about its mean so $${\mathbb{E}}[f(x,y;{\langle}r {\rangle}_m, {\langle}s {\rangle}_m )] \sim f(x,y;{\mathbb{E}}[{\langle}r {\rangle}_m],{\mathbb{E}}[ {\langle}s {\rangle}_m] )$$ where the expectation is taken over interaction times. Defining $\bar{r}_m := {\mathbb{E}}[{\langle}r {\rangle}_m]$ then: $$\begin{aligned}
\label{pfunc1}
\tilde{p}_\omega[r,s](t) & \sim \sum_{x=0}^m \sum_{y=0}^m f(x,y;\bar{r}_m,\bar{s}_m) I_\omega(x,y) \\
&:= p_\omega(\bar{r}_m, \bar{s}_m)
\label{pfunc2}\end{aligned}$$ as $m \rightarrow \infty$. Notice that the quantity $p_\omega(\bar{r}_m, \bar{s}_m)$ is a functional only because the time averages $\bar{r}_m, \bar{s}_m$ are functionals, whereas $p_\omega(\cdot,\cdot)$ is an ordinary function.

Delay equation
--------------

During a short finite time interval $[t,t+\delta t]$, given the history $\{r(u), s(u) | u \leq t\}$, from the learning rule (\[EMA\]) the expected changes in weights will be $$\begin{aligned}
{\mathbb{E}}[\delta r] &= \epsilon [p_R(\bar{r}_m, \bar{s}_m) -r(t)] \delta t \\
{\mathbb{E}}[\delta s] &= \epsilon [p_S(\bar{r}_m, \bar{s}_m) -s(t)] \delta t.\end{aligned}$$ As $L \rightarrow \infty$, these changes become deterministic and the weights obey the following coupled delay equations $$\begin{aligned}
\label{rDelay}
\frac{dr}{dt} &= \epsilon[p_R(\bar{r}_m, \bar{s}_m) - r] \\
\frac{ds}{dt} &= \epsilon[p_S(\bar{r}_m, \bar{s}_m) - s].
\label{sDelay}\end{aligned}$$ Solutions to these equations are shown in Figures \[stable\] and \[hopf\], where we see that they accurately capture the simulated evolution of a system with the same parameter values.

Linear Stability Analysis for Symmetric Case
--------------------------------------------

In order to understand the conditions under which 