Poisson process: Differential-Difference equations

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:

$$\begin{array}{}\text{(Eq:PoissonProc)}& p_n(t)=Pr(N(t)=n)=\frac{(\lambda t{)}^n{e}^{-\lambda t}}{n!},(t\ge 0,n=0,1,2,\cdots )\end{array}$$

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 $$\begin{array}{}\text{(Eq:DD0)}& \frac{d{p}_0(t)}{dt}=-\lambda p_0(t)\end{array}$$ and $$\begin{array}{}\text{(Eq:DDn)}& \frac{d{p}_n(t)}{dt}=\lambda p_{n-1}(t)-\lambda p_n(t)(n\ge 1).\end{array}$$

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 $d{p}_n(t)/dt$, 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:

$$\frac{d{p}_n(t)}{dt}=\underset{\delta t\to 0}{lim}\frac{p_n(t+\delta t)-p_n(t)}{\delta t}.$$

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

$$p_0(t+\delta t)=(1-\lambda \delta t)p_0(t)$$ and $$p_n(t+\delta )=(\lambda \delta t)p_{n-1}(t)+(1-\lambda \delta t)p_n(t)$$

Now, let us define the following conditional probabilities:

$$\begin{array}{rcl}Pr(N(t+\delta t)=n+1|N(t)=n)& =& \lambda \delta t,\\ Pr(N(t+\delta t)=n|N(t)=n)& =& 1-\lambda \delta t.\end{array}$$

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 $\lambda \delta t\le 1$ so that it can be interpreted as a probability. Using these conditional probabilities, the discretized differential-difference equations are expressed as

$$\begin{array}{rcl}\text{(Eq:DD0')}& Pr(N(t+\delta t)=0)& =& Pr(N(t+\delta t)=0|N(t)=0)Pr(N(t)=0),Pr(N(t+\delta )=n)& =& Pr(N(t+\delta t)=n|N(t)=n-1)Pr(N(t)=n-1)\\ \text{(Eq:DDn')}& & & +Pr(N(t+\delta t)=n|N(t)=n)Pr(N(t)=n).\end{array}$$

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: $\lambda \delta t$.