Recent Research Activity

Posted on by
Reply

I’ve spent this week in Luminy, near Marseille, attending a summer school run by ALEA, the organisation of French probabilists. We’ve been staying in CIRM, a dedicated maths research conference centre at the edges of the calanques, the area of mountains and jagged coastal inlets between Marseille and Cassis. The walking possibilities have been excellent, as have the courses and lectures, on a range of topics in probability theory.

Anyway, the time here has been an excellent moment to reflect on my research progress, and try to come up with the sort of fresh ideas that are perhaps slightly inhibited by sitting at a desk with an endless supply of paper on which to try calculations. When I get back, I have to submit a first-year report, so at least for a little while I will have to suppress the desire to make further progress and instead diligently assemble the progress I have made.

The Model

I’ve defined some of these processes in past posts, but I see no harm in doing so again. We take the standard Erdos-Renyi random graph process, where edges are added one-at-a-time uniformly at random between n vertices, and amend it by adding a deletion mechanism. The aim is to arrive at a process which looks in equilibrium more like the critical random graph than either the subcritical or supercritical regimes, where the components are very small, and dominated by one giant component respectively. Rath, Toth and others have studied the process where each vertex is hit by lightning at uniform rate. When this happens, we delete all the edges in the component containing that vertex. Naturally, big components will be hit by lightning more often than small components, and so this acts as a mechanism to prevent the formation of giant components, if scaled correctly.

We take a different approach. We observe that criticality in the original random graph process is denoted by the first appearance of a giant component, but also by the first appearance of a) lots of cycles, and b) large cycles. In particular, it is very unlikely that a giant component could form without containing any cycles. We will therefore use the appearance of a cycle to trigger some form of deletion mechanism.

Our final goal is to treat the so-called ‘Cycle Deletion’ model. Here, whenever a cycle appears, we delete all the edges in that cycle immediately. There are several challenges in treating this model, because the rate at which cycles emerge in a tree is a function of the tree structure. The trees in this model will not be Uniform Spanning Trees (though it is very possible that they will be ‘almost USTs’ in some sense – we need to investigate this further) so it will be hard to make nice statements about the rates. For the standard random graph process, if we are only interested in the sizes of the components, we are actually allowed to ignore the graph structure entirely. The component sizes evolve as a discrete, stochastic version of the multiplicative coalescent (sometimes called a Marcus-Lushnikov process). We would like a deletion mechanism that has a nice interpretation as a fragmentation operation in the same sense. The rate at which a component fragments will be quadratic in the size of the component, since there are O(k^2) possible edges between k vertices forming a component, and adding any of precisely these will create a cycle.

I’ve talked previously about how to overcome the problems with the tree structure in Cycle Deletion with the so-called Uniform Cycle Deleting model. In any case, as a starting point we might consider the Cycle-Induced Forest Fire model. Here, whenever a cycle appears, we delete all the edges, including the new one, in the whole component which contains the cycle.

We suspect this model may resemble the critical random graph at all times. The main characteristic of G(n,1/n) is that the largest component is of size O(n^2/3), and indeed there are arbitrarily many components of this size, with high probability in the limit. Since CIFF is recurrent for any fixed n, meaning that it will visit any state infinitely often (rather than tending to infinity or similar), we should ask what the largest component is typically in the equilibrium distribution. Our aim is to prove that it is O(n^2/3). We might suspect that the typical size of the largest component will be greater in the Cycle Deletion model, since each fragmentation event is less severe there, removing fewer edges.

An Upper Bound

The nice thing about Markov chains is that they have an ergodic property, which means that if you run them for long enough, the proportion of time spent in any state is given by the stationary probability of being in that state. It doesn’t matter whether or not you start in equilibrium, since it will converge anyway. Thus it is meaningful to talk about properties like the average number of isolated vertices as a time-average as well as an average with respect to some distribution.

This quantity is the key to an upper bound. We can equally talk about the average change in the number of isolated vertices in a time-step. This will increase when a component fragments, and will decrease when an isolated vertex coalesces with another component. In particular, the largest possible decrease in the number of isolated vertices in a single time-step is 2, corresponding to an edge appearing between two isolated vertices.

Suppose that with probability \Theta(1) there is a component of size n^\alpha for some \alpha>2/3. Then such a component makes a contribution to the expected change in the number of isolated vertices of

\Theta(1) n^\alpha \left(\frac{n^\alpha}{n}\right)^2. (*)

Where does this come from? Well, we are tracking the contributions from the event that the largest component is of this size and that it fragments, giving n^\alpha new isolated vertices. So the \Theta(1) accounts for the probability that there is such a component to begin with. Then, conditional on that, the probability that it gets fragmented in the next time-step is the probability that both ends of the next edge added lie in that component. Since the edge is chosen uniformly at random, the probability of this is n^\alpha/n. Note that this is under a slightly odd definition of an edge, that allows loops. Basically, I don’t want to have lots of correction terms involving \binom{n}{2} floating around. However, it would make no difference to the orders of magnitude if we to do it with these.

So, this is only one contribution to the typical rate of gain of isolated vertices. Now note that if \alpha>2/3, then this expression is >> 1. This is bad since the negative contributions to this expected flux in the number of isolated vertices is O(1). So this suggests that over time, the number of isolated vertices will keep growing. This is obviously ridiculous since a) we are in equilibrium, so the expected flux should be 0 and b) the number of isolated vertices cannot exceed n, for clear reasons.

This gives us an upper bound of n^2/3 as the typical scale of the largest component. We can come up with a similar argument for the cycle deleting model. The most helpful thing to track there is the number of edges in the graph. Note that since the graph is at all times a forest on n vertices, the number of edges is equal to n minus the number of (tree) components. We use the fact that the typical fragmentation of a component of size k creates O(\sqrt{k}) new components. It is possible to argue via isolated vertices here too, but the estimates are harder, or at least less present in the literature.

Lower Bounds?

The problem with lower bounds is that it is entirely possible that the flux in the number of isolated vertices is not driven by typical behaviour. Suppose for example we had a different rule. We begin a random graph process, and the first time we see a cycle in a component with size larger than n^2/3, we delete all the edges in the whole graph. Then we will see a sequence of random graph processes starting with the empty graph and stopped at some point close to criticality (in fact, with high probability in the *critical window*), and these will all be glued together. So then, most of the time the process will look subcritical, but the gains in isolated vertices will occur only during the critical periods, which are only an asymptotically small proportion of the time.

At the moment, my approach to the lower bound is instead to prove that the upper bound is tight. I mean this in the following sense. Suppose we wanted to be sure that (*) was in fact equal to the average rate of gain of isolated vertices. We would have to check the following:

  • That the total contributions from all other components were similar or smaller than from the component(s) of size roughly n^{\alpha}.
  • That there were only a few components of size n^{\alpha}. In particular, the estimate would be wrong if there were n^\epsilon such components for any \epsilon>0.
  • That it cannot be the case that for example, some small proportion of the time there is a component of size roughly n^{\alpha+\epsilon}, and over a large enough time these make a greater contribution to the average gain in isolated vertices.

A nice way to re-interpret this is to consider some special vertex and track the size of its component in time. It will be involved in repeated fragmentations over the course of time, so it is meaningful to talk about the distribution of the size of the component containing the vertex when it is fragmented. Our aim is to show that this distribution is concentrated on the scaling O(n^\alpha).

So this has turned out to be fairly hard. Rather than try to explain some of the ideas I’ve employed in attempting to overcome this, I will finish by giving one reason why it is hard.

We have seen that the component sizes in random graphs evolve as the multiplicative coalescent, but at a fixed moment in time, we can derive good estimates from an analogy with branching processes. We might like to do that here. If we know what the system looks like most of the time, we might try to ‘grow’ a multiplicative coalescent, viewing it like a branching process, with distribution given by the typical distribution. The problem is that when I do this, I find that the expectation of the offspring distribution is \Theta(1). This looks fine, since 1 is the threshold for extinction with probability 1. However, throughout the analysis, I have only been paying attention to the exponent of n in all the time and size estimates. For example, I view n^\alpha and n^\alpha \log n as the same. This is a problem, as when I say the expectation is \Theta(1), I am really saying it is \sim n^0. This means it could be \frac{1}{\log n} or \log n. Of course, there is a massive difference between these, since a branching process grows expectationally!

