= Poisson thinning
{c}
<Poisson thinning> splits a <Poisson process> by independently marking each arrival as retained or rejected. Let the original <Homogeneous Poisson process> have rate $\lambda>0$, and let $p(s)\in[0,1]$ be a deterministic <measurable function> of time. Attach <independent> <uniform distributions> $U_1,U_2,\ldots$ on $(0,1)$, independent of the entire original <Poisson process>. At arrival time $T_i$, retain the arrival if $U_i\leq p(T_i)$. Given all arrival times, these decisions are <independent> and have the specified retention <probabilities>. Define the two <counting processes> using an <indicator function>:
$$
R_t=\sum_{i:T_i\leq t}\mathbf1_{\{U_i\leq p(T_i)\}},\qquad D_t=N_t-R_t.
$$
The <Poisson thinning theorem> gives <independent> retained and rejected <Inhomogeneous Poisson processes>. Their rates are $\lambda p(s)$ and $\lambda(1-p(s))$, respectively, in the sense of the <intensity function of a point process>. In particular, for $0\leq a<b$,
$$
R_b-R_a\sim\operatorname{Poisson}\!\left(\lambda\int_a^b p(s)\,ds\right),\qquad D_b-D_a\sim\operatorname{Poisson}\!\left(\lambda\int_a^b(1-p(s))\,ds\right).
$$
The <independence> is of the two whole processes: every finite collection of retained counts is <independent> of every finite collection of rejected counts. It is not merely zero <covariance> for one interval. For constant $p$, the streams are <Homogeneous Poisson processes> of rates $\lambda p$ and $\lambda(1-p)$; the endpoints $p=0,1$ include the identically zero stream.<Lectures on the Poisson Process by Günter Last and others>[Definition 5.7, Proposition 5.5, Theorem 5.8 and Corollary 5.9, printed pp. 42-44: location-dependent independent marking, intensities, and independence of the split processes]{ref}
For an exact toy calculation, take $\lambda=2$ per minute and retain with $p(s)=1/4$ on $(0,2]$ and $p(s)=3/4$ on $(2,4]$. In the first two minutes, the retained and rejected <expected values> are $2\cdot2\cdot1/4=1$ and $2\cdot2\cdot3/4=3$. In the next two minutes they are $3$ and $1$. Over all four minutes, $R_4,D_4$ are <independent> <Poisson distributions> with <mean> $4$ each, and their sum $N_4$ has <mean> $8$. Thus
$$
\mathbb P(R_4=0,D_4=0)=e^{-4}e^{-4}=e^{-8},\qquad \mathbb P(R_4=2,D_4=3)=e^{-8}\frac{4^2}{2!}\frac{4^3}{3!}=\frac{256}{3}e^{-8}.
$$
These <probabilities> and <expected values> describe the model, not empirical measurements. In the diagram, the original nine synthetic arrivals split into five retained and four rejected arrivals; those realized counts need not equal their <expected values>.
\Image[/poisson-independent-thinning-expected-counts.png]
{title=Independent marks split a synthetic Poisson arrival path, with exact expected counts for two time-dependent retention probabilities}
The mechanism explains why mutually exclusive labels can yield <independent> counts: randomness of the original total offsets the dependence from allocating a fixed total. For constant $p$ and an interval with original <mean> $m$, the conditional retained count has a <binomial distribution> given $N=n$. The joint <probability generating function> is
$$
\mathbb E[z^Rw^D]=\mathbb E[(pz+(1-p)w)^N]=\exp\{m(pz+(1-p)w-1)\}=e^{mp(z-1)}e^{m(1-p)(w-1)},\qquad 0\leq z,w\leq1.
$$
This factorization proves the two <independent> <Poisson distributions>; disjoint original intervals give <independent> pairs. For varying $p(s)$, the <Independent marking theorem for Poisson point processes> gives the same conclusion with the two integrated intensities. The <joint red and green Poisson count formula> gives the corresponding direct calculation with <probability mass functions>.
Independence here is unconditional. After conditioning on $N=n$, the identity $D=n-R$ gives <covariance> $\operatorname{Cov}(R,D\mid N=n)=-np(1-p)$ for constant $p$. Thus the retained and rejected counts are dependent for $n>0$ and $0<p<1$, even though their unconditional <covariance> is zero. <Poisson thinning> can model independent routing of arrivals, or independent detection of arrivals with a known efficiency schedule; using only a retention fraction without checking the marking mechanism can give a wrong count law.
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