Poisson process: Generating function

 In a previous post, we solved the differential-difference equations for the Poisson process by using the iterative method. In this post, we use the probability generating function (PGF) to solve the same problem.

Let us restate the problem for the sake of completeness.

We consider a stochastic process $\{N(t)\}$ that assumes random non-negative integer values $N(t)=0,1,\cdots$ with continuous time $t\ge 0$. In a previous post, we derived the following differential-difference equations from a few simple assumptions:

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

We would like to solve these equations with the initial condition $$p_0(0)=1.$$ This implies that $p_n(0)=0$ for $n\ge 1$. We can summarize these initial conditions as $$\begin{array}{}\text{(init)}& p_n(0)={\delta }_{0,n},n\ge 0\end{array}$$ where ${\delta }_{0,n}$ is Kronecker's delta.

In general, the PGF is defined as $$\begin{array}{}\text{(Eq:pgf)}& G(s,t)=\sum _{n=0}^{\mathrm{\infty }}{p}_n(t)s^n\end{array}$$ where $s$ is the dummy variable. Note that this PGF also depends on time $t$ (because $p_n(t)$ depends on $t$), so it is a two-variable function. 

Now, multiply both sides of Eq. (DDn) by $s^n$, and sum over $n=0,1,2,\cdots$. We have $$\sum _{n=0}^{\mathrm{\infty }}\frac{d{p}_n(t)}{dt}{s}^n=\lambda \sum _{n=1}^{\mathrm{\infty }}{p}_{n-1}(t)s^n-\lambda \sum _{n=0}^{\mathrm{\infty }}{p}_n(t)s^n.$$

Let's rewrite this equation using the PGF (Eq:pdf). The left-hand side: $$\sum _{n=0}^{\mathrm{\infty }}\frac{d{p}_n(t)}{dt}{s}^n=\frac{\partial }{\partial t}\sum _{n=0}^{\mathrm{\infty }}{p}_n(t)s^n=\frac{\partial G(s,t)}{\partial t}.$$

The first term on the right-hand side:$$\lambda \sum _{n=1}^{\mathrm{\infty }}{p}_{n-1}(t)s^n=\lambda \sum _{m=0}^{\mathrm{\infty }}{p}_m(t)s^{m+1}=sG(s,t).$$

The second term on the right-hand side:$$\lambda \sum _{n=0}^{\mathrm{\infty }}{p}_n(t)s^n=G(s,t).$$

After all, we have the following partial differential equation (PDE):

$$\frac{\partial G(s,t)}{\partial t}=\lambda sG(s,t)-\lambda G(s,t)=\lambda (s-1)G(s,t).$$

Although this is a PDE, it only involves a derivative with respect to $t$. Thus, it can be readily integrated to give 

$$\begin{array}{}\text{(Eq:solA)}& G(s,t)=A(s)e^{\lambda (s-1)t}\end{array}$$

 where $A(s)$ is a function of $s$ alone. With the initial condition Eq. (init) and the definition of the PGF (Eq:pdf), we have 

$$G(s,0)=\sum _{n=0}^{\mathrm{\infty }}{p}_n(0)s^n=1.$$ On the other hand, we have from (Eq:solA) 

$$G(s,0)=A(s).$$

Comparing these, we have $A(s)=1$, and the PGF is $$\begin{array}{rcl}G(s,t)& =& e^{\lambda (s-1)t}=e^{-\lambda t}{e}^{\lambda st}\\ & =& e^{-\lambda t}\sum _{n=0}^{\mathrm{\infty }}\frac{(\lambda st{)}^n}{n!}\\ & =& \sum _{n=0}^{\mathrm{\infty }}\frac{(\lambda t{)}^n{e}^{-\lambda t}}{n!}{s}^n\\ & =& \sum _{n=0}^{\mathrm{\infty }}{p}_n(t)s^n.\end{array}$$

Comparing the coefficients of $s^n$, we have $$p_n(t)=\frac{(\lambda t{)}^n{e}^{-\lambda t}}{n!},(n=0,1,\cdots )$$ as expected.

Related videos