So, this approach appears doomed in its current form. I have some other ideas, but a bit more background may be required before going into those. I’m going to be rather busy with teaching on my return to the office, so unfortunately it is possible that there may be many posts about second year probability and third year applied probability before anything more about CIFF.

The Top-to-Random Shuffle III

Posted on by
Reply

This post concludes my non-exhaustive list of things I think are interesting about the top-to-random shuffle. In previous posts I have talked about the construction and correct sense of convergence to randomness, and that this algorithm does genuinely achieve uniform randomness at some hitting time which is easy to specify. Promising that posts will be short hasn’t worked in the past so I won’t do that again now, but the idea of this post is brief:

When we specified the dynamics of the top-to-random shuffle, we insisted that the top card card could be placed anywhere in the deck with equal probability including back on top. This appears to be doing nothing except slowing down the shuffling process. Why is this important for convergence to randomness?

Fortunately the answer is short: if we do not let the top card be inserted back onto the top, allowing the configuration to stay the same, then we can divide up the set of orderings into two classes, and the pack will alternate between them.

Why is this a problem? Suppose the classes are called X and Y, and X is the class that contains the original ordering 1,2,…,n. Then after k shuffles, the ordering of the deck will be in X if k is even and in Y if k is odd. Remember our definition of ‘close to randomness’ will be the greatest difference in probability of an event between the actual distribution and the uniform distribution. As before, you can think of this by a betting analogy – what proportion profit can you make again someone who thinks it’s uniform by knowing the true distribution?

Well, it will turn out that the sets X and Y have the same size, so in the uniform distribution, the probability that an ordering is in X is 1/2. Whereas if the pack alternates, then so long as we know how many shuffles have occurred, this probability is either 0 or 1. In particular this is far from 1/2. We should remark that if we introduce the notion of sampling at a random time, or taking an average over all large times in some sense, such problems may disappear, but the result obtained may be less useful. See this post on Cesaro Mixing for details presented in a more rigorous style.

So it remains to see why this is true. First a definition. A transposition is when two elements in a permutation are exchanged. Eg 31452 -> 35412 by transposing 1 and 5. It makes sense intuitively that we can get from any permutation to any other permutation by making successive transpositions. Indeed, this is precisely what is happening in the top-to-random shuffle. To avoid continually having to write it out, we call the original permutation 1,2,…,n the identity permutation.

Then the idea is that X is the set of permutations we can obtain by starting with the identity and applying an even number of transpositions, while Y is the set obtained by applying an odd number of transpositions. For this to work, we will need to show that these sets are disjoint. That is, no permutation can be generated by both an odd number and an even number of transpositions. This is important, as a permutation can certainly be generated from transpositions in multiple ways. For example, if the elements are 1,2,3, we can obtain the permutation 2,1,3 by transposing 1 and 2, obviously. However, we could alternatively start by transposing 2 and 3 to get 1,3,2, then 1 and 3 to get 3,1,2, then 2 and 3 again to get 2,1,3. Note that both of these require an odd number of transpositions.

We will call a permutation even if it is generated by an even number of transpositions, and odd otherwise. We also say that its sign (alternatively signature, parity) is +1 or -1 respectively. To prove this is well-defined, we really want to find a different property that is easier to track.

A useful trick is to count how many pairs of elements are not in the correct order. Let’s do this for our previous example: 31452. There are 5 elements so 5 x 4 / 2 = 10 pairs of elements. We list them:

  • 1 and 2 are in the correct order.
  • 1 and 3 are not, as 3 comes before 1 in this permutation.
  • 1 and 4 are correct.
  • 1 and 5 are correct.
  • 2 and 3 are not.
  • 2 and 4 are not.
  • 2 and 5 are not.
  • 3 and 4 are correct.
  • 3 and 5 are correct.
  • 4 and 5 are not.

So 5 pairs are not in the correct order. Since 5 is odd, the claim is that this means 31452 is an odd permutation. To check this, and to confirm that the sign is well-defined, it suffices to check that the number of so-called inversions, or pairs in the wrong order, changes parity every time we apply a transposition.

This is clearly true if we transpose adjacent elements. Then the orderings of all pairs remain the same, apart from the pair we transposed, which changes. Then, if the elements are not adjacent, instead of transposing them directly, we can perform a succession of transpositions of adjacent elements. The easiest way to describe this is again by example. Suppose we want to transpose 3 and 5 in 31452.

31452 -> 13452 -> 14352 -> 14532 -> 15432 -> 51432.

Note that the middle transposition is actually transposing 3 and 5, and the others are symmetric about this middle operation. In particular, there is an odd number of transpositions in total. So we have proved the result for general transpositions, and thus we now know that the sign of a permutation is well-defined. Note also that there are an equal number of odd and even permutations of every n=>2. For every odd permutation, transposing 1 and 2 gives an even permutation, and vice versa, uniquely, giving a bijection.

What’s really going on is that we are able to multiply permutations, by doing one after the other. Unlike multiplying real numbers, the order in which we do this now matters. In this context, the set of permutations is an example of a general structure called a group. The idea of partitioning a group into subsets which are in some sense symmetric and where some other operation jumps between the subsets is a useful motivation point for a whole avenue of interesting theory. Not to be explored now unfortunately…

Related articles

The Top-to-Random Shuffle II

Posted on by
1

In the last post, I introduced the top-to-random shuffle. In particular, we considered why this sort of procedure was important as an alternative to choosing an ordering afresh, and how to we would go about measuring how close we had got to randomness.

In this post, I want to develop the second of these points. The intuition might be that we can get very close to uniform randomness if we repeat the shuffle often enough. Recall this means that even if we choose our bet in a really complicated and careful way, we still couldn’t make much profit by knowing the actual distribution of the ordering. But we might also suspect that the pack will never be exactly random, in the same way that the distribution of the proportion of heads seen on a repeatedly-flipped coin will eventually get very close to 1/2, but will not be exactly 1/2.

This intuition is extremely sensible, and in general is true. It is a nice fact, however, that it fails for the top-to-random shuffle, where we do in fact get to a uniformly random deck. Recall that we approximated how long it would take to get to a state that was roughly random by calculating the time taken for the original bottom card to rise to the top of the deck. This time was:

n+\frac{n}{2}+\frac{n}{3}+\ldots+\frac{n}{n-1},

where n is the total number of cards. A further shuffle is required to send the original card back into the pack somewhere. The claim is that the pack is now uniformly random. Note that if we ever actually use this, we have to be careful because although we can calculate the average time at which this event happens, the time itself is random. Rather than worry about that, let’s see why it is true.

As a motivating example, let’s assume that the pack was originally in the order 1,2,3,…,n, and consider the final relative order of the cards numbered 1 and 2. There is some small probability that on the first go (and then again on the second go possibly also) that the card number 1 stays put. Let’s ignore that possibility and progress to the first time that the 1 has been moved into the interior of the pack, so the 2 is now on top. When we do this, we are choosing a number between 2 and n uniformly at random, and moving the card numbered 1 to this position. Let’s call this number X.

Now, when we try to shuffle the 2, we are choosing a number between 1 and n uniformly at random, and moving the card numbered 2 to this position. Let’s call this number Y. Under exactly what circumstances does 2 end up above 1? The clearest example is if Y=X. Then card 2 has been moved to the position previously occupied by card 1. So card 1 moves up a position (since the card 2 is no longer on the top). The final configuration therefore includes card 1 directly above card 2. So we can say

\mathbb{P}(\text{2 above 1})=\mathbb{P}(Y<X),

where the fact that the inequality in the second probability is strict is important. To calculate this second probability, we want to exploit the symmetry of the situation. The only problems are that the case X=Y is not symmetric as then 1 ends up above 2 as described above, and also that X cannot be 1. So we account for these separately. Note

\mathbb{P}(Y=1)=\frac{1}{n},\quad \mathbb{P}(Y=X)=\frac{1}{n}.

