Superposition Rates from Poisson Count Convolution

Why Independent Poisson Counts Add

Suppose several independent sources generate events, and at a chosen time we count the events from each source. The total count is the sum of the source counts. To find its distribution, we must account for every way the sources can contribute to a given total. The resulting sum is a convolution, and for Poisson counts that convolution has a particularly simple form: the total is Poisson, with parameter equal to the sum of the source parameters.

Throughout, all random variables are defined on a common probability space \((\Omega,\mathcal{F},\mathbb{P})\). A Poisson random variable with parameter \(\theta\ge 0\) has probabilities \(\mathbb{P}(X=k)=e^{-\theta}\theta^k/k!\) for integers \(k\ge0\) when \(\theta>0\); when \(\theta=0\), it equals zero with probability one. Independence is required for the convolution formula below.

The key calculation is for two variables. To have \(X+Y=n\), the first variable can equal any integer \(k\) from \(0\) through \(n\), while the second must then equal \(n-k\). These possibilities are disjoint, so their probabilities add.

Proof. First suppose \(X\) and \(Y\) are independent with Poisson parameters \(a,b\ge0\). If \(a+b=0\), nonnegativity gives \(a=b=0\); both variables are zero with probability one, so their sum is Poisson with parameter zero. Now suppose \(a+b>0\). For each integer \(n\ge0\), independence and the disjoint possibilities \(X=k,Y=n-k\) give \(\mathbb{P}(X+Y=n)=\sum_{k=0}^n\mathbb{P}(X=k)\mathbb{P}(Y=n-k)\). Substituting the Poisson probabilities yields \(e^{-(a+b)}\sum_{k=0}^n a^k b^{n-k}/(k!(n-k)!)\). Since \(1/(k!(n-k)!)=\binom{n}{k}/n!\), the binomial theorem makes this \(e^{-(a+b)}(a+b)^n/n!\). This is the Poisson probability with parameter \(a+b\). The calculation also covers \(a=0\) or \(b=0\): terms with a positive power of zero vanish, and the sum has the stated value. Thus the two-variable claim holds in all cases. For \(m>2\), apply the two-variable result successively. The sum \(X_1+X_2\) is Poisson with parameter \(\theta_1+\theta_2\) and is independent of \(X_3,\ldots,X_m\), because it is a function of \(X_1,X_2\), which are independent of those variables. Repeating this argument gives a Poisson sum with parameter \(\sum_{r=1}^m\theta_r\). If that total parameter is zero, every \(\theta_r=0\), so every count is zero with probability one. \(\square\)

From Parameters to Rates at a Fixed Time

For a Poisson process with rate \(\lambda_r\ge0\), the count \(N^{(r)}_t\) at a fixed time \(t\ge0\) has a Poisson distribution with parameter \(\lambda_r t\). For the result in this tutorial, consider \(m\) such processes on the common probability space, and assume their counts at the time under consideration are independent. Each count has state space \(\{0,1,2,\ldots\}\), and time is continuous, \(t\ge0\). Applying the theorem with \(\theta_r=\lambda_r t\) gives the following fixed-time conclusion.

This is a statement about counts at a fixed time. It does not, by itself, establish that the summed process has independent increments: a claim about counts over several time intervals needs a separate argument about their joint dependence. For a single observation time, however, the convolution result gives the full distribution of the total.

The proof also gives a useful general way to compute the probability of a total count \(n\): sum over all allocations of \(n\) events among the sources. The convolution theorem evaluates that sum without listing every allocation. When the question concerns the total count only, it is usually more efficient to add the parameters first and then use one Poisson probability than to enumerate allocations.

Zero Counts, Means, and Variances

The total-count formula also makes zero-count questions straightforward. If the total count is zero, every source count must be zero. For independent sources, the probability of this joint event is the product of the individual zero-count probabilities. The product becomes the zero-count probability for a Poisson variable whose parameter is the sum.

As another consequence, the total has mean and variance equal to its Poisson parameter. For completeness, if \(Z\) has Poisson parameter \(\theta>0\), direct summation gives \(\mathbb{E}Z=\sum_{k\ge1} k e^{-\theta}\theta^k/k!=\theta\). Also, \(\mathbb{E}[Z(Z-1)]=\sum_{k\ge2} k(k-1)e^{-\theta}\theta^k/k!=\theta^2\). Since \(Z^2=Z(Z-1)+Z\), it follows that \(\operatorname{Var}(Z)=\mathbb{E}Z^2-(\mathbb{E}Z)^2=\theta\). For \(\theta=0\), both mean and variance are zero because \(Z=0\) with probability one. Hence, at fixed time \(t\), the combined count has mean and variance \(t\sum_r\lambda_r\).

Why Independence Matters

Adding Poisson parameters is not a rule for arbitrary dependent counts. The convolution proof multiplies probabilities for pairs of values; that multiplication uses independence. For a concrete failure, let \(X\) and \(Y\) be the same random variable, with \(X=Y\) and each marginally Poisson with parameter \(1\). Then \(X+Y=2X\), so its possible values are even integers. In particular, \(\mathbb{P}(X+Y=0)=\mathbb{P}(X=0)=e^{-1}\), whereas a Poisson variable with the naively added parameter \(2\) would have zero-count probability \(e^{-2}\). These probabilities differ, so the sum is not Poisson with parameter \(2\). The marginals alone do not justify rate addition.

A related common mistake is to infer a process-level property from this one-time calculation. Knowing the distribution of the sum at each fixed time does not establish how the sum behaves jointly across different times. The calculation here identifies fixed-time total-count distributions under independence of the source counts at that time. Questions about independent increments require checking independence across disjoint time intervals as well.