A *Markov process* is a stochastic process such that given the present state of the system, the future behavior is independent of the past history.

Suppose that the size of the population under consideration at time $t$ is represented by a discrete random variable $X(t)$ with probability $$P\{X(t)=n\}=p_n(t),\ n=0,1,2,\cdots$$ At certain instants of time, there will be discrete changes in the population size, due to the loss or death of an individual, or to the appearance or birth of new individuals.

We now consider an example of a stochastic process, which will give rise to a simple model of a Markov process called the *Poisson process*. Suppose that we are measuring the radiation of a certain radioactive substance. Let $X(t)$ be the total number of particles recorded up to time $t$. Also suppose that the chance of a new particle being recorded is any short time interval is independent of not only the previous states but also the present state. Then the chance of a new addition to the total count during a very short time interval $\Delta t$ can be written as $\lambda\Delta t+o(\Delta t)$ where $\lambda$ is a constant and $o(\Delta t)$ is the chance of two or more simultaneous emissions. The chance of no change in the total count would be $1-\Delta t-o(\Delta t)$. Thus, we obtain \begin{equation}\label{eq:poisson}p_n(t+\Delta t)=p_{n-1}(t)\lambda\Delta t+p_n(t)(1-\lambda\Delta t)\end{equation} where terms that are small compared with $\Delta t$ have been disregarded. \eqref{eq:poisson} results in the following differential-difference equation \begin{equation}\begin{aligned}\frac{dp_n(t)}{dt}&=\lim_{\Delta t\to 0}\frac{p_n(t+\Delta t)-p_n(t)}{\Delta t}\\&=\lambda\{p_{n-1}(t)-p_n(t)\},\ n>0\end{aligned}\label{eq:poisson2}\end{equation}

Note that we have $n=0$ at time $t+\Delta t$ only if $n=0$ at time $t$ and no new particles are emitted in $\Delta t$. This means that we have $$p_0(t+\Delta t)=p_0(t)(1-\lambda\Delta t)$$ and subsequently the differential equation $$\frac{dp_0}{dt}=-\lambda p_0(t)$$ whose solution is $$p_0(t)=e^{-\lambda t}$$ with the initial condition $p_0(0)=1$. This initial condition can be obtained by setting the Geiger counter to $0$ in the beginning of the process. Starting off with $p_0(t)$, iterative applications of \eqref{eq:poisson2} will result in $$p_n(t)=\frac{(\lambda t)^ne^{-\lambda t}}{n!},\ n=0,1,2,\cdots$$ which is a Poisson distribution with parameter $\lambda t$.

*References*:

- Norman T. J. Bailey, The Elements of Stochastic Processes with Applications to the Natural Sciences, John Wiley & Sons, Inc., 1964.