The Poisson Probability Mass Function

Why a Count Needs a Probability Model

Suppose a laboratory records how many equipment alerts occur during a shift. The count might be zero, one, two, or another nonnegative whole number. A probability model describes how plausible each of these possible counts is under specified conditions. For count settings that fit a Poisson model, the Poisson probability mass function assigns a probability to each individual count.

The preceding material on recognizing Poisson count settings concerns when this model is suitable. Here, assume the observation window is defined and the count is reasonably modelled by a Poisson distribution: the event rate is stable over that window, and events arise independently. The probability mass function, or PMF, then translates the model’s parameter into probabilities for particular counts.

Definition: A random variable $X$ has a Poisson distribution with parameter $\lambda$, written $X\sim\operatorname{Poisson}(\lambda)$, if $\lambda>0$, its possible values are the nonnegative integers $0,1,2,\ldots$, and its probability mass function is $$ P(X=k)=\frac{e^{-\lambda}\lambda^k}{k!},\qquad k=0,1,2,\ldots. $$ Here $e$ is the base of the natural logarithm, and $k!$ is the factorial of $k$, with $0!=1$. For any value $x$ that is negative or is not an integer, $P(X=x)=0$.

The parameter $\lambda$ is positive and determines the probabilities assigned to the possible counts. In a count setting it represents the model’s expected number of events over the specified observation window. Its value is tied to that window: a parameter for one shift, area, or interval is not automatically the parameter for a different one. How rates and exposure relate is a separate question; here, take $\lambda$ as given for the count being modelled.

The formula is an individual-count formula. In $P(X=k)$, the event is $X=k$: the count equals exactly $k$. For instance, $P(X=2)$ is the probability of exactly two events, not the probability of two or fewer events. A probability involving several possible counts requires combining the probabilities for those individual outcomes.

Reading the Formula

Each part of the PMF has a specific role. The factor $e^{-\lambda}$ depends on the model parameter; $\lambda^k$ depends on both the parameter and the count being evaluated; and $k!$ adjusts for the count. The factorial is the product of all positive integers from 1 through $k$, so $4!=4\cdot3\cdot2\cdot1=24$. The special convention $0!=1$ makes the formula work at zero as well as at positive counts.

At $k=0$, the PMF simplifies because $\lambda^0=1$ and $0!=1$. Thus the probability of a zero count is $e^{-\lambda}$. At a positive count, substitute that integer for $k$, calculate the corresponding power and factorial, and evaluate the resulting expression. The parameter itself need not be an integer.

Worked Example: Probability of No Equipment Alerts

Consider a hypothetical monitoring system for which a shift has been judged suitable for a Poisson model with $\lambda=1.5$ alerts per shift. Find the model probability of no alerts during one shift.

The count $X$ is modelled as $X\sim\operatorname{Poisson}(1.5)$. Zero is a permitted count, so substitute $k=0$ into the PMF:

\begin{aligned} P(X=0) &=\frac{e^{-1.5}(1.5)^0}{0!}\\ &=e^{-1.5}\\ &\approx 0.2231. \end{aligned}

The model assigns probability about 0.2231, or 22.31%, to exactly zero alerts in a shift. This is a probability for the specified one-count outcome; it does not describe all counts up to some threshold.

Worked Example: Probability of Exactly Three Calls

Suppose, hypothetically, that a help desk’s incoming call count during a defined hour is modelled as Poisson with $\lambda=2.4$ calls per hour. Find the probability of exactly three calls in that hour.

The assumptions for using the PMF are that the count in this fixed hour is being treated as Poisson with the stated parameter. With $k=3$:

\begin{aligned} P(X=3) &=\frac{e^{-2.4}(2.4)^3}{3!}\\ &=\frac{e^{-2.4}(13.824)}{6}\\ &=e^{-2.4}(2.304)\\ &\approx (0.09072)(2.304)\\ &\approx 0.2090. \end{aligned}

Thus the model assigns about 20.90% probability to exactly three calls in the hour. The calculation evaluates one term of the PMF; it does not include the probabilities for zero, one, or two calls.

Why the Formula Defines a Probability Distribution

A probability mass function must assign nonnegative probabilities, and the probabilities over all possible outcomes must sum to one. The Poisson expression is nonnegative for every permitted $k$, since $e^{-\lambda}$, $\lambda^k$, and $k!$ are positive. Its total probability is one, as follows from the power series for the exponential function.

