# Lecture 9 – Inhomogeneous random graphs

I am aiming to write a short post about each lecture in my ongoing course on Random Graphs. Details and logistics for the course can be found here.

As we enter the final stages of the semester, I want to discuss some extensions to the standard Erdos-Renyi random graph which has been the focus of most of the course so far. In doing so, we can revisit material that we have already covered, and discover how easily one can extend this directly to more exotic settings.

The focus of this lecture was the model of inhomogeneous random graphs (IRGs) introduced by Soderberg [Sod02] and first studied rigorously by Bollobas, Janson and Riordan [BJR07]. Soderberg and this blog post address the case where vertices have a type drawn from a finite set. BJR address the setting with more general typespaces, in particular a continuum of types. This generalisation is essential if one wants to use IRGs to model effects more sophisticated than those of the classical Erdos-Renyi model G(n,c/n), but most of the methodology is present in the finite-type setting, and avoids the operator theory language which is perhaps intimidating for a first-time reader.

Inhomogeneous random graphs

Throughout, $k\ge 2$ is fixed. A graph with k types is a graph G=(V,E) together with a type function $V\to \{1,\ldots,k\}$. We will refer to a $k\times k$ symmetric matrix with non-negative entries as a kernel.

Given $n\in\mathbb{N}$ and a vector $p=(p_1,\ldots,p_k)\in\mathbb{N}_0^k$ satisfying $\sum p_i=n$, and $\kappa$ a kernel, we define the inhomogeneous random graph $G^n(p,\kappa)$ with k types as:

• the vertex set is [n],
• types are assigned uniformly at random to the vertices such that exactly $p_i$ vertices have type i.
• Conditional on these types, each edge $v\leftrightarrow w$ (for $v\ne w\in [n]$) is present, independently, with probability

$1 - \exp\left(-\frac{\kappa_{\mathrm{type}(v),\mathrm{type}(w)} }{n} \right).$

Notes on the definition:

• Alternatively, we could assign the types so that vertices $\{1,\ldots,p_1\}$ have type 1, $\{p_1+1,\ldots,p_1+p_2\}$ have type 2, etc etc. This makes no difference except in terms of the notation we have to use if we want to use exchangeability arguments later.
• An alternative model considers some distribution $\pi$ on [k], and assigns the types of the vertices of [n] in an IID fashion according to $\pi$. Essentially all the same results hold for these two models. (For example, this model with ‘random types’ can be studied by quenching the number of each type!) Often one works with whichever model seems easier for a given proof.
• Note that the edge probability given is $\approx \frac{\kappa_{\mathrm{type}(v),\mathrm{type}(w)}}{n}$. The exponential form has a more natural interpretation if we ever need to turn the IRGs into a process. Additionally, it avoids the requirement to treat small values of n (for which, a priori, $k/n$ might be greater than 1) separately.

In the above example, one can see that, roughly speaking, red vertices are more likely to be connected to each other than blue vertices. However, for both colours, they are more likely to be connected to a given vertex of the same colour than a vertex of the opposite colour. This might, for example, correspond to the kernel $\begin{pmatrix}3&1\\1&2\end{pmatrix}$.

The definition given above corresponds to a sparse setting, where the typical vertex degrees are $\Theta(1)$. Obviously, one can set up an inhomogeneous random graph in a dense regime by an identical argument.

From an applications point of view, it’s not hard to imagine that an IRG of some flavour might be a good model for many phenomena observed in reality, especially when a mean-field assumption is somewhat appropriate. The friendships of boys and girls in primary school seems a particularly resonant example, though doubtless there are many others.

One particular application is to recover the types of the vertices from the topology of the graph. That is, if you see the above picture without the colours, can you work out which vertices are red, and which are blue? (Assuming you know the kernel.) This is clearly impossible to do with anything like certainty in the sparse setting – how does one decide about isolated vertices, for example? The probabilities that a red vertex is isolated and that a blue vertex is isolated differ by a constant factor in the $n\rightarrow\infty$ limit. But in the dense setting, one can achieve this with high confidence. When studying such statistical questions, these IRGs are often referred to as stochastic block models, and the recent survey of Abbe [Abbe] gives a very rich history of this type of problem in this setting.

