Poisson Distribution

Intuition

Some events happen at a roughly constant average rate: server requests per second, typos per page, radioactive decays per minute. The Poisson distribution counts how many such events occur in a fixed interval of time or space. It works best when events are independent, arrive one at a time, and the average rate does not change across the interval.

Definition

A random variable $X$ follows a Poisson distribution $X\sim \text{Pois}(\mu )$ if it represents the number of events occurring in a fixed interval (time, area, volume, etc.) where events happen independently at a constant average rate.

$X$ can take non-negative integer values $x=0,1,2,\dots$

The parameter $\mu =\lambda t$ is the expected number of events in the interval of length $t$ with rate $\lambda$.

Three assumptions underlie the Poisson model:

1. Independence: the occurrence of one event does not affect the probability of another.
2. Proportionality: for a small sub-interval $\Delta t$, $P(\text{one event})\approx \lambda \Delta t$.
3. No simultaneous events: $P(\text{two or more events in }\Delta t)\approx 0$.

Key Formulas

Probability Mass Function (PMF):

$p(x;\mu )=\frac{e^{-\mu }{\mu }^x}{x!},x=0,1,2,\dots$

Mean:

$E[X]=\mu$

Variance:

$\text{Var}(X)=\mu$

A distinctive (and useful) feature: the mean and variance are equal. This gives you a quick diagnostic - if your observed variance is much larger than the mean, a Poisson model probably doesn’t fit (overdispersion).

Standard Deviation:

$\sigma =\sqrt{\mu }$

Moment Generating Function:

$M(t)=e^{\mu (e^t-1)}$

Poisson approximation to the binomial:

When $n$ is large and $p$ is small, $B(n,p)\approx \text{Pois}(np)$. A common rule of thumb is $n\ge 20$ and $p\le 0.05$.

Example

Aircraft arrive at an airport at an average rate of 2 per hour ($\lambda =2$). What is the probability that exactly 5 arrive in a 2-hour window?

Here $\mu =\lambda t=2\cdot 2=4$.

$P(X=5)=\frac{e^{-4}\cdot 4^5}{5!}=\frac{0.0183\cdot 1024}{120}\approx 0.1563$

There is roughly a 15.6% chance of exactly 5 arrivals in the 2-hour period.

We can also find the probability of no arrivals in a 30-minute window ($\mu =2\cdot 0.5=1$):

$P(X=0)=\frac{e^{-1}\cdot 1^0}{0!}=e^{-1}\approx 0.3679$

So there is about a 37% chance of a quiet half-hour with no aircraft arrivals.

As a binomial approximation: if 1000 components each fail independently with probability $p=0.003$, the expected number of failures is $\mu =np=3$. Rather than computing $\left(\genfrac{}{}{0ex}{}{1000}{x}\right)(0.003{)}^x(0.997{)}^{1000-x}$, we use $P(X=x)\approx e^{-3}\cdot 3^x/x!$.

Why It Matters in CS

The entire field of queueing theory starts with “assume arrivals are Poisson.” The M/M/1 queue, the M/M/c queue, basically every tractable queueing model uses this as the arrival process because it makes the math work out to closed-form solutions for wait times and queue lengths. When you size a buffer on a router or set autoscaling thresholds on a web server, there’s a good chance a Poisson assumption is somewhere in the analysis.

Warning

Real network traffic is often bursty, which violates the independence assumption and produces overdispersion (variance >> mean). The Poisson is a starting point, not gospel. If your observed variance is three times your mean, you need a different model.

The mean-equals-variance property also makes Poisson a natural baseline for anomaly detection. If your error logs normally see $\mu =5$ errors per minute, observing 15 in a single minute is a clear signal - you can compute exactly how unlikely that is without fitting anything more complicated.

Related Notes

• Exponential Distribution - models the continuous time between Poisson events
• Binomial Distribution - the Poisson approximates the binomial for large $n$ and small $p$
• Probability Distributions - broader taxonomy of discrete and continuous distributions