Section 15 Poisson process in infinitesimal time periods

Poisson process in terms of increments in infinitesimal time
Forward equations for the Poisson process

15.1 Definition 3: increments in infinitesimal time

We have seen two definitions of the Poisson process so far: one in terms of increments having a Poisson distribution, and one in therms of holding times having an exponential distribution. Here we will see another definition, by looking at what happens in a very small time period of length $\tau$. Advantages of this approach include that does not require direct assumptions on the distributions involved, so may seem more “natural” or “inevitable”. It also shows links between Markov processes and differential equations. A disadvantage is that it’s normally easier to do calculations with the Poisson or exponential definitions.

Let $(X(t))$ be a stochastic processes counting some arrivals over time. Consider the number of arrivals in a very small time period of length $\tau$; that number of arrivals is $X(t+\tau) - X(t)$. The following (seemingly) weak assumptions seem justified:

It’s quite likely that there will be no arrivals in the very small time period.
It is possible there will be one arrival, and that probability will be roughly proportional to the length $\tau$ of the time period.
It’s extremely unlikely there will be two or more arrivals in such a short time period.

To write this down formally in maths, we will consider $\mathbb P(X(t+\tau) - X(t) = j)$ as the length $\tau$ of the time period tends to zero.

It will be helpful for us to use “little-$o$” notation. Here $o(\tau)$ means a term that is of “lower order” compared to $\tau$. These are terms like $\tau^2$ or $\tau^3$ that are so tiny that we’re safe to ignore them. Formally, $o(\tau)$ stands for a function of $\tau$ that tends to $0$ as $\tau \to 0$ even when divided by $\tau$. That is, \[ f(\tau) = o(\tau) \qquad \iff \qquad \frac{f(\tau)}{\tau} \to 0 \quad \text{as $\tau \to 0$.} \]

Then our suggestions above translate to \[\begin{equation} \mathbb P \big(X(t+\tau) - X(t) = j\big) = \begin{cases} 1 - \lambda \tau + o(\tau) & \text{if $j = 0$,} \\ \lambda\tau + o(\tau) & \text{if $j = 1$,} \\ o(\tau) & \text{if $j \geq 2$.} \end{cases} \tag{15.1} \end{equation}\] as $\tau \to 0$.

It turns out that we have once again defined the Poisson process with rate $\lambda$.

Theorem 15.1 Let $(X(t))$ be a stochastic process with the following properties:

$X(0) = 0$;
infinitesimal increments: $X(t+\tau) - X(t)$ has the structure in (15.1) above, as $\tau \to 0$;
independent increments: $X(t_2) - X(t_1)$ and $X(t_4) - X(t_3)$ are independent for all $t_1 \leq t_2 \leq t_3 \leq t_4$.

Then $(X(t))$ is a Poisson process with rate $\lambda$.

We will prove this in a moment. First let’s see an example of its use.

15.2 Example: sum of two Poisson processes

We mentioned before the sum of two Poisson processes: that if $(X(t))$ is a Poisson process with rate $\lambda$ and $(Y(t))$ is an independent Poisson process with rate $\mu$, then the sum $(Z(t))$ where $Z(t) = X(t) + Y(t)$ is a Poisson process with rate $\lambda + \mu$.

On Problem Sheet 7, Question 3, you proved this from the Poisson increments definition. But it’s perhaps even easier to prove using the infinitesimals definition.

We have \[\begin{align*}\mathbb P\big( Z(t+\tau) - Z(t) = 0 \big) &= \mathbb P\big( X(t+\tau) - X(t) = 0 \big) \, \mathbb P\big( Y(t+\tau) - Y(t) = 0 \big) \\ &=\big(1 - \lambda \tau + o(\tau) \big)\big(1 - \mu \tau + o(\tau) \big) \\ &= 1 - (\lambda + \mu)\tau + o(\tau) , \end{align*}\] since $Z$ does not increase provided neither $X$ nor $Y$ increases.

We also have \[\begin{align*}\mathbb P\big( Z(t+{}&\tau) - Z(t) = 1 \big) \\ &= \mathbb P\big( X(t+\tau) - X(t) = 1 \big) \, \mathbb P\big( Y(t+\tau) - Y(t) = 0 \big) \\ &\qquad {}+ \mathbb P\big( X(t+\tau) - X(t) = 0 \big) \, \mathbb P\big( Y(t+\tau) - Y(t) = 1 \big) \\ &= \big(\lambda \tau + o(\tau) \big)\big(1 - \mu \tau + o(\tau) \big) + \big(1 - \lambda \tau + o(\tau) \big)\big(\mu \tau + o(\tau) \big) \\ &= (\lambda + \mu)\tau + o(\tau) , \end{align*}\] since $Z$ increases by 1 if either $X$ increases by 1 and $Y$ stays fixed or vice versa. We used here that the probability that both $X$ and $Y$ increase is of order $\tau^2 = o(\tau)$.

