Poisson Thinning and Independent Splitting

Why Poisson Counts Split Cleanly

Suppose events arrive in a count \(N\), and each event is independently assigned to one of several categories. A natural question is whether the resulting category counts retain a simple distribution. When \(N\) is Poisson, the answer is especially useful: the category counts are themselves Poisson, and they are mutually independent. The key is to examine all category counts at once through their joint probability-generating function.

The distinction between a random total and a fixed total matters. Before conditioning, the category counts are independent. If the total is fixed at a positive value and at least two categories have positive assignment probabilities, the counts are dependent: an increase in one category leaves fewer events available for the others.

Setup and Joint Probability-Generating Function

Definition: Let \(N\sim\operatorname{Pois}(\lambda)\), where \(\lambda\geq 0\). For each event, independently assign it to exactly one of \(m\) categories, with probabilities \(p_1,\ldots,p_m\), where \(p_j\geq 0\) and \(\sum_{j=1}^m p_j=1\). Assume these assignments are independent of \(N\). Let \(N_j\) be the number assigned to category \(j\). The joint probability-generating function of \((N_1,\ldots,N_m)\) is \[ G(s_1,\ldots,s_m)=E\left[\prod_{j=1}^m s_j^{N_j}\right], \] for \(0\leq s_j\leq 1\).

One way to formalize the assignment mechanism is to generate an independent sequence of category labels, each having probabilities \(p_1,\ldots,p_m\), and use the first \(N\) labels. Given \(N=k\), each of the \(k\) events contributes a factor \(s_j\) with probability \(p_j\). Its expected contribution is therefore \(\sum_{j=1}^m p_js_j\), and independence across events gives the \(k\)th power of this expression.

Theorem: Under the setup above, the category counts \(N_1,\ldots,N_m\) are mutually independent, with \(N_j\sim\operatorname{Pois}(\lambda p_j)\) for each \(j\).

Proof by Joint PGF Factorization

For a Poisson random variable, \(E[a^N]=\exp\{\lambda(a-1)\}\) when \(0\leq a\leq 1\). Conditioning on \(N\) and then applying this identity gives

\begin{aligned} G(s_1,\ldots,s_m) &=E\left[E\left[\prod_{j=1}^m s_j^{N_j}\mid N\right]\right]\\ &=E\left[\left(\sum_{j=1}^m p_js_j\right)^N\right]\\ &=\exp\left\{\lambda\left(\sum_{j=1}^m p_js_j-1\right)\right\}\\ &=\prod_{j=1}^m\exp\{\lambda p_j(s_j-1)\}. \end{aligned}

The univariate PGF of a Poisson random variable with mean \(\lambda p_j\) is \(\exp\{\lambda p_j(s_j-1)\}\). Thus the final expression is the product of the \(m\) Poisson PGFs. To see why this establishes both the marginals and independence, expand each factor as

\exp\{\lambda p_j(s_j-1)\} =e^{-\lambda p_j}\sum_{k=0}^{\infty}\frac{(\lambda p_j)^k}{k!}s_j^k.

Multiplying these series, the coefficient of \(s_1^{k_1}\cdots s_m^{k_m}\) is \[ \prod_{j=1}^m e^{-\lambda p_j}\frac{(\lambda p_j)^{k_j}}{k_j!}. \] This is the joint probability that \(N_j=k_j\) for every \(j\), since a probability-generating function's coefficients give its probabilities. The coefficient is a product of Poisson probabilities, proving the claimed marginals and mutual independence. If \(\lambda p_j=0\), the corresponding count is identically zero, as the same formula indicates. \(\square\)

Worked Examples

Worked Example: Splitting a Count Into Three Categories

Let \(N\sim\operatorname{Pois}(6)\). Assign each event independently to categories 1, 2, and 3 with probabilities \(1/2\), \(1/3\), and \(1/6\). Find the distribution of each category count and calculate \(P(N_1=1,N_2=2,N_3=0)\).

The means are the total mean multiplied by the category probabilities: \[ \lambda p_1=6(1/2)=3,\qquad \lambda p_2=6(1/3)=2,\qquad \lambda p_3=6(1/6)=1. \] Therefore \(N_1,N_2,N_3\) are independent with distributions \(\operatorname{Pois}(3)\), \(\operatorname{Pois}(2)\), and \(\operatorname{Pois}(1)\), respectively. Independence gives \[ \begin{aligned} P(N_1=1,N_2=2,N_3=0) &=P(N_1=1)P(N_2=2)P(N_3=0)\\ &=\left(e^{-3}\frac{3^1}{1!}\right) \left(e^{-2}\frac{2^2}{2!}\right) \left(e^{-1}\frac{1^0}{0!}\right)\\ &=e^{-6}\cdot 3\cdot 2\cdot 1\\ &=6e^{-6}\approx 0.01487. \end{aligned} \] The category counts sum to \(1+2+0=3\), and the event being calculated includes the possibility that the original total was 3; it does not condition the total to equal 3.

Worked Example: A Two-Category Split With Unequal Rates

Let \(N\sim\operatorname{Pois}(12)\), and assign each event to category 1 with probability \(1/3\) and category 2 with probability \(2/3\). Find the joint probability \(P(N_1=2,N_2=5)\).

