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Phase transitions in non-linear urns with interacting types

Marcelo Costa
Universidad de Buenos Aires
[email protected]
Jonathan Jordan
University of Sheffield
[email protected]
Abstract

We investigate reinforced non-linear urns with interacting types, and show that where there are three interacting types there are phenomena which do not occur with two types. In a model with three types where the interactions between the types are symmetric, we show the existence of a double phase transition with three phases: as well as a phase with an almost sure limit where each of the three colours is equally represented and a phase with almost sure convergence to an asymmetric limit, which both occur with two types, there is also an intermediate phase where both symmetric and asymmetric limits are possible. In a model with anti-symmetric interactions between the types, we show the existence of a phase where the proportions of the three colours cycle and do not converge to a limit, alongside a phase where the proportions of the three colours can converge to a limit where each of the three is equally represented.

1 Introduction and definitions

In general terms, an urn model is a system containing a number of particles of different types (often regarded as balls of different colours, for ease of visualisation). At each time step, a set of particles is sampled from the system, whose contents are then altered depending on the sample which was drawn. Pemantle [10] surveys several ways to approach this model framework.

This paper is limited to models with a single urn from which a single ball is drawn, its colour is noted and it is then returned to the urn along with one new ball of that same colour. In addition, we introduce a graph-based interaction according to which the probability of choosing a ball of a given colour is reinforced not only by its own proportion in the urn, but also by the proportions of balls of other colours. Therefore, the interaction arises among balls of different colours, as opposed to the so-called interacting urn models consisting of systems of multiple urns (e.g. Benaïm et al [2], Launay and Limic [7]) in which different urns (each containing balls of different colours) interact with one another. Our model is also different from the graph-based competition described in van der Hofstad et al [11], where the colours correspond to edges of the graph, which compete with, as opposed to being reinforced by, other edges incident on the same vertices.

We now formally define our model. Consider an urn containing balls of dd colours. The vector x(n)=(x1(n),,xd(n))dx(n)=(x_{1}(n),\ldots,x_{d}(n))\in\mathbb{N}^{d} denotes the number of balls of each colour at time n=0,1,2,n=0,1,2,.... The strength of the reinforcement is given by a positive real number β>0\beta>0 and we denote by xβ(n)x^{\beta}(n) the coordinate-wise β\beta power of the column vector x(n)x(n). The interaction is defined as follows. Given a non-negative matrix A=(aij)i,j=1dA=(a_{ij})_{i,j=1}^{d}, define the column vector

u(n):=Axβ(n)u(n):=Ax^{\beta}(n) (1)

Let n\mathcal{F}_{n} be the σ\sigma-algebra generated by the x(m)x(m) for 0mn0\leq m\leq n and write ui(n)u_{i}(n) for the ii-th component of u(n)u(n). The transition probabilities are then

(x(n+1)x(n)=𝐞i|n)=ui(n)j=1duj(n),i=1,,d,\mathbb{P}(x(n+1)-x(n)=\mathbf{e}_{i}\>|\>\mathcal{F}_{n})=\frac{u_{i}(n)}{\sum_{j=1}^{d}u_{j}(n)},\quad i=1,\ldots,d, (2)

where 𝐞i\mathbf{e}_{i} is the unit vector in direction ii. That is, one ball is added to the urn at each time step, and the right hand side of (2) gives the probability that it is of colour ii.

Now, let n0>0n_{0}>0 be the initial number of balls so that at time nn the urn contains n+n0n+n_{0} balls. Then, the proportion of balls of each colour is a process in the (d1)(d-1)-dimensional simplex Δd1\Delta^{d-1} given by the vector

x¯(n)=x(n)/(n+n0).\bar{x}(n)=x(n)/(n+n_{0}). (3)

When AA is a multiple of the identity matrix (therefore, x¯(n)\bar{x}(n) having no interaction) it is well-known (see Oliveira [8]) that the process x¯\bar{x} undergoes a phase transition as follows. For β<1\beta<1, the process converges almost surely to the ‘centre’ of the simplex, that is, the asymptotic proportion of balls of each colour are all the same. For β=1\beta=1, commonly referred to as the Pólya urn model, the process converges almost surely to a non-trivial random variable supported in the interior of the simplex. For β>1\beta>1, the process converges almost surely to one of the corners of the simplex. In this case that a single type dominates was proved by Khanin and Khanin [5] following on from the two-type case which can be covered using Rubin’s Theorem in Davis [4].

For a two-colour urn model d=2d=2 and symmetric interaction,

A=(1aa1),a>0,A=\left(\begin{matrix}1&a\\ a&1\end{matrix}\right),\quad a>0,

it was proved by the first author in Theorem 2.2.1, [3], that there was a phase transition as follows.

(i) if (1a1+a)β1,thenx¯(n)(12,12)a.s.\displaystyle(i)\quad\text{ if }\quad\left(\tfrac{1-a}{1+a}\right)\beta\leq 1,\quad\text{then}\quad\bar{x}(n)\rightarrow(\tfrac{1}{2},\tfrac{1}{2})\quad a.s.
(ii)if (1a1+a)β>1,thenx¯(n)Ψa.s.,\displaystyle(ii)\quad\text{if }\quad\left(\tfrac{1-a}{1+a}\right)\beta>1,\quad\text{then}\quad\bar{x}(n)\rightarrow\Psi\quad a.s.,

where Ψ\Psi is a random vector supported on {(11+r,r1+r),(r1+r,11+r)}\left\{\left(\frac{1}{1+r},\frac{r}{1+r}\right),\left(\frac{r}{1+r},\frac{1}{1+r}\right)\right\} and r:=r(a,β)r:=r(a,\beta) is the unique root in (0,1)(0,1) of 𝒫a,β(z)=azβ+1zβ+za=0.\mathcal{P}_{a,\beta}(z)=az^{\beta+1}-z^{\beta}+z-a=0. In case (ii), [x¯(n)(12,12)]=0.\mathbb{P}[\bar{x}(n)\rightarrow(\frac{1}{2},\frac{1}{2})]=0. Note that for β=1\beta=1, the process (u(n))n0(u(n))_{n\geq 0} is a Friedman’s urn model and statement (i) yields u(n)/(u1(n)+u2(n))(12,12)u(n)/(u_{1}(n)+u_{2}(n))\rightarrow(\frac{1}{2},\frac{1}{2}) a.s.a.s. as expected.

Some similar results appear in Laruelle and Pagès [6]. The definition of the model there allows for a more general skewing function than xβx^{\beta} as above, but in the d=2d=2 case gives phase transitions similar to those in [3], and relates them to the eigenvalues of the generating matrix HH, which is related to our matrix AA. Furthermore they show that, for any dd, for a concave skewing function (corresponding to β<1\beta<1 in our model) and a bi-stochastic generating matrix there is almost sure convergence to the centre of the simplex, and they give conditions, related to the eigenvalues of the generating matrix, under which for a convex skewing function (our β>1\beta>1) there is probability zero of converging to the centre of the simplex.

In this paper, we follow up on the results for d=2d=2 in [3] and those in [6], with the aim to generalise from d=2d=2 to larger values of dd and to see whether more types of behaviour emerge when this is done. We show that this is indeed the case when d=3d=3, where we consider two particular choices of AA. Our results can be seen as extensions of those of [6] in certain specific cases.

First of all, we consider a choice of AA with a symmetric interaction of the same strength aa for each pair of colours. The following theorem shows that in this system there are three phases, as opposed to two when d=2d=2; there are phases where there is almost sure convergence to a symmetric limit and where there is almost sure convergence to one of a number of asymmetric limits, which are analogues of the phases when d=2d=2, but there is also an intermediate phase where both symmetric and asymmetric limits are possible.

Theorem 1.1.

Let AA be the matrix

(1aaa1aaa1),a>0.\begin{pmatrix}1&a&a\\ a&1&a\\ a&a&1\end{pmatrix},\quad a>0. (4)
  1. (i)

    Fix a<1a<1. Then there exists β1(a)\beta_{1}(a) satisfying 1<β1(a)<1+2a1a1<\beta_{1}(a)<\frac{1+2a}{1-a}, with β1(a)\beta_{1}(a) an increasing function of aa satisfying β1(a)\beta_{1}(a)\to\infty as a1a\to 1, and we have the following three phases.

    1. (a)

      Symmetric limit almost surely. If β<β1(a)\beta<\beta_{1}(a), then almost surely x¯(n)(13,13,13)\bar{x}(n)\to\left(\frac{1}{3},\frac{1}{3},\frac{1}{3}\right).

    2. (b)

      Symmetric or asymmetric limit. If β1(a)<β<1+2a1a\beta_{1}(a)<\beta<\frac{1+2a}{1-a} then there exists r2>1r_{2}>1 such that almost surely x¯(n)\bar{x}(n) converges to one of the four points in Δ2\Delta^{2} given by (13,13,13)\left(\frac{1}{3},\frac{1}{3},\frac{1}{3}\right), (r22+r2,12+r2,12+r2)\left(\frac{r_{2}}{2+r_{2}},\frac{1}{2+r_{2}},\frac{1}{2+r_{2}}\right), (12+r2,r22+r2,12+r2)\left(\frac{1}{2+r_{2}},\frac{r_{2}}{2+r_{2}},\frac{1}{2+r_{2}}\right) and (12+r2,12+r2,r22+r2)\left(\frac{1}{2+r_{2}},\frac{1}{2+r_{2}},\frac{r_{2}}{2+r_{2}}\right). All of these points have positive probability of being limits.

