Isolated Vertices and the Second-Moment Method

It’s back-to-school day or week for much of the UK. I’m sure for many, this brings resolutions of better work habits, and while I could always use some of those too as I try to finish my thesis, I also want to start blogging again when possible.

Right now, I’m in Haifa for the Mostly Markov Mixing summer school and workshop hosted at the Technion. The talks have been interesting so far, and the environment stimulating both for the discussion of new problems, and getting work done on existing research.

20150904_104109I wrote much of this post a while ago, after some of the UK olympiad students asked me to tell them about second-moment methods and I politely declined. I was reminded of this by an interesting problem introduced by Elchanan Mossel in the first lecture of his series. I’ll start with this.

Suppose we consider a symmetric inhomogeneous random graph with two types. That is, we divide the vertices into two equally-sized classes, and we connect vertices in the same class with probability p, and vertices in different classes with probability q, all independently. The question is: if we can see the resulting graph structure, with what accuracy can we recover the class division? [Note: this setup with symmetry between the types can be called the block model.]

This is a hard problem, and got us all thinking a great deal. In the most relevant regime, even the most sophisticated techniques do not allow us to identify the partition perfectly with high probability as the number of vertices goes to infinity. For now, though, I only want to use the most uninteresting regime as a starting point for the rest of this post. EM asked: what happens if we take sparse scaling, that is p,q=\Theta(1/N), where N is the number of vertices? Do we have a chance to identify the classes correctly?

Well in this case the answer is easy, because it is ‘no’, and the reason is that in sparse random graphs, there is a positive proportion of isolated vertices. In particular, there is a positive proportion of isolated vertices of each type. And so, when we see an isolated vertex, for large N we can specify accurately the distribution of each type given this extra information, but in doing so, we admit that we certainly cannot partition the isolated vertices into their classes. So for the rest of the talk, EM focused on the regime where the random graph is connected with high probability.

There are two ideas with worthwhile and simple content here. Firstly, the fact that such a random graph has a positive proportion of isolated vertices. I am finishing off a result in a model where the base structure is the inhomogeneous random graph, and have just now proved a short lemma showing exactly this in a slightly more complicated context. It’s a good example of the second-moment method. Secondly, implicit in the final statement of the previous paragraph is that the absence of isolated vertices and connectivity are roughly equivalent in a random graph. This is something I’ve talked about briefly before, and in the interests of keeping the resolution and actually finishing this post, I won’t talk about it here.

Earlier this year, I gave a short talk by request to the UK olympiad students about first-moment methods. In particular, they were hoping that they might occasionally want to apply such approaches to the sort of combinatorics problems you encounter in olympiads. Typically, the best examples in this setting involve demonstrating the existence of a set without certain bad properties, by showing that the probability that all bad properties hold simultaneously is strictly less than one.

The students asked why I wasn’t talking about second-moment methods. Firstly, the models (such as this one) which are the best examples are less familiar to the students, and are really rooted in the randomness. They can’t easily be turned into a combinatorics problem. Secondly, looking for lower bounds in probability suggests we are aiming for convergence in probability, rather than convergence in expectation, and this is a distinction that is unlikely to be appreciated before some undergraduate probability courses have been taken.

Anyway, we want to show that the proportion of isolated vertices in G(n,c/n) is bounded below in probability as n\rightarrow\infty. All the content of the argument is seen in this classical Erdos-Renyi setting. Any inhomogeneous example will merely demand extra notation.

So, first we deal with the expected number of isolated vertices. The event that a given vertex v is isolated demands that the n-1 edges potentially incident to v do not appear. The probability of this is (1-\frac{c}{n})^{n-1} \rightarrow e^{-c}. Thus the expected number of such vertices is ne^{-c}. We now want to show that the fluctuations (standard deviation if you will) of this quantity are small relative to its mean. To bound the variance, we look at pairs of vertices, v and w. Note that the events {v isolated} and {w isolated} are not independent, since knowing that v is isolated tells us that the edge from v to w is not present, which slightly increases the chance of w being isolated. (For a very concrete example, think of the case n=2.)

However, in a large graph, the events are almost independent. That’s a statement which we feel is true, but we need to quantify, so we ask for the probability that {v and w isolated}. This happens precisely if none of the edges incident to either v or w are present. There are 2n-3 such edges, and so the probability of this is (1-\frac{c}{n})^{2n-3}\rightarrow e^{-2c}. So now we can write

\mathbb{E}\left[ \mathbf{1}(1\text{ isolated})+\ldots+\mathbf{1}(n\text{ isolated})\right]^2= \sum_{k=1}^n \mathbb{E}\mathbf{1}(k\text{ isolated})^2 + 2\sum_{j\ne k} \mathbb{E}\mathbf{1}(j,k\text{ isolated}).

An indicator function takes the values 0 and 1, and so in the first sum we can remove the square sign to leave us with the expected number of isolated vertices. Secondly, the vertices are exchangeable and so we can replace each summand with the value we have already established for v and w. We can now calculate the variance of the number of isolated vertices, and we see that it is o(n^2). With a little bit more care about the limits in n, we can check it is actually O(n). In particular, the variance of the proportion of isolated vertices is tends to zero.

For explicit lower bounds in probability on the proportion of isolated vertices we could appeal to Chebyshev’s inequality. However, since the variance vanishes, we have convergence in distribution to a constant, and thus convergence in probability.

Finally, a word on the end of EM’s talk. Having said that the sparse phase is not interesting because it is impossible, we might ask about the dense phase, where p and q are fixed. Just a for concrete example, suppose the probability of connection within the class is ½, and between classes is 1/3. Thus between any pair of vertices in the same class we expect to see roughly N/8+N/18= 13N/72 paths of length 2. The summands correspond to the middle vertex of the path in the same class, and in the opposite class respectively. However, between any pair of vertices in different classes, we expect to see roughly N/6 paths of length 2 for similar reasons. Both of these quantities will be highly concentrated on their means: consider the second-moment as the existence of each possible path is independent. Indeed, the chance that we see a proportion of paths closer to N/6 when it should be 13N/72 is a large deviations event, and so has exponential decay. As a result, the chance that we get the relative positions of any pair of vertices wrong with this method vanishes for large N.

In fact, the condition that (for p<q),

N \mathbb{P}(\text{Bin}(N-1,p)\ge \text{Bin}(N-1,q)) \rightarrow 0

should be enough to identify the partition with high probability, and indeed this is proved by several authors including EM. Note that the dense regime comfortable satisfies this condition, since it holds even without the factor of N. (The sparse regime, completely fails as the probability is roughly constant.) Even closer to the connectivity threshold remains interesting!


1 thought on “Isolated Vertices and the Second-Moment Method

  1. Pingback: Birthday Coincidences and Poisson Approximations | Eventually Almost Everywhere

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