Sum over the support.
The possible outcomes are exactly the integers $k=0,1,2,\ldots$, so add the PMF over those values.

Factor out the constant.
The term $e^{-\lambda}$ does not depend on $k$, so it can be taken outside the sum.

Apply the exponential series.
For any real number $z$, $e^z=\sum_{k=0}^{\infty}z^k/k!$. Set $z=\lambda$ to evaluate the remaining sum.

\begin{aligned} \sum_{k=0}^{\infty}P(X=k) &=\sum_{k=0}^{\infty}\frac{e^{-\lambda}\lambda^k}{k!}\\ &=e^{-\lambda}\sum_{k=0}^{\infty}\frac{\lambda^k}{k!}\\ &=e^{-\lambda}e^\lambda\\ &=1. \end{aligned}

This verifies that the formula allocates all probability across its complete support. The sum is infinite because there is no fixed largest nonnegative integer count in the model, but the exponential series shows that the total probability is exactly one. This justification is about the validity of the PMF, not about how often a particular count will occur in a real sequence of observations.

Worked Example: Probability of Exactly Four Events

Imagine a hypothetical sensor whose event count in a fixed observation window is modelled as Poisson with $\lambda=3$. Find the probability assigned to exactly four events.

Four is a nonnegative integer and therefore lies in the support. Substitute $k=4$, use $4!=24$, and evaluate:

\begin{aligned} P(X=4) &=\frac{e^{-3}(3)^4}{4!}\\ &=\frac{e^{-3}(81)}{24}\\ &=e^{-3}(3.375)\\ &\approx (0.04979)(3.375)\\ &\approx 0.1680. \end{aligned}

The model assigns approximately 0.1680, or 16.80%, probability to exactly four events in the window. The count of four is one possible outcome, and this answer is its individual probability under the stated model.

Support and Common Mistakes

The support—the set of outcomes that can have positive probability—is an essential part of the definition, not an optional detail. A count cannot be negative or fractional. For example, $P(X=-1)=0$ and $P(X=2.5)=0$, even though inserting those numbers into the algebraic expression would produce numbers that are not probabilities for valid count outcomes. The PMF formula is stated for $k=0,1,2,\ldots$; it should not be used as a probability formula for arbitrary real values.

A related mistake is to confuse $P(X=k)$ with a cumulative probability. The expression $P(X=3)$ means exactly three. In contrast, a question about at most three includes four outcomes, $0,1,2,3$, and its probability is the sum of their separate PMF values. Cumulative and tail calculations require that sum and are treated separately from the individual-count formula.

Forgetting the factorial. The denominator is $k!$, not $k$. For $k=3$, use $3!=6$.
Using the wrong count in the exponent. The power is $\lambda^k$; the parameter is not raised to an unrelated value.
Ignoring zero. Zero is in the support, and $0!=1$, so the formula applies directly.
Reporting a cumulative interpretation. A single PMF value refers to exactly one count, not a range of counts.

For example, someone might say, “$P(X=3)$ is the probability of three or fewer events because three is the cutoff.” That reasoning is incorrect: the equality symbol names only the outcome of exactly three. To describe three or fewer, add $P(X=0)+P(X=1)+P(X=2)+P(X=3)$. Keeping the event statement visible while calculating is a reliable way to avoid this mix-up.

Key takeaway: For a Poisson count $X$ with parameter $\lambda>0$, the outcome is a nonnegative integer and its individual probability is $P(X=k)=e^{-\lambda}\lambda^k/k!$. Each permitted $k$ gives one probability; the probabilities over all permitted counts sum to one.

Check Your Understanding

For each question, distinguish the model’s support, parameter, and individual-count probability.

For $X\sim\operatorname{Poisson}(2.2)$, which values are in the support, and what does $P(X=5)$ describe?
Write the PMF expression for $P(X=0)$ when $X\sim\operatorname{Poisson}(\lambda)$. Which factorial convention is needed?
Explain why $P(X=1.7)=0$ for a Poisson count, even though $1.7$ is a real number.
Does $P(X=4)$ include the probabilities of counts zero through three? Explain what event it names.
Use the exponential series to explain why the Poisson PMF probabilities sum to one.

The Poisson Probability Mass Function

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