Event-Scheduling Algorithms with Kalikow Decomposition for Simulating Potentially Infinite Neuronal Networks

Abstract

Event-scheduling algorithms can compute in continuous time the next occurrence of points (as events) of a counting process based on their current conditional intensity. In particular, event-scheduling algorithms can be adapted to perform the simulation of finite neuronal networks activity. These algorithms are based on Ogata’s thinning strategy (Ogata in IEEE Trans Inf Theory 27:23–31, 1981), which always needs to simulate the whole network to access the behavior of one particular neuron of the network. On the other hand, for discrete time models, theoretical algorithms based on Kalikow decomposition can pick at random influencing neurons and perform a perfect simulation (meaning without approximations) of the behavior of one given neuron embedded in an infinite network, at every time step. These algorithms are currently not computationally tractable in continuous time. To solve this problem, an event-scheduling algorithm with Kalikow decomposition is proposed here for the sequential simulation of point processes neuronal models satisfying this decomposition. This new algorithm is applied to infinite neuronal networks whose finite time simulation is a prerequisite to realistic brain modeling.

Appendices

A Link Between Algorithm 2 and the Kalikow Decomposition

To prove that Algorithm 2 returns the desired processes, let us use some additional and more mathematical notation. Note that all the points simulated on neuron i before being accepted or not can be seen as coming from a common Poisson process of intensity M, denoted $\varPi _i$. For any $i \in {\mathbf {I}}$, we denote the arrival times of $\varPi _i$, $(T^i_n)_{n \in \mathbb {Z}}$, with $T^i_1$ being the first positive time.

As in Step 6 of Algorithm 2, we attach to each point of $\varPi ^i$ a stochastic mark X given by

$$\begin{aligned} X^i_{n} = {\left\{ \begin{array}{ll} 1 \quad &{}\text {if} \quad T^i_n \, \text {is accepted in the thinning procedure}\\ 0 \quad &{}\text {otherwise}. \end{array}\right. } \end{aligned}$$

(4)

Let us also define $V^i_n$ the neighborhood choice of $T^i_n$ picked at random and independently of anything else according to $\lambda _i$ and shifted at time $T^i_n$.

In addition, for any $i \in {\mathbf {I}}$, define $N^i= (T^i_n,X^i_n)_{n \in \mathbb {Z}}$ an E-marked point process with $E=\{0;1\}$. In particular, following the notation in Chapter VIII of [2], for any $i \in {\mathbf {I}}$, let

$$\begin{aligned} N^i_t(mark)= \sum _{n \in \mathbb {Z}} \mathbb {1}_{X^i_n =mark} \mathbb {1}_{T^i_n \le t} \quad &\text { for } \quad mark \in E \\ {\mathcal {F}}^{N}_{t^{-}}= \bigvee _{i \in {\mathbf {I}}} \sigma (N^i_s({0}), N^i_s({1});\,\,\,s<t) \quad &\text{ and } \quad {\mathcal {F}}^{N(1)}_{t^{-}}= \bigvee _{i \in {\mathbf {I}}} \sigma (N^i_s({1});\,\,\,s<t). \end{aligned}$$

Moreover note that $(N_t^i(1))_{t\in \mathbb {R}}$ is the counting process associated with the point process $\mathbf{P}$ simulated by Algorithm 2. Let us denote by $\varphi _i(t)$ the formula given by (2) and shifted at time t. Note that since the $\phi _i^v$’s are ${\mathcal {F}}_{0-}^{int}={\mathcal {F}}_{0-}^{N(1)}$, $\varphi _i(t)$ is ${\mathcal {F}}^{N(1)}_{t^{-}}$ measurable. We also denote $\varphi _i^v(t)$ the formula of $\phi _i^v$ shifted at time t.

With this notation, we can prove the following.

Proposition 2

The process $(N^i_t(1))_{t\in \mathbb {R}}$ admits $\varphi _i(t)$ as ${\mathcal {F}}^{N(1)}_{t^{-}}$-predictable intensity.

Proof

Following the technique in Chapter 2 of [2], let us take $C_t$ a non-negative predictable function with respect to (w.r.t) ${\mathcal {F}}^{N^i(1)}_{t}$ that is ${\mathcal {F}}^{N(1)}_{t-}$ measurable and therefore ${\mathcal {F}}^{N}_{t-}$ measurable . We have, for any $i \in {\mathbf {I}}$,

$$\begin{aligned} \mathbb {E}\left( \int \limits _{0}^{\infty } C_t \mathrm{d}N^i_t(1) \right) = \sum _{n=1}^\infty \mathbb {E}\left( C_{T^i_n} \mathbb {1}_{X^i_{n} = 1} \right) \end{aligned}$$

Note that by Theorem T35 at Appendix A1 of [2], any point T should be understood as a stopping time and that by Theorem T30 at Appendix A2 of [2],

$$\begin{aligned} {\mathcal {F}}^{N}_{T-} = \bigvee _j \sigma \{T^j_m, X^j_m \, \text {such that} \, T^j_m < T\} \end{aligned}$$

