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Poisson Distribution



Key Terms
The Probabilistic Experiment Behind Poisson distribution
Poisson Notation
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 FAQ



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.

Binomial(10, 0.4) and Binomial(100, 0.04) against Poisson with lambda = 4ky123456789101112all three have mean 4blue: n = 10, p = 0.4navy: n = 100, p = 0.04amber: Poisson, λ = 4Many trials, small p: the binomial turns into the Poisson
From binomial to Poisson

Three distributions with mean 4. Blue: binomial with n = 10, p = 0.4, visibly narrower. Navy: binomial with n = 100, p = 0.04. Amber: Poisson with λ = 4. The navy and amber bars are almost the same height at every k: with many trials and a small success probability, the binomial count is close to Poisson with λ = np.

The rule of thumb: n large, p small, and λ = np kept moderate.

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.

Poisson Notation

Notation

Poisson Notation

One Greek letter runs this entire page: it fills the declaration's only slot, it is simultaneously the mean and the variance, and it changes size with the window you watch. The marks to fix are the family's spellings, λ's double duty, and a pmf line whose support never closes. All catalogued among the probability symbols.
The declaration tilde and capital/lowercase split come from random variable notation and distribution notation; ee and its exponentials from exponential function notation; n!n! and the counting marks from permutation notation; E[X]E[X] and Var⁡(X)\operatorname{Var}(X) from expected value and variance notation.
X∼Pois⁡(λ)X \sim \operatorname{Pois}(\lambda)
X follows a Poisson distribution with rate lambda
One slot, three spellings: Pois⁡(λ)\operatorname{Pois}(\lambda) standard, Poisson⁡(λ)\operatorname{Poisson}(\lambda) explicit, script P(λ)\mathcal{P}(\lambda) in compact writing. A single parameter carries the whole family — and does double bookkeeping, since E[X]=Var⁡(X)=λE[X] = \operatorname{Var}(X) = \lambda: one number sets both the center and the spread.
CasesThe rare-events bridge is written as an approximation between declarations: Bin⁡(n,p)≈Pois⁡(np)\operatorname{Bin}(n, p) \approx \operatorname{Pois}(np) for large nn, small pp — the binomial's two parameters collapsing into the product that survives.
Do not confuseThe probability operator. The informal X∼P(λ)X \sim P(\lambda) puts capital PP at two jobs in one line — P(X=k)P(X = k) and P(λ)P(\lambda) mean utterly different things a few symbols apart — which is precisely why the script P\mathcal{P} variant exists.
λ\lambda
lambda — the expected number of events per observation window
A rate wearing a count's clothes: λ\lambda is the average number of occurrences in the interval being watched, so it silently carries the window's size — double the window and the model becomes Pois⁡(2λ)\operatorname{Pois}(2\lambda); process texts make the exposure explicit as λt\lambda t.
CasesThe same λ\lambda crosses the discrete/continuous line: counting events per window is Poisson, waiting for the next event is exponential — one rate, two questions, and the letter travels between the two pages unchanged.
Do not confuseA probability. λ=4.2\lambda = 4.2 is perfectly legal — rates are counts, not chances, and live on [0,∞)[0, \infty); capping λ\lambda at 1 imports a constraint from the wrong kind of parameter.
Same glyph elsewhereLinear algebra's eigenvalue — the reserved λ of Av=λvA\mathbf{v} = \lambda\mathbf{v} — is an unrelated tenant of the same letter; context and neighbors decide instantly which is on duty.
P(X=k)=λke−λk!,  k=0,1,2,…P(X{=}k) = \dfrac{\lambda^k e^{-\lambda}}{k!},\ \ k = 0, 1, 2, \ldots
the pmf — with a support that never closes
The trailing list ends in an ellipsis and nothing else: the support is all of N\mathbb{N}, unbounded — any count is possible, however unlikely. Compare the binomial's closed clause k=0,…,nk = 0, \ldots, n: the presence or absence of a final term is itself information.
CasesThe three factors split the labor: λk\lambda^k scales with the count, e−λe^{-\lambda} normalizes the whole family at once, and k!k! divides out orderings — with ee owned by the exponential function and the factorial by permutation notation.
Do not confuseA distribution over k!k! or λk\lambda^k alone. Dropping the ellipsis — or the normalizing e−λe^{-\lambda} — leaves an expression that no longer sums to one; the line works only as a whole, domain clause included.

Parameters


λ\lambda: the average rate (mean number of events), with λ>0\lambda >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 λ\lambda.

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

The single parameter λ\lambda controls both the mean and the variance of the distribution.
Poisson PMFs for lambda = 1, 4 and 10ky24681012141618blue: λ = 1amber: λ = 4navy: λ = 10mean λ, spread √λLarger λ: the peak moves right, widens and evens out
What λ does

Poisson PMFs for λ = 1, 4 and 10. For λ = 1 the bars at 0 and 1 are equal and tallest, and the rest fall away fast. For λ = 4 the peak is at 3 and 4. For λ = 10 the peak sits near 10 and the bars spread wider and more evenly on both sides: the mean is λ and the standard deviation √λ, so one number sets both where the bars sit and how wide they spread.

Mean and variance are both λ: the spread grows with the centre.

