# Inverting the Poisson Process to Find... the Poisson Process

**Prerequisites**: This post assumes knowledge of the basics of probability theory (e.g. independence, densities, law of total probability), and of the Poisson and Exponential distributions. Familiarity with multivariable calculus and the Poisson Process will be helpful.

I’ve taught Berkeley’s standard probability course, Stat 134, a couple of times and one of my favorite topics in this course is the Poisson process. The Poisson process is an example of a stochastic process, i.e., it models the behavior of a sequence of random events over time. It is one of the simplest examples of a stochastic process. Each event occurs completely at random and is identical to, but does not affect the behavior of, any other event; all we know is the average rate at which these events occur.

The Poisson process can be used, for example, to model the lifespan of light bulbs in your home. (In this case, an event would be a light bulb burning out.) Everytime a light bulb burns out, you replace it with an identical bulb; you expect the average number of light bulbs you use each year to be roughly close to the long-run average but naturally, you expect some variation due to how your usage of the light bulb; and finally, there is no real reason to think the lifespan of one bulb should affect the lifespan of any other. The Poisson process is frequently applied across a variety of fields and has been used to model, among other things:

- in queuing theory, phone calls arriving at a call center [1];
- in geology/ seismology, earthquakes in a region of moderate seismic activity [2];
- in economics, companies entering a competitive market [3].

I often describe the Poisson process as being elegant or magical, primarily because of how it so wonderfully balances simplicity and structural wealth. Despite its simple formulation, the Poisson process remains *very* analytically interesting. For example, a large variety of popular distributions (including but not limited to, the Exponential, Gamma, Beta, and of course, Poisson distributions) are intrisically linked to, and can be derived directly from, the Poisson process.

In this post, we will discuss one property of the Poisson process that I am especially fond of: that the traditional definition of the Poisson process can be inverted. (More on what this specifically means later.)

## The Traditional Definition

**Notation**: In the context of Poisson processes, we refer to an event occuring as an arrival and use:

- $X_j$
*to denote the*$j^{th}$*arrival;* - $W_j$
*to denote the waiting time between*$X_{j-1}$*and*$X_{j}$*;* - $T_j$
*to denote the total waiting time until*$X_j$*;* *and*$N_{(s,t)}$*,*$s< t$*to denote the number of arrivals in an interval*$(s,t)$*.*

*See Figure 1 for details.*

Traditionally, in introductory/ intermediate probability courses, the *Poisson Process with rate $\lambda$* is formally defined on the positive real line $(0, \infty)$ as a process where:

- the number of arrivals in any interval of length $t$ has a $Poisson(\lambda t)$ distribution, i.e., $N_{(s,s+t)} \sim Poisson(\lambda t)$;
- and, the numbers of arrivals in disjoint time intervals are independent, i.e., for $s_1 \leq t_1 \leq s_2 \leq t_2$, the number of arrivals in $(s_1, t_1)$ is independent of the number of arrivals in $(s_2, t_2)$.

(This definition may seem rather complicated but it encodes the simple assumptions discussed above: that each event occurs completely at random and is identical to, but does not affect the behavior of, any other event.)

Using this definition, we can show that (1) the time between any two consecutive arrivals (the waiting time) has a $Exponential(\lambda)$ distribution and (2) that all the waiting times are independent of each other.

We now show (1). Consider an arbitrary arrival $X_{j-1}$ that arrives at some time $s$. We will show that $W_j$ is distributed according to an $Exponential(\lambda)$ distribution by showing that they have the same survival function (i.e., $P(W_j > t)$).

As shown in Figure 2, if there are no arrivals in the interval $(s, s+t)$, the waiting time $W_j$ must be greater than $t$. Thus,

We therefore have as desired: $W_j\sim$ $Exponential(\lambda)$.

To build some intuition for why (2) holds, consider Figure 1 again. Look at the total arrival times $T_3$ and $T_4$. $T_3$ being longer necessarily means that $T_4$ will be longer since $T_4$ effectively contains $T_3$. (Note that we can write $T_4 = T_3 + W_4$. So, if $W_4$ is constant, $T_3$ being longer would imply that $T_4$ is longer.) They are clearly not independent since the length of one affects the length of the other. As we can see in the diagram however, there is no obvious relationship between any two waiting times, i.e., there is no clear reason why the length of one waiting time should affect the length of any other. One might therefore intuitively suspect that the waiting times are independent and indeed, they are.

