The Poisson Distribution of a Single Count

A Count in a Fixed Observation Window

A rate describes how quickly events occur, while a count records how many have occurred by a specified time. For a Poisson process, the distribution of the count in a fixed window is determined by one dimensionless quantity: the rate multiplied by the window length. Once that parameter is computed, the probability of each possible count follows from a single formula.

Let \((\Omega,\mathcal{F},\mathbb{P})\) be a probability space, and let \((N_t)_{t\ge0}\) be a continuous-time counting process with state space \(\mathbb{N}_0=\{0,1,2,\ldots\}\). Thus \(N_t\) is the number of events counted from time \(0\) through time \(t\). Suppose the rate is \(\lambda>0\), measured in events per unit time. For an observation duration \(t\ge0\), define \(x=\lambda t\). The units cancel, so \(x\) is a count-scale parameter rather than a rate.

This specification concerns the count at one fixed time. It does not, by itself, describe the joint distribution of counts at several times. For \(t>0\) and \(\lambda>0\), the count probabilities are therefore

In particular, substituting \(k=0\) and \(k=1\) gives the probabilities of no events and exactly one event. The probability of two or more events is obtained by subtracting those two disjoint cases from \(1\). These formulas are exact; they need no approximation for a short or long interval.

Why the Formula Defines a Probability Distribution

Proof. Each \(p_k\) is nonnegative because \(e^{-x}>0\), \(x^k\ge0\), and \(k!>0\), with the stated convention at \(x=0,k=0\). For \(x>0\), the power series for the exponential function gives \(\sum_{k=0}^{\infty}x^k/k!=e^x\). Therefore \(\sum_{k=0}^{\infty}p_k=e^{-x}e^x=1\). For \(x=0\), \(p_0=1\), and \(p_k=0\) for every \(k\ge1\), so the sum is again \(1\). Nonnegative values summing to \(1\) are precisely a probability mass function. \(\square\)

For \(x>0\), consecutive probabilities also have a useful ratio:

Proof. Since \(x>0\), both \(p_k=e^{-x}x^k/k!\) and \(p_{k+1}=e^{-x}x^{k+1}/(k+1)!\) are positive. Dividing the second expression by the first cancels the common factor \(e^{-x}\), leaves one factor \(x\), and uses \((k+1)!=(k+1)k!\). Hence \(p_{k+1}/p_k=x/(k+1)\), as claimed. \(\square\)

The ratio is one way to check adjacent count probabilities, but it does not replace the probability formula: the parameter and the factorial both matter. For the counts most often requested, direct substitution gives

To verify the last line, the events \(N_t=0\), \(N_t=1\), and \(N_t\ge2\) are disjoint and together include every nonnegative integer count. Thus \(\mathbb{P}(N_t\ge2)=1-\mathbb{P}(N_t=0)-\mathbb{P}(N_t=1)\). For any specified integer \(r\ge0\), the exact-count probability is \(e^{-\lambda t}(\lambda t)^r/r!\); this is different from the probability of at least \(r\) events.

Worked Probabilities

Boundary Cases and a Common Mistake

If \(t=0\), the window has length zero and \(x=\lambda t=0\); the count is \(N_0=0\), so \(\mathbb{P}(N_0=0)=1\) and \(\mathbb{P}(N_0=k)=0\) for \(k\ge1\). The same degenerate distribution results if \(\lambda=0\). For the nondegenerate calculations above, both the rate and duration are positive. The formula at \(k=0\) remains \(e^{-x}\), including when \(x=0\).

Common mistake: using the rate as the parameter. Suppose \(\lambda=2\) events per hour and \(t=3\) hours. The incorrect calculation \(P(N_t=0)=e^{-2}\) uses the rate alone. The correct parameter is \(x=\lambda t=2\cdot3=6\), and therefore \(\mathbb{P}(N_t=0)=e^{-6}\). The units expose the error: a rate has units of events per hour, while the argument of an exponential probability must be dimensionless. Multiplying by the duration removes the time unit.

Another distinction to keep explicit is “exactly \(k\)” versus “at least \(k\).” The formula \(e^{-x}x^k/k!\) gives the probability of one value only. For example, at least two means summing the probabilities of every count \(2,3,\ldots\), or more simply taking the complement of counts zero and one. Using the complement gives an exact compact expression without truncating an infinite sum.

The Poisson single-count law is useful whenever events are counted over a fixed exposure period: it converts an estimated rate and a specified duration into probabilities for operationally meaningful outcomes. The key steps are to keep the units consistent, calculate \(x=\lambda t\), and decide whether the question asks for an exact count or a range of counts before substituting.