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Superposition and Thinning with Time-Varying Rates

Learn how superposition and time-dependent thinning change event rates, and verify both results through interval counts and worked calculations.

Advanced 10 min read

What You'll Learn

  • Define nonhomogeneous Poisson processes using their integrated rates and independent increments.
  • Prove the superposition result for independent processes with time-varying rates.
  • Compute event-count distributions after superposing processes.
  • Prove the thinning result for independent retention with probability depending on time.
  • Calculate retained event counts over intervals when rates or retention probabilities vary.
  • Recognize why thinning probabilities must be weighted by the event rate.

Combining and Retaining Events

Suppose several independent event streams operate at once, with each stream becoming more or less active over time. Their combined stream is still Poisson: at each time, the rates add. A related operation keeps each event independently with a probability that can change over time. The retained stream is also Poisson, but its rate is the original rate multiplied by the retention probability at that time. These two rules let us build new time-varying event models from simpler ones.

Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). A counting process \(N(t)\), \(t\ge 0\), is a nonhomogeneous Poisson process with rate function \(\lambda(t)\) if \(\lambda\) is nonnegative and locally integrable, \(N(0)=0\), and counts on disjoint time intervals are independent, with

$$ N(b)-N(a)\sim\operatorname{Poisson}\left(\int_a^b\lambda(u)\,du\right), \qquad 0\le a<b<\infty. $$

Here locally integrable means that the integral is finite on every bounded interval. Write \(\Lambda(a,b)=\int_a^b\lambda(u)\,du\) for the integrated rate on an interval. Rates may vary with time; it is the integrated rate, not the rate at a single instant, that determines the interval count distribution.

Definition: The superposition of counting processes \(N_1,\ldots,N_m\) is the process \(N(t)=\sum_{r=1}^m N_r(t)\), which counts all their events. Time-dependent thinning of \(N\) keeps each event at time \(t\) with probability \(p(t)\), independently from event to event conditional on the event times, where \(p\) is measurable and \(0\le p(t)\le1\).

Why the Rates Add

Assume \(N_1,\ldots,N_m\) are independent nonhomogeneous Poisson processes on the same time domain, with nonnegative, locally integrable rate functions \(\lambda_1,\ldots,\lambda_m\). Independence here means the processes are independent as random objects. On a fixed interval, their counts are therefore independent Poisson random variables. The following result shows that summing the processes preserves the defining properties of a nonhomogeneous Poisson process.

Theorem (Superposition): The process \(N(t)=\sum_{r=1}^m N_r(t)\) is a nonhomogeneous Poisson process with rate \(\lambda(t)=\sum_{r=1}^m\lambda_r(t)\). In particular, on \((a,b]\) its count has Poisson mean \(\int_a^b\sum_{r=1}^m\lambda_r(u)\,du\).

Proof. Fix \(0\le a<b\). For each \(r\), the count \(Y_r=N_r(b)-N_r(a)\) is Poisson with mean \(\theta_r=\int_a^b\lambda_r(u)\,du\), and the variables \(Y_1,\ldots,Y_m\) are independent. For \(0\le z\le1\), independence and the Poisson probability generating function give

$$ \mathbb{E}\left[z^{\sum_{r=1}^mY_r}\right] =\prod_{r=1}^m\mathbb{E}[z^{Y_r}] =\prod_{r=1}^m e^{\theta_r(z-1)} =e^{(\sum_{r=1}^m\theta_r)(z-1)}. $$

This is the probability generating function of a Poisson random variable with mean \(\sum_r\theta_r=\int_a^b\sum_r\lambda_r(u)\,du\), so the superposed count has the required law. For any finite collection of disjoint intervals, the counts of each component process on those intervals are independent. Independence between component processes then implies that all the superposed counts on those intervals are independent as well. Also \(N(0)=0\), and the sum of finitely many nonnegative locally integrable rate functions is nonnegative and locally integrable. Thus the superposition satisfies the definition of a nonhomogeneous Poisson process with the stated rate. \(\square\)

Superposition concerns the combined count. If events also carry a label identifying their source, the labels contain additional information: at time \(t\), the rate contribution from source \(r\) is \(\lambda_r(t)\). When the total rate is positive, the source proportions are therefore \(\lambda_r(t)/\sum_s\lambda_s(t)\). The proportions can vary with time even though the superposed process has a single rate function.

Worked Example: Two Streams with Opposing Time Trends

On \(0\le t\le2\), let two independent processes have rates \(\lambda_1(t)=1+t\) and \(\lambda_2(t)=2-t\). Both rates are nonnegative on this interval. Find the rate and the count distribution of the superposition on \((0,2]\), and calculate the probability of exactly two combined events on \((0,1]\).