The second result holds overall since it holds whenever we condition on a particular value of X. These events are also disjoint. Then

\mathbb{P}(Y<X)=\mathbb{P}(Y=1)+\mathbb{P}(Y<X | Y>1, X\neq Y)\mathbb{P}(Y>1,X\neq Y)

= \frac{1}{n}+\frac{1}{2}(1-\mathbb{P}(Y>1,Y\neq X))

=\frac{1}{n}+\frac{1}{2}(1-\frac{1}{n}-\frac{1}{n}) = \frac{1}{2}.

In summary, the cards 1 and 2 will be in uniformly random order. We might like to extend this idea, but it will get complicated when we add card 3 to the mix, as it is possible (if unlikely) that 1 and 2 will be further mixed before this. This shouldn’t affect the result much, but it will get complicated to define the notation required to carry this sort of argument all the way up to the nth card.

Even using induction is not going to make life substantially easier. Knowing that once we have inserted cards 1,2,…,k into the pack they are in uniformly random order is not enough to make inference about what happens once we put k+1 into the pack. We have to know something about the current positions of 1,2,…,k. For example, if one of these cards is definitely on the bottom of the pack, then the probability that k+1 ends up last among 1,2,…,k+1 is 1/n rather than 1/k+1 as it should be. So in fact we would have to control an annoying amount of information jointly.

In the argument we attempted above, we were looking at the first times some card k got folded back into the pack. Note that this division of time is different to the one we were using for the coupon collector approach to the mixing time in the previous post. Let’s try to use that instead here.

Now we consider the times at which a card is moved below card n. We deliberately decline to say what these cards are. But rather, we want to prove that, conditional on the cards below n being A_k=\{a_1,\ldots,a_k\}, the ordering of these is uniform on S_{A_k}, that is, every possibility is equally likely. Now this is easy to prove by induction. For, by conditioning on A_k and a_{k+1} being the new card to be moved below n, we are conditioning on the set of cards below n now being A_{k+1}=A_k\cup\{a_{k+1}\}. The position of the new card is uniformly random within this, by construction of the top-to-random shuffle, and so the new arrangement is uniformly random on the (k+1)! possibilities.

To see why we have proved the original result we wanted, note that this argument works at the time when the original bottom card is now at the top. So the remaining cards are uniformly randomly ordered. Inserting card n at random gives an arrangement that is uniformly random overall. So as we suggested before, working out how long it takes to get close to randomness in this case reduces to working out how long it is before the original bottom card hits the top and is re-inserted, as at that point, the pack genuinely is uniformly random.

Related articles

The Top-to-Random Shuffle

Posted on by
2

This article is based on a talk I gave to the Maths Society at St Paul’s School on Monday. It may turn into a short series if I have time before I go to ALEA in Luminy near Marseille on Saturday.

My original plan had been to talk about riffle-shuffling, and some of the interesting mixing time themed results one can obtain. As a motivating example, I began by discussing the simpler top-to-random shuffle, and this proved sufficiently interesting to occupy the time I had been allowed (and mea culpa a bit more). It therefore seems worth writing a hopefully moderately accessible blog post on the subject. The aim of this post at least is to discuss the idea that repeatedly shuffling brings a pack of cards close to randomness. We have to settle on a definition of ‘close to randomness’, and find some ways to calculate this.

Suppose we are playing some bizarre card game where it is necessary that three cards labelled, uncontroversially, 1, 2 and 3 need to be placed in a random order. If we are organised, we can write down all the ways to do this in a list:

123, 132, 213, 231, 312, 321.

We want to select each of these with equal probability. We could for example use a dice. Most relevantly, even a computer as ancient as my laptop is very happy simulating a random choice from this set. (Now is not the time to talk about exactly how pseudo-random or otherwise this choice would be.)

Of course, when we play a sensible card game we have not three cards, but fifty-two. So the approach described above still works in theory, but no longer in practice, as the list of possible arrangements now has size 52!. Recall this is defined to be

52!=1\times 2 \times\ldots \times 52.

The reason we get this particular expression is that when we are choosing the first card, we have 52 possible choices. Then, regardless of what this first card actually is, there are precisely 51 cards left from which to choose the second card. So there are 52×51 ways to pick the first two cards in the arrangement, and so on, giving the answer. We can approximate how large 52! is by counting powers of ten rather crudely. It seems reasonable that it should be about 10^{65}. Note that the number of atoms in the universe is *only* about 10^{80}, so if we are going to write down this list, we better have very compact handwriting! But being serious, this number is way too large to realistically compute with, so we have to come up with some cleverer methods.

One way is to spread the cards out on a table then pick them up one at a time, ensuring at all times that the choice of card is uniform among those currently present, and not related to any of the past choices. This is relatively easy for a computer, but hard for a human, and certainly deeply tedious for anyone waiting to receive their hand!

So we seek a different approach, namely an algorithm for shuffling. Our aim is to introduce overall randomness by repeatedly applying some simple but random process. Note we have to be careful about our definition of ‘random’ here. The permutation 123456 is just as ‘random’ as the permutation 361524. That is, if they are fixed, then they are not random at all. Just because it is easier to decribe one of them verbally does not mean it is less random. For example, if I am trying to cheat at poker, then I might be able to if I knew the exact order of the cards in the pack before the dealer dealt. It wouldn’t matter what that order was. I would have to adjust my strategy based on the order, but it wouldn’t affect the fact that I had a massive advantage!

The shuffling algorithm to be discussed here is the top-to-random shuffle. Like all the best things in life, this does exactly what it says on the tin. At a given time, we remove the top card from the deck at present, and insert it at a randomly chosen point in the deck. This could be on the bottom, and it could also be back on the top. It feels like this possibility to remain constant can’t possibly help us, but later we will discuss why we need this.

In any case, it feels natural that if we keep applying this procedure, the arrangement of the deck should start to get more and more random, in the sense that knowing the original arrangement will tell us successively little about the current arrangement as time progresses. But we need to find a way to quantify this if we are to do any mathematics.

When we are talking about real numbers, it is fairly clear what it means if I say that the numbers 2, 1.1, 1.01, 1.001 and so on are getting closer and closer to 1. Indeed we can measure the distance along the number line between each term and 1, using the absolute difference. It is not so clear how to compute the distance between two probability distributions. Bearing in mind the fact that a distribution on the set of permutations of cards is defined to be a set of 52! probabilities that sum to 1, there will be a 52!-1 dimensional space (eg the plane is two-dimensional, the world is three-dimensional, *and so on* – whatever that means) where we have a nice distance formula already.

But this is not what we will choose to use. Rather we return to the cheating-at-poker analogy. Suppose I am playing some sort of game involving the pack of cards with my enemy. He or she thinks the deck is perfectly random, but I know the actual distribution. How big a profit can I make by exploiting this knowledge? This will be our measure of how far a distribution is from uniform. It turns out that this will coincide precisely with the formal definition of total variation distance, but that language belongs to a different level of rigour and is not relevant here.

What is relevant is an explanatory example. Suppose we start with the arrangement 12345678. We are now going to perform one iteration of the top-to-random shuffle. The outcome might, for example, be 23456178, if we insert the 1 between the 6 and the 7. Note there were 8 places for the card to go, so the probability of this particular outcome is 1/8. Now let’s see how I might use my knowledge of the distribution to my advantage. Suppose I suggest the bet that the bottom card is an 8. My enemy thinks the stack is uniformly randomly arranged, so the probability of this is 1/8. On the other hand, I know that the only way the 8 might disappear from the bottom is if I place the 1 under it, which happens with probability 1/8. So in fact, I know the probability of this event is 7/8, which gives me an advantage of 3/4. In fact, I could come up with bets that do even better than this, but they are less simple to describe verbally.

