing vertices. This combination of growth and preferential attachment is key to the scale-free graphs that this Barabasi-Albert model generates. Relaxing either mechanism returns us to the case of exponential tails. However, there are methods in the literature for generating such graphs without the need for a dynamic model, but they are harder to understand and describe. None I have seen so far has a high clustering coefficient.

Hubs are effectively a way to reduce the diameter. Recall the description of Milgram’s experiment where he encouraged randomly chosen people to send a package to Harvard. For the purposes of this model, an undergraduate from Wyoming or a husband from Alabama moving in with his wife in Boston are clear hubs, as for very many people near their previous home, they represent a good connection to Harvard. So it is unsurprising that BA, which reinforces hubs, has a sub-logarithmic small diameter.

Conclusions

I’m not entirely what conclusions I should draw from my reading. Probably the main one is that I should read more as there is plenty of interesting stuff going on in this area. Intuitively, it seems unlikely that there is going to be a single model which unites the descriptions of all relevant real-world networks. As ever, it is pleasant to find structures that are both mathematically interesting in their own right and relevant to applied problems. So it is reassuring to observe how similar many of the models discussed above are to the standard random graph.

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Poisson Tails

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I’ve had plenty of ideas for potential probability posts recently, but have been a bit too busy to write any of them up. I guess that’s a good thing in some sense. Anyway, this is a quick remark based on an argument I was thinking about yesterday. It combines Large Deviation theory, which I have spent a lot of time learning about this year, and the Poisson process, which I have spent a bit of time teaching.

Question

Does the Poisson distribution have an exponential tail? I ended up asking this question for two completely independent reasons yesterday. Firstly, I’ve been reading up about some more complex models of random networks. Specifically, the Erdos-Renyi random graph is interesting mathematical structure in its own right, but the independent edge condition results in certain regularity properties which are not seen in many real-world networks. In particular, the degree sequence of real-world networks typically follows an approximate power law. That is, the tail is heavy. This corresponds to our intuition that most networks contain ‘hubs’ which are connected to a large region of the network. Think about key servers or websites like Wikipedia and Google which are linked to by millions of other pages, or the social butterfly who will introduce friends from completely different circles. In any case, this property is not observed in an Erdos-Renyi graph, where the degrees are binomial, and in the sparse situation, rescale in the limit to a Poisson distribution. So, to finalise this observation, we want to be able to prove formally that the Poisson distribution has an exponential (so faster than power-law) tail.

The second occurrence of this question concerns large deviations for the exploration process of a random graph. This is a topic I’ve mentioned elsewhere (here for the exploration process, here for LDs) so I won’t recap extensively now. Anyway, the results we are interested in give estimates for the rate of decay in probability for the event that the path defined by the exploration process differs substantially from the expected path as n grows. A major annoyance in this analysis is the possibility of jumps. A jump occurs if a set of o(n) adjacent underlying random variables (here, the increments in the exploration process) have O(n) sum. A starting point might be to consider whether O(1) adjacent RVs can have O(n) sum, or indeed whether a single Poisson random variable can have sum of order n. In practice, this asks whether the probability \mathbb{P}(X>\alpha n) decays faster than exponentially in n. If it does, then this is dominated on a large deviations scale. If it decays exactly exponentially in n, then we have to consider such jumps in the analysis.

Approach

We can give a precise statement of the probabilities that a Po(\lambda) random variable X returns a given integer value:

\mathbb{P}(X=k)=e^{-\lambda}\frac{\lambda^k}{k!}.

Note that these are the terms in the Taylor expansion of e^{\lambda} appropriately normalised. So, while it looks like it should be possible to evaluate

\mathbb{P}(X>\alpha n)=e^{-\lambda}\sum_{\alpha n}^\infty \frac{\lambda^k}{k!},

this seems impossible to do directly, and it isn’t even especially obvious what a sensible bounding strategy might be.

The problem of estimating the form of the limit in probability of increasing unlikely deviations from expected behaviour surely reminds us of Cramer’s theorem. But this and other LD theory is generally formulated in terms of n random variables displaying some collective deviation, rather than a single random variable, with the size of the deviation growing. But we can transform our problem into that form by appealing to the three equivalent definitions of the Poisson process.