The rates add pointwise, so \(\lambda(t)=(1+t)+(2-t)=3\) on \([0,2]\). The integrated rates on \((0,2]\) are

$$ \int_0^2(1+t)\,dt=2+\frac{2^2}{2}=4, \qquad \int_0^2(2-t)\,dt=4-\frac{2^2}{2}=2. $$

The two component counts are independent Poisson variables with means \(4\) and \(2\); their sum is Poisson with mean \(4+2=6\). Thus \(N(2)\sim\operatorname{Poisson}(6)\). On \((0,1]\), the integrated combined rate is \(\int_0^1 3\,dt=3\), so

$$ \mathbb{P}(N(1)=2)=e^{-3}\frac{3^2}{2!}=\frac{9}{2}e^{-3}. $$

Check: the component means on \((0,1]\) are \(\int_0^1(1+t)\,dt=1+\tfrac12=\tfrac32\) and \(\int_0^1(2-t)\,dt=2-\tfrac12=\tfrac32\). Their sum is \(3\), matching the combined-rate calculation.

Why Time-Dependent Thinning Multiplies the Rate

Consider a nonhomogeneous Poisson process with rate \(\lambda(t)\), and retain each event at time \(t\) with probability \(p(t)\). The probability is evaluated at the event's actual time: in general, a single average retention probability cannot be used for every interval. For the proof, we use the conditional event-time law: given \(N(b)-N(a)=n\) and \(\Lambda(a,b)>0\), the \(n\) unordered event times in \((a,b]\) have the law of \(n\) independent draws with density \(\lambda(t)/\Lambda(a,b)\) on that interval. This standard result is established in the earlier material on conditional event times.

Theorem (Time-dependent thinning): Let \(N\) be a nonhomogeneous Poisson process with nonnegative, locally integrable rate \(\lambda(t)\), and let \(p(t)\) be measurable with \(0\le p(t)\le1\). If events are retained independently conditional on their times, with retention probability \(p(t)\) at time \(t\), the retained events form a nonhomogeneous Poisson process with rate \(p(t)\lambda(t)\).

Proof. Fix an interval \((a,b]\) and set \(\Lambda=\int_a^b\lambda(t)\,dt\). If \(\Lambda=0\), the original count on that interval is zero almost surely, so the retained count is zero too; also \(\int_a^bp(t)\lambda(t)\,dt=0\). Now suppose \(\Lambda>0\). Conditional on the original count being \(n\), each of its \(n\) event times has density \(\lambda(t)/\Lambda\). After averaging over such a time, an event is retained with probability

$$ q=\int_a^b p(t)\frac{\lambda(t)}{\Lambda}\,dt =\frac{\int_a^b p(t)\lambda(t)\,dt}{\Lambda}. $$

Because the conditional event times are independent and the retention decisions are independent conditional on those times, the retained indicators are independent Bernoulli variables with success probability \(q\). If \(K\) is the retained count on \((a,b]\), then, for \(0\le z\le1\), conditioning on the original count gives

$$ \begin{aligned} \mathbb{E}[z^K] &=\sum_{n=0}^{\infty}e^{-\Lambda}\frac{\Lambda^n}{n!}(1-q+qz)^n\\ &=e^{-\Lambda}\exp\big(\Lambda(1-q+qz)\big)\\ &=\exp\big(\Lambda q(z-1)\big) =\exp\left((z-1)\int_a^b p(t)\lambda(t)\,dt\right). \end{aligned} $$

This is the probability generating function of a Poisson variable with mean \(\int_a^bp(t)\lambda(t)\,dt\). For disjoint intervals, the original restrictions are independent, and the retention decisions on their events are independent; hence the retained counts on those intervals are independent. The function \(p\lambda\) is nonnegative and locally integrable because \(0\le p\le1\) and \(\lambda\) is locally integrable. The retained process therefore has all the defining properties of a nonhomogeneous Poisson process with rate \(p(t)\lambda(t)\). \(\square\)

Worked Example: Retention Probability Changes Halfway Through

Let the original process have rate \(\lambda(t)=2t+2\) for \(t\ge0\). Retain events with probability \(p(t)=\tfrac12\) for \(0\le t<1\), and \(p(t)=\tfrac14\) for \(t\ge1\). Find the retained count distribution on \((0,2]\), including its contributions from the two subintervals.