At what point do I lose this advantage? Well, we said that the probability that the 8 leaves the bottom of the stack is 1/8. And it will continue to be 1/8 on every turn where it is at the bottom. Recalling that the outcomes of successive shuffles are independent, note this is reminiscent of rolling a dice until a six comes up. The number of rolls required to get the six is an example of a geometric random variable. I don’t want to spoil S1 (or whichever module) by going into too much detail, but it turns out that if the probability of an event happening on a single go is p, then the average time we have to wait is 1/p. So 1/(1/8)=8 of course, and this is how long we typically have to wait before the bet I placed before becomes much less effective.

Now seems like a good time to stop talking about 8 cards and start talking about n cards. Obviously, in practice, we will want n to be 52. Anyway, by the same argument as before, it takes on average n iterations before the bottom card leaves the bottom. This is important, because after then, my bet that the bottom card is n is no longer so effective. However, I could equally place a bet that one of the bottom *two* cards is n.

So we consider how long it takes before n is no longer one of the bottom two cards. Well certainly we need to wait until it is no long *the* bottom card, which takes time n on average. Then, once it is second bottom, there is now a 2/n chance that we move the previously top card below it, so by the same argument as before, the time for this to happen is n/2 on average. If we want this effect to disappear, we have to wait until the original bottom card is in fact at the top of the pile for the first time, and by extending our previous argument, the average time for this is

n+\frac{n}{2}+\frac{n}{3}+\ldots+\frac{n}{n-1}.

Fortunately, we have tools for approximating this sort of sum, in particular integration, which is the practice of finding the area under certain curves. It turns out that the answer is roughly n log n. You can think of as log n a measure of the number of digits required to write out n. (This is not the exact definition but it will do for now. In any case, log n gets larger as n gets larger, but not very fast.) There’s a lot more about this in my previous post on the coupon collector problem, from a more technical point of view.

The next question will be to prove that it is actually quite well shuffled by this time, but that’s for another post. The other question to ask is whether this is satisfactory overall? For n=52, the number of operations we have to perform is about 230, which is fine for a computer, but deeply tedious for anyone sitting at a casino table waiting for the next hand. So next time we’ll talk about the riffle shuffle, which seems to introduce a lot of randomness in each go, but we’ll also see that we have to be careful, because the randomness may not be as great as our intuition suggests.

Related articles

The Coupon Collector Problem

Posted on by
Reply

This post talks about a problem which I’ve mentioned in passing a few times previously. A research problem I’ve been thinking about in the past week involves some careful calculations for a related setup, so I thought it would be worth writing some short notes about the classical problem. As an child who has tried to collect football cards or the sets of toys you find inside cereal packets will know, it always takes ages to pick up the last few cards, since you will mainly be picking up types that you already have.

The classical coupon collector problem is exactly a mathematical statement of this situation. We have a finite set, that we might as well take to be [n]. Then we have a collection of IID random variables which are uniformly distributed on [n]. Let’s call these X_1,X_2,\ldots. We are interested in how many samples we need before we have seen each element of [n]. We could define this finishing time as

T=\inf_{k\in\mathbb{N}}\{\{X_1,\ldots,X_k\}=[n]\}.

 We are interested in the distribution of this random variable T. The key initial observation is that we can specify the distribution of the number of samples needed between seeing new objects as geometric. For example, given that we have seen k possible values, the number of Xs until we see a new value is distributed as Geom(n-k/n). This of course has expectation n/(n-k), and so the expectation of T is

\mathbb{E}T=\frac{n}{n}+\frac{n}{n-1}+\ldots+\frac{n}{1}.

We can immediately do a crude approximation of this as

n\int_1^n \frac{dx}{x}=\log n.

We should ask just how approximate this is, and so shortly we will move to a discussion of the harmonic numbers

H_n:=1+\frac12+\ldots+\frac{1}{n}.

First though, we should settle the concentration question. Note that this is concretely different to the question of approximating the harmonic numbers. Concentration asks how tightly the distribution is packed around its mean. Later, we will ask exactly how close the mean is to n log n.

The most crude measure is variance. Note that in the above calculation we used linearity of expectation, which holds generally. We also have a result concerning linearity of variance, which Wikipedia asserts is commonly called Bienayme’s formula, provided the summand random variables are independent. And that is the case here. Recall that the variance of a Geom(p) random variable is \frac{q}{p^2}, where p+q=1. So we obtain

\mathrm{Var}(T)= \sum_{k=1}^n \frac{n(n-k)}{k^2}

\le \sum_{k=1}^n \frac{n^2}{k^2}\le n^2\frac{\pi^2}{6}.

So the standard deviation is an order of magnitude less than the expectation, hence by Chebyshev or similar, we can get some concentration bounds.

Harmonic Numbers

We return now to the question of how good an estimate log n is for

H_n:=1+\frac12+\ldots+\frac{1}{n}.

First, we note that log n must be an underestimate. In the following picture

DSC_1814 - Copy

log n is the area under the black curve between 1 and n, whereas H_n also includes the block between n and n+1, and the red area above the curve. So, in the limit, the estimate is off by at least the total red area (as the first error term decays to zero). The value of this area is defined to be the Euler-Mascheroni constant, a name attributed to the two men who first made a serious attempt to approximate it.

Why should this be finite? Well note that if we translated each red bit so they lay one on top of the other, they would not overlap (apart from boundaries) and be contained within the block with width [1,2] and height 1. So the constant is bounded above by 1. Since the function 1/x is strictly convex, approximating each red sector by a triangle with the same boundary points would be an underestimate, so the constant \gamma>1/2. The value turns out to be about 0.577.

So thus far we have the approximation

H_n\approx \log n + \gamma.

In fact a strong asymptotic statement is

H_n=\log n +\gamma+\frac{1}{2n}+o\left(\frac{1}{n}\right).

Rather than address that, I will show that, asymptotically:

\log n+\gamma< H_n<\log n +\gamma+\frac{1}{n}.

We treat the upper bound first. Note that \log(n+1)=\log n + \log (1+\frac{1}{n})=\log n +\frac{1}{n}+O(\frac{1}{n^2}). So, just by bounding with areas, noting that we need to take the upper limit of the integral to be (n+1), we get:

H_n < \int_1^{n+1} \frac{dx}{x} + \gamma=\log(n+1)+\gamma\leq \log n + \frac{1}{n}+\gamma.

For the lower bound, it is required to show that

\int_n^{n+1}\frac{dx}{x}

is strictly greater than the red area contributing to \gamma that lies to the right of x=n+1 But for this we can use the same argument as before, by translating all the red areas as far to the left as possible so they all lie on top of each other, and in particular within a box of dimensions 1 x 1/n+1, which is less than the integral.

We conclude this article by relating these harmonic numbers to the Stirling numbers of the first kind, which are a particular example of Bell polynomials. The easiest way to state the result is

H_n=\frac{1}{n!}\#\{\text{permutations of }[n+1]\text{ with 2 cycles}\}.

To see why this might be, consider the number of permutations of [n+1] with two cycles of length k and n+1-k, where for now we assume these are not equal. Then the number of such permutations is

\binom{n+1}{k}(k-1)! (n-k)!=\frac{(n+1)!}{k\cdot(n-k+1)}=n!\left[\frac{n+1}{k\cdot(n-k+1)}\right]= n!\left[\frac{1}{k}+\frac{1}{n-k+1}\right].

After checking the case k=n+1/2 if applicable, and making sure all the bounds match up correctly, the result follows.

Related articles

Responding to the Monty Hall Problem

Posted on by
5

It’s been a while since I got round to writing anything, so I’m easing myself back in by avoiding anything too mathematically strenuous. Shortly after my last post, there was a nice clip on Horizon featuring Marcus du Sautoy explaining the famous Monty Hall Problem to the ever-incredulous Alan Davies. The BBC article here shows the clip, and then an excellent account by probabilist John Moriarty explaining the supposed paradox.