So

$$\begin{aligned} \mathbb {E}\left( \int \limits _{0}^{\infty } C_t dN^i_t(1) \right)= & {} \sum _{n=1}^\infty \mathbb {E}\left( C_{T^i_n} \mathbb {E}( \mathbb {1}_{X^i_{n} = 1} | {\mathcal {F}}^{N}_{{T^i_n}^{-}}, {V}^i_n)\right) \\= & {} \sum _{n=1}^\infty \mathbb {E} \left( C_{T^i_n} \dfrac{\varphi _i^{V^i_n}(T^i_n)}{M} \right) . \end{aligned}$$

Let us now integrate with respect to the choice $V^i_n$, which is independent of anything else.

$$\begin{aligned}&\mathbb {E}\left( \int \limits _{0}^{\infty } C_t \mathrm{d}N^i_t(1) \right) \\&\quad = \sum _{n=1}^\infty \mathbb {E}\left( C_{T^i_n} \frac{\lambda _i(\emptyset ) \varphi _i^{\emptyset } + \sum \nolimits _{v \in {\mathcal {V}}, v\ne \emptyset }\lambda _i(v) \times \varphi _i^{v}(T^i_n)}{M} \right) \\&\quad = \mathbb {E}\left( \int \limits _0^{\infty } C_t \frac{\varphi _i(t)}{M} \mathrm{d}\varPi ^i(t)\right) . \end{aligned}$$

Since $\varPi ^i$ is a Poisson process with respect to $({\mathcal {F}}_t^N)_t$ with intensity M, and since $C_t \frac{\varphi _i(t)}{M}$ is ${\mathcal {F}}_{t-}^N$ measurable, we finally have that

$$\begin{aligned} \mathbb {E}\left( \int \limits _{0}^{\infty } C_t \mathrm{d}N^i_t(1) \right) = \mathbb {E}\left( \int \limits _0^{\infty }C_t\varphi _i(t) \mathrm{d}t\right) , \end{aligned}$$

which ends the proof.$\square$

B Proof of Proposition 1

Proof

We do the proof for the backward part, starting with $T=T_{next}$ as the next point after $t_0$ (Step 4 of Algorithm 3), the proof being similar for the other $T_{next}$ generated at Step 23. We construct a tree with root (i, T). For each point $(j_{T'},T')$ in the tree, the points which are simulated in $V_{T'}$ (Step 12 of Algorithm 3) define the children of $(j_{T'}, T')$ in the tree. This forms the tree $\tilde{{\mathcal {T}}}$.

Let us now build a tree $\tilde{{\mathcal {C}}}$ with root (i, T) (that includes the previous tree) by mimicking the previous procedure in the backward part, except that we simulate on the whole neighborhood even if it has a part that intersects with previous neighborhoods (if they exist) (Steps 11–12 of Algorithm 3). By doing so, we make the number of children at each node independent of anything else.

If the tree $\tilde{{\mathcal {C}}}$ goes extinct then so does the tree $\tilde{{\mathcal {T}}}$ and the backward part of the algorithm terminates.

But if one only counts the number of children in the tree $\tilde{{\mathcal {C}}}$, we have a marked branching process whose reproduction distribution for the mark i is given by

• no children with probability $\lambda _i(\emptyset )$

• Poissonian number of children with parameter l(v)M if v is the chosen neighborhood with probability $\lambda _i(v)$

This gives that the average number of children issued from a node with the mark i is

$$\begin{aligned} \zeta _i=\lambda _i(\emptyset ) \times 0+\sum _{v\in {\mathcal {V}}, v \ne \emptyset } \lambda _i(v) l(v)M. \end{aligned}$$

If we denote $\tilde{{\mathcal {C}}}^k$ as the collection of points in the tree $\tilde{{\mathcal {C}}}$ at generation k, and by $K_{T'}$ the set of points generated independently as a Poisson process of rate M inside $V_{T'}$, we see recursively that

$$\begin{aligned} \tilde{{\mathcal {C}}}^{k+1} = \bigcup _{T'\in \tilde{{\mathcal {C}}}^{k}} K_{T'} \end{aligned}$$

But

$$\begin{aligned} \mathbb {E}(|K_{T'}| | T')= \zeta _{j_{T'}}. \end{aligned}$$

Therefore, if we denote the total number of sites in $\tilde{{\mathcal {C}}}^k$ by $Z^{(k)}$, we have

$$\begin{aligned} \mathbb {E}(Z^{(k+1)}|\tilde{{\mathcal {C}}}^k) \le Z^{(k)} \sup _{i\in I} \zeta _i. \end{aligned}$$

One can then conclude by recursion that

$$\begin{aligned} \mathbb {E}(Z^{(k)}) \le (\sup _{i\in I} \zeta _i)^{k} <1. \end{aligned}$$

The last inequality uses the sparsity neighborhood assumption. Then, we deduce that the mean number of children in each generation goes to 0 as k tends to infinity. So using classical branching techniques in [15], we conclude that the tree $\tilde{{\mathcal {C}}}$ will go extinct almost surely. This also implies that the backward steps end a.s. $\square$