Poisson multitype branching processes

As in the case of the classical random graph G(n,c/n), we learn a lot about the IRG by studying its local structure. Let’s assume from now on that we are given a sequence of IRGs $G^n(p^n,\kappa)$ for which $\frac{p^n}{n}\rightarrow \pi$, where $\pi=(\pi_1,\ldots,\pi_k)\in[0,1]^k$ satisfies $||\pi||_1=1$.

Now, let $v^n$ be a uniformly-chosen vertex in [n]. Clearly $\mathrm{type}(v^n)\stackrel{d}\rightarrow \pi$, with the immediate mild notation abuse of viewing $\pi$ as a probability distribution on [k].

Then, conditional on $\mathrm{type}(v^n)=i$:

• when $j\ne i$, the number of type j neighbours of $v^n$ is distributed as $\mathrm{Bin}\left(p_j,1-\exp\left(-\frac{\kappa_{i,j}}{n}\right)\right)$.
• the number of type i neighbours of $v^n$ is distributed as $\mathrm{Bin}\left( p_i-1,1-\exp\left(-\frac{\kappa_{i,i}}{n}\right)\right)$.

Note that $p_j\left[1-\exp\left(-\frac{\kappa_{i,j}}{n}\right)\right]\approx \frac{p_j\cdot \kappa_{i,j}}{n} \approx \kappa_{i,j}\pi_j$, and similarly in the case j=i, so in both cases, the number of neighbours of type j is distributed approximately as $\mathrm{Poisson}(\kappa_{i,j}\pi_j)$.

This motivates the following definition of a branching process tree, whose vertices have k types. Continue reading

# Perturbation of Eigenvectors

In the previous post, I talked about eigenvalues, and some alternative characterisations which could be useful in some circumstances. Recently, I’ve been interested in controlling how eigenvalues and eigenvectors change as the matrix is varied. My particular example concerns positive matrices, which have a well-defined largest eigenvalue (or Perron root), and a unique (up to normalising in some way) principal eigenvector.

We might expect that perturbing a matrix slightly does not change the eigenvectors very much, since any original eigenvector is still almost an eigenvector, in the sense that its image under the action of the perturbed matrix is almost equal to a multiple of itself. But how to make this precise? And when does it go wrong?

Eigenvalues – The non-multiple case

Throughout, we assume we have a k x k matrix. We might want to allow the entries to be complex, but for now, real entries are perfectly interesting enough.

It makes sense to start with eigenvalues, since it’s easy to define these through the characteristic equation of the matrix. The coefficients of this polynomial are well-behaved (indeed polynomial) functions of the entries of the matrix. So we are really asking how the set of roots of a finite polynomial evolves as the (k+1) coefficients of the polynomial evolve. It is fairly clear that, under any sensible choice of topology on the space of k-(multi)-subsets of $\mathbb{C}$, the multiset of roots is continuous in the coefficients of the polynomial.

To say anything more precise, we have to introduce some notation.

Let $\chi_{A}(z)=z^k+\gamma_{k-1}(A)z^{k-1}+\ldots+\gamma_1(A)z+\gamma_0(A)$ be the characteristic polynomial of A. Each $\gamma_i$ is a polynomial of degree $k-i$ in the entries of A. Let’s consider now a matrix-valued function A(t), and we assume that the entries of A(t) are all differentiable with respect to t. So each $\gamma_i(A(t))$ is also differentiable with respect to t.

At this point, let’s make the assumption that t lies in some interval [r,s] for which the eigenvalues of A(t) are distinct. Let $\lambda(t)$ be some eigenvalue of A(t), chosen such that $\lambda$ is a continuous function of t. For example, we might take $\lambda(t)=\Lambda_1(t)$, the eigenvalue with largest absolute value (with some canonical tie-breaking mechanism). Then $\chi_{A(t)}(\lambda(t))=0$, and so differentiating with respect to $\gamma_i$:

$0=\chi'_{A(t)}(\lambda(t)) \frac{\partial \lambda}{\partial \gamma_i} \Big|_{A(t)} + \lambda(t)^i.$

Because we deliberately demanded that the eigenvalues were disjoint, we have $\chi'_{A(t)}(\lambda(t))\ne 0$, and so $\frac{\partial \lambda(t)}{\partial \gamma_i}=-\lambda(t)^i / \chi'_{A(t)}(\lambda(t))$. In particular, $\lambda(t)$ is differentiable with respect to the coefficients of the characteristic polynomial, and thus with respect to t also.