The formal proof of (2) is not terribly difficult, but is rather notationally dense. Thus, in the interest of brevity and clarity, I have elected to not include it here. However, it is a good exercise in proof writing and a good test of your understanding of Poisson processes to attempt the proof yourself. Feel free to post in the comments or reach out to me directly via email if you need any help getting started/ if you get stuck.

## Inverting the Definition

We now know that any system where (1) the number of arrivals in any time interval is Poisson also has the property that (2) the time between arrivals are independent Exponentials; i.e., (1) implies (2). A natural question to ask is, “Is the converse true?”, or “Does (2) imply (1) back?”. In other words, what happens if we invert the definition of the Poisson process and start with (2) instead of (1): if we start with a system where the time between any two arrivals are independent Exponentials, are the number of arrivals in any time interval Poisson? As suggested by the title of this post, the answer is yes!

We will prove a version of this result that retains the essence of the problem while being simpler and more intuitive. Specifically, we will show that the number of arrivals in any time interval beginning at 0 (as opposed to any time interval) is distributed according to a $Poisson(\lambda t)$ distribution. I do this, once again, in the interest of brevity and clarity but will outline at the end of this post, how to prove the full result.

Formally, we define the system as follows: given a sequence of independent and identically distributed $Exponential(\lambda)$ random variables $W_1, W_2, W_3, \ldots$, let $T_j = W_1 + W_2 + \ldots + W_j$ denote the time of the $j^{th}$ arrival. Let $N_t = N_{(0,t)} = \max_{T_j < t} j$ be the number of arrivals in $(0,t)$. We will show that $N_t \sim Poisson(\lambda t)$.

We will proceed slowly to build some intuition. We will first find the probability that there are 0, 1, and 2 arrivals in the interval $(0,t)$, and show that it is equivalent that a $Poisson(\lambda)$ distribution takes values 0, 1, and 2 respectively; finally, leveraging the insight from these cases, we will show the desired, general result.

### Case 0

Consider some arbitrary interval $(0,t)$. First, let’s find the probability that there are no arrivals in $(0,t)$, i.e. that $N_t = 0$. Note that the only information we are given, and therefore allowed to use, is the fact that the waiting times are independent and identically distributed $Exponential(\lambda)$ random variables; we must therefore reduce this problem into one dealing with waiting times. We do so as follows: if there are no arrivals in $(0,t)$, then the first arrival *must* have arrived after $t$ and therefore the first waiting time $W_1$ *must* be greater than $t$. Thus,
This result is what we would expect if $N_t\sim Poisson(\lambda t)$. Great, we’re on the right track.

### Case 1

* Note*:

*Recall that the probability that a continuous random variable $X$ takes any specific value is 0, i.e. $P(X = x) = 0$. (This is (sorta) because there are an infinite number of real numbers in any continuous interval: since the probabilities associated with each of these numbers must sum to one, they must be effectively infiticeminal to begin with.) To overcome this, we will compute $P(X \in \Delta x) = P(X \in (x, x+dx))$ and take the limit as $dx \rightarrow 0$. (We henceforth use $\Delta x$ refers to the interval $(x, x+dx)$.) Recall that for small $dx$, we have $P(X \in \Delta x) = P(X \in (x, x+dx)) = f_X(x) \cdot dx$.*

Now, let’s find the probability that there is exactly one arrival in $(0,t)$. This means that the time of the first arrival $T_1$ (and therefore the first waiting time $W_1$) must be at some point less than $t$; for now, let’s say that the first arrival is near $x$ (i.e. $x\in \Delta x = (x, x+dx)$). The second arrival must however be at some point greater than $t$; as shown in Figure 3, this means that the second waiting time $W_2$ must be greater that $t-x$. (Technically speaking, we require that $W_2 > t-x-dx$ but we can ignore the $dx$ term since it will vanish when we take the limit $dx\rightarrow 0$.)

Now, what is this mysterious $x$ value that $W_1$ takes? The answer is that $x$ could be anything between $0$ and $t$; as long as $x$ falls within this interval and the $W_2$ is greater than $t-x$, we will have exactly one arrival in this interval. Thus, by the Law of Total Probability, we have to compute the probability that $W_1$ is near $x$ and $W_2$ is greater than $t-x$ for *every possible value of $x$* in the interval $(0,t)$, and then sum all these probabilities up to arrive at the desired answer: the total probability that we have exactly one arrival in this interval. Since we are dealing with continuous probabilities, we replace the sum with an integral and proceed as follows:

As suggested by my final line of work, this is again what we would expect if $N_t\sim Poisson(\lambda t)$.