Recall the situation. A game show is hosted by Monty Hall. At some point, to decide the grand prize, or whatever, a contestant plays the following lottery. There are three doors, and three prizes, one behind each. One of the prizes is worthwhile, normally a car, while the other two are rubbish. For reasons I don’t understand, it is normally said that behind the other two doors are goats. Anyway, MH knows where the car is from the beginning. Then the contestant picks a door, but doesn’t open it yet. MH then dramatically opens a different door, revealing a goat. The contestant then has the option to stick with his original choice, or switch to the third door. What should he or she do?

So the natural error to make is that since one possibility has been eliminated, since the remaining two were a priori equally likely, they should still be equally likely, and hence each occur with probability 1/2. So it makes no difference whether you stay or switch. But, of course, as well as removing information (ruling out one door), we have also gained information. The fact that MH chose to reveal that door, rather than the potential third door is important.

In my opinion, the best way to think about this is to imagine you are the host. The nice thing about being in this position is that then you can imagine you know everything about the configuration in advance, so are less likely to fall foul of counter-intuitive conditional probabilities. Anyway, there’s loads of symmetry, so from my point of view as the host, I only care whether the contestant initially selects the right door or the wrong door. If he selects the wrong door, then I need to open the ONLY OTHER wrong door. In other words, I have no choice in what I do next. I open the other wrong door. Therefore in this case the contestant should switch. If the contestant in fact selects the right door initially, I then have a choice about which door to open. However, it doesn’t really matter which I pick, by symmetry, and in either case the contestant would do better to stay with his original choice. Note that we have defined the outcome entirely by the initial choice. And as the host, I knew where the car was at the start, so from my point of view, unless he had somehow cheated, the contestant always had a 1/3 chance of guessing correctly. So it is better to switch 2/3 of the time.

The BBC’s weekly collation feature, The Loop, printed a couple of responses to the article. This is one:

“I believe that the problem is a game of two halves. In the first instance you have a 1:3 chance of getting the prize. Once you’ve made your initial choice, even though the ‘box’ contents remain the same, one ‘false’ choice is removed and you are then offered the chance to choose again. There is a play on words here, perhaps, with the option to ‘switch’. Effectively you are being asked to make a choice of 1:2. Now, if this is always going to happen then the odds of you winning the prize in the first instance is 1:2, as it doesn’t really matter whether or not you choose correctly the first time – one ‘false’ is going to be removed and you’ll be presented with just two options. It’s mind trickery. The first choice is totally irrelevant because you will always end up having to pick one option from two in order to attain the prize. In my humble opinion.”

I highlight this not to inspire contempt, since it is an easy mistake to make, even after reading as nice an explanation as Moriarty’s. Rather it is an excellent example of how using too many words can convince you an argument is complete before you’ve really made any concrete statement at all. Note that the sentences beginning ‘Effectively…’, ‘Now, if this…’ and ‘The first choice…’ all say exactly the same thing, which is, to use inappropriately formal terminology, exactly what the writer is trying to prove. The padding of ‘mind trickery’ and ‘humble opinions’ is as close as a maths argument can ever really come to an ad hominem.

So I was trying to come to a clean conclusion about exactly why the Monty Hall problem does such a good job at generating confusion. I think the following aspects of the set-up are important:

1) In general, people often get confused about the meaning of ‘random’. In some sense, the sequence 416235 is more random than the sequence 123456. However, typically, we use ‘random’ to mean ‘not determined in advance’. The key is that the bit where MH reveals a goat is not a random action. Whatever you do initially, you know that this will happen. So the fact that it does actually happen should not add extra information.

2) This fact is compounded by the time-steps at which decisions have to be made. It feels like the arc of the narrative is directing everything towards this final choice, which feels very separate from the initial choice. You can imagine the dramatic music that undoubtedly plays as MH opens the appropriate door, and the camera pans back to the contestant making their decision. All of which is an excellent way to suggest that something complicated has happened, when in fact in terms of what we are interested in overall, nothing has happened.

A point I hadn’t thought about before was that it is important that MH knows what happens in advance. Let’s develop this point and consider what happens if MH in fact has no knowledge of where the car is, and chooses a door at random. We will still assume that he must choose a different door to the one that the contestant originally chose.

Then, 1/3 of the time the contestant picks a goat, and MH picks the car, so he gets a goat whether he stays or switches. 1/3 of the time, the contestant picks a goat, and MH picks the other goat, so he should switch. And 1/3 of the time, the contestant originally picks the car, and then by default MH picks a goat, so he should stay. So, overall, it is equally preferable to stay as to switch.

I guess the morals of this story are: “be careful with conditional probabilities; sometimes apparently new information doesn’t change anything; and that it’s probably not a good idea to get into a discussion about the Monty Hall problem with a skeptic unless you have a piece of paper to hand on which to draw a probability table.”

Related articles

PS. A final moral is that, based on the links WordPress suggested, at least 10% of the internet is devoted to explanations of the Monty Hall problem…

The Secretary Problem

Posted on by
Reply

This post explores in a direction rather different to my research and recent problems I’ve been talking about. We discuss a problem of optimal stopping theory. In less politically correct times, there were various other names for this including (according to Wikipedia) the marriage problem, the Sultan’s dowry problem and the fussy suitor problem. A much more appealing situation might be someone running an auction or selling a house, who wants the maximum bid, but has to accept or reject bids immediately. In any case, we shall quickly reduce it to a context-free setting.

The problem is as follows. A manager needs to hire a secretary and has n applicants, who are interviewed one at a time. I can’t think of a good reason why this should be the case in this context, but the manager’s only concern is to maximise the probability of picking the best of the n candidates. This would be easy if the choice were made at the end, but in fact the decision whether to accept or reject an applicant has to be made straight after the interview. What is the optimal strategy?

It must be clarified what is going on. In a practical setting, there would be some underlying numerical attribute which could be compared, and so the problem would reduce to what I’ve previously called the Passport Photograph problem. In this setting, after the kth candidate, we only know the relative ranks of the first k candidates to be seen. This in turn massively reduces the possibilities for a strategy. The strategy must be of the form:

Accept the kth candidate if they are the best candidate yet seen with some probability that is a function of k and n. Reject otherwise. In particular, this probability cannot depend on the history of the process – that is of basically no relevance at all since we only know the relative ranks at any given time.

There are two ideas that I want to justify informally for now. It seems reasonable to assume that this probability will either be 0 or 1 for each k and n. For example, one could work backwards in k, from n back to 1 working out the probabilities, and proving at each stage inductively that there is no advantage to a mixed strategy.

Secondly, suppose we decide we will accept the kth candidate if they are the best so far. Well if they are not the best, we pass to the next candidate. I claim that if we are using an optimal strategy, we should also accept the (k+1)st candidate if they are the best so far. This property is called monotonicity for the algorithm. In many senses this is the meat of the problem. If we can prove this, then everything else will be relatively straightforward. For now though, we assume it is true. This means that our optimal strategy for n is precisely defined by a threshold value of k. So we rewrite the strategy as follows:

We observe the first (k-1) candidates. After that, as soon as a candidate is the best so far, we accept them.

Note that this allows for the possibility that we do not choose anyone, so if this feels like a problem, we could always demand we choose the last candidate, even if we know they are not optimal. The question that remains is: what value should k take? Let’s make life easy and take n to be large, so we can assume k=cn for some constant c.

The probability that the second best was in the first cn, but the first was not is c(1-c). On that event, this strategy will definitely give us the best candidate. The probability that the third best was in the first cn, but the first and second were not is c(1-c)^2. Our strategy gives us the best candidate overall if and only if the first candidate appears before the second candidate in what remains of the process. Since we are just sampling from a random permutation, these are equally likely, so the probability that we get the best candidate is

\frac{1}{2} c(1-c)^2.

Continuing, the probability that the best candidate in the first cn is the fourth best overall, and the strategy gives us the best candidate is:

\frac{1}{3}c(1-c)^3.

So to find the optimal c, we have to maximise the sum

\sum_{k\geq 1}\frac{1}{k}c(1-c)^k

=-c\log c.

