4.6 Poisson Distribution (Optional)

Poisson Distribution

The Poisson distribution models how often rare events occur within a fixed interval of time or space. Think of counts like the number of car accidents at an intersection per week, typos per page, or radioactive decays per second. Where the binomial asks "how many successes in $n$ trials?", the Poisson asks "how many events in this interval?" with no fixed number of trials at all.

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Poisson Distribution for Fixed Intervals

The Poisson distribution applies when you're counting events that meet three conditions:

• Events occur independently of each other (one accident doesn't make another more or less likely)
• Events occur at a constant average rate throughout the interval
• Events are rare relative to the size of the interval (you can't have infinitely many events piling up)

The single parameter $\lambda$ (lambda) represents the average number of events per interval. It completely defines the distribution. If a hospital emergency room averages 4.2 patient arrivals per hour, then $\lambda = 4.2$ for a one-hour interval. For a two-hour window, you'd use $\lambda = 8.4$.

Unlike the binomial, there's no cap on the number of events. $X$ can technically take any non-negative integer value (0, 1, 2, 3, ...), though very large values become extremely unlikely.

Probability Calculations with Poisson

The probability mass function (PMF) gives the probability of observing exactly $x$ events:

$P(X = x) = \frac{e^{-\lambda}\lambda^x}{x!}$

where:

• $x$ = the specific count you're finding the probability for (0, 1, 2, ...)
• $\lambda$ = the average number of events in the interval
• $e \approx 2.71828$ (Euler's number)
• $x!$ = $x$ factorial

Step-by-step example: A bookstore averages 3 online orders per hour ($\lambda = 3$). What's the probability of exactly 5 orders in a given hour?

1. Identify $\lambda = 3$ and $x = 5$
2. Plug into the PMF: $P(X = 5) = \frac{e^{-3} \cdot 3^5}{5!}$
3. Compute the pieces: $e^{-3} \approx 0.04979$, $3^5 = 243$, $5! = 120$
4. Result: $P(X = 5) = \frac{0.04979 \times 243}{120} \approx 0.1008$

So there's about a 10.1% chance of exactly 5 orders in that hour.

For cumulative probabilities like $P(X \leq 2)$, sum the individual PMF values:

$P(X \leq 2) = P(X=0) + P(X=1) + P(X=2)$

In practice, your calculator's poissonpdf and poissoncdf functions handle this. Use poissonpdf(λ, x) for exact probabilities and poissoncdf(λ, x) for cumulative (at most $x$) probabilities.

Poisson as a Binomial Approximation

The Poisson distribution can stand in for the binomial when direct binomial calculations become unwieldy. This works well when:

1. The number of trials $n$ is large (typically $n > 20$)
2. The probability of success $p$ is small (typically $p < 0.05$; some texts allow $p < 0.1$)
3. You set $\lambda = n \times p$

Under these conditions, $\text{Binomial}(n, p) \approx \text{Poisson}(\lambda = np)$.

Why this works: When $n$ is large and $p$ is tiny, most trials result in failure, and the few successes are scattered independently across the trials. That pattern matches exactly what the Poisson models.

Example: Suppose 1 in 500 manufactured chips is defective ($p = 0.002$), and you inspect a batch of 1000 chips ($n = 1000$). Finding the exact binomial probability of 3 or fewer defects requires heavy computation. Instead, use $\lambda = 1000 \times 0.002 = 2$ and calculate $P(X \leq 3)$ with the Poisson PMF. The approximation will be very close to the true binomial answer.

Properties of the Poisson Distribution

• Mean and variance are both equal to $\lambda$. This is a distinctive feature. If you see count data where the sample mean and sample variance are roughly equal, the Poisson may be a good model.
• The standard deviation is $\sqrt{\lambda}$.
• The distribution is right-skewed for small $\lambda$ and becomes more symmetric as $\lambda$ increases.
• It's a discrete distribution since it models whole-number counts.
• The Poisson is related to the exponential distribution, which models the time between consecutive Poisson events. If events arrive at rate $\lambda$ per hour, the waiting time between events follows an exponential distribution with the same parameter.