Poisson Thinning and Independent Counts

Concept & Intuition

Suppose total arrivals N in a unit interval follow a Poisson(λ) distribution, and each arrival is marked independently as retained with probability p or rejected with probability 1 - p. Conditional on a fixed total N = n, the retained count R and rejected count J = n - R are negatively correlated (since their sum is fixed at n).

Remarkably, when N itself is random and Poisson-distributed, the randomness in total arrivals exactly cancels this negative conditional correlation, making R and J unconditionally independent.

Mathematical Formulation

For any nonnegative integers r and j, the joint probability that R = r and J = j requires N = r + j total arrivals and exactly r retained marks:

P(R = r, J = j) = [e-λ λr+j / (r + j)!] · [((r + j)! / (r! j!)) pr (1 - p)j] = [e-λp (λp)r / r!] · [e-λ(1-p) (λ(1 - p))j / j!].

Thus R ~ Poisson(λp) and J ~ Poisson(λ(1 - p)) are independent, with Cov(R, J) = 0.

Worked Example & Boundary Behavior

Let arrivals follow a Poisson distribution with mean λ = 4, and each arrival is retained independently with probability p = 1/4 (retention quarters = 1). We want to find the expected number of retained and rejected arrivals, as well as the joint probability of R = 1 and J = 2.

1. Marginal Distributions: The retained arrivals R follow a Poisson distribution with mean λR = λp = 4(1/4) = 1. The rejected arrivals J follow a Poisson distribution with mean λJ = λ(1 - p) = 4(3/4) = 3.
2. Probability of R = 1: Using the Poisson PMF, P(R = 1) = e-1(1)1 / 1! ≈ 0.36788.
3. Probability of J = 2: Similarly, P(J = 2) = e-3(3)2 / 2! ≈ 0.22404.
4. Joint Probability: Since thinning a Poisson process yields two independent Poisson processes, the joint probability is the product of their marginals:
   P(R = 1, J = 2) = P(R = 1) · P(J = 2) ≈ 0.36788 · 0.22404 ≈ 0.08242.

Interactive Experiment

This interactive experiment allows you to visually and computationally verify that splitting a Poisson process yields two independent Poisson processes.

1. Open the Poisson Thinning module.
2. Set the Poisson mean to 4 and Retention quarters to 1 (p = 1/4).
3. Observe that the Marginal Means update to 1 for Retained and 3 for Rejected.
4. Set the Retained count to 1 and the Rejected count to 2, and verify that Joint P(R=1, J=2) equals the product of the marginals while Cov(R, J) remains 0.

Interpretation & Limitations

Independence of thinned counts holds only when the total count N is Poisson-distributed and marks are independent of N and of each other. The displayed 0..6 × 0..6 grid reports the exact unnormalized joint probabilities together with the explicit outside-grid tail probability P(R ≥ 7 ∪ J ≥ 7).

References

These references provide the formal foundation and proofs for the Poisson splitting mechanism:

• Random Services: Splitting a Poisson Process — Derives the split rates λp and λ(1 - p) and proves the independence of the resulting sub-processes.