Probability Mass Function (PMF) and Support (Range)


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

Poisson PMF
P(X=k)=λke−λk!,k=0,1,2,…P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!}, \quad k = 0, 1, 2, \ldots
Learn more about this formula: Poisson PMF →


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:

Probability Mass Function (PMF) and Support (Range)
∑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.

0.0000.0500.1000.1500.2000.250024681012141618Value (k)Probability P(X = k)
Poisson PMF, lambda = 3

The bars peak at 2 and 3, the integers around lambda, and fall off in a right tail that never quite ends; each height is lambda to the k times e to the minus lambda over k factorial. Small lambda piles the mass near zero, large lambda makes the pattern almost symmetric. Change lambda and watch the peak follow it on the PMF visualizer.

The CDF adds these bars from the left.

Cumulative Distribution Function (CDF)


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

Cumulative Distribution Function (CDF)
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:

Cumulative Distribution Function (CDF)
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:

Cumulative Distribution Function (CDF)
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":

Cumulative Distribution Function (CDF)
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.
0.00.20.40.60.81.0024681012141618Value (k)Cumulative Probability F(k)
Poisson CDF, lambda = 3

The event of at most k occurrences has probability equal to the sum of the first k plus 1 bars, and the staircase shows that sum passing one half between 2 and 3 for lambda equal to 3. The final steps are tiny because the tail bars are small. Read P(X ≤ k) at any step on the discrete CDF explorer.

The mean and variance both equal lambda.

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:

Expected Value (Discrete)
E[X]=∑xx⋅P(X=x)E[X] = \sum_{x} x \cdot P(X = x)
Learn more about this formula: Expected Value (Discrete) →


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

Formula


Poisson Mean
E[X]=λE[X] = \lambda
Learn more about this formula: Poisson Mean →


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:

Expected Value (Mean)
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:

Expected Value (Mean)
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:

Expected Value (Mean)
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:

Expected Value (Mean)
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.

To compute the expected value of a Poisson distribution for your own parameters, step by step, use the Poisson expected value calculator.

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:

Variance and Standard Deviation
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:

Variance and Standard Deviation
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


Poisson Variance
Var(X)=λ\mathrm{Var}(X) = \lambda
Learn more about this formula: Poisson Variance →


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!}:

Variance and Standard Deviation
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:

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

Applying the shortcut formula:

Variance and Standard Deviation
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


Variance and Standard Deviation
σ=λ\sigma = \sqrt{\lambda}

Example


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

Variance and Standard Deviation
Var(X)=16\mathrm{Var}(X) = 16

Variance and Standard Deviation
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:

Mode and Median
∑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.

Poisson PMF with lambda = 5: the bars at 4 and 5 are equal, two modesky1234567891011121314P(X = 4) = P(X = 5) ≈ 0.175ratio of neighbours: λ/kk = 5: ratio 5/5 = 1λ = 5, an integer: two equal tallest bars at 4 and 5
Two modes when λ is an integer

The Poisson PMF for λ = 5. Each bar is λ/k times the bar before it, so the bars grow while k < 5 and shrink once k > 5. At k = 5 the ratio is exactly 1, so P(X = 5) = P(X = 4) ≈ 0.175: two equal tallest bars. For a non-integer λ such as 4.7 the ratio never equals 1 and the single mode is ⌊4.7⌋ = 4.

The ratio λ/k decides the mode: bars rise while it is above 1 and fall once it is below 1.

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.

Poisson variables also add: independent Poisson counts combine into another Poisson count whose rate is the sum of the rates.

Poisson Sum (Independent)
X∼Poisson(λX),  Y∼Poisson(λY)  ⟹  X+Y∼Poisson(λX+λY)X \sim \text{Poisson}(\lambda_X),\; Y \sim \text{Poisson}(\lambda_Y) \implies X + Y \sim \text{Poisson}(\lambda_X + \lambda_Y)
Learn more about this formula: Poisson Sum (Independent) →


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

Poisson Distribution FAQ

Why does the Poisson variance equal the mean?

+
Because a single parameter fixes the whole distribution. λ is the only input, so it has to determine both the center and the spread — E[X] = Var(X) = λ. That gives a practical test: if count data has variance far above its mean the data are overdispersed and Poisson is the wrong model; far below, underdispersed.Read more →

How is the Poisson distribution related to the binomial?

+
Poisson is the limit of the binomial for rare events. When n is large and p small, Bin(n, p) ≈ Pois(np): the two binomial parameters collapse into the single product that survives. A common rule of thumb is n > 20 with p < 0.05. Counting rare successes among many trials behaves like a Poisson count.Read more →

Can λ be greater than 1?

+
Yes — λ = 4.2 is perfectly legal. λ is a rate, not a probability: it counts expected occurrences and lives on [0, ∞), with no upper bound. Capping it at 1 imports a constraint from the wrong kind of parameter. A λ of 4.2 simply means about four events are expected in the window being watched.Read more →

What happens to λ if you change the time window?

+
It scales with the window. λ is the average count over the interval being watched, so it silently carries that interval's size: double the window and the model becomes Pois(2λ). Process texts make the exposure explicit by writing λt. Quoting a λ without saying over what period is an incomplete specification.Read more →