Poisson process: Differential-Difference equations

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Read Poisson process: Introduction first.

Recall that the Poisson process \(\{N(t)\}\) with parameter \(\lambda\) is a stochastic process characterized by the following probability distribution:
\begin{equation} p_n(t) = \Pr(N(t) = n) = \frac{(\lambda t)^ne^{-\lambda t}}{n!}, ~~ (t \geq 0, n = 0, 1, 2,\cdots) \tag{Eq:PoissonProc}\end{equation}

Differential-Difference Equations

To understand the meaning of a stochastic process, it is often useful to find the differential(-difference) equations satisfied by the probability distribution. 
By differentiating (Eq:PoissonProc) with respect to time \(t\), we have \begin{equation} \frac{dp_0(t)}{dt} = -\lambda p_0(t) \tag{Eq:DD0}\end{equation} and \begin{equation} \frac{dp_n(t)}{dt} = \lambda p_{n-1}(t) - \lambda p_{n}(t) ~ (n \geq 1). \tag{Eq:DDn}\end{equation}
These are the differential-difference equations satisfied by the sequence of probabilities \(\{p_0(t), p_1(t), \cdots \}\). It is "differential" because of the derivatives \(dp_n(t)/dt\), and "difference" because of the sequence of probabilities \(p_0(t), p_1(t), \cdots\).

Exercise. Derive the above differential-difference equations.

So, how do these differential-difference equations help us understand the Poisson process? To see that, let us discretize the equations. Recall the definition of derivatives:
\begin{equation} \frac{dp_n(t)}{dt} = \lim_{\delta t\to 0}\frac{p_n(t + \delta t) - p_n(t)}{\delta t}. \end{equation}
That is, we can approximate the derivative on the left-hand side by the right-hand without the limit. Then (Eq:DD0) and (Eq:DDn) can be approximated as
\begin{equation} p_0(t + \delta t) = (1 - \lambda\delta t)p_0(t)\end{equation} and \begin{equation} p_n(t+\delta) = (\lambda\delta t)p_{n-1}(t) + (1 - \lambda \delta t)p_n(t) \end{equation}
Now, let us define the following conditional probabilities:
$$\begin{eqnarray} \Pr(N(t+\delta t) = n+1| N(t) = n) & = & \lambda \delta t, \\ \Pr(N(t+\delta t) = n | N(t) = n) & = & 1 -\lambda\delta t. \end{eqnarray}$$
That is, given \(N(t) = n\), after the time interval of \(\delta t\), the value of \(N(t+\delta t)\) either increase by 1 (with probability \(\lambda \delta t\)) or stays the same (with probability \(1 - \lambda \delta t\)). We should assume that \(\lambda \delta t \leq 1\) so that it can be interpreted as a probability. Using these conditional probabilities, the discretized differential-difference equations are expressed as
$$\begin{eqnarray} \Pr(N(t + \delta t) = 0) &=& \Pr(N(t+\delta t) = 0 | N(t) = 0)\Pr(N(t)=0),\tag{Eq:DD0'}\\ \Pr(N(t+\delta)=n) &=& \Pr(N(t+\delta t) = n | N(t) = n-1)\Pr(N(t) = n-1) \nonumber\\ &&+ \Pr(N(t+\delta t) = n | N(t) = n)\Pr(N(t)=n). \tag{Eq:DDn'} \end{eqnarray}$$
Note that these are in the form of the partition theorem. From these equations, we can see that the Poisson process is a Markov process because each \(p_n(t+\delta t)\) depends only on the quantities at time \(t\) such as \(p_{n-1}(t)\) and \(p_{n}(t)\), not on those before \(t\) such as  \(p_{m}(t - \delta t), p_{m}(t-2\delta t), p_{m}(t - 3\delta t), \cdots \) for any \(m = 0, 1, \cdots\).
(Eq:DD0') says:
  • \(N(t+\delta t) = 0\) at time \(t + \delta t\) with probability \(p_0(t+\delta t)\) if \(N(t) = 0\) at time \(t\) with probability \(p_0(t)\), and \(N(t)\) stays the same with probability \(1 - \lambda \delta t\).
Similarly, (Eq:DDn') says
  • \(N(t+\delta t) = n\) with probability \(p_n(t + \delta t)\) if either
    • \(N(t) = n-1\) with probability \(p_{n-1}(t)\) and \(N(t)\) increases by 1 with probability \(\lambda \delta t\),
    • or \(N(t) = n\) with probability \(p_n(t)\)and \(N(t)\) stays the same with probability \(1 - \lambda \delta t\). 
Thus, the Poisson process is a process where the value of \(N(t)\) increases by at most 1 within a short time interval \(\delta t\) with probability proportional to the interval: \(\lambda\delta t\).



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