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

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

\[\begin{eqnarray} \frac{dp_0(t)}{dt} &=& -\lambda p_0(t),\tag{DD0}\\ \frac{dp_n(t)}{dt} &=& \lambda p_{n-1}(t) - \lambda p_n(t), ~~ (n \geq 1).\tag{DDn} \end{eqnarray}\]

Now, let's solve these equations with the initial condition \[p_0(0) = 1. \tag{Init}\] This initial condition means that \(N(0) = 0\) almost surely. It also means that \[p_n(0) = 0, ~~ (n \geq 1).\]

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

\[p_0(t) = e^{-\lambda t}. \tag{Sol0}\]

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 \[p_1(t) = \lambda t e^{-\lambda t}.\tag{Sol1}\]

Similarly, for \(n = 2\), Eq. (DDn) can written as \[ \frac{\mathrm{d}p_2(t)}{\mathrm{d}t} + \lambda p_{2}(t) = \lambda p_1(t) = \lambda^2 te^{-\lambda t}.\]

Using the same technique as with \(n=1\), we have 

\[\frac{\mathrm{d}}{\mathrm{d}t}[e^{\lambda t}p_2(t)] = \lambda^2t.\]

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

\[p_2(t) = \frac{(\lambda t)^2}{2!}e^{-\lambda t}.\tag{Sol2}\]

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 \(n = k\),

\[p_k(t) = \frac{(\lambda t)^k}{k!}e^{-\lambda t}.\tag{Solk}\]

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

\[p_n(t) = \frac{(\lambda t)^n e^{-\lambda t}}{n!}, ~~ (n \geq 0).\]

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

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