Small Leakage

We now relax the constraint of zero leakage and suppose that $\gamma$ is non-vanishing. To facilitate the analysis, we consider the case for which $\gamma \ll \sigma$. With this assumption, the $n^{th}$ population fraction, $\rho _{n}(v,t)$, although no longer a delta function, is still tightly concentrated about $v_{n}=n\epsilon$. Furthermore, we shall presume that any leakage backwards through the threshold is insignificant (otherwise, we'd have to take care to impose the boundary condition $\rho (1,t)=0$). In other words, the likelihood of a neuron that has fired leaking backwards through threshold is small since it receives input events so rapidly.

To generate the interspike interval distribution, we once again ignore the reset flux in equation (4). The population equation is even more tractable if we approximate $v\approx n\epsilon$ in the equation for $\rho _{n}(v,t),$ which is reasonable if $\gamma \ll \sigma$. We then solve the equations

$$\frac{\partial \rho _{n}}{\partial t}=\gamma \epsilon n\frac{\par... ...{n}}{\partial v}+\sigma \left[\rho _{n-1}(v-\epsilon ,t)-\rho _{n}(v,t)\right].$$ (18)

The distribution $\rho _{0}(v,t)$, which is the fraction of the population that has received zero input events, may be solved for without the above approximation:

$$\frac{\partial \rho _{0}}{\partial t} = -\sigma \rho _{0}$$

$$\rho _{0}(v,0) = \delta (v)\, .$$

The solution is

$$\rho _{0}(v,t)=\delta (v)e^{-\sigma t}.$$ (19)

We solve the remaining equations using a Fourier transform in both space and time,

$$\rho _{n}(v,t) = \frac{1}{2\pi }\int _{-\infty }^{\infty }dk\int _{-\infty }^{\infty }d\omega \, \tilde{\rho }_{n}(k,\omega )e^{i(kx+\omega t)}$$ (20)

$$\tilde{\rho }_{n}(k,\omega ) = \frac{1}{2\pi }\int _{-\infty }^{\infty }dx\int _{-\infty }^{\infty }dt\, \rho _{n}(x,t)e^{-i(kx+\omega t)}.$$ (21)

In Fourier space (18) becomes

$$i\omega \tilde{\rho }_{n}=i\gamma \epsilon nk\tilde{\rho }_{n}+\sigma e^{-ik\epsilon }\tilde{\rho }_{n-1}-\sigma \tilde{\rho }_{n}\, ,$$

which gives

$$\tilde{\rho }_{n}=\frac{e^{-ik\epsilon }}{1+\frac{i}{\sigma }(\omega -\gamma \epsilon nk)}\tilde{\rho }_{n-1}.$$ (22)

Using (22) and the fact that

$$\tilde{\rho }_{0}(k,\omega )=\frac{1}{2\pi \sigma \left(1+\frac{i}{\sigma }\omega \right)}\, ,$$

we find

$$\tilde{\rho }_{n}(k,\omega )=\frac{(-i\sigma )^{n+1}e^{-ik\epsilo... ...}{2\pi \sigma }\prod _{m=0}^{n}\frac{1}{\omega -(i\sigma +m\gamma \epsilon k)}.$$ (23)

Our task now is to invert $\tilde{\rho }_{n}(k,\omega )$. There are a total of $n+1$ simple poles associated with the $\omega$-integration. We will see that there are no poles with respect to $k$.