    3. (c)

      Asymmetric limit almost surely. If β>1+2a1a\beta>\frac{1+2a}{1-a} then there exists r+>1r_{+}>1 such that almost surely x¯(n)\bar{x}(n) converges to one of the three points in Δ2\Delta^{2} given by (r+2+r+,12+r+,12+r+)\left(\frac{r_{+}}{2+r_{+}},\frac{1}{2+r_{+}},\frac{1}{2+r_{+}}\right), (12+r+,r+2+r+,12+r+)\left(\frac{1}{2+r_{+}},\frac{r_{+}}{2+r_{+}},\frac{1}{2+r_{+}}\right) and (12+r+,12+r+,r+2+r+)\left(\frac{1}{2+r_{+}},\frac{1}{2+r_{+}},\frac{r_{+}}{2+r_{+}}\right).

  2. (ii)

    Fix a1a\geq 1. Then almost surely x¯(n)(13,13,13)\bar{x}(n)\to\left(\frac{1}{3},\frac{1}{3},\frac{1}{3}\right).

Theorem 1.1 presents the results in terms of phase transitions in β\beta with aa fixed. However, because both β1(a)\beta_{1}(a) and 1+2a1a\frac{1+2a}{1-a} are increasing functions of aa which converge to 11 as a0a\to 0 and to \infty as a1a\to 1, it is also possible to see them as phase transitions in aa with β>1\beta>1 fixed: if a<β12+βa<\frac{\beta-1}{2+\beta} then we will be in case (c), if β12+β<a<β11(β)\frac{\beta-1}{2+\beta}<a<\beta_{1}^{-1}(\beta) then we will be in case (b), and if a>β11(β)a>\beta_{1}^{-1}(\beta) we will be in case (a).

Theorem 1.1 can be seen as an extension of the results of Proposition 2.15 of Laruelle and Pagès [6] in the case where the matrix H=11+2a(1aaa1aaa1)H=\frac{1}{1+2a}\begin{pmatrix}1&a&a\\ a&1&a\\ a&a&1\end{pmatrix} in that their result shows the non-convergence to (13,13,13)(\frac{1}{3},\frac{1}{3},\frac{1}{3}) in the case (i)(c), as the second largest eigenvalue of HH is 1a1+2a\frac{1-a}{1+2a}. We can also state the condition β>1+2a1a\beta>\frac{1+2a}{1-a} in terms of the eigenvalues of AA: the right hand side can be seen as the ratio of the two largest eigenvalues. It might be reasonable to conjecture that an extension to d>3d>3 of Theorem 1.1 might involve a similar condition on the eigenvalues; however we note that the other phase transition does not appear to be related to the eigenvalues of AA. We discuss the question of what happens with d>3d>3 further in Section 5.4.

We also consider a system where each colour is reinforced by itself and by one other, in a cyclic way. For this system, the following theorem shows the existence of a phase transition between a phase with convergence with positive probability to a symmetric limit and a phase where there is no convergence to a limit and there is cycling behaviour.

Theorem 1.2.

Let AA be the matrix

(110011101).\begin{pmatrix}1&1&0\\ 0&1&1\\ 1&0&1\end{pmatrix}.
  • When β<4\beta<4 there is positive probability that X¯(n)(1/3,1/3,1/3)\bar{X}(n)\to(1/3,1/3,1/3).

  • When β>4\beta>4, almost surely X¯(n)\bar{X}(n) fails to converge, and the limit set is either a periodic orbit or a connected union of periodic orbits.

In Section 2 we discuss the stochastic approximation methods we use in the proofs, while the proofs themselves are in Section 3 for Theorem 1.1 and Section 4 for Theorem 1.2. In the final Section 5, we illustrate the results with some examples and simulations, including some examples beyond those covered by Theorems 1.1 and 1.2.

2 Stochastic approximation approach

In this section we introduce some of the stochastic approximation ideas which appear in our proofs.

The type of stochastic approximation process we will be interested in is a Robbins-Monro algorithm in d\mathbb{R}^{d}, following section 4.2 of Benaïm [1]. This is a stochastic process (y(n))n(y(n))_{n\in\mathbb{N}}, with natural filtration (n)(\mathcal{F}_{n}), taking values in d\mathbb{R}^{d} which satisfies

y(n+1)y(n)=γn(F(y(n))+ξn+1),y(n+1)-y(n)=\gamma_{n}(F(y(n))+\xi_{n+1}), (5)

where F:ddF:\mathbb{R}^{d}\to\mathbb{R}^{d} is a deterministic vector field, γn\gamma_{n} is a step size satisfying certain conditions, and 𝔼(ξn+1|n)=0\mathbb{E}(\xi_{n+1}|\mathcal{F}_{n})=0 with ξn\xi_{n} being n\mathcal{F}_{n}-measurable. For most results it is required that γn0\gamma_{n}\to 0 as nn\to\infty but that n=1γn=\sum_{n=1}^{\infty}\gamma_{n}=\infty, and some conditions are also needed on the noise term ξn+1\xi_{n+1}, typically that it is bounded in LqL^{q} for some qq which depends on the γn\gamma_{n}: see for example Proposition 4.2 of Benaïm [1].

The above sequence can be thought as a numerical approximating method with varying step size γn\gamma_{n} for solving the ODE dx/dt=F(x)dx/dt=F(x). Under the conditions discussed above, the asymptotic behavior of (x¯n)n(\bar{x}_{n})_{n\in\mathbb{N}} and the underlying ODE are closely connected: as described in [1], define an interpolated version (X(t))t0(X(t))_{t\geq 0} by defining τ0=0\tau_{0}=0 and for nn\in\mathbb{N} τn=i=1nγi\tau_{n}=\sum_{i=1}^{n}\gamma_{i}, then defining X(τn+s)=x¯n+x¯n+1xn¯γn+1X(\tau_{n}+s)=\bar{x}_{n}+\frac{\bar{x}_{n+1}-\bar{x_{n}}}{\gamma_{n+1}} for 0s<γn+10\leq s<\gamma_{n+1}. Then the interpolated process (X(t))t0(X(t))_{t\geq 0} is an asymptotic pseudotrajectory for the ODE. This is called the ODE method or the dynamical system approach, which alongside some probabilistic techniques, is applied to examine almost sure dynamics of stochastic approximation processes.

Let Φ:+×dd\Phi:\mathbb{R}_{+}\times\mathbb{R}^{d}\to\mathbb{R}^{d} be the semiflow induced by the vector field FF, so that (Φ(t,x))t0)(\Phi(t,x))_{t\geq 0}) is the trajectory of FF started at xx. A useful situation for analysis of stochastic approximation processes is where there is a Lyapunov function for the vector field FF: a function V:ddV:\mathbb{R}^{d}\to\mathbb{R}^{d} such that on a trajectory (Φ(t,x))t0(\Phi(t,x))_{t\geq 0} of the vector field, V(Φ(t,x))V(\Phi(t,x)) is strictly decreasing in tt except where Φ(t,x)\Phi(t,x) is a stationary point of FF. If this holds, then under mild conditions then the Robbins-Monro algorithm will converge almost surely to a (possibly random) limit, which will be a stationary point of FF. See section 6.2 of Benaïm [1], and in particular Corollary 6.6 therein.

We now show that our process (x¯n)(\bar{x}_{n}) can be put into Robbins-Monro form, and that the assumptions of Proposition 4.2 of Benaïm [1] are satisfied, allowing the theory of asymptotic pseudotrajectories to be used. For a general matrix AA and a given configuration of balls x(n)x(n) at time nn, let in+1{1,,d}i_{n+1}\in\{1,\ldots,d\} be the random colour of the ball to be added in the urn at time n+1n+1. Then, note that

x¯(n+1)\displaystyle\bar{x}(n+1) =x(n)+𝐞in+1n0+n+1=(n0+n)x¯(n)+𝐞in+1n0+n+1\displaystyle=\frac{x(n)+{\bf e}_{i_{n+1}}}{n_{0}+n+1}=\frac{(n_{0}+n)\bar{x}(n)+{\bf e}_{i_{n+1}}}{n_{0}+n+1}
=(11n0+n+1)x¯(n)+𝐞in+1n0+n+1,\displaystyle=\left(1-\frac{1}{n_{0}+n+1}\right)\bar{x}(n)+\frac{{\bf e}_{i_{n+1}}}{n_{0}+n+1}, (6)

implying

x¯(n+1)x¯(n)=1n0+n+1(𝐞in+1x¯(n)).\bar{x}(n+1)-\bar{x}(n)=\frac{1}{n_{0}+n+1}({\bf e}_{i_{n+1}}-\bar{x}(n)). (7)

To put (7) into Robbins-Monro form, we rearrange the right-hand side into a deterministic part and a zero mean “noise”. More specifically, let

F(x¯(n)):=𝔼[𝐞in+1|n]x¯(n),F(\bar{x}(n)):=\mathbb{E}[{\bf e}_{i_{n+1}}\,|\,\mathcal{F}_{n}]-\bar{x}(n), (8)

and

ξn+1:=𝐞in+1𝔼[𝐞in+1|n].\xi_{n+1}:={\bf e}_{i_{n+1}}-\mathbb{E}[{\bf e}_{i_{n+1}}\,|\,\mathcal{F}_{n}]. (9)

By setting γn=1/(n0+n+1)\gamma_{n}=1/(n_{0}+n+1), we obtain

x¯(n+1)x¯(n)=γn(F(x¯(n))+ξn+1).\bar{x}(n+1)-\bar{x}(n)=\gamma_{n}(F(\bar{x}(n))+\xi_{n+1}). (10)

We note that the components of ξn+1\xi_{n+1} are uniformly bounded in modulus by 22, meaning that the conditions of Proposition 4.2 of Benaím [1] are satisfied, meaning that the interpolated process (X(t))t0(X(t))_{t\geq 0} is indeed an asymptotic pseudotrajectory of FF.