Recall that the Poisson process is the canonical description of, say, an arrivals process, where events in disjoint intervals are independent, and the expected number of arrives in a fixed interval is proportional to the width of the interval, giving a well-defined notion of ‘rate’ as we would want. The two main ways to define the process are: 1) the times between arrivals are given by i.i.d. Exponential RVs with parameter \lambda equal to the rate; and 2) the number of arrivals in interval [s,t] is independent of all other times, and has distribution given by Po(\lambda(t-s)). The fact that this definition gives a well-defined process is not necessarily obvious, but let’s not discuss that further here.

So the key equivalence to be exploited is that the event X>n for X\sim \text{Po}(\lambda) is a statement that there are at least n arrivals by time 1. If we move to the exponential inter-arrival times definition, we can write this as:

\mathbb{P}(Z_1+\ldots+Z_n<1),

where the Z’s are the i.i.d. exponential random variables. But this is exactly what we are able to specify through Cramer’s theorem. Recall that the moment generating function of an exponential distribution is not finite everywhere, but that doesn’t matter as we construct our rate function by taking the supremum over some index t of:

I(x)=\sup_t (xt-\log \mathbb{E}e^{tZ_1})=\sup_t(xt-\log(\frac{\lambda}{\lambda-t})).

A simple calculation then gives

I(x)=\lambda x-1 - \log \lambda x.

\Rightarrow I(x)\uparrow \infty\text{ as }x\downarrow 0.

Note that I(1) is the same for both Exp(\lambda) and Po(\lambda), because of the PP equality of events:

\{Z_1+\ldots+Z_n\leq n\}=\{\text{Po}(\lambda n)=\text{Po}(\lambda)_1+\ldots+\text{Po}(\lambda)_n> n\},

similar to the previous argument. In particular, for all \epsilon>0,

\mathbb{P}(\text{Po}(\lambda)>n)=\mathbb{P}(\frac{Z_1+\ldots+Z_n}{n}<\frac{1}{n})<\mathbb{P}(\frac{Z_1+\ldots+Z_n}{n}<\epsilon),\text{ for large }n.

\mathbb{P}(\text{Po}(\lambda)>n)=O(e^{-nI(\epsilon)}),\text{ for all }\epsilon.

Since we can take I(\epsilon) as large as we want, we conclude that the probability decays faster than exponentially in n.

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Delayed Connectivity in Random Graphs

Aside

I presented a poster at the Oxford SIAM Student Chapter Conference on Friday. It was nice to win the prize for best poster, but mainly I enjoyed putting it together. For once it meant ignoring the technical details and anything requiring displayed formulae, and focusing only on aspects that could be conveyed with bullet points and images. Anyway, this is what I came up with. The real thing is sitting safely in a tube in my office, ready for the next time it is needed in a hurry!

Delayed Connectivity Poster

Mixing Times 5 – Cesaro Mixing

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We have just finished discussing chapters 11 and 12 of Markov Chains and Mixing Times, the end of the ‘core material’. I thought that, rather than addressing some of the more interesting but technical spectral methods that have just arisen, it would be a good subject for a quick post to collate some of the information about Cesaro mixing, which is spread throughout this first section.

Idea

A main result in the introductory theory of Markov chains is that for an irreducible aperiodic chain X, the distribution of X_t\rightarrow \pi, the (unique) equilibrium distribution. The mixing time gives the rate at which this first mode of convergence takes place. We have freedom over the initial state, so we typically consider the ‘worst case scenario’, ie the slowest convergence. The most appropriate metric is given by the total variation distance, which is defined in previous posts. The most important point to note is that the mixing time should be thought of as the correct timescale for convergence, rather than some threshold. In particular, the time at which the chain is within 1/4 of the equilibrium distribution in the TV metric has the same order magnitude (in n, some parameter controlling the number of states) as the time at which it is within 1/20 of the equilibrium distribution.