The category means are \(12(1/3)=4\) and \(12(2/3)=8\). The theorem gives independent \(N_1\sim\operatorname{Pois}(4)\) and \(N_2\sim\operatorname{Pois}(8)\). Hence \[ \begin{aligned} P(N_1=2,N_2=5) &=\left(e^{-4}\frac{4^2}{2!}\right) \left(e^{-8}\frac{8^5}{5!}\right)\\ &=e^{-12}\left(\frac{16}{2}\right) \left(\frac{32768}{120}\right)\\ &=e^{-12}(2184.5333\ldots)\\ &\approx 0.01342. \end{aligned} \] The exponents combine to \(e^{-12}\), and the two numerical factors multiply to approximately \(2184.5333\), giving the stated probability.

What Changes When the Total Is Fixed?

The PGF argument proves independence without conditioning on the total. Conditional on \(N=n\), however, the allocation has a different familiar form: it is multinomial. This conditional law follows by counting the category-label sequences that produce a specified vector of counts.

Result: For nonnegative integers \(k_1,\ldots,k_m\) summing to \(n\), \[ P(N_1=k_1,\ldots,N_m=k_m\mid N=n) =\frac{n!}{\prod_{j=1}^m k_j!}\prod_{j=1}^m p_j^{k_j}. \] If the counts do not sum to \(n\), this conditional probability is zero.

For a count vector summing to \(n\), there are \(n!/\prod_j k_j!\) distinct sequences of category labels with exactly those counts. Each such sequence has probability \(\prod_j p_j^{k_j}\), by independence of the event assignments. Multiplying the number of sequences by the probability of each sequence proves the formula.

For two distinct categories \(i\) and \(j\), define \(I_{ri}\) to be 1 when event \(r\) is assigned to category \(i\), and 0 otherwise. Given \(N=n\), \(N_i=\sum_{r=1}^n I_{ri}\) and \(N_j=\sum_{r=1}^n I_{rj}\). For a single event, \(I_{ri}I_{rj}=0\), since it cannot be assigned to both categories. Thus \[ \operatorname{Cov}(I_{ri},I_{rj}\mid N=n) =0-p_ip_j=-p_ip_j. \] Assignments for different events are independent, so their cross-event covariances are zero. Adding the covariances over the \(n\) events gives \[ \operatorname{Cov}(N_i,N_j\mid N=n)=-np_ip_j. \] When \(n>0\) and both \(p_i\) and \(p_j\) are positive, this covariance is nonzero, so those counts are dependent. If \(n=0\), all counts are constant zero and are independent. Likewise, if only one category has positive probability, its count is fixed at \(n\) and the other counts are zero; all are constant and independent.

Worked Example: Conditional Dependence Despite Unconditional Independence

Suppose \(N\sim\operatorname{Pois}(4)\), with each event independently assigned to category 1 with probability \(1/4\) and category 2 with probability \(3/4\). Compare the category counts unconditionally and conditional on \(N=4\).

Unconditionally, \(N_1\sim\operatorname{Pois}(4(1/4))=\operatorname{Pois}(1)\) and \(N_2\sim\operatorname{Pois}(4(3/4))=\operatorname{Pois}(3)\), independently. Conditional on \(N=4\), the two counts satisfy \(N_2=4-N_1\), and their covariance is \[ \operatorname{Cov}(N_1,N_2\mid N=4) =-4(1/4)(3/4)=-3/4. \] That negative, nonzero covariance confirms conditional dependence. For instance, \[ P(N_1=1,N_2=3\mid N=4) =\frac{4!}{1!3!}(1/4)(3/4)^3 =4\cdot\frac14\cdot\frac{27}{64} =\frac{27}{64}. \] Fixing the total removes the freedom for the two counts to vary independently.

Thinning, Interpretation, and a Common Pitfall

The two-category case is often called Poisson thinning. If each event is retained with probability \(p\), the retained count is \(\operatorname{Pois}(\lambda p)\), and the discarded count is \(\operatorname{Pois}(\lambda(1-p))\); these counts are independent. The result extends directly to any finite number of categories, including categories interpreted as event types, destinations, or outcomes.

A common mistake is to reason that because \(N_1+N_2=N\), the category counts must always be dependent. That identity does not make the counts dependent when \(N\) itself is random: the joint PGF factors, so the counts are unconditionally independent. The mistake is treating the random total as though it had already been fixed. Once one conditions on \(N=n\), the sum is fixed and, for \(n>0\) with at least two positive category probabilities, the counts are dependent. The degenerate cases \(n=0\) or only one positive category probability are exceptions because all category counts are then constant.

Key takeaway: Independent assignment of Poisson events with mean \(\lambda\) to categories of probabilities \(p_j\) produces mutually independent counts \(N_j\sim\operatorname{Pois}(\lambda p_j)\). Conditioning on the total instead gives a multinomial allocation and generally introduces dependence.

Check Your Understanding

Use the definitions and results above to answer the following questions.

If \(N\sim\operatorname{Pois}(15)\) and each event is assigned to category 1 with probability \(0.2\), what is the distribution of the category 1 count and of the count assigned elsewhere?
For three assignment probabilities \(p_1,p_2,p_3\), write the joint PGF of the category counts and show how its factorization identifies their means.
What is the conditional distribution of the category counts given \(N=n\), and what feature of its probability formula counts the possible event-label sequences?
When \(n>0\), why are two conditional category counts dependent if both corresponding probabilities are positive?
Why does the identity \(N_1+N_2=N\) not contradict the unconditional independence of \(N_1\) and \(N_2\)?