Poisson Additivity

Why Poisson Rates Add

A Poisson variable models a count over a specified interval or exposure. If two independent sources contribute counts to the same total, it is natural to ask whether that total is still Poisson and, if so, what its rate is. The answer is exact: the rates add. The independence condition matters, and transform methods make both the conclusion and that condition transparent.

We prove the identity in two ways. A probability-generating function (PGF) is particularly suited to nonnegative integer-valued variables; a characteristic function (CF) gives the same identification using a transform defined for every real argument. These methods also illustrate a general principle from transform products for independent sums: independence turns the transform of a sum into a product.

Poisson Transforms

For a nonnegative integer-valued variable, its PGF is \(G_X(z)=E[z^X]\) for \(|z|\leq 1\), with the convention \(z^0=1\). Its CF is \(\phi_X(t)=E[e^{itX}]\) for real \(t\), where \(i^2=-1\). For a Poisson variable, both transforms follow directly from the probability mass function.

The series calculation is valid for every \(|z|\leq1\): the exponential power series converges absolutely, and the Poisson probabilities sum to one. Substituting \(e^{it}\), whose absolute value is one, gives the CF formula. In particular, a Poisson PGF is determined by its rate, and so is its CF.

Poisson Additivity by Probability-Generating Functions

The additivity statement concerns a finite collection of independent counts. The rate of each count may be zero; such a variable is simply zero with probability one. A finite collection ensures that the sum of the rates is finite and that no issue about infinite sums arises.

Proof using PGFs. Set \(\Lambda=\sum_{j=1}^{m}\lambda_j\). Since the \(X_j\) are independent and \(z^{X_j}\) is bounded in absolute value by one for \(|z|\leq1\), expectations factor: $$ G_S(z)=E\left[z^{\sum_{j=1}^{m}X_j}\right] =E\left[\prod_{j=1}^{m}z^{X_j}\right] =\prod_{j=1}^{m}E[z^{X_j}]. $$ Using the Poisson PGF for each factor gives $$ G_S(z)=\prod_{j=1}^{m}e^{\lambda_j(z-1)} =e^{(\sum_{j=1}^{m}\lambda_j)(z-1)} =e^{\Lambda(z-1)}. $$ This is the PGF of a \(\operatorname{Poisson}(\Lambda)\) variable. A PGF determines the probabilities of a nonnegative integer-valued variable: its power-series coefficient of \(z^k\) is \(P(S=k)\). Therefore \(S\) has that Poisson distribution. If \(\Lambda=0\), the same formula is \(G_S(z)=1\), corresponding to \(S=0\) with probability one. \(\square\)

The coefficient argument also gives the resulting mass function explicitly. Expanding the final PGF, $$ G_S(z)=e^{-\Lambda}e^{\Lambda z} =e^{-\Lambda}\sum_{k=0}^{\infty}\frac{\Lambda^k z^k}{k!}, $$ shows that \(P(S=k)=e^{-\Lambda}\Lambda^k/k!\) for each nonnegative integer \(k\). Thus additivity identifies the whole distribution, not merely its mean.

The Same Identification by Characteristic Functions

The CF route uses the same hypotheses but a different uniqueness result. Characteristic functions determine probability distributions, so equality of the CFs is enough to establish equality in distribution. Here the transforms can be calculated for every real \(t\).

Proof using CFs. Independence gives the product formula for the CF of a sum: $$ \phi_S(t)=\prod_{j=1}^{m}\phi_{X_j}(t). $$ For each \(j\), the Poisson CF derived above is \(\phi_{X_j}(t)=e^{\lambda_j(e^{it}-1)}\). Therefore $$ \phi_S(t) =\prod_{j=1}^{m}e^{\lambda_j(e^{it}-1)} =e^{(\sum_{j=1}^{m}\lambda_j)(e^{it}-1)}. $$ This is exactly the CF of \(\operatorname{Poisson}(\Lambda)\), where \(\Lambda=\sum_{j=1}^{m}\lambda_j\). By uniqueness of characteristic functions, \(S\sim\operatorname{Poisson}(\Lambda)\). This proves the same additivity statement by the CF method. \(\square\)

The two proofs agree, but their tools emphasize different features. The PGF calculation directly concerns probabilities on the nonnegative integers and displays the mass function through coefficients. The CF proof uses uniqueness of characteristic functions and fits into the broader transform approach to distributional identities. Neither proof says that matching just a few moments would be enough to identify the distribution.

Interpreting Rates and Recognizing the Independence Condition

A rate is tied to the count being modeled: it is the parameter in that count's Poisson law. When independent components are combined, their rates add because their PGFs and CFs multiply. For example, if counts from two disjoint sources are independent and have rates \(3\) and \(5\) over the same observation window, their combined count is Poisson with rate \(8\). The calculation does not require the component rates to be equal.

Independence cannot be dropped merely because the separate variables have Poisson distributions. A shared source of randomness or another dependence structure can change the law of the total. The following example makes this failure explicit.

A useful practical habit is to check the model before applying the theorem: identify the component counts, verify that each has a Poisson distribution, and justify their mutual independence. If these checks hold, the combined count has the sum of the component rates. If independence is uncertain, the marginal Poisson distributions alone do not determine the distribution of the total.

Common Mistake

• Wrong reasoning: “Both counts are Poisson, so their sum is Poisson with the rates added.” This omits the condition that the variables be independent.
• Correction: State and verify independence before multiplying transforms. With independence, the PGF or CF of the sum is the product of the component transforms; without it, that product identity need not hold.
• Another overclaim: The theorem does not assert that arbitrary weighted sums are Poisson. It concerns the unweighted sum of independent Poisson variables.