But this isn’t the only result about convergence in distribution of functionals of Markov chains. Perhaps more intuitive is the ergodic theorem which asserts that the proportion of time spent in a particular state also converges to the equilibrium probability as time advances. We might write:

\frac{1}{t}\sum_t \mathbf{1}(X_t=x)\rightarrow \pi(x),\quad \forall x\in \Omega.

Note that if the state space \Omega is finite, then we can also assume that this occurs uniformly in x. We can also think of the LHS of this convergence as a measure on \Omega varying in time

\frac{1}{t}\sum_t \mathbf{1}(X_t=\cdot),

and the mixing time for this sequence of measures is defined as for the conventional mixing time, and is called the Cesaro mixing time, at least in the Levin / Peres / Wilmer text.

Advantages

There are some obvious advantages to considering Cesaro mixing. Principally, a main drawback of conventional mixing is that we are unable to consider periodic chains. This property was the main content of the previous post, but to summarise, if a chain switches between various classes in a partition of the state space in a deterministic periodic way, then the distribution does not necessarily converge to equilibrium. The previous post discusses several ways of resolving this problem in specific cases. Note that this problem does not affect Cesaro mixing as the ergodic theorem continues to hold in the periodic case. Indeed the form of the distribution (which we might call an occupation measure in some contexts) confirms the intuition about viewing global mixing as a sort of sum over mixing modulo k in time.

Other advantages include the fact that the dependence on the initial state is weaker. For instance, consider the occupation measures for a chain started at x which moves first to y, versus a chain which starts at y then proceeds as the original. It requires very little thought to see that for O(1) values of t, this difference in occupation measures between these chains becomes small.

Another bonus is that we can use so-called stationary times to control Cesaro mixing. A stationary time is a stopping time such that X_\tau\stackrel{d}{=}\pi. It is clear that if we wait until \tau, then run the chain for a further \alpha\tau, the chain will have spent \frac{\alpha}{1+\alpha} of its duration in the equilibrium distribution, and so using Markov’s inequality and bounding the total variation distance between occupation measure and \pi by 1 up until the stationary time, we can get good bounds for the Cesaro mixing time in terms of \mathbb{E}\tau.

Why does this fail to work for normal mixing? The key to the above argument was that by taking an average over time up to some T>>\tau, the dependence on the actual value of X_\tau was suppressed. Consider the deterministic walk on the cycle \mathbb{Z}_n, which advances by 1 modulo n on each go. Now sample independently a random variable Z distributed uniformly on \mathbb{Z}_n. By definition, the random hitting time \tau_Z is a stationary time, but in fact the chain’s distribution does not converge. The condition we actually require for normal mixing is that \tau be a strong stationary time, meaning that X_\tau\stackrel{d}{=}\pi, and the value of X_\tau is independent of \tau. With this definition we can proceed with a similar result for normal mixing. An example of a strong stationary time would be for shuffling a pack of cards by repeatedly inserting the top card into a random place in the rest of the pile. Then consider the moment at which the original bottom card first reaches the top of the pile. It does not take too much to reassure oneself that after the next move, we have a strong stationary time, since every card has been randomised at least once, and the position of the other cards is independent of how long it took the original bottom card to rise to the top.

Disadvantages

So why do we not consider Cesaro mixing rather than the conventional variety? Well, mainly because of how we actually use mixing times. The Metropolis algorithm gives a way to generate chains with a particular equilibrium distribution, including ones for which it is hard to sample directly. Mixing time theory then gives a quantitative answer to the question of how long it is necessary to run such a chain for before it gives a good estimate to the equilibrium distribution. In many cases, such a random walk on a large unknown network, the main aim when applying such Monte Carlo procedures is to minimise the difficulty of calculation. For Cesaro mixing, you have to store all the information about path states, while for conventional mixing you only care about your current location.

The other phenomenon that is lacking in Cesaro mixing is cutoff. This is where the total variation distance

d_n(t)=||P^t(x,\cdot)-\pi||_{TV}

converges to 0 suddenly. More formally, there is some timescale f(n) such that

\lim_{n\rightarrow\infty}d_n(cf(n))=\begin{cases}1& c<1\& c>1,\end{cases}

so in the n-limit, the graph of d looks like a step-function. Several of the shuffling chains exhibit this property, leading to the statements like “7 shuffles are required to mix a standard pack of cards”. Cesaro mixing smooths out this effect on an f(n) timescale.