Multiple Eigenvalues

It gets more complicated when the characteristic equation has multiple roots. Typically we will be interested in the evolution of the eigenvalue with some extremal property, probably the largest one. Let’s restrict to the real, symmetric case, where the set of eigenvalues is complete and real. Suppose we have $t_0$ such that $A(t_0)$ has a repeated eigenvalue. Then, in a small enough region of $t_0$, we can define eigenvalues $\lambda(t),\mu(t)$ continuously such that $\lambda(t_0)=\mu(t_0)$ while $\lambda(t)\ne \mu(t)$ for $t=\ne t_0$. Then, if the entries of A(t) are analytic functions of t, then so are $\lambda(t),\mu(t)$.

But then $\max(\lambda(t),\mu(t))$ will in general not be analytic, as the maximum of two smooth functions is in general Lipschitz.

This effect is most obvious in the case of a diagonal matrix $A(t)=\begin{pmatrix}t&0\\0&-t\end{pmatrix}$, for which the largest eigenvalue is $|t|$.

Eigenvectors

When the matrix A is real and symmetric, we know it has real eigenvalues, and an orthogonal basis of eigenvectors. Then the Rayleigh quotient characterises the eigenvector as well as the eigenvalue. Recall that for any $x\in\mathbb{R}^k$ with $||x||_2=1$, we have

$\lambda_1\ge x^T A x \ge \lambda_k,$

with equality precisely at the respective eigenvectors. So if we perturb A slightly, keeping it real and symmetric, we can control the principal eigenvector quite well by this method.

If A is not diagonalisable, we can still say something about this principal eigenvector, via large powers of A, sometimes called the Van Mises iteration. This says that for large N, $A^N v$ should have direction close to that of the eigenvector, for any test vector v. The rate of convergence depends on the ratio of the largest eigenvalue to the second largest eigenvalue, though if the matrix is not diagonalisable, it is not completely trivial to quantify this convergence. We have to be careful though, since A maps the subspace orthogonal to the eigenvector to itself, so the magnitude of the projection of v onto the eigenvector determines the speed of convergence. Indeed, if v is orthogonal to the eigenvector, it won’t converge towards the principal eigenvector at all. (But if there is a well-defined ‘second eigenvector’ then it will converge towards that.)

Continuity of Eigenvectors

The reason why I ended up reading about some of these topics was that I wanted to show that the Perron eigenvector of a positive matrix (that is, the unique eigenvector corresponding to the Perron root) was Lipschitz continuous as a function of the entries of the positive matrix. Since for such a matrix, the largest eigenvalue is simple, we are able to make some progress.

In general, the condition that $v$ is an eigenvector of matrix A with eigenvalue $\lambda$ is described by the relation:

$(A-\lambda I)v=0,\quad ||u||_1=1,$ (*)

or whatever the most appropriate normalising condition appears. This describes an implicit relation between A and the eigenvalue-eigenvector pair $(\lambda,v)$. So given a matrix $A_0$ with eigenvalue $\lambda_0$ corresponding to eigenvector $v_0$, in a neighbourhood of $A_0$ in $\mathbb{R}^{k\times k}$ we can use the implicit function theorem to comment on the differentiability of $(\lambda,v)$ with respect to A in this neighbourhood.

Precisely, we require the matrix of partial derivatives from (*)

$\begin{pmatrix} A_0-\lambda_0 I&v_0 \\ \mathbf{1}&0\end{pmatrix}$

to have non-zero determinant. But if $\lambda_0$ is not simple, then if we apply this matrix from the left to one of the other eigenvectors (with a zero appended) we can see that it has non-trivial kernel. With a bit more work, we can show the converse too, and conclude that $(\lambda,v)$ are smooth with respect to A in some neighbourhood of $A_0$.

Finally, we observe that when the eigenvalues are not simple, we can’t even guarantee continuity of the eigenvectors. This is unsurprising really, since for a multiple eigenvalue, a) we might not know how many LI eigenvectors exists; and b) we might have complete freedom over the choice of eigenvectors. Think about the identity matrix! Indeed the eigenvectors of $\begin{pmatrix}1+\epsilon &0\\ 0&1+\epsilon\end{pmatrix}$ are (1,0) and (0,1), while the eigenvectors of $\begin{pmatrix}1&\epsilon\\ \epsilon&1\end{pmatrix}$ are (1,1), (1,-1). So no continuous choice of eigenvectors is possible here.