Visual Tools
Calculators
Tables
Mathematical Keyboard
Converters
Other Tools


Poisson Distribution



Key Terms
The Probabilistic Experiment Behind Poisson distribution
Notation Used
Parameters
Probability Mass Function (PMF) and Support (Range)
Cumulative Distribution Function (CDF)
Expected Value (Mean)
Variance and Standard Deviation
Mode and Median
Applications and Examples
Interactive Calculator
Poisson at a Glance



Poisson Distribution: Event Counts Over an Interval


The Poisson distribution models an experiment where events occur randomly over a continuous interval of time or space at a constant average rate. There are no discrete trials and no fixed number of opportunities; instead, the experiment observes how many events happen within a given interval. The defining assumptions are independence of events and stability of the average rate, making this distribution suitable for modeling counts of rare or spontaneous occurrences.

Key Terms

Poisson Distribution— counts events in a fixed interval at constant rate λ\lambda
Expected Value— E[X]=λE[X] = \lambda
Variance— Var⁡(X)=λ\operatorname{Var}(X) = \lambda (equals the mean)
Exponential Distribution— models the waiting time between Poisson events

See All Probability Definitions →


The Probabilistic Experiment Behind Poisson distribution


The Poisson distribution models the number of events occurring in a fixed interval of time or space, when events happen independently at a constant average rate. Unlike other discrete distributions, it does not rely on repeated Bernoulli trials or success–failure experiments.

The defining idea is event intensity, not trial structure. Events are assumed to occur randomly but with a stable long-run average frequency. The random variable represents the number of events that occur in any given interval of time, regardless of when they occur within it.

The Poisson distribution is especially effective for representing rare or spontaneous events, and it often arises as an approximation to the binomial distribution when the number of trials is large and the probability of success is small.


Example:

Counting the number of emails received by a support desk in one hour when emails arrive randomly but at a stable average rate. The exact timing does not matter — only the total count within the hour.

Notation Used


X∼Poisson(λ)X \sim \text{Poisson}(\lambda) or X∼P(λ)X \sim \mathcal{P}(\lambda) — distribution of the random variable.

Poisson(λ)\text{Poisson}(\lambda) — used to denote the distribution itself (not the random variable).

P(λ)\text{P}(\lambda) — sometimes used informally, especially in compact notation.

P(X=k)=λke−λk!,for k=0,1,2,…P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!}, \quad \text{for } k = 0, 1, 2, \ldots — probability mass function (PMF), where:

λ\lambda — average rate of occurrence (expected number of events in the interval)

kk — number of events observed

e≈2.71828e \approx 2.71828 — Euler's number (base of natural logarithm)

Key properties:

E[X]=λE[X] = \lambda — expected value (mean)

Var(X)=λ\text{Var}(X) = \lambda — variance (equal to the mean)

Relationship to binomial distribution:

Poisson(λ)≈Binomial(n,p)\text{Poisson}(\lambda) \approx \text{Binomial}(n, p) where λ=np\lambda = np, when nn is large and pp is small (rare events approximation)


See All Probability Symbols and Notations →

Parameters


𝜆𝜆: the average rate (mean number of events), with 𝜆>0𝜆>0

The Poisson distribution models the number of events occurring in a fixed interval of time or space, assuming events happen independently and at a constant average rate 𝜆𝜆.

It describes counts: 0,1,2,...,0, 1, 2, ..., with probabilities determined by how large or small 𝜆𝜆 is.

The single parameter 𝜆𝜆 controls both the mean and the variance of the distribution.

Probability Mass Function (PMF) and Support (Range)


The probability mass function (PMF) of a Poisson distribution is given by:

P(X=k)=λke−λk!,k=0,1,2,…P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!}, \quad k = 0, 1, 2, \ldots


Counting Rare Events: The Poisson distribution models the number of events occurring in a fixed interval of time or space when events occur independently at a constant average rate.

Support (Range of the Random Variable):
* The random variable XX can take on values 0,1,2,3,…0, 1, 2, 3, \ldots (all non-negative integers).
* X=kX = k means exactly kk events occur in the interval.
* The support is thus a countably infinite set.