A Further Example

Perhaps the best example where Cesaro mixing happens faster than normal mixing is in the case of a lazy biased random walk on \mathbb{Z}_n. (Ex. 4.10 in MTMC) Here, we stay put with probability 1/2, otherwise move clockwise with probability p>1/4 and counter-clockwise with probability 1/2 – p < 1/4. This chain is not reversible, as we can determine the direction (or arrow) of time by examining a path. Roughly speaking, the chain will drift clockwise at rate 2p – 1/2 > 0. In particular, at some time Kn, where K is large, we would expect to have completed ~ \frac{K}{2p-\frac12} circuits of the vertices, and so the occupation measure will be close to the uniform equilibrium distribution if we choose K large enough.

On the other hand, the distribution of X at time Kn is still fairly concentrated. If we assume we are instead performing the random walk on \mathbb{Z}, the distribution after Kn i.i.d. increments is

X_t\sim N(Kn(2p-\frac12), Kn\sigma^2).

That is, the standard deviation is O(\sqrt{n})<<n. So, even once we return to considering the random walk on the cyclic group, if we view it as a circle, we expect most of the probability mass to be concentrated near Kn(2p-\frac12) \mod n. By an identical heuristic argument, we see that the mixing time is achieved when the variance has order n, that is when time has order n^2.

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Mixing Times 4 – Avoiding Periodicity

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A Markov chain is periodic if you can partition the state space such it is possible to be in a particular class only at certain, periodic times. Concretely, suppose we can find a decomposition into classes \Omega=V_1\cup\ldots\cup V_k such that conditional on X_t\in V_i, we have \mathbb{P}(X_{t+1}\in V_{i+1})=1, where the indices of the Vs are taken modulo k. Such a chain is called periodic with period k. In most cases, we would want to define the period to be the maximal such k.

Why is periodicity a problem? It prevents convergence to equilibrium. The distribution at time t has some fairly strong dependence on the initial distribution. For example, if the initial distribution is entirely supported on V_1 as defined above, then the distribution at time t will be entirely supported on V_i, where i\equiv t \mod k. In particular, this cannot converge to some equilibrium.

Aperiodicity thus becomes a necessary condition in any theorem on convergence to equilibrium. Note that by construction this is only relevant for chains in discrete time. In an first account of Markov chains, most of the examples will either have a small state space, for which the transition matrix will have to contain lots of zeros before it stands a chance of being periodic, or obviously aperiodic birth-death or queue type processes. But some of the combinatorially motivated chains we consider for interesting mixing properties are more likely to be periodic. In particular, for a random walk on a group say, the generator measure may well be supported only on a small subset of the whole group, which is completely natural (eg transpositions as a subset of the symmetric group). Then it becomes more plausible that periodicity might arise because of some underlying regularity or symmetry in the group structure.

My first claim is that periodicity is not a disaster for convergence properties of Markov chains. Firstly, by the definition above, P^k(x,y) for x,y\in V_1 is an irreducible (aperiodic if k is maximal) transition matrix on V_1, and so we have convergence to some equilibrium distribution on V_1 of (X_{kt+a}) or similar. An initial distribution mixed between classes gives a mix of such equilibria. Alternatively, we could think about large-time ergodic properties. By taking an average over all distributions up to some large t, the periodic problems get smoothed out. So, for mixing on a periodic chain, it might be possible to make headway with Cesaro mixing, which looks at the speed of convergence of the ergodic average distribution.

In most cases, though, we prefer to alter the chain directly to remove periodicity, or even any chance of periodicity. The preferred method in many contexts is to replace the transition matrix P with \frac12 (P+I). This says that at every time t, we toss an independent fair coin, and with probability 1/2 make the transition suggested by P, and with probability 1/2 we stay where we are. Note that if a chain is irreducible, and some P(x,x)>0, then it is definitely aperiodic, as x cannot be in more than one class as per the definition of periodicity.