Finally, since probabilities must add up to 1, we must have \[ \mathbb P\big( Z(t+\tau) - Z(t) \geq 2 \big) = o(\tau) . \]

But what we have written down is precisely the infinitesimals definition of a Poisson process with rate $\lambda + \mu$, so we are done.

15.3 Forward equations and proof of equivalence

If we compare Theorem 15.1 to the definition of the Poisson process in terms of Poisson increments, we see that properties 1 and 3 are the same. We only need to check property 2, that $X(t+s) - X(s)$ is a Poisson distribution with rate $\lambda t$.

Since the increments as defined in (15.1) are time homogeneous, it will suffice to consider $s = 0$. So we need to show that $X(t) \sim \text{Po}(\lambda t)$.

Write $p_j(t) = \mathbb P(X(t) = j)$. Then for $j \geq 1$ we have \[ p_j(t + \tau) = \big(1 - \lambda\tau + o(\tau)\big)p_j(t) + \big(\lambda\tau + o(\tau)\big) p_{j-1}(t) + o(\tau) , \] since we get to $j$ either by staying at $j$ or by moving up from $j-1$, with all other highly unlikely possibilities being absorbed into the $o(\tau)$. In order to deal with increments, it will be convenient to take a $p_{j}(t)$ to the left-hand side and rearrange, to get the increment \[ p_j(t + \tau) - p_j(t) = -\lambda\tau p_j(t) + \lambda\tau p_{j-1}(\tau) + o(\tau) , \] where “constant times $o(\tau)$” is itself $o(\tau)$.

The only way to deal with the $o(\tau)$ term, given its definition, is to divide everything by $\tau$ and send $\tau \to 0$. Dividing by $\tau$ gives \[ \frac{p_j(t + \tau) - p_j(t)}{\tau} = -\lambda p_j(t) + \lambda p_{j-1}(t) + \frac{o(\tau)}{\tau} . \] Sending $\tau$ to 0, the term $o(\tau)/\tau$ tends to 0 and vanishes, while we recognise the limit of the left-hand side as being the derivative $\mathrm d p_j(t) / \mathrm d t = p'_j(t)$. This leaves us with the differential equation \[ p'_j(t) = -\lambda p_j(t) + \lambda p_{j-1}(t) . \] We also have the initial condition $p_j(0) = 0$ for $j \geq 1$, since we start at 0, not at any $j \geq 1$.

We have to deal with the case $j = 0$ separately. This is similar but a bit easier. We have \[ p_0(t + \tau) = \big(1 - \lambda\tau + o(\tau)\big)p_0(t) , \] since we must previously been at 0, then seen no arrivals. After rearranging to get an increment and sending $\tau \to 0$, as before, we get the differential equation \[ p'_0(t) = -\lambda p_0(t) \] with the initial condition $p_0(0) = 1$, since we always start at 0.

In summary, we have derived the following equations: \[\begin{align*} p'_0(t) &= -\lambda p_0(t) & p_0(0) & = 1 ,\\ p'_j(t) &= -\lambda p_j(t) + \lambda p_{j-1}(t) & p_j(0) &= 0 \quad \text{for $j \geq 1$} . \end{align*}\] These are called the Kolmogorov forward equations.

We claim that the solution to these equations is the Poisson distribution \[ p_j(t) = \mathrm{e}^{-\lambda t} \frac{(\lambda t)^j}{j!} . \] This would prove the theorem.

We must check that the claim holds. For $j = 0$, the Poisson probability is \[ p_0(t) = \mathrm{e}^{-\lambda t} \frac{(\lambda t)^0}{0!} = \mathrm{e}^{-\lambda t} . \] This has \[ p'_0(t) = -\lambda \mathrm{e}^{-\lambda t} = -\lambda p_0(t) , \] as desired, and we also have the correct initial condition $p_0(0) = \mathrm{e}^{0} = 1$.

For $j \geq 1$, the left-hand side is \[\begin{align*} p'_j(t) &= \frac{\mathrm d}{\mathrm d t} \mathrm{e}^{-\lambda t} \frac{(\lambda t)^j}{j!} \\ & = \frac{1}{j!} \left(-\lambda \mathrm{e}^{-\lambda t}(\lambda t)^j + \mathrm{e}^{-\lambda t} \lambda j(\lambda t)^{j-1} \right) \\ &= - \lambda \mathrm{e}^{-\lambda t} \frac{(\lambda t)^j}{j!} + \lambda \mathrm{e}^{-\lambda t} \frac{(\lambda t)^{j-1}}{(j-1)!} \\ &= -\lambda p_j(t) + \lambda p_{j-1}(t) \end{align*}\] as desired. We also have the initial condition $p_j(0) = \mathrm{e}^{0} 0^j/j! = 0$. The claim is proven, and we are (finally) done.

In the next section, we two new types of process that are similar to the Poisson process.