Poisson process: Solving differential-difference equations by an iterative method

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:

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

Now, let's solve these equations with the initial condition $$\begin{array}{}\text{(Init)}& p_0(0)=1.\end{array}$$ This initial condition means that $N(0)=0$ almost surely. It also means that $$p_n(0)=0,(n\ge 1).$$

We first solve them by the iterative method.

For $n=0$, Eq. (DD0) is easy to solve. Together with the initial condition Eq. (Init), we immediately have

$$\begin{array}{}\text{(Sol0)}& p_0(t)=e^{-\lambda t}.\end{array}$$

For $n=1$, we have

$$\frac{\mathrm{d}{p}_1(t)}{\mathrm{d}t}=\lambda p_0(t)-\lambda p_1(t),$$

or, using Eq. (Sol0),

$$\frac{\mathrm{d}{p}_1(t)}{\mathrm{d}t}+\lambda p_1(t)=\lambda e^{-\lambda t}.$$

This is an inhomogeneous linear differential equation and can be solved by using, for example, the method of variation of parameters. But here, we employ a more heuristic approach. Multiply both sides of the above equation by $e^{\lambda t}$, and we have

$$e^{\lambda t}\frac{\mathrm{d}{p}_1(t)}{\mathrm{d}t}+\lambda e^{\lambda t}{p}_1(t)=\lambda ,$$

which can be rearranged into

$$\frac{\mathrm{d}}{\mathrm{d}t}[e^{\lambda t}{p}_1(t)]=\lambda .$$

Solving this, we have $$e^{\lambda t}{p}_1(t)=\lambda t+C$$ where $C$ is a constant. With the initial condition $p_1(0)=0$, we have $$\begin{array}{}\text{(Sol1)}& p_1(t)=\lambda t{e}^{-\lambda t}.\end{array}$$

Similarly, for $n=2$, Eq. (DDn) can written as $$\frac{\mathrm{d}{p}_2(t)}{\mathrm{d}t}+\lambda p_2(t)=\lambda p_1(t)={\lambda }^{2}t{e}^{-\lambda t}.$$

Using the same technique as with $n=1$, we have 

$$\frac{\mathrm{d}}{\mathrm{d}t}[e^{\lambda t}{p}_2(t)]={\lambda }^{2}t.$$

Solving this with the initial condition $p_2(0)=0$, we have

$$\begin{array}{}\text{(Sol2)}& p_2(t)=\frac{(\lambda t{)}^2}{2!}{e}^{-\lambda t}.\end{array}$$

Why the factorial (2!) rather than just 2? To answer this, we need to continue the same process for $n=3,4,\cdots$, and then we begin to see some patterns.

For $n=3$, applying the same method, we have

$$p_3(t)=\frac{(\lambda t{)}^3}{3!}{e}^{-\lambda t}.$$

Now, you get the idea. We prove the general solution by mathematical induction. Suppose that we have, for $n=k$,

$$\begin{array}{}\text{(Solk)}& p_k(t)=\frac{(\lambda t{)}^k}{k!}{e}^{-\lambda t}.\end{array}$$

Then, Eq. (DDn) for $n=k+1$ becomes

$$\frac{\mathrm{d}{p}_{k+1}(t)}{\mathrm{d}t}+\lambda p_{k+1}(t)=\lambda \frac{(\lambda t{)}^k}{k!}{e}^{-\lambda t},$$

which can be rearranged into

$$\frac{d}{dt}[e^{\lambda t}{p}_{k+1}(t)]={\lambda }^{k+1}\frac{t^k}{k!}.$$

Solving this with the initial condition $p_{k+1}(0)=0$, we have

$$p_{k+1}(t)=\frac{(\lambda t{)}^{k+1}}{(k+1)!}{e}^{-\lambda t}.$$

Therefore, the general solution is given by

$$p_n(t)=\frac{(\lambda t{)}^n{e}^{-\lambda t}}{n!},(n\ge 0).$$

This is indeed the Poisson distribution (with parameter $\lambda t$).

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