We can write the components of FF as

Fi(x)=uij=1dujxi,i=1,,d,F_{i}(x)=\frac{u_{i}}{\sum_{j=1}^{d}u_{j}}-x_{i},\quad i=1,\ldots,d,

where uiu_{i} has the same relationship to xx as ui(n)u_{i}(n) to x(n)x(n).

For a stochastic approximation heuristic, and to apply some of the results we use, it is necessary to classify the stationary points, for which we use the following terminology.

Definition 1.

Consider a stationary point pp of FF. We will say that pp is stable if it is an attractor for the vector field FF, meaning that there exists a neighbourhood WW of pp where for xWx\in W the trajectory (Φ(t,x))t0(\Phi(t,x))_{t\geq 0} satisfies Φ(x,t)p\Phi(x,t)\to p as tt\to\infty, uniformly in WW. Furthermore, if all eigenvalues of DF(p)DF(p) have negative real part, pp is said to be linearly stable, while if some eigenvalue has positive real part, pp is said to be linearly unstable. If all eigenvalues have positive real part, then pp is said to be a source.

Typically convergence happens with positive probability to a stable stationary point but with probability zero to a linearly unstable one. For the former, Theorem 7.3 of Benaïm [1], which holds under the assumptions on the noise of Proposition 4.2, shows that a stable stationary point has positive probability of being a limit as long as its basin of attraction WW contains points which are attainable in the sense that the process has positive probability of being indefinitely close to them at indefinitely large times, a condition which is usually satisfied for all points in the simplex for urn type processes under mild conditions; see Example 7.2 of [1].

To show that linearly unstable stationary points have zero probability of being limits it is usually necessary to check that there is expectation bounded away from zero of the positive part of the component of the noise in any given direction; see Pemantle [9], Theorem 1. This condition ensures that the process has zero probability of being trapped on a stable manifold and converging to the unstable point. Theorem 1 of [9] also requires the noise term ξn+1\xi_{n+1} (in our notation) to be uniformly bounded; as noted above this is satisfied in our process.

3 Proofs for the symmetric case

In this section we prove Theorem 1.1.

Throughout this section we let AA be the matrix

(1aaa1aaa1),a>0.\begin{pmatrix}1&a&a\\ a&1&a\\ a&a&1\end{pmatrix},\quad a>0. (11)

In this case the vector field FF is given by

F1(x1,x2,x3)\displaystyle F_{1}(x_{1},x_{2},x_{3}) =\displaystyle= x1β+ax2β+ax3β(1+2a)(x1β+x2β+x3β)x1\displaystyle\frac{x_{1}^{\beta}+ax_{2}^{\beta}+ax_{3}^{\beta}}{(1+2a)(x_{1}^{\beta}+x_{2}^{\beta}+x_{3}^{\beta})}-x_{1} (12)
F2(x1,x2,x3)\displaystyle F_{2}(x_{1},x_{2},x_{3}) =\displaystyle= ax1β+x2β+ax3β(1+2a)(x1β+x2β+x3β)x2\displaystyle\frac{ax_{1}^{\beta}+x_{2}^{\beta}+ax_{3}^{\beta}}{(1+2a)(x_{1}^{\beta}+x_{2}^{\beta}+x_{3}^{\beta})}-x_{2} (13)
F3(x1,x2,x3)\displaystyle F_{3}(x_{1},x_{2},x_{3}) =\displaystyle= ax1β+ax2β+x3β(1+2a)(x1β+x2β+x3β)x3\displaystyle\frac{ax_{1}^{\beta}+ax_{2}^{\beta}+x_{3}^{\beta}}{(1+2a)(x_{1}^{\beta}+x_{2}^{\beta}+x_{3}^{\beta})}-x_{3} (14)

The stochastic approximation approach indicates that the possible limits of our process will be stationary points of FF, so we start by identifying these. Noting that the lines x1=x2x_{1}=x_{2}, x1=x3x_{1}=x_{3} and x2=x3x_{2}=x_{3} are each invariant under FF, define the function

𝒫a,β(z)=azβ+1zβ+(1+a)z2a,\mathcal{P}_{a,\beta}(z)=az^{\beta+1}-z^{\beta}+(1+a)z-2a, (15)

which we will see is related to the dynamics restricted to one of these lines. The following result shows that all stationary points of FF are located on at least one of these lines and expresses them in terms of solutions to 𝒫a,β(z)=0\mathcal{P}_{a,\beta}(z)=0.

Proposition 3.1.

All stationary points (x1,x2,x3)(x_{1},x_{2},x_{3}) of FF have at least two of x1,x2,x3x_{1},x_{2},x_{3} equal, and are of one of the forms (rr+2,1r+2,1r+2)(\frac{r}{r+2},\frac{1}{r+2},\frac{1}{r+2}), (1r+2,rr+2,1r+2)(\frac{1}{r+2},\frac{r}{r+2},\frac{1}{r+2}) or (1r+2,1r+2,rr+2)(\frac{1}{r+2},\frac{1}{r+2},\frac{r}{r+2}), with rr a solution of 𝒫a,β(z)=0\mathcal{P}_{a,\beta}(z)=0 in +\mathbb{R}^{+}.

Furthermore, there are at most three possible values of rr, one of which is always 11, corresponding to the stationary point (13,13,13)(\frac{1}{3},\frac{1}{3},\frac{1}{3}).

Proof.

We start off by showing that any stationary point has at least two co-ordinates equal. We do this by writing the stationary point in the form (x,rx,sx)(x,rx,sx) and showing that one of r=1r=1, s=1s=1 or r=sr=s must hold.

Rearranging (12), (13) and (14) at (x,rx,sx)(x,rx,sx) gives

x\displaystyle x =\displaystyle= 1+a(rβ+sβ)(2a+1)(1+rβ+sβ)\displaystyle\frac{1+a(r^{\beta}+s^{\beta})}{(2a+1)(1+r^{\beta}+s^{\beta})} (16)
rx\displaystyle rx =\displaystyle= rβ+a(1+sβ)(2a+1)(1+rβ+sβ)\displaystyle\frac{r^{\beta}+a(1+s^{\beta})}{(2a+1)(1+r^{\beta}+s^{\beta})} (17)
sx\displaystyle sx =\displaystyle= sβ+a(1+sβ)(2a+1)(1+rβ+sβ).\displaystyle\frac{s^{\beta}+a(1+s^{\beta})}{(2a+1)(1+r^{\beta}+s^{\beta})}. (18)

It follows that

rβ+a(1+sβ)=r(1+a(rβ+sβ))\displaystyle r^{\beta}+a(1+s^{\beta})=r(1+a(r^{\beta}+s^{\beta})) (19)
sβ+a(1+rβ)=s(1+a(rβ+sβ)).\displaystyle s^{\beta}+a(1+r^{\beta})=s(1+a(r^{\beta}+s^{\beta})). (20)

Take the linear combination (s+1a)×(s+\frac{1}{a})\times(19)(r+1)×-(r+1)\times(20). This eliminates sβs^{\beta} and sβ+1s^{\beta+1}, giving

(rβ+a)sar(1+rβ)=(arβ+1)s(a(1+rβ)+rarβa+rβ+11),(r^{\beta}+a)s-ar(1+r^{\beta})=(ar^{\beta}+1)s-\left(a(1+r^{\beta})+\frac{r}{a}-\frac{r^{\beta}}{a}+r^{\beta+1}-1\right), (21)

which can be rearranged to give

s(rβ1)(a1)=(rβ+11)(1a)+(rβr)(a1a).s(r^{\beta}-1)(a-1)=(r^{\beta+1}-1)(1-a)+(r^{\beta}-r)\left(a-\frac{1}{a}\right). (22)

Assuming a1a\neq 1, (22) gives r=1r=1 or

s=a(rβ+1)(1r)+rβra(rβ1)=𝒫a,β(r)arβ+aa(1rβ).s=\frac{a(r^{\beta}+1)(1-r)+r^{\beta}-r}{a(r^{\beta}-1)}=\frac{\mathcal{P}_{a,\beta}(r)-ar^{\beta}+a}{a(1-r^{\beta})}. (23)

Using this form for ss in (20) gives (if r1r\neq 1)

sβ=arβ+1+rβ+ara(r1)=𝒫a,β(r)a(1r)+1.s^{\beta}=\frac{-ar^{\beta+1}+r^{\beta}+a-r}{a(r-1)}=\frac{\mathcal{P}_{a,\beta}(r)}{a(1-r)}+1. (24)

Combining (23) and (24) tells us that either r=1r=1 or s=1s=1 or

sβ1s1=rβ1r1,\frac{s^{\beta}-1}{s-1}=\frac{r^{\beta}-1}{r-1},

and the latter case implies r=sr=s. Hence any stationary point has two co-ordinates equal.

We now assume, without loss of generality, that the stationary point is of the form (x,x,rx)(x,x,rx) or equivalently (1r+2,1r+2,rr+2)\left(\frac{1}{r+2},\frac{1}{r+2},\frac{r}{r+2}\right). That the stationary point equations for a point of this form imply 𝒫a,β(r)=0\mathcal{P}_{a,\beta}(r)=0 is easy to check, and it is also easy to check that 𝒫a,β(1)=0\mathcal{P}_{a,\beta}(1)=0 for any a,β>0a,\beta>0.

