= Dependence in Poisson thinning
{c}
Independent marking is the hypothesis that guarantees the <Poisson thinning theorem>. Equal marginal retention <probabilities> do not guarantee it. Consider one <Bernoulli random variable> $B$ with success <probability> $p\in(0,1)$, independent of a <Poisson process>, and retain every arrival when $B=1$ and none when $B=0$. Each arrival is retained with <probability> $p$, but the marks share the same coin. In an interval whose original count has <Poisson distribution> of <mean> $m>0$, the retained count is $R=BN$ and the rejected count is $D=(1-B)N$, so
$$
\mathbb E[R]=pm,\qquad \operatorname{Var}(R)=pm+p(1-p)m^2,\qquad \operatorname{Cov}(R,D)=-p(1-p)m^2.
$$
The extra <variance> shows that $R$ is not <Poisson distributed>; the negative <covariance> proves dependence. Its no-arrival <probability> is $1-p+pe^{-m}$, rather than $e^{-pm}$. For the toy values $p=1/2,m=2$, both rules have retained <mean> $1$, but the common-coin rule has <variance> $2$ and no-arrival <probability> $(1+e^{-2})/2\approx0.567668$, compared with <variance> $1$ and no-arrival <probability> $e^{-1}\approx0.367879$ for independent marking. These formulas follow by conditioning on $B$ and using the <law of total variance>.
An arrival-dependent or adversarial retention rule can change the law just as strongly. Keeping every second arrival gives $R_t=\lfloor N_t/2\rfloor$. Its successive retained gaps are sums of two <independent> <exponential random variables>, hence have the <Erlang distribution> of shape two and rate $\lambda$, not an <exponential distribution>. The resulting <renewal process> and its count <variance> are analyzed in <Every second Poisson arrival has underdispersed counts>, and the two dependent streams in <alternating split of a Poisson process>. As an even more restrictive rule, keeping only the first arrival ever gives $R_t=\mathbf1_{\{N_t\geq1\}}$, a <Bernoulli distribution> at each $t>0$, so it cannot have a positive-mean <Poisson distribution>. A thinning rule must specify how decisions depend on arrivals and on one another; the independent-marking conclusion cannot be applied automatically to arbitrary deletion.
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