If you want to know about the mixing time of the original chain, note that this so-called lazy chain moves at half the speed of the original, so to get exact asymptotics (eg in the case of cutoff, that is mixing speed faster than the scale of the mixing time) you must multiply by 2. Also note that all of the eigenvalues of \frac12 (P+I) are non-negative, and in fact, the eigenvalues are subject to a linear transform in the construction of the lazy transition matrix \lambda\mapsto \frac12(1+\lambda).

Note that choosing 1/2 as the parameter is unnecessary. Firstly, it would suffice to take some P(x,x)=\epsilon and rescale the rest of the row appropriately. Also, in some cases, a different constant gives a more natural interpretation of the underlying mechanism. For example, one model worth considering is the Random Transposition Random Walk on the symmetric group, where at time t we multiply (ie compose) with a transposition chosen uniformly at random. This model is interesting partly because the orbits of an element resemble, at least initially, the component size process of a Erdos-Renyi random graph, on the grounds that when the number of transpositions is small, they don’t interact too much, so can be viewed as independent edges. Anyway, some form of laziness is needed in RTRW, otherwise the chain will alternative between odd and even permutations. In this case, 1/2 is not the most natural choice. The most sensible way to sample random transpositions is to select the two elements of [n] to be transposed uniformly and independently at random. Thus each transposition is selected with probability \frac{2}{n^2}, while the identity, which corresponds to ‘lazily’ staying at the current state in the random walk, is selected with probability 1/n.

The lazy chain is also useful when the original chain has a lot of symmetry involved. In particular, if the original chain involves ‘switching’ say one coordinate. The best example is the random walk on the vertices of the n-hypercube, but there are others. Here, the most helpful way to visualise the configuration is to choose a coordinate uniformly at random and then flip its value (from 0 to 1 or 1 to 0). Now the lazy chain can be viewed similarly, but note that the dependence on the current value of the coordinate is suppressed. That is, having chosen the coordinate to be affected, we set it to be 0 with probability 1/2 and 1 with probability 1/2, irrespective of the prior value at that coordinate. Thus instead of viewing the action on coordinates to be ‘stay or switch’, we can view the action on the randomly chosen coordinate to be ‘randomly resample’, to use statistical terminology. This is ideal for coupling, because from the time coordinate j is first selected, the value at that coordinate is independent of the past, and in particular, the initial value or distribution. So we can couple arbitrary initial configurations or distributions, and we know that as soon as all coordinates have been selected (a time that can be described as a coupon collector problem), the chains are well coupled, that is, the values are the same.

Note that one way we definitely get periodicity is if the increment distribution for random walk on a group is supported entirely in a single coset of a normal subgroup. Why? Well if we take H\lhd G to be the normal subgroup, and gH to be the relevant coset, then P^t(g',\cdot) is supported entirely on g^tg'H, so is periodic with period equal to the order of gH in the quotient group G/H. Note that if the coset is the normal subgroup itself, then it might well include support on the identity, which immediately makes the chain aperiodic. However, there will be then be no transitions between cosets, so the chain is not irreducible on G.

The previous paragraph is the content of Remark 8.3 in the book we are reading. My final comment is that normality is precisely what is needed for this to hold. The key idea is that the set of subsets {gH, gHgH, gHgHgH, … } forms a partition of the group. This is certainly true if H is normal and gH generates G/H. If the latter statement is not true, then the set of subsets still forms a partition, but of some subset of G. The random walk is then neither irreducible nor aperiodic on the reduced state space. If H is not normal, then there are no such restrictions. For example, gHgH might be equal to the whole group G. Then the random walk is aperiodic, as this would imply we can move between any pair of states in two steps, and so by extension between any pair of states in three steps. (2,3)=1, hence the chain is aperiodic. As a concrete example, consider

\tau=\langle (1 2)\rangle \leq S_3,

the simplest example of a non-normal subgroup. Part of the problem is that cosets are different in the left-case and the right-case. Consider the left coset of \tau given by \sigma\tau=\{(1 2 3),(2 3)\}. These elements have order three and two respectively, and so by a similar argument to the general one above, this random walk is aperiodic.

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