### Case 2

Let’s finally look at one slightly harder iteration of this problem before doing the general case; now, we will find the probability that there are exactly two arrivals in $(0,t)$. As with before, we require that that the time of the first arrival, $T_1$ (and therefore also $W_1$), is near some $t_1$, where $t_1 < t$ (we change the name here for notational convenience). We similarly require that $T_2$ is near some $t_2$, where $0 < t_1 < t_2 < t$; this means that we require that $W_2$ is near $(t_2 - t_1)$. We finally require that $T_3$ be greater than $t$; therefore, we require that $W_3$ be greater than $(t-t_2)$. (See Figure 4 for a visual description of this.) We will once again integrate over all the possible values of $t_1$ and $t_2$ to compute the total probability. (So yes, we do have a double integral this time around!)

The bounds of this integral are a little tricky. We will first integrate over $t_1$, considering $t_2$ to be some fixed value. (This is important; the problem becomes annoyingly difficult if you choose to integrate in some other order.) $t_1$ can take any value between $0$ and $t_2$, so these are the bounds we will integrate $t_1$ over. Now, onto $t_2$. Recall that $t_2$ is bounded as follows: $0< t_1 < t_2 < t$, i.e., $t_2$ is bounded between $t_1$ and $t$. However, since we have marginalized $t_1$ already, we need no longer consider it when we integrate over $t_2$; it has, for all intents and purposes, been dealt with. Thus, we take the next tightest bound for $t_2$ and find that $t_2$ can take on values from $0$ to $t$. We thus proceed as follows:

Voila! We get exactly what we expect again. Using what we learnt from these examples, we can go ahead and find $P(N_t = n)$ for some arbitrary $n$ to show that $N_t\sim Poisson(\lambda t)$.

### The General Case

If there are exactly $n$ arrivals in $(0,t)$, we must have that the time of the first $n$ arrivals, $T_1, \ldots, T_n$ is near $t_1, \ldots, t_n$ respectively, and that the $(n+1)^{th}$ arrival is after $t$, i.e. that $T_{n+1} > t$. Translating this into the language of waiting times and integrating over bounds derived similarly to above, we find:

Thus, we have as desired, that $N_t$ has a $Poisson(\lambda t)$ distribution.

There you go! We have shown that the definition of the Poisson process can be inverted. Well, almost. To complete the proof, we need to show that (1) the above result holds for any arbitrary interval $(s, s+t)$ instead of only intervals of the form $(0,t)$ and (2) the number of arrivals in disjoint intervals are independent. Both of these results can be shown true by exploiting the memoryless property of the Exponential distribution and the result above. (Once again, I think this is worth doing if you have not done this before.)

## Wrapping Up

This result shows that there are actually multiple, *equivalent* ways of defining the Poisson process. You can define the Poisson process as a process where either (1) the number of arrivals in any time interval are Poisson or (2) the time between intervals is Exponential. (There is in fact a third equivalent definition. See the Appendix for details.) However, it does not matter which definition you choose: the characteristics of the resulting process are the same. i.e., whichever definition you start with, you get the properties specified by the other definition(s) for free!

As I’ve said before, the Poisson process is a wonderfully elegant stochastic model and there are plenty more of these beautiful results hidden within it. I highly recommend exploring the Poisson process on your own if you haven’t already but just in case you want somewhere concrete to get started, here’s another fun one: conditional on the fact there are $n$ arrivals in the interval (0, 1), what is the distribution of the time of the $k^{th}$ arrival, for $1\leq k \leq n$? What if we condition on there being $n$ arrivals in some arbitrary interval $(s, s+t)$ instead?

Please do leave your thoughts and questions in the comments section below, or send them to me directly via email. I will try to get back to you as soon as possible. Your feedback is very much appreciated. Until next time :)

## Appendix

** Bonus:** There is in fact a third equivalent definition of the Poisson process: as a special case of a pure birth process.

Consider a process where (1) in any small interval $dt$, there may be at most one arrival; (2) this arrival occurs with some constant probability $\lambda\ dt$; (3) and this arrival is independent of all other arrivals outside this small interval. Formally, let $N_t$ denote the number of arrivals in $(0,t)$. $N_t$ is such that (1)

and (2) for $s< t$, the increment $N_t - N_s$ is independent of all arivals prior to $s$.

(This definition is similar to that presented in Probability and Random Process by Grimmett and Stirzaker.)

This process is also (surprise, surprise) a Poisson process! We can show that this definition is in fact equivalent to both the other definitions presented in this blog post; this is left as an exercise to the reader :)

*Thank you to Alvin Wan and Chin Chang for reading early drafts of this post and offering helpful suggestions.*