Poisson process: Derivation from simple assumptions

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 In a previous post, we have seen that the Poisson process satisfies a set of differential-difference equations. By reversing that argument, we can "derive" the Poisson process from a few simple assumptions. 

Let's pretend we don't know the Poisson process. We want to model the stochastic process \(\{N(t)\}\) that assumes non-negative integer values \(N(t) = 0, 1, 2, \cdots\) and depends on non-negative continuous time \(t \in [0, \infty)\). Our first assumption is the following:

(1) The probability that \(N(t)\) increases by 1 within the time interval of \(\delta t\) is approximately proportional to \(\delta t\). In other words,

\[\Pr(N(t + \delta t) = n + 1 | N(t) = n) = \lambda \delta t + o(\delta t)\]

where \(\lambda\) is a positive constant. "\(o(\delta t\)" in the "little o" notation that represents any term that converges to 0 faster than \(\delta t\) when \(\delta t \to 0\). That is,

\[\lim_{\delta t\to 0}\frac{o(\delta t)}{\delta t} = 0.\]

Our next assumption is:

(2) For a sufficiently small \(\delta t\), \(N(t+\delta t) - N(t)\) is no more than 1. That is, \(N(t)\) increases by 1, or it doesn't increase at all within \(\delta t\). Accordingly, we have

\[\Pr(N(t + \delta t) = n | N(t) = n) = 1 - \Pr(N(t+\delta t) = n+1| N(t) = n) = 1 - \lambda \delta t + o(\delta t).\]

Note that we are implicitly assuming that \(\{N(t)\}\) is a Markov process.

Let \(p_n(t) = \Pr(N(t) = n)\) denote the probability that \(N(t) = n\) at time \(t\). 

By the above assumptions, we have

\[p_0(t + \delta t) = (1 - \lambda \delta t + o(\delta t))p_0(t),\]

and, for \(n \geq 1\),

\[p_n(t+\delta t) = (\lambda \delta t + o(\delta t))p_{n-1}(t) + (1 - \lambda \delta t + o(\delta t))p_n(t).\]

By rearranging these, we have

\[\begin{eqnarray} \frac{p_0(t + \delta t) - p_0(t)}{\delta t} &=& -(\lambda + o(\delta t)/\delta t)p_0(t),\\ \frac{p_n(t + \delta t) - p_n(t)}{\delta t} &=& (\lambda + o(\delta t)/\delta t)p_{n-1}(t) - (\lambda + o(\delta t)/\delta t)p_n(t). \end{eqnarray}\]

With the limit \(\delta t \to 0\), we have the (familiar) differential-difference equations:

\[\begin{eqnarray} \frac{dp_0(t)}{dt} &=& -\lambda p_0(t),\\ \frac{dp_n(t)}{dt} &=& \lambda p_{n-1}(t) - \lambda p_n(t) ~~ (n \geq 1). \end{eqnarray}\]

By solving these equations with the initial condition \(p_0(0) = 1\), we can show that

\[p_n(t) = \frac{(\lambda t)^n e^{-\lambda t}}{n!}.\]

That is, \(N(t)\) is a Poisson process. We will see how to solve the above differential-difference equations in the following posts.

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