The function 𝒫a,β\mathcal{P}_{a,\beta} satisfies 𝒫a,β(0)<0\mathcal{P}_{a,\beta}(0)<0 and 𝒫a,β(z)\mathcal{P}_{a,\beta}(z)\to\infty as zz\to\infty; furthermore it is concave for z<β1a(β+1)z<\frac{\beta-1}{a(\beta+1)} and convex for z>β1a(β+1)z>\frac{\beta-1}{a(\beta+1)}, which indicates that it has either one root or three in +\mathbb{R}^{+}, counting multiplicity. This completes the proof.

We now show that the process will indeed, almost surely, converge to one of the stationary points identified in Proposition 3.1.

Lemma 3.2.

The limit set of the process (x¯(n))n(\bar{x}(n))_{n\in\mathbb{N}}, defined in (3) with given matrix (11), will, almost surely, be a single point which is a stationary point of FF.

Proof.

Let

L(x1,x2,x3)=(x1+x2+x3)12a+1[alog(x1x2x3)1β(a1)log(x1β+x2β+x3β)].L(x_{1},x_{2},x_{3})=(x_{1}+x_{2}+x_{3})-\frac{1}{2a+1}\left[a\log(x_{1}x_{2}x_{3})-\frac{1}{\beta}(a-1)\log(x_{1}^{\beta}+x_{2}^{\beta}+x_{3}^{\beta})\right]. (25)

Then LL is a strict Lyapunov function for FF. In fact, straightforward differentiation gives

Lxi=1xiFi.\frac{\partial L}{\partial x_{i}}=-\frac{1}{x_{i}}F_{i}.

Then, denoting an trajectory of FF by x(t)=(x1(t),x2(t),x3(t))x(t)=(x_{1}(t),x_{2}(t),x_{3}(t)), we obtain

d(Lx)dt=i=13Lxidxidt=i=13xi(Lxi)20,\frac{d(L\circ x)}{dt}=\sum_{i=1}^{3}\frac{\partial L}{\partial x_{i}}\frac{dx_{i}}{dt}=-\sum_{i=1}^{3}x_{i}\left(\frac{\partial L}{\partial x_{i}}\right)^{2}\leq 0,

where the equality holds in the above inequality if and only if F(x)=0F(x)=0. Hence L(x(t))L(x(t)) is decreasing in tt, and strictly decreasing except where x(t)x(t) is a stationary point of FF.

We now apply Corollary 6.6 of Benaïm [1] to show that the limit set of (x¯(n))n(\bar{x}(n))_{n\in\mathbb{N}} will almost surely be a stationary point of FF. As Δ2\Delta^{2} is precompact and we have identified a strict Lyapunov function, the only condition which needs to be checked is that FF has countably many stationary points, which follows from Proposition 3.1. In fact, Proposition 3.1 shows that FF has no connected sets of stationary points other than single points, so the limit set must be a single point, which is one of the stationary points of FF. ∎

Note that the Lyapunov function LL generalises in an obvious way to more than three types, as long as all off-diagonal entries are equal.

Proposition 3.3.

Consider 𝒫a,β(z)=azβ+1zβ+(1+a)z2a\mathcal{P}_{a,\beta}(z)=az^{\beta+1}-z^{\beta}+(1+a)z-2a for z+z\in\mathbb{R}^{+}, with a,β>0a,\beta>0.

  1. (i)

    For a given value of a>1a>1, 𝒫a,β\mathcal{P}_{a,\beta} has only one root at z=1z=1.

  2. (ii)

    For a given value of a<1a<1, there exists β1(a)\beta_{1}(a) satisfying 1+2a1a>β1(a)>1\frac{1+2a}{1-a}>\beta_{1}(a)>1 such that

    1. (a)

      If β>1+2a1a\beta>\frac{1+2a}{1-a} then 𝒫a,β(1)<0\mathcal{P}_{a,\beta}^{\prime}(1)<0 and we have that 𝒫a,β\mathcal{P}_{a,\beta} has three roots in +\mathbb{R}^{+}, 11, rr_{-} and r+r_{+}, labelled so that r<1<r+r_{-}<1<r_{+}. As functions of β\beta for fixed aa, r+r_{+} is increasing and rr_{-} is decreasing.

    2. (b)

      If β1(a)<β<1+2a1a\beta_{1}(a)<\beta<\frac{1+2a}{1-a} then 𝒫a,β(1)>0\mathcal{P}_{a,\beta}^{\prime}(1)>0 and 𝒫a,β\mathcal{P}_{a,\beta} has three roots in +\mathbb{R}^{+}, 11, r1r_{1} and r2r_{2}, labelled so that 1<r1<r21<r_{1}<r_{2}. As functions of β\beta for fixed aa, r2r_{2} is increasing and r1r_{1} is decreasing.

    3. (c)

      If β<β1(a)\beta<\beta_{1}(a) then 𝒫a,β(1)>0\mathcal{P}_{a,\beta}^{\prime}(1)>0 and the only root of 𝒫a,β\mathcal{P}_{a,\beta} in +\mathbb{R}^{+} is 11.

    Furthermore β1(a)\beta_{1}(a) is an increasing function of aa with β1(a)\beta_{1}(a)\to\infty as a1a\to 1.

Proof.

We start by observing that 𝒫a,β(0)=2a\mathcal{P}_{a,\beta}(0)=-2a and that 𝒫a,β(z)\mathcal{P}_{a,\beta}(z)\to\infty as zz\to\infty, indicating that 𝒫a,β\mathcal{P}_{a,\beta} has an odd number of roots in +\mathbb{R}^{+}, counting multiplicity. Differentiating with respect to zz, we have

𝒫a,β(z)=a(β+1)zββzβ1+1+a\mathcal{P}_{a,\beta}^{\prime}(z)=a(\beta+1)z^{\beta}-\beta z^{\beta-1}+1+a (26)

and

𝒫a,β′′(z)=zβ2(aβ(β+1)zβ(β1)).\mathcal{P}_{a,\beta}^{\prime\prime}(z)=z^{\beta-2}(a\beta(\beta+1)z-\beta(\beta-1)). (27)

Because 𝒫a,β′′(z)\mathcal{P}_{a,\beta}^{\prime\prime}(z) is increasing on z+z\in\mathbb{R}^{+} there cannot be more than three roots of 𝒫a,β\mathcal{P}_{a,\beta} in +\mathbb{R}^{+}. (i) First, if β<1\beta<1 we have that 𝒫a,β′′(z)>0\mathcal{P}_{a,\beta}^{\prime\prime}(z)>0 in +\mathbb{R}^{+} implying that 𝒫a,β(z)\mathcal{P}_{a,\beta}^{\prime}(z) is strictly increasing. Moreover, 𝒫a,β(z)\mathcal{P}_{a,\beta}^{\prime}(z) goes from -\infty to ++\infty when zz ranges from 0 to ++\infty and 𝒫a,β(1)>0\mathcal{P}_{a,\beta}^{\prime}(1)>0. Then 𝒫a,β(z)\mathcal{P}_{a,\beta}^{\prime}(z) changes sign only once at some z<1z^{*}<1. Now, since 𝒫a,β(0)=2a<0\mathcal{P}_{a,\beta}(0)=-2a<0 and 𝒫a,β(z)\mathcal{P}_{a,\beta}(z) is decreasing for z<z<1z<z^{*}<1 and increasing otherwise, it follows that 𝒫a,β(z)\mathcal{P}_{a,\beta}(z) crosses the z=0z=0 line only once at z=1z=1. Second, the same happens for β>1\beta>1 since 𝒫a,β(z)>0\mathcal{P}_{a,\beta}^{\prime}(z)>0 in +\mathbb{R}^{+} and so 𝒫a,β(z)\mathcal{P}_{a,\beta}(z) is strictly increasing. The case β=1\beta=1 is trivial.

(ii) If β>1+2a1a\beta>\frac{1+2a}{1-a} we have 𝒫a,β(1)=1+2aβ(1a)<0\mathcal{P}_{a,\beta}^{\prime}(1)=1+2a-\beta(1-a)<0, indicating that in this case 𝒫a,β\mathcal{P}_{a,\beta} must have three roots. Note that if β=1+2a1a\beta=\frac{1+2a}{1-a} then 𝒫a,β(1)=0\mathcal{P}_{a,\beta}^{\prime}(1)=0 but that 𝒫a,β′′(1)<0\mathcal{P}_{a,\beta}^{\prime\prime}(1)<0, showing that this is a double root, not a triple root, and so there must be another root in that case for larger zz. If β<1+2a1a\beta<\frac{1+2a}{1-a} then 𝒫a,β(1)>0\mathcal{P}_{a,\beta}^{\prime}(1)>0 and 𝒫a,β\mathcal{P}_{a,\beta} has no root less than 11. Then, there must be either none, one double, or two distinct additional roots greater than 1.

