Multicolor Thinning into Independent Poisson Streams

From One Poisson Stream to Several

Suppose arrivals to a system occur according to a Poisson process, and each arrival is independently assigned to one of several categories. Examples include calls routed to different services, inspected items classified by outcome, or customers assigned to different groups. Each category receives only some of the original arrivals. The key fact is that, when assignments are independent from arrival to arrival and use fixed probabilities, the resulting category streams are themselves Poisson processes—and are independent of one another.

Let \((\Omega,\mathcal{F},\mathbb{P})\) be a probability space. Let \((N_t)_{t\ge0}\) be a Poisson process with rate \(\lambda\ge0\), so \(N_0=0\), its increments over disjoint time intervals are independent, and \(N_{t+h}-N_t\) has a Poisson distribution with mean \(\lambda h\) for \(t,h\ge0\). For each arrival number \(k\ge1\), let \(Z_k\) be its category label in \(S=\{1,\ldots,m\}\), where \(m\) is a positive integer. Assume the labels are independent and identically distributed, independent of the Poisson process, and satisfy \(\mathbb{P}(Z_k=r)=p_r\), where \(p_r\ge0\) and \(\sum_{r=1}^m p_r=1\).

The result is stronger than saying that each category’s count has a Poisson distribution at a fixed time. It says that counts from different categories are independent, and that each category has independent increments over time. To establish both claims at once, consider a finite collection of pairwise disjoint time intervals \(I_1,\ldots,I_q\), with lengths \(h_1,\ldots,h_q\). Write \(C_{r\ell}\) for the number of category-\(r\) arrivals in interval \(I_\ell\).

The Joint Distribution and Independence

Proof. First suppose \(\lambda>0\); the same calculation below also applies when \(\lambda=0\). Let \(K_\ell\) be the total number of original arrivals in \(I_\ell\). The variables \(K_1,\ldots,K_q\) are independent, and \(K_\ell\) is Poisson with mean \(\lambda h_\ell\). For numbers \(z_{r\ell}\in[0,1]\), condition on the total arrivals in these intervals. Each arrival in \(I_\ell\) independently contributes a factor \(z_{r\ell}\) with probability \(p_r\). Thus the conditional probability generating function of the category counts is \[ \mathbb{E}\!\left[\prod_{\ell=1}^q\prod_{r=1}^m z_{r\ell}^{C_{r\ell}}\,\middle|\,K_1,\ldots,K_q\right] =\prod_{\ell=1}^q\left(\sum_{r=1}^m p_r z_{r\ell}\right)^{K_\ell}. \] Taking expectations and using the probability generating function \(\mathbb{E}[a^{K_\ell}]=\exp(\lambda h_\ell(a-1))\) of a Poisson variable gives \[ \mathbb{E}\!\left[\prod_{\ell=1}^q\prod_{r=1}^m z_{r\ell}^{C_{r\ell}}\right] =\prod_{\ell=1}^q \exp\!\left(\lambda h_\ell\left(\sum_{r=1}^m p_r z_{r\ell}-1\right)\right) =\prod_{\ell=1}^q\prod_{r=1}^m \exp\!\left(\lambda p_r h_\ell(z_{r\ell}-1)\right). \] The last expression is a product of the probability generating functions of Poisson variables with means \(\lambda p_r h_\ell\). Probability generating functions determine the probabilities of integer-valued vectors by their power-series coefficients. The factorization therefore proves the asserted distributions and the mutual independence of all the counts \(C_{r\ell}\). This holds also when a mean is zero, in which case that count equals zero with probability one.

Since the intervals were arbitrary, each category process starts at zero, has independent increments, and has Poisson increments with mean \(\lambda p_r h\) on an interval of length \(h\). These are the defining properties of a Poisson process of rate \(\lambda p_r\). The factorization across categories and intervals gives independence for every finite collection of increments. Equivalently, the category processes are independent as processes: their finite-dimensional cylinder events are independent, and the standard \(\pi\)-\(\lambda\) argument extends this factorization to the sigma-fields generated by their paths. If \(\lambda=0\), there are no arrivals and every category process is identically zero, so the conclusions hold directly as well. \(\square\)

The calculation identifies why the result works. In one interval of length \(h\), the joint generating function factors as \[ \prod_{r=1}^m \exp\!\left(\lambda p_r h(z_r-1)\right). \] Each factor is the generating function of one category count. Independent assignments alone would not give this conclusion for every arrival process: the Poisson distribution of the original count is what makes the generating function exponential and produces the factorization.

Rates Add Back to the Original Rate

Proof. Each arrival by time \(t\) receives exactly one label in \(\{1,\ldots,m\}\). In the sum \(\sum_r N_r(t)\), it is therefore counted exactly once, proving the pathwise identity. For the rates, \(\sum_r\lambda p_r=\lambda\sum_r p_r=\lambda\), using the assumption that the category probabilities sum to one. \(\square\)

This gives a useful consistency check: the independent streams, when superposed, account for all original arrivals. The identity is pathwise, not merely an equality of expected counts. It also shows why rates can be calculated category by category without changing the overall arrival rate.

A Common Independence Pitfall

It is important to distinguish unconditional independence of the streams from independence after fixing the original number of arrivals. For example, take a rate-\(2\) Poisson process over one time unit and assign each arrival to either of two categories with probability \(\tfrac12\). If the original count is fixed at \(1\), the two category counts cannot both equal \(1\). In fact, each category count has conditional probability \(\tfrac12\) of being \(1\), while the conditional probability that both are \(1\) is \(0\), not \(\tfrac12\cdot\tfrac12=\tfrac14\). Thus the category counts are not independent under that conditioning.

Wrong calculation: Treating the category counts as independent after fixing the total at \(1\) gives \(\mathbb{P}(N_1=1,N_2=1\mid N_1+N_2=1)=\tfrac14\). This contradicts the fact that their sum is \(1\), so the conditional calculation is wrong.

Correction: The theorem asserts independence without conditioning on the original count. For instance, the unconditional probability that each category count is exactly \(1\) is \(e^{-1}e^{-1}=e^{-2}\), since each count is Poisson with mean \(1\) and the two streams are independent. Directly, this event requires exactly two original arrivals, which has probability \(e^{-2}2^2/2!=2e^{-2}\), followed by one arrival in each category, which has probability \(2(\tfrac12)^2=\tfrac12\). The product is \(e^{-2}\), agreeing with the independent-stream calculation.

For applications, this result lets each category be analysed as its own Poisson process with rate \(\lambda p_r\), while retaining independence between categories. Before using it, check that every arrival receives exactly one category, that the category probabilities are fixed, and that labels are independent across arrivals and independent of the original process. If those assumptions change, the conclusion need not hold.