We use the residue theorem to effect the $\omega$-integration. First define

$$\widehat{\rho }_{n}(k,t)=\frac{1}{\sqrt{2\pi }}\int _{-\infty }^{\infty }d\omega \, e^{i\omega t}\tilde{\rho }_{n}(k,\omega )\, ,$$ (24)

so that

$$\widehat{\rho }_{n}(k,t)=\frac{(-i\sigma )^{n+1}}{2\pi \sigma \sq... ...{\infty }d\omega \, e^{i\omega t}\prod _{m=0}^{n}\frac{1}{\omega -\omega _{m}},$$ (25)

where we have defined the poles

$$\omega _{m}=i\sigma +m\gamma \epsilon k\, .$$ (26)

The residue calculation involves the integral

$$I_{n}(t)\equiv \int _{-\infty }^{\infty }d\omega \, e^{i\omega t}... ... -\omega _{m}}=2\pi i\sum _{l=0}^{n}Res\left(g_{l}(\omega );\omega _{l}\right),$$ (27)

where

$$g_{l}(\omega )=e^{i\omega t}\prod _{\begin{array}{c} p=0\\ p\neq l\end{array} }^{n}\frac{1}{\omega -\omega _{p}}\, .$$ (28)

The residues are given by

$$Res\left(g_{l}(\omega );\omega _{l}\right)=(-1)^{n-l}e^{i\omega _... ...{array} }^{n}\left\vert\frac{1}{\omega -\omega _{p}}\right\vert\right)^{-1}\, .$$ (29)

Further simplifications are possible using the binomial theorem. Putting (29) together with (25) yields

$$\widehat{\rho }_{n}(k,t)=\frac{e^{-\sigma t}}{n!\sqrt{2\pi }}\lef... ...ma k}\right)^{n}\sum _{l=0}^{n}(-1)^{l}{{n \choose l}}e^{il\epsilon \gamma kt},$$ (30)

where we have used the fact that

$$\prod _{\begin{array}{c} p=0\\ p\neq l\end{array}}^{n}\left\vert\frac{1}{\omega -\omega _{p}}\right\vert=l!(n-l)!\, .$$

Next, we apply the binomial theorem

$$(x-y)^{n}=(-1)^{n}\sum _{l=0}^{n}(-1)^{l}{{n \choose l}}x^{l}y^{n-l}$$

to arrive at the simplification

$$(-1)^{n}\left(e^{i\epsilon \gamma kt}-1\right)^{n}=\sum _{l=0}^{n}(-1)^{l}{{n \choose l}}e^{il\epsilon \gamma kt}\, .$$

Finally we have

$$\widehat{\rho }_{n}(k,t)=\frac{(-i\sigma )^{n}t^{n}e^{-\sigma t}}... ...psilon n}\left(\frac{e^{i\epsilon \gamma kt}-1}{\epsilon \gamma kt}\right)^{n}.$$ (31)

We then note that

$$\left(\frac{e^{i\epsilon \gamma kt}-1}{\epsilon \gamma kt}\right)... ...a }\right)^{n}t^{-n}\left(\int _{0}^{\epsilon \gamma t}e^{ikq}\, dq\right)^{n},$$

which gives a cleaner expression for $\widehat{\rho }_{n}(k,t)$,

$$\widehat{\rho }_{n}(k,t)=\frac{\left(\frac{\sigma }{\epsilon \gam... ... t}e^{-ik\epsilon n}\left(\int _{0}^{\epsilon \gamma t}e^{ikq}\, dq\right)^{n}.$$

A final Fourier inversion integral gives $\rho _{n}(v,t)$:

$$\rho _{n}(v,t)=\frac{\left(\frac{\sigma }{\epsilon \gamma }\right... ...i }\int _{-\infty }^{\infty }dk\, e^{ik(v-\epsilon n+\sum _{j=1}^{n}q_{j})}\, ,$$

which after the integration with respect to $k$ is

$$\rho _{n}(x,t)=\frac{\left(\frac{\sigma }{\epsilon \gamma }\right... ...ilon \gamma t}dq_{n}\, \delta \left(q_{1}+\ldots +q_{n}+v-\epsilon n\right)\, .$$ (32)

This last expression can be written in a more illuminating form as repeated convolutions of the boxcar function, $B(\xi )$, which here we define to be

$$B(\xi )=\left\{ \begin{array}{ccc} 1 & & 0<\xi <1\\ 0 & & \xi \notin (0,1)\end{array} \right..$$ (33)