The derivative of 𝒫a,β(z)\mathcal{P}_{a,\beta}(z) with respect to β\beta, for fixed aa and zz, is (az1)(logz)zβ(az-1)(\log z)z^{\beta}. Thus, for any z(1,1/a)z\in(1,1/a), 𝒫a,β(z)\mathcal{P}_{a,\beta}(z) is decreasing in β\beta, and it follows that if there are roots of 𝒫a,β\mathcal{P}_{a,\beta} in this range for a particular value of β\beta there must also be for any larger β\beta. As 𝒫a,β(z)>0\mathcal{P}_{a,\beta}(z)>0 if z1/az\geq 1/a, this shows that there exists β1(a)[1,1+2a1a]\beta_{1}(a)\in[1,\frac{1+2a}{1-a}] such that there is one root of 𝒫a,β\mathcal{P}_{a,\beta} when β<β1(a)\beta<\beta_{1}(a) and three when β>β1(a)\beta>\beta_{1}(a).

Let β0(a)>21a1>1\beta_{0}(a)>\frac{2}{1-a}-1>1 be the unique solution to

1+a=((12β+1)1a)β1.1+a=\left(\left(1-\frac{2}{\beta+1}\right)\frac{1}{a}\right)^{\beta-1}. (28)

(It can be seen that (28) has a unique solution for fixed a<1a<1, as in that case the right hand side is increasing in β\beta if the right hand side is greater than 11, is equal to 11 at β=1\beta=1 and tends to \infty as β\beta\to\infty. That β0(a)>21a1\beta_{0}(a)>\frac{2}{1-a}-1 can be seen by noting that if β\beta is a solution of (28) we must have (12β+1)1a>1\left(1-\frac{2}{\beta+1}\right)\frac{1}{a}>1.) Then if β<β0(a)\beta<\beta_{0}(a) we have 𝒫a,β(z)>0\mathcal{P}_{a,\beta}^{\prime}(z)>0 for all zz and hence 𝒫a,β\mathcal{P}_{a,\beta} is increasing in zz and so z=1z=1 is the only root. This shows that β1(a)β0(a)>1\beta_{1}(a)\geq\beta_{0}(a)>1.

Now, the fact that as mentioned above there is a root greater than 11 when β=1+2a1a\beta=\frac{1+2a}{1-a} together with the continuity of 𝒫a,β(z)\mathcal{P}_{a,\beta}(z) in β\beta ensures that there remains a root greater than 11 for β(1+2a1aϵ,1+2a1a)\beta\in\left(\frac{1+2a}{1-a}-\epsilon,\frac{1+2a}{1-a}\right) for some ϵ>0\epsilon>0, so β1(a)<1+2a1a\beta_{1}(a)<\frac{1+2a}{1-a}.

The claims that r+r_{+} and r2r_{2} are increasing functions of β\beta and that r1r_{1} is a decreasing function of β\beta also follow from the negative derivative of 𝒫a,β(z)\mathcal{P}_{a,\beta}(z) with respect to β\beta for z(1,1/a)z\in(1,1/a). Similarly the claim that rr_{-} is a decreasing function of β\beta follows from the derivative of 𝒫a,β(z)\mathcal{P}_{a,\beta}(z) with respect to β\beta being positive on (0,1)(0,1).

To see that β1(a)\beta_{1}(a) is an increasing function of aa, note that the derivative of 𝒫a,β(z)\mathcal{P}_{a,\beta}(z) with respect to aa is zβ+1+z2z^{\beta+1}+z-2, and for fixed z>1z>1 this is positive, meaning that if we are in case (c) for particular choices of aa and β\beta we will also be in case (c) for the same value of β\beta and any larger value of aa. That β1(a)\beta_{1}(a)\to\infty as a1a\to 1 follows from β0(a)>21a1\beta_{0}(a)>\frac{2}{1-a}-1. ∎

We now investigate the stability of these roots, for which recall the terminology in Definition 1.

Proposition 3.4.

If a stationary point for FF is of the form (x,x,rx)(x,x,rx) or (x,rx,x)(x,rx,x) or (rx,x,x)(rx,x,x), then it is linearly stable if 𝒫a,β(r)>0\mathcal{P}_{a,\beta}^{\prime}(r)>0 and rβ+2r+2>β(1a)2a+1\frac{r^{\beta}+2}{r+2}>\frac{\beta(1-a)}{2a+1}, and linearly unstable if either 𝒫a,β(r)<0\mathcal{P}_{a,\beta}^{\prime}(r)<0 or rβ+2r+2<β(1a)2a+1\frac{r^{\beta}+2}{r+2}<\frac{\beta(1-a)}{2a+1}.

Proof.

Without loss of generality we focus on the case (x,x,rx)(x,x,rx).

We note that the differential equation driven by FF keeps the line x1=x2x_{1}=x_{2} invariant, so we consider it restricted to this line; the equation for F3F_{3} gives

F3(1x32,1x32,x3)=2a(1x32)β+x3β(1+2a)(2(1x32)β+x3β)x3.F_{3}\left(\frac{1-x_{3}}{2},\frac{1-x_{3}}{2},x_{3}\right)=\frac{2a\left(\frac{1-x_{3}}{2}\right)^{\beta}+x_{3}^{\beta}}{(1+2a)\left(2\left(\frac{1-x_{3}}{2}\right)^{\beta}+x_{3}^{\beta}\right)}-x_{3}.

Let x3=z/(z+2)x_{3}=z/(z+2) so that x1=x2=1/(z+2)x_{1}=x_{2}=1/(z+2). Then

F3(1x32,1x32,x3)=2𝒫a,β(z)(1+2a)(2+zβ),F_{3}\left(\frac{1-x_{3}}{2},\frac{1-x_{3}}{2},x_{3}\right)=\frac{-2\mathcal{P}_{a,\beta}(z)}{(1+2a)(2+z^{\beta})},

and so is positive when 𝒫a,β(z)\mathcal{P}_{a,\beta}(z) is negative and vice versa. Hence a stationary point (x,x,rx)(x,x,rx) is stable in this direction if 𝒫a,β(r)>0\mathcal{P}_{a,\beta}^{\prime}(r)>0 and linearly unstable in this direction if 𝒫a,β(r)<0\mathcal{P}_{a,\beta}^{\prime}(r)<0.

Because FF is symmetric in x1x_{1} and x2x_{2}, the other direction in which we need to consider stability will be perpendicular to this one. Hence we consider

F1(x+ϵ,xϵ,rx)=(x+ϵ)β+(xϵ)β+arβxβ(2a+1)((x+ϵ)β+(xϵ)β+rβxβ)xϵ=F1(x,x,rx)+ϵ(1+β(1a)(2a+1)(2+rβ)x)+o(ϵ)=F1(x,x,rx)+ϵ(1+β(1a)(r+2)(2a+1)(2+rβ))+o(ϵ).\begin{split}F_{1}(x+\epsilon,x-\epsilon,rx)&=\frac{(x+\epsilon)^{\beta}+(x-\epsilon)^{\beta}+ar^{\beta}x^{\beta}}{(2a+1)((x+\epsilon)^{\beta}+(x-\epsilon)^{\beta}+r^{\beta}x^{\beta})}-x-\epsilon\\ &=F_{1}(x,x,rx)+\epsilon\left(-1+\frac{\beta(1-a)}{(2a+1)(2+r^{\beta})x}\right)+o(\epsilon)\\ &=F_{1}(x,x,rx)+\epsilon\left(-1+\frac{\beta(1-a)(r+2)}{(2a+1)(2+r^{\beta})}\right)+o(\epsilon).\end{split} (29)

It follows that (x,x,rx)(x,x,rx) is a stable stationary point in the direction perpendicular to the line x1=x2x_{1}=x_{2} if rβ+2r+2>β(1a)2a+1\frac{r^{\beta}+2}{r+2}>\frac{\beta(1-a)}{2a+1} and linearly unstable in that direction if the reverse inequality applies. ∎

We shall henceforth restrict ourselves to the case a<1a<1 since Propositions 3.1, 3.3(1) and 3.4 imply that if a>1a>1, (13,13,13)(\frac{1}{3},\frac{1}{3},\frac{1}{3}) is the only stable stationary point for FF and by Lemma 3.2, (x¯(n))n(\bar{x}(n))_{n\in\mathbb{N}} must converge to it. The case a=1a=1 has the probabilities of each colour being 1/31/3 regardless of x¯(n)\bar{x}(n) and so it is easily seen that x¯(n)(1/3,1/3,1/3)\bar{x}(n)\to(1/3,1/3,1/3) almost surely.

Corollary 3.5.

Assume a<1a<1.

  1. (i)

    If β<β1(a)\beta<\beta_{1}(a), then the stationary point (13,13,13)(\frac{1}{3},\frac{1}{3},\frac{1}{3}) is stable, and is the limit with probability 11.

  2. (ii)

    If β1(a)<β<1+2a1a\beta_{1}(a)<\beta<\frac{1+2a}{1-a}, then the stationary points (13,13,13)(\frac{1}{3},\frac{1}{3},\frac{1}{3}) and (1r2+2,1r2+2,r2r2+2)\left(\frac{1}{r_{2}+2},\frac{1}{r_{2}+2},\frac{r_{2}}{r_{2}+2}\right) (and its permutations) are stable, while the stationary point (1r1+2,1r1+2,r1r1+2)\left(\frac{1}{r_{1}+2},\frac{1}{r_{1}+2},\frac{r_{1}}{r_{1}+2}\right) and its permutations are linearly unstable.