Logic Behind the Formula:
* λk\lambda^k: Represents the rate parameter raised to the power of the number of events
* e−λe^{-\lambda}: The exponential decay factor ensuring probabilities sum to 1
* k!k!: Accounts for the number of ways kk events can be ordered
* The total probability sums to 1:

∑k=0∞P(X=k)=∑k=0∞λke−λk!=e−λ∑k=0∞λkk!=e−λ⋅eλ=1\sum_{k=0}^{\infty} P(X = k) = \sum_{k=0}^{\infty} \frac{\lambda^k e^{-\lambda}}{k!} = e^{-\lambda} \sum_{k=0}^{\infty} \frac{\lambda^k}{k!} = e^{-\lambda} \cdot e^{\lambda} = 1


This uses the Taylor series expansion: eλ=∑k=0∞λkk!e^{\lambda} = \sum_{k=0}^{\infty} \frac{\lambda^k}{k!}

Poisson Distribution

Rare events over time interval (rate λ)

Explanation

The Poisson distribution models the number of events occurring in a fixed interval when events happen at a constant average rate. The probability mass function is P(X=k)=λke−λk!P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!}, where λ\lambda is the average rate of events. Both the expected value and variance equal λ\lambda: E[X]=λE[X] = \lambda and Var(X)=λ\text{Var}(X) = \lambda. Common applications include customer arrivals per hour, website hits per minute, radioactive decay events, and other scenarios where rare events occur independently at a constant average rate.


Cumulative Distribution Function (CDF)


The cumulative distribution function (CDF) of a Poisson distribution is given by:

FX(k)=P(X≤k)=∑i=0kλie−λi!F_X(k) = P(X \leq k) = \sum_{i=0}^{k} \frac{\lambda^i e^{-\lambda}}{i!}


Where:
λ\lambda = average rate of occurrence (expected number of events, λ>0\lambda > 0)
kk = number of events observed (where k=0,1,2,3,…k = 0, 1, 2, 3, \ldots)
ee = Euler's number (approximately 2.71828)

Intuition Behind the Formula


Definition: The CDF gives the probability of observing kk or fewer events in a fixed interval of time or space.

Summation of Probabilities:
We sum the individual probabilities from 0 events up to kk events:

P(X≤k)=P(X=0)+P(X=1)+P(X=2)+⋯+P(X=k)P(X \leq k) = P(X=0) + P(X=1) + P(X=2) + \cdots + P(X=k)


Alternative Formulation via Incomplete Gamma Function:
The CDF can be expressed using the regularized incomplete gamma function:

FX(k)=Γ(k+1,λ)k!=Q(k+1,λ)F_X(k) = \frac{\Gamma(k+1, \lambda)}{k!} = Q(k+1, \lambda)


This relationship is often used in statistical software for efficient computation, especially for large values of kk.

Complementary Probability:
For "more than kk events":

P(X>k)=1−FX(k)P(X > k) = 1 - F_X(k)


Infinite Support: Unlike finite discrete distributions, the Poisson distribution has infinite support (kk can be arbitrarily large), though probabilities decrease rapidly for k≫λk \gg \lambda.

Poisson Distribution CDF

CDF for rare events over time

CDF Explanation

The Poisson CDF is F(k)=P(X≤k)=∑i=0kλie−λi!F(k) = P(X \leq k) = \sum_{i=0}^{k} \frac{\lambda^i e^{-\lambda}}{i!} for k≥0k \geq 0. This gives the probability of observing kk or fewer events in the fixed interval. The CDF starts at F(0)=e−λF(0) = e^{-\lambda}, which is the probability of observing zero events. As kk increases, the CDF approaches 1.0, with the rate of convergence depending on λ\lambda. For larger values of λ\lambda, the CDF increases more gradually across a wider range of kk values, while smaller λ\lambda values lead to faster convergence near k=0k = 0.

Expected Value (Mean)


As explained in the general case for calculating expected value, the expected value of a discrete random variable is computed as a weighted sum where each possible value is multiplied by its probability:

E[X]=∑xx⋅P(X=x)E[X] = \sum_{x} x \cdot P(X = x)


For the Poisson distribution, we apply this general formula to the specific probability mass function of this distribution.