On \((0,1]\), the original integrated rate is \(\int_0^1(2t+2)\,dt=[t^2+2t]_0^1=3\). The retained mean is \(\tfrac12\cdot3=\tfrac32\). On \((1,2]\), the original integrated rate is \([t^2+2t]_1^2=8-3=5\), and the retained mean is \(\tfrac14\cdot5=\tfrac54\). The retained counts on these disjoint intervals are independent, so the total retained count is Poisson with mean \(\tfrac32+\tfrac54=\tfrac{11}{4}\). In particular, the probability of no retained events on \((0,2]\) is \(e^{-11/4}\).

Check: integrating the retained rate directly gives \(\int_0^1(t+1)\,dt=\tfrac32\) and \(\int_1^2(\tfrac12t+\tfrac12)\,dt=[\tfrac14t^2+\tfrac12t]_1^2=\tfrac54\). Their sum is again \(\tfrac{11}{4}\).

Worked Example: Superpose First, Then Thin

Let two independent processes on \(t\ge0\) have rates \(\lambda_A(t)=1\) and \(\lambda_B(t)=t\). Retain every event with probability \(\tfrac12\), independently. Find the retained count distribution on \((0,2]\), and the probability of exactly one retained event from process \(A\) and exactly two from process \(B\).

The superposed rate is \(1+t\). Its integrated rate on \((0,2]\) is \(\int_0^2(1+t)\,dt=2+2=4\). Thinning halves this rate, giving retained rate \(\tfrac12(1+t)\) and total retained mean \(\tfrac12\cdot4=2\). Thus the total retained count has distribution \(\operatorname{Poisson}(2)\).

For the source-specific calculation, thinning process \(A\) gives mean \(\tfrac12\int_0^2 1\,dt=1\); thinning process \(B\) gives mean \(\tfrac12\int_0^2t\,dt=\tfrac12\cdot2=1\). The source-specific retained counts are independent because the original processes and their retention decisions are independent. Therefore the requested probability is

$$ \left(e^{-1}\frac{1^1}{1!}\right) \left(e^{-1}\frac{1^2}{2!}\right) =\frac12e^{-2}. $$

Check: adding the two retained means gives \(1+1=2\), as required by the total-rate calculation.

A Common Pitfall: Averaging by Time Instead of by Events

When the retention probability varies with time, its ordinary time average need not be the probability that a randomly selected event is retained. Events are more likely to occur at times when the rate is higher, so the relevant weighting is by \(\lambda(t)\). For the rate \(\lambda(t)=2t+2\) and retention rule in the previous example, the total original mean on \((0,2]\) is \([t^2+2t]_0^2=8\). The ordinary time average of \(p(t)\) over this interval is \((\tfrac12+\tfrac14)/2=\tfrac38\). Multiplying \(8\) by \(\tfrac38\) gives \(3\), but that is the wrong retained mean: it treats all times as equally likely locations of an event.

The correct calculation weights the retention probability by the rate. The retained mean is

$$ \int_0^2p(t)\lambda(t)\,dt =\frac12\int_0^1(2t+2)\,dt+\frac14\int_1^2(2t+2)\,dt =\frac12(3)+\frac14(5) =\frac{11}{4}. $$

This differs from \(3\). The exact weighting matters whenever the event rate and retention probability both vary with time. In contrast, if \(p(t)=p\) is constant, then the retained mean on \((a,b]\) is \(p\int_a^b\lambda(t)\,dt\), as the constant-probability case suggests.

Takeaway: For independent streams, add the rate functions before calculating interval counts. For independent time-dependent retention, multiply the original rate by the retention probability at each time, then integrate over the interval. The resulting count is Poisson, and retained counts on disjoint intervals remain independent.

Check Your Understanding

In each question, use integrated rates over the specified interval and distinguish a time average from an event-weighted retention probability.

  1. Two independent processes have rates \(2\) and \(t\) on \((0,2]\). What is the rate of their superposition, and what is its count distribution on that interval?
  2. A process has rate \(\lambda(t)=t+1\) on \((0,1]\). Each event is retained with probability \(p(t)=\tfrac12\). Find the retained count distribution.
  3. A process has rate \(2\) on \((0,1]\), and each event is retained with probability \(p(t)=t\). Compute the retained mean and explain why it is not found by using a constant retention probability of \(\tfrac12\) without integrating.
  4. Why does the thinning proof check the case \(\int_a^b\lambda(t)\,dt=0\) separately?
  5. Two independent processes have rates \(\lambda_1(t)\) and \(\lambda_2(t)\), and a common retention probability \(p(t)\) is used. State the rate of the retained superposition and the condition needed for the thinning rule.

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