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 \( \Delta t \), \( p(1,\Delta t) \), is \( \lambda \Delta t \) for small \( \Delta t \) and \( \lambda \in \mathbb{R}^+ \).
2. The probability of \( \gt 1 \) success happening in an interval \( \Delta t \) is negligible so that
\( p(0,\Delta t) + p(1, \Delta t) = 1 \)
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}