Retained Counts Conditional on the Original Count

Fixing the Original Count Changes the Distribution

Suppose events arrive according to a Poisson process, and each event is independently retained with a fixed probability. The retained count has two different descriptions, depending on what is known. If the original number of arrivals is fixed at \(n\), the retained count is binomial: it counts successes in \(n\) independent retention trials. If the original count is allowed to vary, averaging over its Poisson distribution gives a Poisson retained count. These statements concern different probability laws, not competing answers to the same conditioning question.

Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). Let \((N_t)_{t\ge0}\) be a Poisson process with rate \(\lambda\ge0\). Write \(S_k\) for its \(k\)th arrival time, and let \(I_k\) be the indicator that its \(k\)th arrival is retained. Assume that the \(I_k\) are independent Bernoulli random variables with \(\mathbb{P}(I_k=1)=p\), where \(0\le p\le1\), and that the entire sequence of marks is independent of the Poisson process. The retained count by time \(t\) is \(R(t)=\sum_{k=1}^{N_t} I_k\), with an empty sum equal to zero. For definiteness, the information available through time \(t\) is the filtration \(\mathcal{F}_t=\sigma(N_s,I_k\mathbf{1}_{\{S_k\le s\}}:0\le s\le t,\ k\ge1)\). Set \(\theta=\lambda t\). When \(\theta>0\), \(N_t\) has the Poisson distribution with mean \(\theta\), and every event \(\{N_t=n\}\), \(n\ge0\), has positive probability.

Proof. On the event \(N_t=n\), the retained count is the sum \(I_1+\cdots+I_n\). Independence of the marks from \(N_t\) means that conditioning on \(N_t=n\) leaves these \(n\) marks independent, each with probability \(p\) of being one. For any particular set of \(r\) indices to be retained and the other \(n-r\) indices to be rejected, the conditional probability is \(p^r(1-p)^{n-r}\). There are \(\binom nr\) such sets, and their events are disjoint. Summing their probabilities gives the displayed formula. If \(r<0\) or \(r>n\), no such set exists. When \(p=0\), all \(n\) indicators equal zero with probability one; when \(p=1\), all equal one. These direct cases also cover the endpoint interpretations of the formula. \(\square\)

This is a conditional distribution: it describes uncertainty in the marks after the original count has been fixed. In particular, the number of possible retained counts is limited to \(0,\ldots,n\). Its conditional mean and variance follow from the same sum of indicators.

Proof. On \(\{N_t=n\}\), write \(R(t)=\sum_{k=1}^n I_k\). Each indicator has conditional mean \(p\) and conditional variance \(p(1-p)\), because the marks are independent of \(N_t\). Linearity of conditional expectation gives \(\mathbb{E}[R(t)\mid N_t=n]=\sum_{k=1}^n p=np\). Conditional independence makes the covariance between distinct indicators zero, so the variance of the sum is \(\sum_{k=1}^n p(1-p)=np(1-p)\). If \(n=0\), the sum is empty and both moments are zero, as the formulas state. \(\square\)

Allowing the Original Count to Vary

Without conditioning on \(N_t\), the retained count is a mixture over all possible original counts. Each term in that mixture uses the conditional binomial law, weighted by the probability of the corresponding Poisson count. The resulting distribution is not binomial with a fixed number of trials: the number of trials itself is random.

Proof. First take \(\theta>0\) and \(0<p<1\). For \(r\ge0\), the event \(R(t)=r\) can occur only when \(N_t=n\) for some \(n\ge r\). Conditioning on \(N_t\), and using the Poisson probability for \(N_t=n\) and the conditional binomial probability, gives \(\mathbb{P}(R(t)=r)=\sum_{n=r}^{\infty}e^{-\theta}\frac{\theta^n}{n!}\binom nr p^r(1-p)^{n-r}\). Set \(m=n-r\), so \(n!=(m+r)!\) and \(\binom{m+r}{r}=(m+r)!/(r!m!)\). Substitution and cancellation yield \[ \mathbb{P}(R(t)=r) =e^{-\theta}\frac{(\theta p)^r}{r!} \sum_{m=0}^{\infty}\frac{(\theta(1-p))^m}{m!} =e^{-\theta}\frac{(\theta p)^r}{r!}e^{\theta(1-p)} =e^{-\theta p}\frac{(\theta p)^r}{r!}. \] The series identity used here is the power-series expansion of the exponential function. If \(\theta=0\), there are no arrivals by time \(t\), so \(R(t)=0\). If \(p=0\), no arrival is retained. If \(p=1\), every arrival is retained, so \(R(t)=N_t\), which has mean \(\theta=\theta p\). These cases establish the result at the boundaries as well. \(\square\)

The conditional and unconditional laws have different supports and parameters. Given \(N_t=n\), the count cannot exceed \(n\), and its mean is \(np\). Unconditionally, there is no fixed upper bound on the retained count when \(\theta p>0\); its Poisson mean is \(\theta p\). The unconditional result follows by averaging over the original count, not by replacing \(n\) in the conditional formula with its mean.

Reversing the Conditioning

Conditioning on the original count and conditioning on the retained count are not interchangeable. The first leaves \(n\) Bernoulli retention decisions. The reverse question asks how many original arrivals could have produced the observed retained count, and it allows any number of rejected arrivals. The next calculation makes that distinction explicit.

A Common Error and Its Correction

A frequent mistake is to treat the conditional retained count as Poisson with mean \(np\). For instance, with \(N_t=2\) fixed and \(p=\tfrac12\), the correct conditional probability of no retained events is \(\mathbb{P}(R(t)=0\mid N_t=2)=(1-\tfrac12)^2=\tfrac14\). If one incorrectly substitutes the conditional mean \(np=1\) into a Poisson formula, one obtains \(e^{-1}\) instead. The error is that a Poisson law with mean \(np\) describes neither the fixed-\(n\) retention experiment nor its conditional distribution. The correct conditional law is binomial; the Poisson law arises only after averaging over a Poisson-distributed original count.

When solving a retention problem, first identify what is being conditioned on. If the original count is specified as \(n\), use the binomial probability and restrict the retained count to \(0\le r\le n\). If no original count is specified, average the binomial probabilities over the Poisson law, or use the resulting Poisson distribution with mean \(\lambda t p\). Keeping those two steps separate prevents confusing a conditional law with its mixture.