Dependence in Poisson thinning

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 $D=(1-B)N$, soThe extra variance shows that $R$ is not Poisson distributed; the negative covariance proves dependence. Its no-arrival probability is $1-p+p{e}^{-m}$, 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 $(1+e^{-2})/2\approx 0.567668$, compared with variance $1$ and no-arrival probability $e^{-1}\approx 0.367879$ 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 $R_t=1_{\{N_t\ge 1\}}$, 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.