We can derive the poisson process for which the number of successes in a finite interval is a poisson distribution.
Assumptions.
1. The probability of success happening in a time interval Δt, p(1,Δt), is λΔt for small Δt and λ∈R+.
2. The probability of >1 success happening in an interval Δt is negligible so that
p(0,Δt)+p(1,Δt)=1
3. The number of successes in one interval is independent of the number of successes in another.
Derivation.
The probability of no ‘successes’ until time t, denoted by P(t), is equal to the probability that no ‘successes’ occur until t – \Delta t followed by no successes in the interval \Delta t
\begin{align}
P(t) &= P(t – \Delta t) p(0, \Delta t) \\
&= P(t – \Delta t) ( 1 – \lambda \Delta t)
\end{align}
so
\begin{align}
\frac{P(t) – P(t – \Delta t)}{\Delta t} = – \lambda P(t -\Delta t)
\end{align}
Let \Delta t \rightarrow 0 gives the differential equation
\begin{align}
\frac{d P}{dt} = -\lambda P
\end{align}
The solution to this is
P(t) = P(0) e^{-\lambda t} and P(0) = 1 since P(t) is the probability of no successes until time t and P(0) = \lim_{t \rightarrow 0}p(0, \Delta t) = 1
Consider the probability k ‘successes’ occur.
P(k,t+\Delta t) = P(k,t)p(0,\Delta t) + P(k-1,t)p(1,\Delta t)
\begin{align}
P(k,t + \Delta t) = P(k,t)(1- \lambda \Delta t) + P(k-1,t) \lambda \Delta t
\end{align}
which implies
\begin{align}
\frac{P(k,t+\Delta t) -P(k,t)}{\Delta t} = – \lambda P(k,t) + \lambda P(k-1,t)
\end{align}
Let \Delta t \rightarrow 0 yields
\begin{align}
\frac{dP(k)}{dt} = – \lambda P(k,t) + \lambda P(k-1,t)
\end{align}
multiplying by the integrating factor e^{\lambda t} and integrating
\begin{align}
e^{\lambda t}P(k) = \int_0^t \lambda e^{\lambda s} P(k-1) ds + C
\end{align}
C = 0 since P(k,0) = 0
from earlier we know P(0,s) = e^{-\lambda s} so
\begin{align}
e^{\lambda t}P(1) = \int_0^t \lambda ds = \lambda t
\end{align}
and
\begin{align}
e^{\lambda t}P(2) = \int_0^t \lambda^2 s ds = \frac{(\lambda t)^2}{2}
\end{align}
and in general
\begin{align}
e^{\lambda t}P(k) = \int_0^t \lambda^k \frac{s^{k-1}}{(k-1)!} ds = \frac{(\lambda t)^k}{k!}
\end{align}
so that we get the final result
\begin{align}
P(k) = e^{-\lambda t}\frac{(\lambda t)^k}{k!}
\end{align}