Formula


E[X]=λE[X] = \lambda


Where:
λ\lambda = average rate of occurrence (expected number of events per interval)

Derivation and Intuition


Starting from the general definition and substituting the PMF P(X=k)=λke−λk!P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!} for k=0,1,2,…k = 0, 1, 2, \ldots:

E[X]=∑k=0∞k⋅λke−λk!E[X] = \sum_{k=0}^{\infty} k \cdot \frac{\lambda^k e^{-\lambda}}{k!}


The k=0k = 0 term vanishes, so we start from k=1k = 1:

E[X]=∑k=1∞k⋅λke−λk!=e−λ∑k=1∞k⋅λkk!=e−λ∑k=1∞λk(k−1)!E[X] = \sum_{k=1}^{\infty} k \cdot \frac{\lambda^k e^{-\lambda}}{k!} = e^{-\lambda} \sum_{k=1}^{\infty} \frac{k \cdot \lambda^k}{k!} = e^{-\lambda} \sum_{k=1}^{\infty} \frac{\lambda^k}{(k-1)!}


Substituting j=k−1j = k - 1:

E[X]=e−λλ∑j=0∞λjj!=e−λλ⋅eλ=λE[X] = e^{-\lambda} \lambda \sum_{j=0}^{\infty} \frac{\lambda^j}{j!} = e^{-\lambda} \lambda \cdot e^{\lambda} = \lambda


The result E[X]=λE[X] = \lambda has a particularly elegant interpretation: the parameter λ\lambda is both the rate of occurrence and the expected value. The Poisson distribution is parameterized directly by its mean.

Example


Consider phone calls arriving at a call center at an average rate of λ=12\lambda = 12 calls per hour:

E[X]=12E[X] = 12


The expected number of calls in one hour is exactly 12, which is the defining parameter of the distribution. This self-referential property makes the Poisson distribution especially natural for modeling count data.

Variance and Standard Deviation


The variance of a discrete random variable measures how spread out the values are around the expected value. It is computed as:

Var(X)=E[(X−μ)2]=∑x(x−μ)2P(X=x)\mathrm{Var}(X) = \mathbb{E}[(X - \mu)^2] = \sum_{x} (x - \mu)^2 P(X = x)


Or using the shortcut formula:

Var(X)=E[X2]−μ2\mathrm{Var}(X) = \mathbb{E}[X^2] - \mu^2


For the Poisson distribution, we apply this formula to derive the variance.

Formula


Var(X)=λ\mathrm{Var}(X) = \lambda


Where:
λ\lambda = average rate of occurrence (expected number of events per interval)

Derivation and Intuition


Starting with the shortcut formula, we need to calculate E[X2]\mathbb{E}[X^2].

We know from the expected value section that μ=λ\mu = \lambda.

Using the PMF P(X=k)=λke−λk!P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!}:

E[X2]=∑k=0∞k2⋅λke−λk!\mathbb{E}[X^2] = \sum_{k=0}^{\infty} k^2 \cdot \frac{\lambda^k e^{-\lambda}}{k!}


Through algebraic manipulation using properties of the exponential series:

E[X2]=λ2+λ\mathbb{E}[X^2] = \lambda^2 + \lambda


Applying the shortcut formula:

Var(X)=(λ2+λ)−λ2=λ\mathrm{Var}(X) = (\lambda^2 + \lambda) - \lambda^2 = \lambda


The result Var(X)=λ\mathrm{Var}(X) = \lambda reveals a remarkable property: the Poisson distribution's variance equals its mean. This makes the Poisson distribution unique among common distributions. The single parameter λ\lambda completely determines both the center and the spread of the distribution.

This property provides a practical test: if observed count data has variance approximately equal to its mean, the Poisson distribution may be an appropriate model.

Standard Deviation


σ=λ\sigma = \sqrt{\lambda}


Example


Consider phone calls arriving at a rate of λ=16\lambda = 16 calls per hour:

Var(X)=16\mathrm{Var}(X) = 16


σ=16=4\sigma = \sqrt{16} = 4


The variance equals the expected value (16), and the standard deviation of 4 indicates that in most hours, the number of calls will fall within roughly 12 to 20 calls (within one standard deviation of the mean).

Mode and Median

Mode


The mode is the value of kk (number of events) with the highest probability—the peak of the PMF.

