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Poisson distribution
You will learn: Connect event counts, rates, windows, and the conditions of a Poisson model.
Start with: Binomial distribution
New to the notation? Start with the connected foundation for the underlying definitions and a worked example.
Suppose requests arrive independently at a constant average rate of 2.5 per minute. In one minute, what is the probability of receiving no more than three requests? The Poisson distribution turns an expected count into probabilities for each possible count.
Its parameter λ is the mean count in the chosen window. If the rate is r events per minute and the window lasts t minutes, λ = rt. Always specify the window before entering a number.
Experiment: count probabilities
The accent bars show a selected event. Choose “at most,” “exactly,” or “more than” and compare their probabilities. Changing the sample count changes the noise in the simulated bars; it does not change the theoretical distribution.
Probability: 43.347%. Accent bars show the selected event.
The count has no upper limit. Probability beyond the plotted k = 20 is 1.92e-7%.
The model at a glance
| Property | Meaning |
|---|---|
| Support | Nonnegative integers: 0, 1, 2, … |
| Parameter | λ ≥ 0, the expected count in the window |
| Mean and variance | Both equal λ |
| Standard deviation | √λ |
| Zero events | exp(−λ) |
| CDF | The sum of probabilities from zero through the threshold |
A Poisson process assumes independent counts in disjoint intervals, a constant rate, and no simultaneous batches. Individual events need not be rare over the whole window: a large λ simply means many arrivals are expected. Rarity describes the chance within a sufficiently short interval.
Worked example: no more than three
“No more than three” includes 0, 1, 2, and 3. With λ = 2.5:
The terms sum to 0.757576, about 75.76%. Exactly three has probability about 21.38%. More than three is the complement, about 24.24%. These are different events even though the same threshold appears in each question.
from math import exp, factorial
expected_count = 2.5
probability = sum(
exp(-expected_count) * expected_count**j / factorial(j)
for j in range(4)
)
print(f"{100 * probability:.2f}%") # 75.76%Make a prediction
At the same rate, a two-minute window has twice the chance of zero arrivals. True or false?
Show a hint
Double λ, then compare exp(−λ) with exp(−2λ).
Explore the answer
False. The expected count doubles, but the zero-arrival probability becomes exp(−5) ≈ 0.67%, down from exp(−2.5) ≈ 8.21%. Probabilities are not generally proportional to the window length.
The same events, viewed as waiting times
Count the ticks in a row below. Then look at the gaps between them. Under a constant-rate Poisson process, counts are Poisson and the time to the next event is exponential, with mean 1/r minutes.
Across 100 synthetic windows, mean count = 3.67, sample variance = 3.33. The model mean is 4.00; its variance is 4.00.
Switch to quiet-or-busy windows. Each window is still conditionally Poisson, but its rate is now random. Pooling those windows produces a mixture with variance larger than its mean. A single Poisson parameter cannot describe both features simultaneously.
Where it comes from
Divide a window into n tiny opportunities, each with event probability p. The count is approximately binomial. As n grows and p shrinks while np stays at λ, its probabilities approach the Poisson probabilities.
The crucial limit holds the mean fixed. Increasing n while keeping p fixed makes the mean grow instead. At large λ, a normal approximation can also be useful, but accuracy depends on the event and tail being computed; the exact Poisson probabilities are preferable when available.
Binomial: n=20, p=0.1250. Both means equal 2.5; binomial variance is 2.1875, versus Poisson variance 2.5.
| Binomial (filled, left) | Poisson (outline, right) | Absolute difference |
|---|---|---|
| 0.765332 | 0.757576 | 0.007756 |
When the model misses
A constant-rate model can fail when arrivals come in groups, one event triggers another, or rates vary across observation windows. Excess variance is a clue to investigate those mechanisms, not proof of a particular alternative.
A negative-binomial model can represent one kind of heterogeneous Poisson rate. Scheduled arrivals can be less variable than Poisson. A count is not automatically Poisson merely because it is a nonnegative integer.
Continue exploring
Use exponential waiting times for the next arrival and gamma for time to the k-th arrival. Return to binomial when there is a fixed number of independent opportunities.
References
- NIST: Poisson distribution, for the PMF, CDF, and parameter properties.
- The timeline and quiet/busy comparison above are synthetic experiments defined on this page. They are not measurements of a real service.
Compare the assumptions and event probabilities in Poisson vs binomial, or use the distribution chooser.
Follow the arrivals
Continue to the Poisson process to follow counts, waiting times, and independently retained events on one shared path. Total variance explains the extra spread in a mixture of rates.