# Homogeneous Poisson Process of spike train

In Theoretical Neuroscience by Dayan and Abbott page 25,

the probability $$P[t_1, t_2, ..., t_n]$$ that a sequence of n spikes occurs with spike $$i$$ falling between times $$t_i$$ and $$t_i+\Delta t$$ for $$i=1,2,...,n$$ is given in terms of this by the relation $$P[t_1, t_2, ..., t_n]=p[t_1,...,t_n](\Delta t)^n$$... the probability $$P[t_1, t_2, ..., t_n]$$ can be expressed in terms of another probability function $$P_T[n]$$, which is the probability that an arbirary sequence of exactly n spikes occurs within a trial of duration T. Assuming that the spike times are ordered so that $$0 \le t_1 \le t_2 \le ... \le t_n \le T$$, the relationship is $$P[t_1, t_2, ..., t_n] = n!P_T[n]\big({\Delta t \over T}\big)^n$$

I get that $$P_T[n]$$ is the probability function of $$n$$ spikes occurring(which later is proved to be a Poisson distribution). I'm guessing $$n!$$ is different ways of re-arranging the $$n$$ spikes. What exactly is $$\big({\Delta t \over T}\big)^n$$?

First of all you should notice the assumption that spikes are independent and that every sequence of spikes over an interval has the same probability. That's why you can express $$P[t_{1}, t_{2}, ..., t_{n}]$$ in terms of $$P_{T}[n]$$. Second, note that $$p[t_{1}, t_{2}, ..., t_{n}]$$ is a probability density. So you can express the probability as the integral of $$p[t_{1}, t_{2}, ..., t_{n}] dt_{1}dt_{2}...dt_{n}$$
$$\Delta t / T$$ is the probability that one spike occurs in that time interval. If $$\Delta t$$ is small enough, only one spike is within that interval, so just let $$\Delta t \rightarrow dt_{i}$$. Then $$p[t_{1}, t_{2}, ..., t_{n}] dt_{1}dt{2}...dt{n} = n!P_{T}[n]dt_{1}dt_{2}...dt_{n}(\frac{1}{T})^{n}$$