A Mean Field Limit

We consider a system consisting of N interacting objects. As we let the number of objects increase, we can characterize the limiting behaviour of the system.

We define our model as follows:

  • The are N objects moving in discrete time and each of which taking states in the finite set {\mathcal X}.
  • We let X_n^N(t) be the state of the n-th object at time t.
  • We let M_x^N(t) give the proportion of objects in state x\in{\mathcal X} at time t. Specifically,
  • X^N_n preforms transitions that depend on its previous state and the state of the system as recorded through a vector S^N(t) \in {\mathbb R}^d. Specifically, there is a Transition matrix P^N_{xy}(r) such that

  • Further the state of the system S^N evolves according to the mean state vector M^N and the previous state of the system:

We now take a limit where the number of objects in the system, N, is sent to infinity. We make the following assumption

  • We assume that P_{xy}^N(\cdot) converges uniformly to some probability transition matrix P_{xy}(\cdot).

The following results shows that for M^N and S^N a sensible limit comes from our analysis

Thrm [MF:Thrm] Provided

for probability distribution \mu (0) and vector \sigma(0) then, almost surely,

for each t\in{\mathbb Z}_+ where \mu(t) and \sigma(t) satisfy the equations

For the proof we will require the following elementary lemma followed by a proposition.

Prop [MF:Bern] Let B^N be an integer in \{0,1,...,N\} such that \lim_{N\rightarrow\infty} B^N/N = \beta>0. Let Y^N_b b=1,...,B be iid Bernoulli such that {\mathbb E} Y^N_b = y and we define

with Z^N=0 when B^N=0. Then, almost surely,

Proof By a Chernoff bound, there exists \alpha >0 such that

for each \epsilon >0. Thus

By the Borel-Cantelli Lemma

But we know that B^N/N > b eventually for any b<\beta so we can see that (almost surely) eventually \limsup_{N\rightarrow \infty} | Z^N-y| <\epsilon. Since \epsilon is arbitrary, it must be that

as required. \square

To prove the result we compare the chain X^N(t) with \tilde{X}^N_n(t), the chain where we replace S^N(t) with its limit process \sigma(t). That is,

and X^N_n(0)=\tilde{X}^N_n(0). Further, we note that since our state-space is finite we can assume the elements of {\mathcal X} are ordered (with ordering x\leq y) and we can achieve a jump in our chain X^N_n(t) to X^N_n(t+1) with S^N(t)=s [or \tilde{X}^N_n(t) to \tilde{X}^N_n(t+1) with \sigma(t)=s) by taking a uniform [0,1] random variable U and setting

We will assume that the versions of X^N and \tilde{X}^N are achieved by this process and that we use the same uniform random variables the jumps of X^N_n(t) and X^N_n(t+1) respectively.

We now prove the convergence of \tilde{X}:

Prop [MF:Prop] Provided M^N(0) \rightarrow \mu(0) for probability distribution \mu(0) then, almost surely, \tilde{M}^N(t) \rightarrow \mu(t) for each t\in{\mathbb Z}_+ where \mu(t) satisfies the equation \mu(t+1) = \mu(t) P(\sigma(t)).

Proof. We prove the result by induction on t, where we assume that the convergence result holdsup until time t. The result trivally holds at time 0.

Now, assuming that the result holds until time t. We let I_{xy}^n be the indicator function if the n-th object X^N_n performs a jump from x to y at step t.

Focusing on each term

We see that it counts up \tilde{M}^N_x(t) Bernoulli random variables and so Lemma [MF:Bern] applies and

provided \lim_{N\rightarrow\infty}\tilde{M}^N_x(t) >0. If \lim_{N\rightarrow\infty}\tilde{M}^N_x(t) >0 the above term, which is bounded above by \tilde{M}^N_n(t), must converge to zero. \square

We now prove the main result. The main idea is to show, by induction, that \tilde{X}^N and X^N exhibit the same behaviour as N goes to infinity.

Proof [Proof of Theorem [MF:Thrm]] One could think of this as an extension of Proposition [MF:Prop]. We preform induction on t with the hypothesis that

The hypothesis trivially holds for t=0. Assuming that the induction hypothesis holds until time t, we show that the result holds for time t+1. Note that


Since the final term above converges to zero, it is sufficient to prove each A_x^N converges to zero. Note that both the random variables X^N_n(t) and \tilde{X}^N_n(t) respectively are governed by the same condition with s equal to S^N(t) and \sigma(t) respectively. Letting L_{xy} and V_{xy} respectively equal the minimum and minimum of \

We see that X^N_n(t) and \tilde{X}^N_n(t) are not equal when

We let G^N(z) be the empircal distribution of our uniform random variables, that is

We note that

\sup_z | G^N(z) -z | converges to zero by the Glivenko-Cantelli Theorem. Thus holds at time t+1.

At this point we see the M^N_n(t+1)\rightarrow \mu(t+1) since holds at time t+1 and since \tilde{M}^N_n(t)\rightarrow\mu(t). Further by continuity of g(\cdot,\cdot) and the induction hypothesis, we have that

This proves the induction hypothesis and the theorem. \square


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