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Limit of an infinite chain of first-order exponential smoothers

First-order exponential smoother

The simplest model how the voltage $x$ at a synapse responds to input $u$ is a first-order filter:

$$\tau \dot x = -x + u.$$

This corresponds to convolving signal $u(t)$ with exponential filter $\operatorname H(t) \exp(-t/\tau)$, where $\operatorname H(\cdot)$ is the Heaviside step function:

$$\begin{aligned}x(t) &= h(t) * u(t)\\h(t)&=\operatorname H(t) \exp(-t/\tau).\end{aligned}$$

The alpha function

A first-order filter has a discontinuous jump in response to an abrupt inputs (like spikes). A more realistic response is the "alpha function"  $t\cdot \exp(-t)$. The alpha function can be obtained by convolving two first decay functions (i.e. chaining together two first-order filters):

$$\begin{aligned}\tau \dot x_1 &= -x_1 + u\\\tau \dot x_2 &= -x_2 + x_1.\end{aligned}$$

This is sometimes written in the compact notation

$$\left(\tau \frac{d }{dt} + 1\right)^2 x = u.$$

Higher orders

You can repeat this operation many times, obtaining responses with increasing smoothness. The family $t^n\cdot \exp(-t)$ reflects $n+1$ feed-forward variables coupled by exponential decay $\dot x_n=x_{n-1}-x_n$. The integral of $t^n \exp(-t)$ grows with $n$. To normalize, divide by $n! = \Gamma(n+1)$: 

$$h(t) = \operatorname H(t)  \frac {t^n}{\Gamma(n+1)}e^{-t}.$$

The response of $t^n\exp(-t)$ peaks at time $t=n$. Rescale time with $t\gets nt$ to get a peak response at $t=1$. To keep the integral of the response normalized when rescaling, multiply by $n$.

$$h(t) = \operatorname H(t) n \frac {(nt)^n}{\Gamma(n+1)}e^{-nt}$$

This is equivalent to choosing a time constant $\tau=1/n$ for each of the $n$ filtering stages. To place the peak response at time $t_0$, set $\tau = t_0/n$. 

This corresponds to a gamma distribution with $k=n+1$ and $\theta=1/n$. For large $n$ this approximates a Gaussian with $\mu=\frac {n+1}{n}$ and $\sigma^2=\frac{n+1}{n^2}$. As $n\to\infty$ this converges to a Dirac delta (impulse) distribution. 

To stabilize the variance instead of time-to-peak, rescale time by $1/\sqrt{n+1}$. This corresponds to a gamma distribution with $k=n+1$ and $\theta=1/\sqrt{k}$. The time-to-peak in this case diverges as $n\to\infty$.