The Poisson Mean-Variance Relationship

Why the Poisson Mean and Variance Matter

The Poisson distribution is a standard model for nonnegative integer counts. Its defining mean-variance relationship is unusually simple: the variance equals the mean. That equality is not an extra assumption added after choosing the distribution; it follows from the Poisson probability mass function and, in particular, from the derivatives of its log-partition function.

This connection matters in generalized linear models because the mean describes the response's expected count, while the variance describes its variability around that mean. Deriving both from the same function makes clear which part of the relationship is fixed by the Poisson family and why its dispersion is called one.

Canonical Form and Log-Partition Function

For a Poisson random variable \(Y\) with mean parameter \(\mu>0\), the probability mass function is

Set the natural parameter to \(\theta=\log\mu\), so that \(\mu=e^\theta\) and \(\theta\) ranges over all real numbers. Rewriting the mass function gives

In the canonical exponential-family form \(\exp\{y\theta-b(\theta)+c(y)\}\), the log-partition function is \(b(\theta)=e^\theta\), and the data-only term is \(c(y)=-\log(y!)\). The support, the nonnegative integers, does not depend on \(\theta\).

The normalizing role of \(b\) can be checked by summing over the support:

Deriving the Mean and Variance

For a regular one-parameter exponential family with a fixed support and a differentiable log-partition function, differentiating \(b\) once gives the expectation, and differentiating twice gives the variance. Here the support is fixed and the relevant series can be differentiated term by term: on any bounded interval of \(\theta\), the series of derivatives is dominated by convergent exponential series. We can also verify the result directly from the Poisson mass function.

Proof. Let \(Z(\theta)=\sum_{y=0}^{\infty}e^{y\theta}/y!\). For any bounded interval \(|\theta|\le M\), the \(k\)th derivative terms are bounded in absolute value by \(y^k e^{My}/y!\). The series of these bounds converges for each fixed nonnegative integer \(k\), by the ratio test. Thus the series may be differentiated term by term there. Since \(Z(\theta)=e^{e^\theta}\), its log is \(b(\theta)=\log Z(\theta)=e^\theta\).

To connect these derivatives to moments, differentiate \(Z\). The probability of \(Y=y\) is \(e^{y\theta}/(y!Z(\theta))\), so

Differentiating the ratio once more yields

Because \(\log Z=b\), these identities say \(E[Y]=b'(\theta)\) and \(\operatorname{Var}(Y)=b''(\theta)\). Substituting \(b'(\theta)=b''(\theta)=e^\theta=\mu\) proves the result. \(\square\)

The variance conclusion can also be checked through the second factorial moment. For \(\mu>0\),

Since \(Y^2=Y(Y-1)+Y\), it follows that \(E[Y^2]=\mu^2+\mu\). Therefore

This second derivation checks the same relationship from the probability mass function rather than relying only on the log-partition derivative rule.

Worked Calculations

Why the Poisson Family Has Unit Dispersion

In an exponential-dispersion representation, the response variance is written as \(\operatorname{Var}(Y)=\phi V(\mu)\), where \(\phi\) is a dispersion factor and \(V(\mu)\) is the variance function. For the standard Poisson family, the mass function has no additional free scale parameter: its canonical form is \(\exp\{y\theta-b(\theta)+c(y)\}\), with no nontrivial divisor of the canonical term. Thus the standard family fixes \(\phi=1\). Since the variance derived above is \(\mu\), its variance function is \(V(\mu)=\mu\), and the variance law is \(1\cdot\mu=\mu\).

“Unit dispersion” does not mean the response variance is one. It means the multiplicative dispersion factor is one; the response variance still changes with the mean. For example, a Poisson response with mean \(4\) has variance \(4\), not variance \(1\). The value one belongs to the dispersion factor, not to every response variance.

This distinction is useful when interpreting a Poisson model: the family links the expected count to its variance. It also explains why checking the variance relative to the mean is meaningful when assessing whether the standard Poisson assumption is plausible. The present derivation establishes the model's implication; it does not assert that every real count dataset must follow it.

Common Mistake

A common error is to reason, “The Poisson dispersion is one, so the variance must be one.” This confuses the dispersion multiplier with the variance itself. The correct calculation is

so a mean of \(0.5\) gives variance \(0.5\), and a mean of \(4\) gives variance \(4\). A related error is to say that an observed count \(y\) has variance \(\mu\). Variance describes the distribution of the random variable \(Y\), not the fixed value observed in one realization. Once \(Y=y\) has been observed, \(y\) is a number; the model variance remains a property of the distribution that generated it.