  3. (iii)

    If β>1+2a1a\beta>\frac{1+2a}{1-a}, then there are three stationary points of FF of the form (x,x,rx)(x,x,rx) corresponding to the three solutions r<1<r+r_{-}<1<r_{+} of 𝒫a,β(z)=0\mathcal{P}_{a,\beta}(z)=0 in +\mathbb{R}^{+}. The stationary points (13,13,13)(\frac{1}{3},\frac{1}{3},\frac{1}{3}) and (1r+2,1r+2,2r+2)\left(\frac{1}{r_{-}+2},\frac{1}{r_{-}+2},\frac{2}{r_{-}+2}\right) (and its permutations) are linearly unstable, while (1r++2,1r++2,r+r++2)\left(\frac{1}{r_{+}+2},\frac{1}{r_{+}+2},\frac{r_{+}}{r_{+}+2}\right) and its permutations are stable.

Proof.

(i) Stability follows from Proposition 3.4, and almost sure convergence from Lemma 3.2.

(ii) The shape of 𝒫a,β\mathcal{P}_{a,\beta} as discussed in the proof of Proposition 3.3 ensures that r1,r2>1r_{1},r_{2}>1 and that 𝒫a,β(r1)<0\mathcal{P}_{a,\beta}^{\prime}(r_{1})<0 and 𝒫a,β(r2)>0\mathcal{P}_{a,\beta}^{\prime}(r_{2})>0, showing that (1r1+2,1r1+2,r1r1+2)\left(\frac{1}{r_{1}+2},\frac{1}{r_{1}+2},\frac{r_{1}}{r_{1}+2}\right) is linearly unstable, and that for the other two stationary points we just need to check the stability perpendicular to the line x1=x2x_{1}=x_{2}. But r2β+2r2+2>1>β(1a)2a+1\frac{r_{2}^{\beta}+2}{r_{2}+2}>1>\frac{\beta(1-a)}{2a+1} by our assumption on β\beta, so the condition from Propostion 3.4 is satisfied and so (13,13,13)(\frac{1}{3},\frac{1}{3},\frac{1}{3}) and (1r2+2,1r2+2,r2r2+2)\left(\frac{1}{r_{2}+2},\frac{1}{r_{2}+2},\frac{r_{2}}{r_{2}+2}\right) are stable. (iii) That 𝒫a,β(z)=0\mathcal{P}_{a,\beta}(z)=0 has three solutions follows from Proposition 3.3. As 𝒫a,β(1)<0\mathcal{P}_{a,\beta}^{\prime}(1)<0 it follows from Propostion 3.4 that (13,13,13)(\frac{1}{3},\frac{1}{3},\frac{1}{3}) is linearly unstable, and as r<1r_{-}<1 we have rβ+2r+2<1<β(1a)2a+1\frac{r_{-}^{\beta}+2}{r_{-}+2}<1<\frac{\beta(1-a)}{2a+1}, so (1r+2,1r+2,rr+2)\left(\frac{1}{r_{-}+2},\frac{1}{r_{-}+2},\frac{r_{-}}{r_{-}+2}\right) is also linearly unstable. The global minimum of the Lyapunov function on Δ2\Delta^{2} must be a stable stationary point of FF, so the remaining stationary points, (1r++2,1r++2,r+r++2)\left(\frac{1}{r_{+}+2},\frac{1}{r_{+}+2},\frac{r_{+}}{r_{+}+2}\right) and its permutations, must be stable. ∎

The following result completes the proof of Theorem 1.1.

Proposition 3.6.

Let l(x¯)l(\bar{x}) denote the limit set of the process (x¯(n))n(\bar{x}(n))_{n\in\mathbb{N}} defined in (10) and recall the stability criteria in Proposition 3.4. Then we have

  1. (i)

    [l(x¯)={p}]>0\mathbb{P}[l(\bar{x})=\{p\}]>0 for stable points pp of FF.

  2. (ii)

    [l(x¯)={p}]=0\mathbb{P}[l(\bar{x})=\{p\}]=0 for linearly unstable points pp of FF.

Proof.

Without loss of generality we focus on the case (x,x,rx)(x,x,rx).

(i) Let us now show that the process x¯\bar{x} in fact converges with positive probability toward a given stable fixed point. Of course, it is necessary that the process has positive probability of being arbitrarily close to the attractor at arbitrarily large times. That is, a point pp is said to be attainable by a process XX if for each t>0t>0 we have that [st:X(s)Np]>0\mathbb{P}[\exists\leavevmode\nobreak\ s\geq t\leavevmode\nobreak\ :\leavevmode\nobreak\ X(s)\in N_{p}]>0 for every neighborhood NpN_{p} of pp. It turns out that if the function F+IdF+Id associated with an urn process XX maps the simplex into its interior, it follows that every point of the simplex is attainable by XX. This is indeed the case for our process x¯\bar{x}. Finally, Benaïm [1] Theorem 7.3 ensures that a given attainable attractor pp with non-empty basin of attraction is such that [l(x¯)={p}]>0\mathbb{P}[l(\bar{x})=\{p\}]>0.

(ii) Let pp be a linearly unstable critical point in the interior of Δ2\Delta^{2} and 𝒩pΔ2\mathcal{N}_{p}\subset\Delta^{2} a neighborhood of pp. The simplex is considered as a differential manifold by identifying its tangent space at any point with the linear subspace TΔ2={x3:ixi=0}T\Delta^{2}=\{x\in\mathbb{R}^{3}\>:\>\sum_{i}x_{i}=0\}. We need to check Pemantle’s non-convergence criteria (Theorem 1 in [9]). As we have γn=1/(n0+n+1)\gamma_{n}=1/(n_{0}+n+1) and we have bounded noise, the only condition we need to check is condition (6): that the expectation of the positive part of the component of the noise in any given direction is uniformly bounded away from zero. Formally, we need that whenever x¯n𝒩p\bar{x}_{n}\in\mathcal{N}_{p}, there is a constant κ\kappa such that 𝔼[max{ξn+1θ,0}|n]κ\mathbb{E}[\max\{\xi_{n+1}\cdot\theta,0\}\>|\>\mathcal{F}_{n}]\geq\kappa for every unit vector θ=(θ1,θ2,θ3)TΔ2\theta=(\theta_{1},\theta_{2},\theta_{3})\in T\Delta^{2}, where ξn+1\xi_{n+1} is the noise term from (9). For notational simplicity, write u~i=ui(n)/(juj(n))\tilde{u}_{i}=u_{i}(n)/(\sum_{j}u_{j}(n)) and note that

𝔼[max{ξn+1θ,0}|n]=\displaystyle\mathbb{E}[\max\{\xi_{n+1}\cdot\theta,0\}\>|\>\mathcal{F}_{n}]= u~1max{θ1(1u~1)θ2u~2θ3u~3,0}+\displaystyle\tilde{u}_{1}\max\{\theta_{1}(1-\tilde{u}_{1})-\theta_{2}\tilde{u}_{2}-\theta_{3}\tilde{u}_{3},0\}+
u~2max{θ1u~1+θ2(1u~2)θ3u~3),0}+\displaystyle\tilde{u}_{2}\max\{-\theta_{1}\tilde{u}_{1}+\theta_{2}(1-\tilde{u}_{2})-\theta_{3}\tilde{u}_{3}),0\}+
u~3max{θ1u~1θ2u~2+θ3(1u~3),0}.\displaystyle\tilde{u}_{3}\max\{\theta_{1}\tilde{u}_{1}-\theta_{2}\tilde{u}_{2}+\theta_{3}(1-\tilde{u}_{3}),0\}. (30)

Now, write θ=(θ1,θ2,θ3)\theta=(\theta_{1},\theta_{2},\theta_{3}) with θ1+θ2+θ3=0\theta_{1}+\theta_{2}+\theta_{3}=0 and θ12+θ22+θ32=1\theta_{1}^{2}+\theta_{2}^{2}+\theta_{3}^{2}=1, and suppose that θi<0\theta_{i}<0 for exactly two coordinates, without loss of generality i=2,3i=2,3. Then minθ2,θ316\min{\theta_{2},\theta_{3}}\leq-\frac{1}{\sqrt{6}}, as otherwise (θ2θ3)2+θ22+θ32<1(-\theta_{2}-\theta_{3})^{2}+\theta_{2}^{2}+\theta_{3}^{2}<1. Then we can write

θ1(1u~1)θ2u~2θ3u~3=θ2(1u~1+u~2)θ3(1u~1+u~3)2a(1+2a)6,\theta_{1}(1-\tilde{u}_{1})-\theta_{2}\tilde{u}_{2}-\theta_{3}\tilde{u}_{3}=-\theta_{2}(1-\tilde{u}_{1}+\tilde{u}_{2})-\theta_{3}(1-\tilde{u}_{1}+\tilde{u}_{3})\geq\frac{2a}{(1+2a)\sqrt{6}},

as u~111+2a\tilde{u}_{1}\leq\frac{1}{1+2a}, meaning that the first term on the right hand side of (3) is at least 2a2(1+2a)26\frac{2a^{2}}{(1+2a)^{2}\sqrt{6}}. On the other hand, suppose that θi<0\theta_{i}<0 for exactly one coordinate; without loss of generality assume θ3<0\theta_{3}<0. Then if θ1>θ2\theta_{1}>\theta_{2} we can write

θ1(1u~1)θ2u~2θ3u~3=θ1(1u~1+u~3)θ2(u~2u~3).\theta_{1}(1-\tilde{u}_{1})-\theta_{2}\tilde{u}_{2}-\theta_{3}\tilde{u}_{3}=\theta_{1}(1-\tilde{u}_{1}+\tilde{u}_{3})-\theta_{2}(\tilde{u}_{2}-\tilde{u}_{3}).

