Mean and Variance of a Poisson Count

From the Poisson Probability Mass Function to Its Moments

A Poisson count is described by one parameter, but that parameter has two useful interpretations: it is both the average count and the variance of the count. These facts are not part of the definition; they follow from summing the probabilities in the Poisson distribution. The derivation also gives a practical check on calculations and a way to quantify how much counts typically fluctuate.

Let \((N_t)_{t\ge0}\) be a Poisson counting process on a probability space \((\Omega,\mathcal{F},\mathbb{P})\), with state space \(\{0,1,2,\ldots\}\) and continuous time \(t\ge0\). At a fixed time \(t\), suppose the rate is \(\lambda\ge0\), and write \(\theta=\lambda t\). The single-count distribution is

For \(\theta=0\), this distribution means \(\mathbb{P}(N_t=0)=1\) and \(\mathbb{P}(N_t=k)=0\) for \(k\ge1\). For \(\theta>0\), the displayed formula applies directly. In either case the count is finite almost surely and has finite first and second moments. We derive those moments below. No conditioning on a filtration is needed: each calculation concerns the distribution of the count at one fixed time.

The factorial moment is convenient because multiplying the Poisson probability at \(k\) by \(k\), or by \(k(k-1)\), cancels part of the factorial in the denominator. The resulting sums are exponential series. We first establish both sums in a form that will give the mean and variance.

Proof. If \(\theta=0\), then \(N_t=0\) almost surely, so both equalities hold. Suppose \(\theta>0\). Using the probability mass function and the fact that the \(k=0\) term contributes zero to the first moment,

In the second line, \(k/k!=1/(k-1)!\), and in the third line we set \(r=k-1\). The exponential series sums to \(e^\theta\), which cancels the factor \(e^{-\theta}\). For the second factorial moment, the terms \(k=0,1\) vanish, and for \(k\ge2\), \(k(k-1)/k!=1/(k-2)!\). Therefore

Both sums are finite, so the moments used in these calculations exist. This proves the proposition. \(\square\)

Proof. The proposition gives \(\mathbb{E}[N_t]=\theta\) and \(\mathbb{E}[N_t(N_t-1)]=\theta^2\). Since \(N_t^2=N_t(N_t-1)+N_t\), taking expectations gives \(\mathbb{E}[N_t^2]=\theta^2+\theta\). Thus

When \(\theta=0\), the count is zero almost surely, so its mean and variance are both zero, as asserted. \(\square\)

What the Two Moments Tell You

Because \(\theta=\lambda t\), a larger rate or a longer observation time increases both the expected count and its variance in direct proportion. The standard deviation, the square root of the variance, is \(\sqrt{\lambda t}\). For \(\lambda t>0\), the coefficient of variation is the standard deviation divided by the mean, so it is \(1/\sqrt{\lambda t}\). Thus the standard deviation increases with the expected count, while the relative size of typical fluctuations decreases as the parameter grows. At \(\lambda t=0\), the count is constant; its coefficient of variation is not defined because its mean is zero.

The equality of mean and variance is a feature of the Poisson distribution, not a universal rule for counts. It is useful in two ways. First, the mean provides a quick check on a probability calculation or a proposed expected count. Second, comparing a count model’s variance with its mean helps assess whether the Poisson distribution is a plausible description. A discrepancy can flag a problem with the model or calculation, but agreement alone does not prove that the data follow a Poisson distribution.

A Common Variance Error

A frequent mistake is to treat the second factorial moment as if it were the second raw moment. They are not equal: \(N_t^2=N_t(N_t-1)+N_t\). For example, when \(\theta=2\), the second factorial moment is \(4\), while the second raw moment is \(4+2=6\). A wrong calculation that substitutes the factorial moment directly into the variance formula gives \(4-2^2=0\). The correction is to add the mean first: \(\mathbb{E}[N_t^2]=4+2=6\), then \(\operatorname{Var}(N_t)=6-2^2=2\).

When checking a proposed answer, keep the parameter and the moment distinct: calculate \(\theta=\lambda t\) with consistent time units, then compare the stated mean and variance with \(\theta\). A variance that differs from the mean is not compatible with this Poisson model at the specified rate and time. A match is a useful consistency check, not a proof that the Poisson assumptions are appropriate.