Poisson process: Derivation from simple assumptions

 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 $t\in [0,\mathrm{\infty })$. 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,

$$\underset{\delta t\to 0}{lim}\frac{o(\delta t)}{\delta t}=0.$$

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 $n\ge 1$,

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

By rearranging these, we have

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

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

$$\begin{array}{rcl}\frac{d{p}_0(t)}{dt}& =& -\lambda p_0(t),\\ \frac{d{p}_n(t)}{dt}& =& \lambda p_{n-1}(t)-\lambda p_n(t)(n\ge 1).\end{array}$$

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.

Related videos