To find the maximum, we set the derivative equal to zero. Solving, we get c=1/e.

It turns out the most natural way to generalise this problem is to consider n independent random variables taking the values 0 and 1. The aim is to find a strategy that maximises the probability of picking the final 1. An informal analysis proceeds much as in the secretary problem. In this setting, we will work with fixed finite n rather than in some limit. Let us call the {0,1}-valued RVs, I_1,I_2,\ldots,I_n, and set

p_j=\mathbb{P}(I_j=1),\quad q_j=1-p_j,\quad r_j=\frac{p_j}{q_j},

in keeping with Bruss’s original paper dealing with this so-called ‘Odds problem’. Then, as before, we want to find the threshold for k. We should remark that the argument given below is shorter than Bruss’s proof, which uses generating functions, but again we will not address the matter of proving that the optimal strategy should have this threshold form.

First we explain why the secretary problem is a special case of this. Observe that, given no information about the first (k-1) candidates apart from their relative ranks, the probability that the kth candidate is the best so far is independent of the history of the process, and is equal to 1/k. So taking p_k=1/k in the odds problem precisely replicates the secretary problem, since the best candidate is, by construction, the final candidate who had the property of being the best so far.

Now we work out how to calculate the threshold for k. We could define k by:

k=\arg\max_{1\leq j \leq n}\left\{\mathbb{P}(I_k+\ldots+I_n=1)\right\}.

By summing over the possibilities for which of the I_j’s is 1, we get

\mathbb{P}(I_k+\ldots+I_n)=q_kq_{k+1}\ldots q_n\big[\frac{p_k}{q_k}+\ldots+\frac{p_n}{q_n}\big]=q_k\ldots q_n[r_k+\ldots+r_n].

So we ask, for which values of k is:

q_k\ldots q_n[r_k+\ldots+r_n]\geq q_{k+1}\ldots q_n[r_{k+1}+\ldots+r_n] ?

This is equivalent to

q_k\ldots q_n \cdot r_k \geq (1-q_k)q_{k+1}\ldots d_n[r_{k+1}+\ldots+r_n],

and we recall that (1-q_k)=p_k=q_kr_k, hence

\iff q_{k+1}\ldots q_n\ge q_{k+1}\ldots q_n[r_{k+1}+\ldots+r_n],

\iff 1\geq r_{k+1}+\ldots+r_n.

So we conclude that the correct optimal value for the threshold k is the largest value of j such that

r_j+\ldots+r_n < 1.

We can check that this agrees with our answer for the secretary problem since

\frac{1}{cn}+\frac{1}{cn+1}+\ldots+\frac{1}{n}\approx \int_{cn}^n \frac{dx}{x}=-\log c

which is approximately 1 precisely when c is roughly 1/e.

Finally a word on generalisations. Dendievel’s paper which appeared on arXiv yesterday considers a problem of Richard Weber where the RVs instead take values {-1,0,1}, and you want a strategy that successfully selects either the last time the value -1 appears or the last time +1 appears. The case where the probabilities of the event 1 and the event -1 are 1) equal and 2) constant is relatively straightforward by backwards induction as here, and the author deals with the two natural generalisations by removing 1) and 2) in turn. It seems natural that such a method should adapt for RVs taking any finite number of values.

REFERENCES

Bruss – Sum the odds to one and stop (2000)

Bruss – A note on bounds for the odds theorem of optional stopping (2003)

Dendievel – Weber’s optimal stopping problem and generalisations (2013) http://arxiv.org/abs/1309.2860

Related articles

Random Mappings for Cycle Deletion

Posted on by
1

In previous posts here and here, I’ve talked about attempts to describe a cycle deleting process. We amend the dynamics of the standard random graph process by demanding that whenever a cycle is formed in the graph we delete all the edges that lie on the cycle. The aim of this is to prevent the system growing giant components, and perhaps give a system that displays the characteristics of self-organised criticality. In the posts linked to, we discuss the difficulties caused by the fact that the tree structure of components in such a process is not necessarily uniform.

Today we look in the opposite direction. It gives a perfectly reasonable model to take a multiplicative coalescent with quadratic fragmentation (this corresponds to cycle deletion, since there are O(n^2) edges which would give a cycle if added to a tree on n vertices) and a fragmentation kernel corresponding to adding an extra edge to a uniform spanning tree on n vertices then deleting the edges of the unique cycle. The focus of the rest of this post, we consider this fragmentation mechanism, in particular thinking about how we would sample from it most practically. Not least, without going through Prufer codes or some other clever machinery, it is not trivial to sample a uniform spanning tree.

First, we count the number of unicyclic graphs on n labelled vertices. If we know that the vertices on the cycle are v_1,\ldots,v_k, then the number of cycles with an identified edge is

u_1=1,\quad u_k=\frac{k!}{2},\, k\ge 2.

If we know that the tree coming off the cycle from vertex v_i has size m, say, then each of the possible rooted labelled trees with size m is equally likely. So taking w_j=j^{j-1}, the number of rooted trees on j labelled vertices, we get B_n(u_\bullet,w_\bullet) for the number of such unicyclic graphs on [n]. Recall B_n is the nth Bell polynomial, which gives the size of a compound combinatorial structure, where we have some structure on blocks and some other structure within blocks. Then the random partition of [n] given by the tree sizes has the distribution \text{Gibbs}_n(u_\bullet,w_\bullet).

Consider now a related object, the so-called random mapping digraph. What follows is taken from Chapter 9 of Combinatorial Stochastic Processes. We can view any mapping M_n:[n]\rightarrow[n] as a digraph where every vertex has out-degree 1. Each such digraph contains a collection of directed cycles, supported on those elements x for which M_n^k(x)=x for some k. Such an element x is called a cyclic point. Each cyclic point can be viewed as the root of a labelled tree.

In an identical manner to the unicyclic graph, the sizes of these directed trees in the digraph decomposition of a uniform random mapping is distributed as \text{Gibbs}_n(\bullet !,w_\bullet). So this is exactly the same as the cycle deletion kernel, apart from in the probability that the partition has precisely one block. In practice, for large n, the probability of this event is very small in both cases. And if we wanted to sample the cycle deletion kernel exactly, we could choose the trivial partition with some probability p, and otherwise sample from the random mapping kernel, where p is chosen such that

p+\frac{1-p}{B_n(\bullet !, w_\bullet)}=\frac{1}{B_n(u_\bullet,w_\bullet)}.

At least we know from the initial definition of a random mapping, that B_n(\bullet !,w_\bullet)=n^n. The number of unicyclic graphs with an identified edge is less clear. It turns out that the partition induced by the random mapping has a nice limit, after rescaling, as the lengths of excursions away from 0 in the standard Brownian bridge on [0,1].

The time for a fuller discussion of this sort of phenomenon is in the context of Poisson-Dirichlet distributions, as the above exchangeable partition turns out to be PD(1/2,1/2). However, for now we remark that the jumps of a subordinator give a partition after rescaling. The case of a stable subordinator is particularly convenient, as calculations are made easier by the Levy-Khintchine formula.

A notable example is the stable-1/2 subordinator, which can be realised as the inverse of the local time process at zero of a Brownian motion. The jumps of this process are then the excursion lengths of the original Brownian motion. A calculation involving the tail of the w_j’s indicates that 1/2 is the correct parameter for a subordinator to describe the random mappings. Note that the number of blocks in the partition corresponds to the local time at zero of the Brownian motion. (This is certainly not obvious, but it should at least be intuitively clear why a larger local time roughly indicates more excursions which indicates more blocks.)

So it turns out, after checking some of the technicalities, that it will suffice to show that the rescaled number of blocks in the random mapping partition \frac{|\Pi_n|}{\sqrt{n}} converges to the Raleigh density, which is a size-biased Normal random variable (hence effectively first conditioned to be positive), and which also is the distribution of the local time of the standard Brownian bridge.

