No Leakage

In this section we suppose that the membrane discharge rate, $\gamma$, is negligible compared to the input spike rate $\sigma$. With $\gamma =0$ and with no reset of cells which have spiked, the dynamical population equation is

$$\frac{\partial \rho }{\partial t}=\sigma \left[\rho (v-\epsilon ,t)-\rho (v,t)\right]\, .$$ (7)

In this case, the population density changes solely in response to the randomly-distributed synaptic input events. Because each neuron's voltage is bumped forward by $\epsilon$ upon each synaptic event, the distribution function is non-vanishing only at integer multiples of $\epsilon$. It therefore makes sense to decompose $\rho (v,t)$ into a sum of distributions for neurons that have received no input events, one input event, and so forth, so that

$$\rho (v,t)=\sum _{n=0}^{N}\rho _{n}(v,t)\, ,$$ (8)

where $\rho _{n}(v,t)$ is to be interpreted as the probability density for that fraction of the population which has received exactly $n$ input spikes. We assume that it takes $N+1$ input events to drive a cell past threshold, so that the following inequality holds:

$$N\epsilon <1<(N+1)\epsilon \, .$$ (9)

Owing to the discrete nature of the voltage jumps, and no leakage, we may write

$$\rho _{n}(v,t)=\delta \left(v-n\epsilon \right)P_{n}(t)\, .$$ (10)

Upon inserting (8) and (10) into (7) and equating delta functions, one derives the initial value problem for $P_{n}(t)$:

$$\frac{dP_{n}}{dt} = \sigma \left(P_{n-1}-P_{n}\right)$$ (11)

$$P_{n}(0) = \left\{ \begin{array}{ccc} 1 & & n=0\\ & & \\ 0 & & n\neq 0\end{array}\right.,$$ (12)

where $P_{k}(t)\equiv 0$ for $k<0$.

This system may be solved using the Z-transform. After multiplying the equation involving $\frac{dP_{n}}{dt}$ by $Z^{n}$ we obtain

$$Z^{0}\frac{dP_{0}}{dt} = -Z^{0}\sigma P_{0}$$

$$Z^{1}\frac{dP_{1}}{dt} = Z^{1}\sigma \left(P_{0}-P_{1}\right)$$

$$\ldots = \ldots$$

$$Z^{N}\frac{dP_{N}}{dt} = Z^{N}\sigma \left(P_{N-1}-P_{N}\right),$$

which when summed over all equations yields

$$\frac{d\tilde{P}}{dt} = -\sigma \tilde{P}+\sigma Z\tilde{P}$$ (13)

$$\tilde{P}(Z,t) \equiv \sum _{n=0}^{N}P_{n}(t)Z^{n}\, .$$ (14)

Equation (13), with the initial condition $\tilde{P}(Z,0)=1$, gives

$$\tilde{P}(Z,t)=e^{-\sigma t}e^{\sigma tZ}=e^{-\sigma t}\sum _{n=0}^{\infty }\frac{(\sigma t)^{n}}{n!}Z^{n}\, .$$ (15)

The functions $P_{n}(t)$ are thus determined by the Taylor coefficients of $\tilde{P}(Z,t)$,

$$P_{n}(t)=\frac{(\sigma t)^{n}}{n!}e^{-\sigma t}.$$ (16)

The interspike interval distribution, $p(\tau )$, identical to the firing rate $r(\tau )$ determined by (5), is

$$p(\tau ) = \sigma \int _{1-\epsilon }^{1}dv\, \rho (v,\tau )$$

$$= \sigma \int _{1-\epsilon }^{1}dv\, \sum _{n=0}^{N}\delta (v-n\epsilon )P_{n}(\tau )$$

$$= \sigma P_{N}(\tau )\, ,$$

so we have

$$p(\tau )=\frac{\sigma (\sigma \tau )^{N}}{N!}e^{-\sigma \tau }\, .$$ (17)

This last expression follows from the fact that $\rho _{N}(v,t)=\delta (v-N\epsilon )P_{N}(t)$ is the only portion of the population that contributes to the firing rate at time $t$.