Integrated Problems with Superposition, Thinning, and Arrival Times

Combining Streams Without Losing the Conditioning

Problems involving several arrival streams often have more than one layer of randomness. An arrival may come from one of several sources; it may be retained or rejected; and its location in time may be observed or conditioned upon. A reliable calculation keeps these layers separate until the relevant conditioning is clear. The key ingredients are the rates of the streams that remain after retention, the distribution of source labels given a total count, and the conditional law of arrival locations given that count.

Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). Let \(N^{(\ell)}_t\), for \(\ell=1,\dots,r\), be independent homogeneous Poisson processes on \(t\ge0\), with rates \(\lambda_\ell\ge0\). Independently for each arrival of source \(\ell\), retain it with probability \(a_\ell\in[0,1]\). All retention decisions are independent of the arrival processes and of one another. Write \(R^{(\ell)}_t\) for the number of retained arrivals of source \(\ell\) by time \(t\). The combined retained count is \(R_t=\sum_{\ell=1}^r R^{(\ell)}_t\). The retained rate of source \(\ell\) is \(r_\ell=\lambda_\ell a_\ell\), and the total retained rate is \(\Lambda=\sum_{\ell=1}^r r_\ell\).

Two Facts That Organize the Calculations

The first fact identifies the processes to use after thinning and combining. The second describes what can be said about labels and arrival locations once the total count is fixed. Both will be proved here so that the later calculations have explicit foundations.

Proof. Consider a time interval of length \(h\). For one source with rate \(\lambda_\ell\), let \(K\) be its original count on that interval, and let \(J\) be its retained count. Conditional on \(K\), each of its \(K\) arrivals is retained independently with probability \(a_\ell\). For \(0\le z\le1\), conditioning on \(K\) gives \(\mathbb{E}[z^J\mid K]=(1-a_\ell+a_\ell z)^K\). Since \(K\) has a Poisson distribution with mean \(\lambda_\ell h\), \(\mathbb{E}[z^J]=\exp\{\lambda_\ell h((1-a_\ell+a_\ell z)-1)\}=\exp\{r_\ell h(z-1)\}\). This is the probability generating function of a Poisson random variable with mean \(r_\ell h\). Disjoint intervals have independent original counts and independent retention decisions, so retained counts in disjoint intervals are independent. Independent sources also have independent retained counts. Thus the retained processes are independent Poisson processes. Their sum has independent increments, and on an interval of length \(h\) its count has probability generating function \(\prod_{\ell=1}^r\exp\{r_\ell h(z-1)\}=\exp\{\Lambda h(z-1)\}\). This is Poisson with mean \(\Lambda h\), proving the assertion about the sum.

For the conditional statement, suppose first that \(\Lambda>0\), and consider \(T>0\). For an ordered sequence of source labels \((\ell_1,\dots,\ell_n)\), the joint density of ordered arrival times \(0<t_1<\cdots<t_n<T\) and that label sequence is \(e^{-\Lambda T}\prod_{k=1}^n r_{\ell_k}\). This follows from independent Poisson increments: each specified arrival contributes its source rate, and the probability of no other arrivals on the interval contributes the exponential factor. Summing over all label sequences gives the time density \(e^{-\Lambda T}\Lambda^n\) on the ordered region. The volume of that region is \(T^n/n!\), so \(\mathbb{P}(R_T=n)=e^{-\Lambda T}(\Lambda T)^n/n!\). Dividing the joint density by this probability gives the conditional density \(n!/T^n\prod_{k=1}^n(r_{\ell_k}/\Lambda)\). The factor \(n!/T^n\) is the density of \(n\) ordered independent uniforms; the remaining factor is the probability of the label sequence for independent labels with probabilities \(r_\ell/\Lambda\). This factorization proves both the stated conditional laws and their independence. For \(n=0\), there are no times or labels, and the assertion is vacuous. \(\square\)

The conditional label probabilities use the retained rates, not the original rates. Also, conditioning on the total count does not fix the arrival locations: their order statistics remain random. Those two points are central in the examples below.

Combining Labels with Arrival-Time Events

When the total count on \([0,T]\) is fixed, the theorem permits a useful separation: a condition on the ordered times can be calculated from uniform order statistics, and a condition on the labels can be calculated from the categorical label probabilities. Their probabilities multiply when the two conditions are considered together, because the times and labels are conditionally independent. This multiplication requires that both probabilities refer to the same conditioning event, \(R_T=n\).

Conditioning on Both the Original Count and Retention

A different conditional question fixes the number of original arrivals and the number retained. In that setting, it is not correct to treat retained arrivals as a fresh Poisson count without regard to the conditioning. Given \(m\) original arrivals and exactly \(k\) retained, the retained subset is uniformly distributed over the \(\binom{m}{k}\) subsets of arrival indices: before conditioning, every particular subset of size \(k\) has the same probability \(a^k(1-a)^{m-k}\), provided \(0<a<1\). The locations and retention decisions are independent, so the selected indices do not favor early or late locations.

A Common Rate Error and a Practical Checklist

Common mistake. In the first example, using the original rates to calculate the conditional source probability gives \(\mathbb{P}(\text{A label}\mid\text{retained arrival})=2/(2+1)=\tfrac23\). That is wrong: it ignores that A arrivals are retained only half the time. After thinning, the relevant rates are \(2\cdot\tfrac12=1\) for A and \(1\cdot\tfrac23=\tfrac23\) for B. The corrected probability is \(1/(1+\tfrac23)=\tfrac35\). Substituting \(\tfrac35\) into the five-arrival calculation gives \(\binom{5}{3}(\tfrac35)^3(\tfrac25)^2=\tfrac{216}{625}\), rather than the result obtained from the unthinned rates.

Before solving a multi-step problem, identify which count is fixed: the original count, the retained count, or the combined count. Then calculate the relevant retained rates and check that the conditioning event has positive probability. Given a fixed combined count, use uniform order statistics for locations and the retained-rate proportions for source labels. If the original count and the number retained are both fixed, instead reason about a uniformly selected subset of the original arrival indices. These are different conditional setups, even though both involve retention.

• For a source-specific retained count, use mean \(\lambda_\ell a_\ell T\) on \([0,T]\).
• For a combined retained count, use total rate \(\Lambda=\sum_\ell\lambda_\ell a_\ell\).
• Given the combined count, use label probabilities \(\lambda_\ell a_\ell/\Lambda\), provided \(\Lambda>0\).
• Separate label and time events only under the fixed-total-count conditioning where their conditional independence applies.
• When both the original count and retained count are fixed, account for the selected subset rather than substituting an unconditional Poisson law.