After that very approximate description, we conclude by showing that the distribution of the number of blocks does indeed converge as we require. Recall Cayley’s formula kn^{n-k-1} for the number of labelled forests on [n] with a specified set of k roots. We also need to know how many labelled forests there are with any set of roots. Suppose we introduce an extra vertex, labelled 0, and connect it only to the roots of a rooted labelled forest on [n]. This gives a bijection between unlabelled trees on {0,1,…,n} and labelled forests with a specified set of roots on [n]. So we can use Cayley’s original formula to conclude there are (n+1)^{n-1} such forests. We can do a quick sanity check that these are the same, which is equivalent to showing

\sum_{k=1}^n k n^{-k-1}\binom{n}{k}=\frac{1}{n}(1+\frac{1}{n})^{n-1}.

This odd way of writing it is well-motivated. The form of the LHS is reminiscent of a generating function, and the additional k suggests taking a derivative. Indeed, the LHS is the derivative

\frac{d}{dx}(1+x)^n,

evaluated at \frac{1}{n}. This is clearly the same as the RHS.

That said, having established that the random mapping partition is essentially the same, it is computationally more convenient to consider that instead. By the digraph analogy, we again need to count forests with k roots on n vertices, and multiply by the number of permutations of the roots. This gives:

\mathbb{P}(|\Pi_n|=k)=\frac{kn^{n-k-1}\cdot k! \binom{n}{k}}{n^n}=\frac{k}{n}\prod_{i=1}^{k-1}\left(1-\frac{i}{n}\right).

Now we can consider the limit. Being a bit casual with notation, we get:

\lim \mathbb{P}(\frac{|\Pi_n|}{\sqrt{n}}\in dl)\approx \sqrt{n}dl \mathbb{P}(|\Pi_n|=l\sqrt{n}).

Since the Raleigh distribution has density l\exp(-\frac12 l^2)dl, it suffices for this informal verification to check that

\prod_{i=1}^{l\sqrt{n}}(1-\frac{i}{n})\approx \exp(-\frac12 l^2). (*)

We take logs, so the LHS becomes:

\log(1-\frac{1}{n})+\log(1-\frac{2}{n})+\ldots+\log(1-\frac{l\sqrt{n}}{n}).

If we view this as a function of l and differentiate, we get

d(LHS)=\sqrt{n}dl \log (1-\frac{l}{\sqrt{n}})\approx \sqrt{n}dl \left[-\frac{l}{\sqrt{n}}-\frac{l^2}{2n}\right]\approx -ldl.

When l is zero, the LHS should be zero, so we can obtain the desired result (*) by integrating then taking an exponential.

Related articles

The Configuration Model

Posted on by
2

In the past, I’ve talked about limitations of the Erdos-Renyi model of homogeneous random graphs for applications in real-world networks. In a previous post, I’ve discussed a dynamic model, the Preferential Attachment mechanism, that ‘grows’ a graph dynamically by adding edges from new vertices preferentially to existing vertices with high degree. The purpose of this adjustment is to ensure that the distribution of the degrees is not concentrated around some fixed value (which would be c in G(n,c/n) ) but rather exhibits a power-law tail such as observed in many genuine examples.

In this post, we introduce some aspects of the configuration model, which achieves this property more directly. This idea probably first arose in the guise of regular graphs. Recall a regular graph has all degrees equal. How would we construct a random d-regular graph on a large number of vertices?

What we probably want to do is to choose uniformly at random from the set of such graphs, but it is not clear even how large this set is, let alone how one would order its elements to make it possible to make this uniform choice. Instead, we try the following. Assign to each vertex d so-called stubs, which will end up being ‘half-edges’. We then choose two stubs uniformly at random, and glue them together. More formally, we construct an edge between the host vertices, and then delete the chosen stubs. We then continue.

The construction makes no reference to the distribution of stubs, so we are free to choose this as we please. We could for example specify some sequence of degrees which approximates a power-law, so we could sample a random sequence of degrees in some way. So long as we have a sequence of stub set sizes before we start building the edges of the graph we will be able to use the above algorithm.

So what might go wrong? There seem to me to be three potential problems that might arise with this construction.

Firstly, there might be a stub left over, if the sum of the stub set sizes is odd. Recall that in a graph the sum of the degrees is twice the sum of the number of edges, and so in particular the sum of the degrees should be even. But this is a small problem. When the degree sequence is deterministic we can demand that it have even sum, and if it is random, we will typically be working in a large N regime, and so deleting the solitary stub, if such a thing exists, will not affect the sort of properties of the graph we are likely to be interested in.

The second and third objections are perhaps more serious. If we glue together stubs naively, we might end up with loops, that is, edges that ‘begin’ and ‘end’ at the same vertex. These are not allowed in the standard definition of a graph. Alternatively, we might end up with more than one edge between the same pair of vertices.

Our overall aim is that this mechanism gives a convenient way of simulating the uniform distribution on simple graphs with a given degree sequence. At present we have the uniform distribution on potential multigraphs, with a weighting of 1/k! for every multi-edge with multiplicity k, and a weighting of 1/2 for every loop. The latter can be seen because there is an initial probability proportional to d(v_i)d(v_j) that vertices v_i and v_j will be joined, whereas a probability proportional (with the same constant) to d(v_i)^2 that v_i will receive a loop. The multi-edge weighting justification is similar.

However, conditional on getting a simple graph, the distribution is uniform on the set of simple graphs with that degree sequence. So it remains to investigate the probability that a graph generated in this way is simple. So long as this probability does not tend to 0 as n grows, we will probably be happy.

The strongest results on this topic are due to Janson. First observe that if the sum of the degrees grows faster than the number of vertices n, we fail to get a graph without loops with high probability. Heuristically, note that on the first pass, we are taking two picks from the set of vertices, biased by the number of stubs. By Cauchy-Schwarz, Rearrangement Inequality or just intuition, the probability of getting the same vertex is greater than if we picked uniformly from the set of vertices without biasing. So the probability of getting no loop on the first pass is \le (1-\frac{1}{n}). Take some function a(n) that grows faster than n, but slower than the sum of the degrees. Then after a(n) passes, the degree distribution is still roughly the same. In particular, the sum of the degrees is still an order of magnitude greater than n. So we obtain:

\mathbb{P}(\text{no loops})\leq (1-\frac{1}{n})^{a(n)}\approx e^{-\frac{a(n)}{n}}\rightarrow 0.

So, since isolated vertices have no effect on the simplicity or otherwise, we assume the sum of the degrees is \Theta(n). Then, Janson shows that the further condition

\sum_{i=1}^n d_i^2=O(n),

is essentially necessary and sufficient for simplicity. We can see why this might be true by looking at the probability that the first edge added is a loop, which is roughly

\frac{d_1^2+d_2^2+\ldots+d_n^2}{2(\sum d_i)^2}.

We have to consider O(\sum d_i) edges, so if the above expression is much larger than this, we can perform a similar exponential estimate to show that the probability there are no loops is o(1). The technical part is showing that this probability doesn’t change dramatically as the first few stubs disappear.

Note that in both cases, considering only loops is sufficient for simplicity. Although it looks like loop appearance is weaker than multiplicity of edges, in fact they have the same threshold. It should also be pointed out that, like the uniform random forests, an alternative approach is simply to count the number of simple graphs and multigraphs with a given degree sequence. Good asymptotics can then be found for the probability of simplicity.

In the case of G(n,c/n), we were particularly interested in the emergence of the giant component at time c=1. While first-moment methods can be very effective in demonstrating such results, a branching process local limit representation is probably easiest heuristic for this phase transition.

So long as the degree sequences converge in a natural way, we can apply a similar approach to this configuration model. Concretely, we assume that the proportion of vertices with degree i is \lambda_i in the limit. Although the algebra might push through, we should be aware that this means we are not explicitly specifying how many vertices have degree, eg \Theta(n^{1/2}). For now assume the \lambda_is sum to 1, so specify a probability distribution for degree induced by choosing a vertex uniformly at random.