For the Poisson distribution, the mode depends on the parameter λ\lambda:

If λ\lambda is an integer:
The distribution has two modes: λ−1\lambda - 1 and λ\lambda

If λ\lambda is not an integer:
The mode is ⌊λ⌋\lfloor \lambda \rfloor

Intuition: The mode sits at or just below the expected value λ\lambda, representing the most likely number of events. The Poisson distribution models rare events, and the mode indicates the count that balances the competing effects of increasing event likelihood with decreasing probability of higher counts.

Example:
For λ=4.7\lambda = 4.7:

Mode = ⌊4.7⌋=4\lfloor 4.7 \rfloor = 4

Getting exactly 4 events is more likely than any other outcome.

Example:
For λ=5\lambda = 5:

Modes = 4 and 5 (both equally likely)

Median


The median is the value mm such that P(X≤m)≥0.5P(X \leq m) \geq 0.5 and P(X≥m)≥0.5P(X \geq m) \geq 0.5.

For the Poisson distribution, there is no simple closed-form expression for the median, but it can be found numerically by solving:

∑k=0mλke−λk!≥0.5\sum_{k=0}^{m} \frac{\lambda^k e^{-\lambda}}{k!} \geq 0.5


Properties of the median:
• The median is approximately λ+13−0.02λ\lambda + \frac{1}{3} - \frac{0.02}{\lambda} for large λ\lambda
• For integer λ\lambda, median ≈λ\approx \lambda
• The distribution approaches symmetry as λ\lambda increases

Example:
For λ=5\lambda = 5:
Mean = 5

The median is approximately 5 (nearly symmetric)

Example:
For λ=2\lambda = 2:
Mean = 2

The median is approximately 2 (found numerically)

Unlike continuous distributions where finding the median requires integration, for discrete distributions, the median is found by summing probabilities until reaching 0.5.


Applications and Examples


### Practical Example

Suppose a call center receives an average of λ=4\lambda = 4 calls per hour. The probability of receiving exactly k=6k = 6 calls in a given hour is:

P(X=6)=46e−46!=4096⋅0.0183720≈0.104P(X = 6) = \frac{4^6 e^{-4}}{6!} = \frac{4096 \cdot 0.0183}{720} \approx 0.104

This means there's about a 10.4% chance of receiving exactly 6 calls in an hour.

Note: The Poisson distribution is often used as an approximation to the binomial distribution when nn is large and pp is small, with λ=np\lambda = np.

Interactive Calculator


This interactive calculator computes probabilities for the Poisson distribution, which models the number of events occurring in a fixed interval of time or space. Enter the average rate (λλ) of events and choose whether you want the full distribution or specific probabilities like P(X=k)P(X=k) or P(X≤k)P(X≤k) to analyze event frequencies. Essential for modeling customer arrivals, website traffic, defect rates, or any situation involving rare events occurring at a constant average rate.

1. Select probability type: 'All values' for full distribution, or choose a specific query
2. Enter λ (lambda - average rate) - average number of events per time/space interval
3. For specific queries, enter k (number of events) - the target event count
4. Click Calculate to see probabilities and distribution

Poisson Distribution Calculator

Calculate probabilities and distribution properties

Average number of events per interval

,

Poisson at a Glance

The table below collects the full anatomy of the Poisson distribution into a single reference card — its rate parameter, support, PMF and CDF, the mean and variance (both equal to λ), mode and median behavior, the binomial approximation, and a canonical example.
Aspect Formula / statement Note / example
Parameter λ > 0 — average rate (events per interval) events occur independently at a constant rate
Support k ∈ {0, 1, 2, ...} countably infinite — no upper limit on count
PMF P(X = k) = λk e−λ / k! normalized via the Taylor series eλ = ∑ λk/k!
CDF F(k) = ∑i=0k λi e−λ / i! no simple closed form; computed via the regularized incomplete gamma
Expected value E[X] = λ the rate parameter is the mean
Variance Var(X) = λ — equal to the mean unique to Poisson; sample mean ≈ sample variance is a diagnostic for fit
Mode and median mode = ⌊λ⌋ (or λ − 1 and λ when λ is an integer); median ≈ λ + 1/3 − 0.02/λ distribution approaches symmetry as λ grows
Binomial approximation Poisson(λ) ≈ Binomial(n, p) with λ = np, large n, small p the "law of rare events"; rule of thumb n > 20, p < 0.05
Canonical example call center receiving 12 calls/hour → λ = 12 E[X] = 12, Var(X) = 12, σ ≈ 3.46