If u~3u~2\tilde{u}_{3}\geq\tilde{u}_{2} then this is at least θ1(1u~1+u~3)2a(1+2a)6\theta_{1}(1-\tilde{u}_{1}+\tilde{u}_{3})\geq\frac{2a}{(1+2a)\sqrt{6}}, while if u~3<u~2\tilde{u}_{3}<\tilde{u}_{2} it is at least θ1(1u~1u~2+2u~3)\theta_{1}(1-\tilde{u}_{1}-\tilde{u}_{2}+2\tilde{u}_{3}), which is equal to 3θ1u~33\theta_{1}\tilde{u}_{3} because u~1+u~2+u~3=1\tilde{u}_{1}+\tilde{u}_{2}+\tilde{u}_{3}=1. Hence in that case θ1(1u~1)θ2u~2θ3u~33a(1+2a)6\theta_{1}(1-\tilde{u}_{1})-\theta_{2}\tilde{u}_{2}-\theta_{3}\tilde{u}_{3}\geq\frac{3a}{(1+2a)\sqrt{6}}, so that again the first term on the right hand side of (3) is at least 3a2(1+2a)26\frac{3a^{2}}{(1+2a)^{2}\sqrt{6}}. If θ2θ1\theta_{2}\geq\theta_{1} similar arguments show the the same but for the second term. Hence we can set κ=2a2(1+2a)26>0\kappa=\frac{2a^{2}}{(1+2a)^{2}\sqrt{6}}>0, and apply Theorem 1 of [9] to conclude the result. ∎

4 Proofs for the cyclic case

4.1 Introduction

In this section we prove Theorem 1.2. Here AA is the matrix

(110011101),\begin{pmatrix}1&1&0\\ 0&1&1\\ 1&0&1\end{pmatrix},

and we have

F1(x1,x2,x3)\displaystyle F_{1}(x_{1},x_{2},x_{3}) =\displaystyle= x1β+x2β2(x1β+x2β+x3β)x1\displaystyle\frac{x_{1}^{\beta}+x_{2}^{\beta}}{2(x_{1}^{\beta}+x_{2}^{\beta}+x_{3}^{\beta})}-x_{1}
F2(x1,x2,x3)\displaystyle F_{2}(x_{1},x_{2},x_{3}) =\displaystyle= x2β+x3β2)(x1β+x2β+x3β)x2\displaystyle\frac{x_{2}^{\beta}+x_{3}^{\beta}}{2)(x_{1}^{\beta}+x_{2}^{\beta}+x_{3}^{\beta})}-x_{2}
F3(x1,x2,x3)\displaystyle F_{3}(x_{1},x_{2},x_{3}) =\displaystyle= x3β+x1β2(x1β+x2β+x3β)x3\displaystyle\frac{x_{3}^{\beta}+x_{1}^{\beta}}{2(x_{1}^{\beta}+x_{2}^{\beta}+x_{3}^{\beta})}-x_{3}

First, we note that for any choice of β\beta, (13,13,13)(\frac{1}{3},\frac{1}{3},\frac{1}{3}) is a stationary point of FF. The following two results give information on its stability and show that it is in fact the only stationary point.

Lemma 4.1.

For the vector field FF, the stationary point at (13,13,13)(\frac{1}{3},\frac{1}{3},\frac{1}{3}) is stable if β<4\beta<4, and a linearly unstable source if β>4\beta>4.

Proof.

As we are working with xΔ2x\in\Delta^{2}, write x3=1x1x2x_{3}=1-x_{1}-x_{2}. Routine calculus then shows that the Jacobian matrix at (13,13,13)(\frac{1}{3},\frac{1}{3},\frac{1}{3}) is

(β21β2β21).\begin{pmatrix}\frac{\beta}{2}-1&\frac{\beta}{2}\\ -\frac{\beta}{2}&-1\end{pmatrix}.

The eigenvalues of this Jacobian are then the roots λ\lambda of

λ2+λ(2β2)(β21)+(β2)2\lambda^{2}+\lambda\left(2-\frac{\beta}{2}\right)-\left(\frac{\beta}{2}-1\right)+\left(\frac{\beta}{2}\right)^{2}

which are

(β41)±i3β4.\left(\frac{\beta}{4}-1\right)\pm\frac{i\sqrt{3}\beta}{4}.

Since the real part of both eigenvalues is positive if β>4\beta>4 and negative if β<4\beta<4, the result follows. ∎

Lemma 4.2.

The only stationary point of FF in Δ2\Delta^{2} is (13,13,13)(\frac{1}{3},\frac{1}{3},\frac{1}{3}).

Proof.

For x=(x1,x2,x3)Δ2x=(x_{1},x_{2},x_{3})\in\Delta^{2} with F1(x)=F2(x)=F3(x)=0F_{1}(x)=F_{2}(x)=F_{3}(x)=0, we have x1=x1β+x2βx2β+x3βx2x_{1}=\frac{x_{1}^{\beta}+x_{2}^{\beta}}{x_{2}^{\beta}+x_{3}^{\beta}}x_{2} and similarly x2=x2β+x3βx3β+x1βx3x_{2}=\frac{x_{2}^{\beta}+x_{3}^{\beta}}{x_{3}^{\beta}+x_{1}^{\beta}}x_{3} and x3=x3β+x1βx1β+x2βx1x_{3}=\frac{x_{3}^{\beta}+x_{1}^{\beta}}{x_{1}^{\beta}+x_{2}^{\beta}}x_{1}. Using this,

x1x2=x2x1βx3βx2β+x3β,x_{1}-x_{2}=x_{2}\frac{x_{1}^{\beta}-x_{3}^{\beta}}{x_{2}^{\beta}+x_{3}^{\beta}},

indicating that (if x2>0x_{2}>0) if x1>x2x_{1}>x_{2} then also x1>x3x_{1}>x_{3}, while if x1<x2x_{1}<x_{2} then x1<x3x_{1}<x_{3}. Similarly, if x3>0x_{3}>0 then the signs of x2x3x_{2}-x_{3} and x2x1x_{2}-x_{1} are the same, and if x1>0x_{1}>0 then the signs of x3x1x_{3}-x_{1} and x3x2x_{3}-x_{2} are the same. Hence the only stationary point of FF in the interior of Δ2\Delta^{2} is (13,13,13)(\frac{1}{3},\frac{1}{3},\frac{1}{3}).

It is also easy to check that if x1=0x_{1}=0 then x2=0x_{2}=0, and similarly that if x2=0x_{2}=0 then x3=0x_{3}=0 and if x3=0x_{3}=0 then x1=0x_{1}=0. Hence there are no stationary points of FF on the boundary of Δ2\Delta^{2}. ∎

We can now complete the proof of Theorem 1.2.

That we only have one stationary point, and that it is never a saddle, restricts the possibilities for chain transitive sets. In two dimensions Theorem 6.12 of Benaïm [1] states that chain transitive sets must be unions of stationary points, periodic orbits and cyclic orbit chains. However, with only one stationary point which is not a saddle cyclic orbit chains are impossible. We can thus conclude that the limit set must be a connected union of periodic orbits and stationary points.

By Lemma 4.1, if β<4\beta<4 then by Lemma 4.1 the stationary point (13,13,13)(\frac{1}{3},\frac{1}{3},\frac{1}{3}) is stable, and hence is an attractor for the flow given by FF. As in the proof of Proposition 3.6, by Theorem 7.3 of Benaïm [1], to show that there is positive probability of convergence to (13,13,13)(\frac{1}{3},\frac{1}{3},\frac{1}{3}) it is enough to show that it is an attainable point, that is that we have that [st:X(s)N]>0\mathbb{P}[\exists\leavevmode\nobreak\ s\geq t\leavevmode\nobreak\ :\leavevmode\nobreak\ X(s)\in N]>0 for every neighborhood NN of (13,13,13)(\frac{1}{3},\frac{1}{3},\frac{1}{3}). This is straightforward to show: for any ϵ>0\epsilon>0 there will be points of the form (n1n,n2n,n3n)\left(\frac{n_{1}}{n},\frac{n_{2}}{n},\frac{n_{3}}{n}\right) with n1,n2,n3n_{1},n_{2},n_{3} integers satisfying n1+n2+n3=nn_{1}+n_{2}+n_{3}=n and max{n1n13,n2n13,n3n13}<ϵ\max\left\{\frac{n_{1}}{n}-\frac{1}{3},\frac{n_{2}}{n}-\frac{1}{3},\frac{n_{3}}{n}-\frac{1}{3}\right\}<\epsilon for arbitrarily large nn. Because even if only one colour is present in the urn initially, there will be positive probability that a second colour is chosen at the first time step, and once two colours are present in the urn all three colours have positive probability of being chosen at each step, there will be positive probability of any such point being reached, so (13,13,13)(\frac{1}{3},\frac{1}{3},\frac{1}{3}) is indeed attainable.

If β>4\beta>4, then by Lemma 4.1 (13,13,13)(\frac{1}{3},\frac{1}{3},\frac{1}{3}) is linearly unstable, so as in Proposition 3.6 it will follow that it is a limit with probability zero if we have positive expectation of the positive part of the component of the noise in any given direction. In fact, the same argument as in Proposition 3.6 will work here, except that the explicit bounds for the u~i\tilde{u}_{i} must be replaced by the fact that in a neighbourhood of (13,13,13)(\frac{1}{3},\frac{1}{3},\frac{1}{3}) there will exist δ\delta such that δ<u~i<1δ\delta<\tilde{u}_{i}<1-\delta for each ii, giving κ=δ2/6\kappa=\delta^{2}/\sqrt{6}. So we can conclude that x¯(n)\bar{x}(n) has probability zero of converging to a stationary point, and it follows that the limit set must be a periodic orbit or a connected union of periodic orbits, completing the proof of Theorem 1.2.