We simplify the bounds of integration using the definition

$$q_{j}=\epsilon \gamma tQ_{j}\, ,$$

which yields

$$\rho _{n}(v,t)=\frac{(\sigma t)^{n}}{n!}e^{-\sigma t}\int _{0}^{1... ...t[\sum _{j=1}^{n}Q_{j}-\frac{\epsilon n-v}{\epsilon \gamma t}\right]\right)\, .$$

We now use the following definition and observation,

$$P_{n}(t) \equiv \frac{(\sigma t)^{n}}{n!}e^{-\sigma t}$$

$$\int _{0}^{1}dx\, \delta (kx-a) = \frac{1}{k}B\left(\frac{a}{k}\right)\, ,$$

and thus

$$\rho _{n}(v,t)=\frac{P_{n}(t)}{\epsilon \gamma t}\int _{0}^{1}dQ_... ...B\left(\frac{\epsilon n-v}{\epsilon \gamma t}-\sum _{j=1}^{n-1}Q_{j}\right)\, .$$

Using the recursion definition

$$B*B_{n-1}\equiv B_{n}(\xi )=\int _{-\infty }^{\infty }dy\, B(y)B_{n-1}(\xi -y)\, ,$$

we see that

$$\rho _{n}(v,t) = \frac{P_{n}(t)}{\epsilon \gamma t}\int _{0}^{1}dQ_{1}\dots \int _... ...ght)B\left(\frac{\epsilon n-v}{\epsilon \gamma t}-\sum _{j=1}^{n-1}Q_{j}\right)$$

$$\equiv \frac{P_{n}(t)}{\epsilon \gamma t}\int _{0}^{1}dQ_{1}\dots \int _... ...B_{2}\left(\frac{\epsilon n-v}{\epsilon \gamma t}-\sum _{j=1}^{n-2}Q_{j}\right)$$

$$= \dots$$

$$= \frac{P_{n}(t)}{\epsilon \gamma t}\int _{0}^{1}dQ_{1}\, B_{n-1}\left(\frac{\epsilon n-v}{\epsilon \gamma t}-Q_{1}\right)$$

$$= \frac{P_{n}(t)}{\epsilon \gamma t}\int _{-\infty }^{\infty }dQ_{1}\, B(Q_{1})B_{n-1}\left(\frac{\epsilon n-v}{\epsilon \gamma t}-Q_{1}\right)\, .$$

This shows that $\rho _{n}(v,t)$ involves the $n$-fold convolution of the boxcar function:

$$\rho _{n}(v,t)=\frac{P_{n}(t)}{\epsilon \gamma t}B*B_{n-1}\left(\frac{\epsilon n-v}{\epsilon \gamma t}\right).$$ (34)

For $n=1$ we get the moving boxcar solution

$$\rho _{1}(v,t)=\frac{P_{n}(t)}{\epsilon \gamma t}B\left(\frac{\ep... ...psilon [1-\gamma t],0)<v<\epsilon \\ & & \\ 0 & & otherwise\end{array} \right..$$ (35)

It turns out that $\rho _{n}(v,t)$ is non-vanishing in the interval $v\in (\max (\epsilon [n-\gamma t],0),n\epsilon )$. As $n\rightarrow \infty$, the convolution approaches a Gaussian.

Assuming that $N$ input events puts us just below threshold, and the $(N+1)^{th}$ event takes a cell past threshold, then the interspike interval distribution is determined, up to the order of the small-$\gamma$ approximation, by

$$p(\tau )=\sigma \int _{1-\epsilon }^{1}dv\, \sum _{j=1}^{N}\rho _... ...}dv\, \sum _{j=1}^{N}B*B\left(\frac{\epsilon n-v}{\epsilon \gamma t}\right)\, .$$ (36)