So we start at a vertex, and look at its neighbours. The expected number of neighbours of this root vertex is \sum i\lambda i. Thereafter, when we consider a child vertex, based on how the stubs are paired up (and in particular the fact that the order of the operations does not matter – the choice of partner of a given stub is chosen uniformly at random), we are really choosing a stub uniformly at random. This corresponds to choosing a vertex at random, biased by the number of stubs available. The quantity of interest is how many additional stubs (other than the one that led to the vertex) are attached to this vertex. We assume we don’t need to worry too much about repeating vertices, in a similar way to G(n,c/n). So the expected number of additional stubs is

\frac{1}{\sum i\lambda_i}\sum i\lambda_i(i-1).

For an infinite component, we required the expectation to be > 1, which is equivalent to

\sum \lambda_i i(i-2)>0.

This was proven by Molloy and Reed (95), then with fewer conditions by Janson (07). The latter also shows how to use this construction to derive the giant component for G(n,c/n) result.

REFERENCES

Janson – A New Approach to the Giant Component Problem

Molloy, Reed – A Critical Point for Random Graphs with a Given Degree Sequence

Janson – The Probability that  Random Multigraph is Simple

Related articles

Characterisations of Geometric Random Graphs

Posted on by
Reply

Continuing the LMS-EPSRC summer school on Random Graphs, Geometry and Asymptotic Structure, we’ve now had three of the five lectures by Mathew Penrose on Geometric Random Graphs.

The basic idea is that instead of viewing a graph entirely abstractly, we now place the vertices in the plane, or some other real space. In many network situations, we would expect connectivity to depend somehow on distance. Agents or sites which are close together might be considered more likely to have the sort of relationship indicated by being connected with an edge. In the model discussed in this course, this dependence is deterministic. We have some parameter r, and once we have chosen the location of all the vertices, we connect a pair of vertices if the distance between them is less than r.

For the purposes of this, we work in a compact space [0,1]^d, and we are interested in the limit as the number of vertices n grows to infinity. To avoid the graph getting too connected, as in the standard random graph model, we take r to be a decreasing function of n. Anyway, we place the n points into the unit hypercube uniformly at random, and then the edges are specified by the adjacency rule above. In general, because r_n will be o(1), we won’t have to worry too much above boundary effects. The number of vertices within r_n of the boundary of the cube will be o(1). For some results, this is a genuine problem, when it may be easier to work on the torus.

In G(n,p), the order of np in the limit determines the qualitative structure of the graph. This is the expected degree of a given fixed vertex. In this geometric model, the relevant parameter is nr_n^d, where d is the dimension of the hypercube. If this parameter tends to 0, we say the graph is sparse, and dense if it tends to infinity. The intermediate case is called a thermodynamic limit. Note that the definition of sparse here is slightly different from G(n,p).

Much of the content of the first three lectures has been verifying that the distributions of various quantities in the graph, for example the total number of edges, are asymptotically Poisson. Although sometimes arguments are applicable over a broad spectrum, we also sometimes have to use different calculations for different scaling windows. For example, it is possible to show convergence to a Poisson distribution for the number of edges in the sparse case, from which we get an asymptotic normal approximation almost for free. In the denser regimes, the argument is somewhat more technical, with some substantial moment calculations.

A useful tool in these calculations are some bounds derived via Stein’s method for sums of ‘almost independent’ random variables. For example, the presence or non-presence of an edge between two pairs of vertices are independent in this setting if the pairs are disjoint, and the dependence is still only mild if they share a vertex. An effective description is via a so-called dependency graph, where we view the random variables as the vertices of a graph, with an edge between them if there is some dependence. This description doesn’t have any power in itself, but it does provide a concise notation for what would otherwise be very complicated, and we are able to show versions of (Binomials converge to Poisson) and CLT via these that are exactly as required for this purpose.

In particular, we are able to show that if E_n is the total number of edges, under a broad set of scaling regimes, if \lambda_n is the expected total number of edges, then d_{TV}(E_n,\mathrm{Po}(\lambda_n))\rightarrow 0, as n grows. This convergence in total variation distance is as strong a result as one could hope for, and when the sequence of \lambda_n is O(1), we can derive a normal approximation as well.

At this point it is worth discussing an alternative specification of the model. Recall that for a standard homogenous random graph, we have the choice of G(n,m) and G(n,p) as definitions. G(n,m) is the finer measure, and G(n,p) can be viewed as a weighted mix of G(n,m). We can’t replicate this directly in the geometric setting because the edges and non-edges are a deterministic function of the vertex locations. What we can randomise is the number of vertices. Since we are placing the vertices uniformly at random, it makes sense to consider as an alternative a Poisson Point Process with intensity n. The number of vertices we get overall will be distributed as Po(n), which is concentrated near n, in the same manner as G(n,c/n).

As in G(n,p), this is a less basic model because it is a mixture of the fixed-vertex models. Let’s see if how we would go about extending the total variation convergence result to this slightly different setting without requiring a more general version of the Poisson Approximation Lemma. To avoid having to define everything again, we add a ‘ to indicate that we are talking about the Poisson Point Process case. Writing d(.,.) for total variation distance, the result we have is:

\lim_{n\rightarrow\infty} d(E_n,\mathrm{Po}(\lambda_n))=0.

We want to show that

\lim_{n\rightarrow\infty}d(E_n',\mathrm{Po}(\lambda_n'))=0,

which we can decompose in terms of expectations in the original model by conditioning on N_n

\leq \lim_{n\rightarrow\infty}\mathbb{E}\Big[\mathbb{E}[d(E_{N_n},\mathrm{Po}(\lambda_n')) | N_n]\Big],

where the outer expectation is over N. The observation here, is that the number of points given by the Poisson process induces a measure on distributions, the overwhelming majority of which look quite like Poisson distributions with parameter n. The reason we have a less than sign is that we are applying the triangle inequality in the sum giving total variation distance:

d(X,Y)=\sum_{k\geq 0}|\mathbb{P}(X=k)-\mathbb{P}(Y=k)|.

From this, we use the triangle inequality again:

\lim_{n\rightarrow\infty} \mathbb{E}\Big[\mathbb{E}[d(E_{N_n},\mathrm{Po}(\lambda_{N_n})) | N_n]\Big]

+\lim_{n\rightarrow\infty}\mathbb{E}\Big[\mathbb{E}[d(\mathrm{Po}(\lambda_{N_n}),\mathrm{Po}(\lambda_n')) | N_n]\Big].

Then, by a large deviations argument, we have that for any \epsilon>0, \mathbb{P}(|N_n-n|\geq \epsilon n)\rightarrow 0 exponentially in n. Also, total variation distance is, by definition, bounded above by 1. In the first term, the inner conditioning on N_n is irrelevant, and we have that E_{N_n} converges to the Poisson distribution for any fixed N_n\in (n(1-\epsilon),n(1+\epsilon)). Furthermore, we showed in the proof of the non-PPP result that this convergence is uniform in this interval. (This is not surprising – the upper bound is some well-behaved polynomial in 1/n.) So with probability 1- e^{-\Theta(n)} N_n is in the region where this convergence happens, and elsewhere, the expected TV distance is bounded below 1, so the overall expectation tends to 0. With a similar LD argument, for the second term it suffices to prove that when \lambda\rightarrow\mu, we must have d(\mathrm{Po}(\lambda),\mathrm{Po}(\mu))\rightarrow 0. This is ‘obviously’ true. Formally, it is probably easiest to couple the distributions \mathrm{Bin}(n,\lambda/n),\mathrm{Bin}(n,\mu/n) in the obvious way, and carry the convergence of TV distance as the parameter varies through the convergence in n.

That all sounded a little bit painful, but is really just the obvious thing to do with each term – it’s only the language that’s long-winded!

Anyway, I’m looking forward to seeing how the course develops. In particular, when you split the space into small blocks, the connectivity properties resemble those of (site) percolation, so I wonder whether there will be concrete parallels. Also, after reading about some recent results concerning the metric structure of the critical components in the standard random graph process, it will be interesting to see how these compare to the limit of a random graph process which comes equipped with metric structure for free!