5 Examples and simulations

In this section we consider some examples, including some where exact calculations are possible, and some simulations. We also consider some examples which go beyond the cases covered by Theorems 1.1 and 1.2.

5.1 The symmetric case with β=2\beta=2

In the case of Theorem 1.1 with β=2\beta=2, the possible limits and the phase transitions can be explicitly identified. We find that 𝒫a,2(z)=(z1)(az2+(a1)+2a)\mathcal{P}_{a,2}(z)=(z-1)(az^{2}+(a-1)+2a), with roots given by z=1z=1 and z=1a±12a7a22az=\frac{1-a\pm\sqrt{1-2a-7a^{2}}}{2a}. If a<817a<\frac{\sqrt{8}-1}{7}, then these are real and positive, and letting λ1=3a+112a7a24(2a+1),λ2=3a+1+12a7a24(2a+1),λ3=a+112a7a22(2a+1)\lambda_{1}=\frac{3a+1-\sqrt{1-2a-7a^{2}}}{4(2a+1)},\lambda_{2}=\frac{3a+1+\sqrt{1-2a-7a^{2}}}{4(2a+1)},\lambda_{3}=\frac{a+1-\sqrt{1-2a-7a^{2}}}{2(2a+1)} and λ4=a+1+12a7a22(2a+1)\lambda_{4}=\frac{a+1+\sqrt{1-2a-7a^{2}}}{2(2a+1)}, we obtain linearly stable stationary points (λ1,λ1,λ4)(\lambda_{1},\lambda_{1},\lambda_{4}), (λ1,λ4,λ1)(\lambda_{1},\lambda_{4},\lambda_{1}) and (λ4,λ1,λ1)(\lambda_{4},\lambda_{1},\lambda_{1}), and linearly unstable stationary (except at a=14a=\frac{1}{4}) points (λ2,λ2,λ3)(\lambda_{2},\lambda_{2},\lambda_{3}), (λ2,λ3,λ2)(\lambda_{2},\lambda_{3},\lambda_{2}), (λ3,λ2,λ2)(\lambda_{3},\lambda_{2},\lambda_{2}). In addition, the stationary point (13,13,13)(\frac{1}{3},\frac{1}{3},\frac{1}{3}) is linearly stable if a>14a>\frac{1}{4}, and linearly unstable if a<14a<\frac{1}{4}.

If a>817a>\frac{\sqrt{8}-1}{7}, then (13,13,13)(\frac{1}{3},\frac{1}{3},\frac{1}{3}) is the only stationary point, and is stable. In the notation of Theorem 1.1, we have β1(817)=2\beta_{1}\left(\frac{\sqrt{8}-1}{7}\right)=2, and the three phases are as follows:

Figure 1: 20 simulations of the symmetric model for β=2\beta=2
Refer to caption
i a=0.2a=0.2
Refer to caption
ii a=0.26a=0.26
Refer to caption
iii a=0.5a=0.5
  • When a<14a<\frac{1}{4}, (13,13,13)(\frac{1}{3},\frac{1}{3},\frac{1}{3}) is linearly unstable; there are three other stationary points, (λ1,λ1,λ4)(\lambda_{1},\lambda_{1},\lambda_{4}) and permutations, placed symmetrically, which are stable. For example when a=15a=\frac{1}{5}, (0.1847,0.1847,0.6306)(0.1847,0.1847,0.6306) and permutations are stable. Almost surely, one of these three points will be the limit. A simulation of 20 trajectories of the stochastic process in this case appears in Figure 1i.

  • For 14<a<817\frac{1}{4}<a<\frac{\sqrt{8}-1}{7}, (13,13,13)(\frac{1}{3},\frac{1}{3},\frac{1}{3}) is now stable but there are also stable stationary points elsewhere, near (14,14,12)(\frac{1}{4},\frac{1}{4},\frac{1}{2}). In this case, both symmetric and asymmetric limits have positive probability. For example, at a=0.26a=0.26, there are stable stationary points at (0.2792,0.2792,0.4416)(0.2792,0.2792,0.4416) and permutations as well as (13,13,13)(\frac{1}{3},\frac{1}{3},\frac{1}{3}). A simulation of 20 trajectories of the stochastic process in this case appears in Figure 1ii.

  • For a>817a>\frac{\sqrt{8}-1}{7}, (13,13,13)(\frac{1}{3},\frac{1}{3},\frac{1}{3}) is the only stationary point, and is stable, and will be the limit almost surely. A simulation of 20 trajectories of the stochastic process in the a=12a=\frac{1}{2} case appears in Figure 1iii.

At the critical value a=14a=\frac{1}{4}, (13,13,13)=(λ2,λ2,λ3)(\frac{1}{3},\frac{1}{3},\frac{1}{3})=(\lambda_{2},\lambda_{2},\lambda_{3}) has zeros as eigenvalues of its Jacobian and so is neither linearly stable nor linearly unstable, while there are stable stationary points at (14,14,12)(\frac{1}{4},\frac{1}{4},\frac{1}{2}) and permutations; similarly at the critical value a=817a=\frac{\sqrt{8}-1}{7} the stationary point (λ2,λ2,λ3)=(λ1,λ1,λ4)(\lambda_{2},\lambda_{2},\lambda_{3})=(\lambda_{1},\lambda_{1},\lambda_{4}) is neither linearly stable nor linearly unstable.

5.2 The symmetric case with β=3\beta=3

It is also possible to do some explicit calculations when β=3\beta=3. In this case (13,13,13)(\frac{1}{3},\frac{1}{3},\frac{1}{3}) is linearly stable when a>25a>\frac{2}{5} and linearly unstable when a<25a<\frac{2}{5}, and we have

𝒫a,3(z)=(z1)(az3+(a1)z2+(a1)z+2a),\mathcal{P}_{a,3}(z)=(z-1)(az^{3}+(a-1)z^{2}+(a-1)z+2a),

where the cubic factor has one real root (which is negative) when a>ac=1166(3.(2)1/2.(3)1/4+24.(3)1/2+13.(2)1/2.(3)3/420)=0.4160306a>a_{c}=\frac{1}{166}(3.(2)^{1/2}.(3)^{1/4}+24.(3)^{1/2}+13.(2)^{1/2}.(3)^{3/4}-20)=0.4160306 and three real roots (one of which is negative) when a<aca<a_{c}. (In the notation of Theorem 1.1, β1(ac)=3\beta_{1}(a_{c})=3.) Hence for a>aca>a_{c} we get almost sure convergence to (13,13,13)(\frac{1}{3},\frac{1}{3},\frac{1}{3}), for a<25a<\frac{2}{5} we get almost sure convergence to one of three asymmetric stationary points, and for 25<a<ac\frac{2}{5}<a<a_{c} the process may converge either to (13,13,13)(\frac{1}{3},\frac{1}{3},\frac{1}{3}) or to an asymmetric stationary point, each with positive probability.

5.3 The cyclic model

Figure 2i shows 20 simulations of the cyclic model when β=3\beta=3, showing convergence to (13,13,13)(\frac{1}{3},\frac{1}{3},\frac{1}{3}). Figure 2ii shows 20 simulations with β=6\beta=6, showing apparent convergence to a single limit cycle.

Figure 2: 20 simulations of the cyclic model
Refer to caption
i β=3\beta=3
Refer to caption
ii β=6\beta=6

5.4 The symmetric case with more than three types

It is natural to extend Section 3 to d>3d>3 types, letting AA be the d×dd\times d matrix with aii=1a_{ii}=1 for each ii and aij=aa_{ij}=a for iji\neq j. It is straightforward to extend the Lyapunov function (25) to this case, meaning that Lemma 3.2 applies. However, the later calculations, starting with Proposition 3.1, do not apply. It thus may be interesting to investigate whether more complex patterns of phases can occur in this case than when d=3d=3; however, numerical solution of the stationary point equations for particular values of aa when d=4,5,6d=4,5,6 suggests that the behaviour is in fact very similar to the d=3d=3 case, with three phases which parallel those found in Theorem 1.1.

5.5 A more general cyclic case

It is natural to extend Section 4 by allowing the matrix AA to take the form

(1a001aa01),\begin{pmatrix}1&a&0\\ 0&1&a\\ a&0&1\end{pmatrix},

allowing for different strengths of the cyclic reinforcement. It is straightforward to extend Lemma 4.1 to this case, showing that the stationary point at (13,13,13)(\frac{1}{3},\frac{1}{3},\frac{1}{3}) is linearly stable if a2a\geq 2 or if a<2a<2 and β<2(1+a)2a\beta<\frac{2(1+a)}{2-a}, and linearly unstable if a<2a<2 and β>2(1+a)2a\beta>\frac{2(1+a)}{2-a}. However Lemma 4.2 does not apply for general aa and there may be other stationary points.

Numerical investigation when β=2\beta=2 suggests that there are three phases in aa: in addition to a phase with almost sure convergence to (13,13,13)(\frac{1}{3},\frac{1}{3},\frac{1}{3}) and a phase with convergence to a limit cycle, there is a phase up to a0.25057a\approx 0.25057 where there are stable stationary points other than (13,13,13)(\frac{1}{3},\frac{1}{3},\frac{1}{3}) and that the process usually converges to one of these.

Acknowledgements

M.C.’s research was supported by CONICET [grant number 10520170300561CO]. The authors are grateful to Andrew Wade for suggestions that improved the presentation.

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