Proving Independence of Retained and Rejected Streams

Why the Two Thinned Streams Are Independent

Suppose arrivals occur according to a Poisson process of rate \(\lambda\), and each arrival is independently retained with probability \(p\) or rejected with probability \(1-p\). The retained stream and rejected stream have the expected rates \(p\lambda\) and \((1-p)\lambda\). A stronger fact is also true: the two streams are independent processes. The key is to calculate their joint counts before making any claim about independence.

Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). Let \((N_t)_{t\ge0}\) be a Poisson process of rate \(\lambda\ge0\): \(N_0=0\), its increments on disjoint time intervals are independent, and \(N_{t+h}-N_t\) has a Poisson distribution with mean \(\lambda h\). Give each arrival an independent Bernoulli mark, independent also of the original process, with probability \(p\in[0,1]\) of being retained. Write \(N_t^R\) and \(N_t^J\) for the numbers of retained and rejected arrivals by time \(t\). Both are counting processes with state space \(\mathbb{N}_0\), indexed by continuous time \(t\ge0\).

We first calculate the two counts on a single interval of length \(h\). Assume \(\lambda h>0\) and \(0<p<1\) for the calculation; the endpoint cases will be addressed after the result. Let \(R\) and \(J\) be the retained and rejected counts on that interval. For nonnegative integers \(r,s\), the original count must be \(r+s\). Conditional on this total, there are \(\binom{r+s}{r}\) ways to assign \(r\) of its arrivals to the retained stream, and each such assignment has probability \(p^r(1-p)^s\). Therefore

The final line is a product of a Poisson probability with mean \(p\lambda h\) and a Poisson probability with mean \((1-p)\lambda h\). This factorization for every \(r,s\in\mathbb{N}_0\) proves that \(R\) and \(J\) are independent on this interval, with those respective Poisson distributions. The argument uses the unconditional joint probability; conditioning on the total alone does not make the two counts independent.

Proof. First take finitely many pairwise disjoint intervals \(I_1,\ldots,I_m\) with lengths \(h_1,\ldots,h_m\), and let \(R_\ell,J_\ell\) denote the retained and rejected counts on \(I_\ell\). The original counts on these intervals are independent. The marks assigned to arrivals in one interval are independent of the marks and counts in the other intervals. Consequently, the pairs \((R_\ell,J_\ell)\) are independent across \(\ell\). On each interval, the calculation above gives, for \(r_\ell,s_\ell\in\mathbb{N}_0\),

Multiplying these probabilities over \(\ell=1,\ldots,m\) shows that the entire vector of retained increments is independent of the entire vector of rejected increments. It also shows that increments within either stream on pairwise disjoint intervals are independent. Summing over the other count in the one-interval law gives the Poisson marginal laws: for example, \(\mathbb{P}(R=r)=e^{-p\lambda h}(p\lambda h)^r/r!\), because the probabilities for \(J=s\) sum to \(1\). The same argument gives the rejected marginal.

It remains to pass from increments to independence of the processes. Any finite vector of counts at times \(0\le t_1<\cdots<t_k\) can be written as sums of increments on the disjoint intervals \((0,t_1],(t_1,t_2],\ldots,(t_{k-1},t_k]\). Thus the finite-dimensional vector of retained counts is independent of any finite-dimensional vector of rejected counts, even if the selected times in the two vectors differ. The cylinder events determined by finite vectors of counts form generating classes for the two sigma-fields in the definition. Applying the standard \(\pi\)-\(\lambda\) theorem to extend independence from these generating classes to the generated sigma-fields proves that the processes are independent. The \(\pi\)-\(\lambda\) theorem is a standard measure-theoretic result; its proof is not needed for this count calculation. If \(p=0\), the retained process is identically zero; if \(p=1\), the rejected process is identically zero. If \(\lambda=0\), both processes are identically zero. A constant random process is independent of every process, so these degenerate cases satisfy the result as well. \(\square\)

Worked Count Calculations

What Independence Does—and Does Not—Say

Independence here is between the two process-generated sigma-fields. It does not say that arbitrary increments of one process are independent of each other. For example, with \(\lambda=1\) and \(p=\tfrac12\), the events \(N_1^R=0\) and \(N_2^R=0\) are nested: if there are no retained arrivals by time \(2\), then there are none by time \(1\). Their joint probability is \(\mathbb{P}(N_2^R=0)=e^{-1}\), whereas the product of their probabilities is \(e^{-1/2}e^{-1}=e^{-3/2}\). These differ. The independent-increment assertion applies to disjoint intervals, not to overlapping windows such as \((0,1]\) and \((0,2]\).

Common mistake: treating the streams as independent after conditioning on the total. If the original count on an interval is fixed to be \(1\), then exactly one of \(R,J\) equals \(1\) and the other equals \(0\). For \(p=\tfrac12\), \(\mathbb{P}(R=1,J=0\mid N=1)=\tfrac12\), but the conditional marginal probabilities are both \(\tfrac12\), whose product is \(\tfrac14\). Thus the two counts are not conditionally independent given the original total. The correct proof first computes the unconditional joint law: the random Poisson total produces the factorization into independent Poisson probabilities. Conditioning on a fixed total instead gives a binomial allocation with a fixed-sum constraint.

The factorization is useful beyond checking the two rates. It supplies the full joint distribution on each interval, and the independent-increment structure extends it to any finite family of disjoint intervals. When checking a claim about the streams, specify whether it concerns the two process sigma-fields, counts on a particular interval, or increments on disjoint intervals. Those are related statements, but they are not interchangeable.