Poisson process: Introduction

 We consider stochastic processes with continuous time. For example,

• The number of births (or deaths) in a population by time $t$.
• The number of phone calls at an office by time $t$.
• The number of radioactive particles detected by a Geiger counter by time $t$.

The Poisson process is a basic stochastic process that can be used to model above phenomena. Let's define it.

The Poisson process

Definition

Let $N(t)$ be a time-dependent random variable of integer values, and $\lambda >0$ a constant. Here, $t$ represents time. Suppose the probability mass function of $N(t)$ is given by

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

A stochastic process $\{N(t)\}$ characterized by the probability distribution (Eq:PoissonProc) is called a Poisson process.

Example: Let $N(t)$ be the number of phone calls at an office by time $t$. Then $N(t)$ can be modeled as a Poisson process. (The same applies to other examples above.)

Why is it called the "Poisson" process? Because the probability distribution $p_n(t)$ is the Poisson distribution with parameter (the mean of $N(t)$) $\lambda t$. You should check that $$\begin{array}{}\text{(Eq:Mean)}& \mu (t)=\mathbb{E}[N(t)]=\sum _{n=0}^{\mathrm{\infty }}n{p}_n(t)=\lambda t\end{array}$$ where $\mathbb{E}[\cdot ]$ represents the expectation value of the random variable in the argument. The variance is $$\begin{array}{}\text{(Eq:Var)}& \mathbb{V}[N(t)]=\mathbb{E}[(N(t)-\mu (t){)}^2]=\lambda t.\end{array}$$

What can we learn from this?

• According to (Eq:Mean), the mean of $N(t)$ is proportional to time $t$. Thus, $N(t)$ is a non-decreasing function of time, on average.
• According to (Eq:Var), the variance of $N(t)$ is also proportional to time $t$. Thus, for a larger $t$, the difference between different samples of $N(t)$ becomes larger, on average.

Here is a sample path of a Poisson process (with $\lambda =0.1$).

We can see step-wise increases of $N(t)$ at random intervals.

Simulating Poisson processes

(See this page for details.)

How can we simulate a Poisson process with parameter $\lambda$, such as in the above figure? Here, I explain a rough approximate method without details. In later posts, we will see why this method correctly (yet approximately) simulates the Poisson process and why it is a rough approximation. Since we cannot represent continuous variables on a computer, we must discretize them. In the present case, it is the time variable $t$ that needs to be discretized. So, we consider discrete time steps:

$$t=0,\delta t,2\delta t,3\delta t,\cdots$$ where $\delta t$ is some small positive real number (floating-point number). As the initial condition, we assume that $$N(0)=0.$$

Now, suppose we have $N(k\delta t)=n$ for some $k=0,1,\cdots$. We determine the value of $N((k+1)\delta t)$ in the following manner:

1. Generate a random number $r$ from the uniform distribution in [0, 1) (i.e., between 0 and 1, including 0, excluding 1).
2. If $r<\lambda \delta t$, then we set $N((k+1)\delta t)=n+1$, otherwise we set $N((k+1)\delta t)=n$.

By this procedure, the value of $N((k+1)\delta t)$ is incremented by 1 with probability $\lambda t$, and stays the same ($N((k+1)\delta t)=n$) with probability $1-\lambda \delta t$. Let's see an example.

1. Suppose we have $\lambda =0.1$ and $\delta t=0.1$. Thus, $\lambda \delta t=0.01$.
2. We start from $N(0)=0$.
3. Generate a random number from [0, 1). Say, we obtained $r=0.2358$. Since this is greater than $\lambda \delta t=0.01$, we set $N(\delta t)=N(0)+0=0$.
4. Generate another random number. Say we obtained $r=0.0052103<\lambda \delta t$. Then, we set $N(2\delta t)=N(\delta t)+1=1$.
5. Generate another random number. Say we obtained $r=0.025139>\lambda \delta t$. Then, we set $N(3\delta t)=N(2\delta t)+0=1$.
6. We repeat this procedure over and over.

Since the quantity $\lambda \delta t$ serves as a probability,  this method is applicable only when $\lambda \delta t\le 1$ (otherwise $\lambda \delta t(>1)$cannot be interpreted as a probability). For example, if we have $lambda=20,\delta t=0.1$, then $\lambda \delta t=2>1$ so that $r<\lambda \delta t$ for any random number $r\in [0,1)$. This means $N(k\delta t)$ is always incremented by 1 regardless of the random numbers, which is nonsense. To use a larger $\lambda$ value, we replace $r<\lambda \delta t$ with $$r<1-e^{-\lambda \delta t}(<1)$$ in Step 2. Note that when $\lambda \delta t$ is sufficiently small, the Maclaurin expansion gives

$$1-e^{-\lambda \delta t}=\lambda \delta t+o(\delta t^2).$$

That is, $\lambda \delta t$ can be obtained as the first-order approximation of $1-e^{-\lambda \delta t}$. In a later post, we will see why $1-e^{-\lambda \delta t}$ is indeed the correct probability of increment. (Hint: $1-e^{-\lambda \delta t}$ is the cumulative distribution function of the exponential distribution of $\delta t$ with parameter $\lambda$.)

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