Lecture 15: Poisson Process II

What this lecture covers

This lecture finishes the course's treatment of the Poisson process, moving from definitions to problem-solving technique. It reviews the Poisson PMF and the exponential and Erlang distributions for arrival times, then works a fishing problem that combines several of these ideas, before revisiting and extending merging, introducing splitting (thinning), and closing with the counterintuitive random incidence phenomenon.

Building on the previous lecture's derivation of the Poisson process, this session shows how memorylessness, merging, and splitting let you solve seemingly hard problems, such as the expected time until the last of several light bulbs burns out, without calculus. It ends with the random incidence paradox: why a person arriving at a bus stop at a random time tends to observe longer-than-average gaps between buses.

Key ideas

• Poisson PMF recap: the number of arrivals in an interval of length tau follows P(k,tau) = (lambda*tau)^k / k! * e^(-lambda*tau), with mean and variance both lambda*tau.
• Erlang distribution: the time until the k-th arrival in a Poisson process is a sum of k independent exponentials, giving a PDF known as the Erlang distribution with parameters k and lambda.
• Used is as good as new: because of memorylessness, the remaining time until the next Poisson arrival does not depend on how long you've already waited.
• Merging Poisson processes: two independent Poisson processes with rates lambda1 and lambda2 merge into one Poisson process with rate lambda1 + lambda2, and the origin of each merged arrival is independent of the origins of other arrivals.
• Splitting (thinning): sending each arrival of a Poisson process to one of several streams by independent coin flips produces new Poisson processes, each with a rate scaled by the corresponding probability.
• Maximum of independent exponentials via merging: the expected time until the last of several independent, exponentially distributed lifetimes ends can be computed by summing the expected inter-arrival times of successively smaller merged processes.
• Random incidence: an interval observed by arriving at a random time is, on average, longer than a "typical" interarrival interval, because random arrival times are more likely to land inside longer intervals.

Walkthrough

Reviewing the Poisson process and Erlang distribution (0:00)

The lecture restates the Poisson process as a continuous-time limit of the Bernoulli process, recaps the Poisson PMF for arrival counts, and reviews that the time to the first arrival is exponential while the time to the k-th arrival follows the Erlang distribution, a sum of k independent exponentials.

The fishing problem (11:52)

A worked example models fish being caught as a Poisson process at 0.6 per hour, with a rule to fish for two hours and then stop once the first fish is caught (or keep fishing until one is caught if none arrive in the first two hours). The lecture computes the probability of fishing longer than two hours, of fishing between two and five hours, of catching at least two fish, the expected number of fish caught, and the expected total fishing time, using the exponential distribution, memorylessness, and the Erlang PDF.

Merging Poisson processes again (21:36)

The lecture reviews merging two independent Poisson processes into one with combined rate lambda1 + lambda2, and formalizes that the origin of each arrival in the merged process is an independent coin flip with probability lambda1/(lambda1+lambda2) of coming from the first process.

Maximum of three exponential lifetimes (27:11)

Using the light-bulb example, the lecture finds the expected time until all three of three independent, identically distributed exponential lifetimes have ended, by viewing each exponential as the first arrival of an embedded Poisson process, merging the three processes, and summing the expected times between successive burnouts: 1/(3*lambda) + 1/(2*lambda) + 1/lambda.

Splitting a Poisson process (33:36)

The lecture introduces splitting, or thinning: routing each arrival of a Poisson process to one of two streams via an independent coin flip with probability p. Each resulting stream is itself a Poisson process, with rate scaled by p or 1-p.

Random incidence at the bus stop (36:02)

Using a story about a bus company claiming Poisson arrivals at a rate of four per hour, the lecture shows that a rider arriving at a random time and measuring the gap between the previous and next bus observes an average gap of 30 minutes, not the 15-minute average interarrival time, because the gap splits into a forward and a backward exponential piece, each with expectation 1/lambda.

Why random incidence happens (41:34)

The lecture explains the paradox by distinguishing between picking a bus at random and picking a time at random: since random times are more likely to fall inside longer intervals, the observed interval is biased toward longer lengths. A simplified example with only 5- and 10-minute gaps, and analogies to family size and crowded buses, illustrate how the choice of sampling procedure changes what "typical" means.

Before you watch

• Watch the previous lecture on the definition and basic properties of the Poisson process.
• Review the exponential distribution and its memorylessness property.
• Be comfortable with the idea of merging independent Poisson processes from the end of the prior lecture.

Check your understanding

1. Why can the expected time until the last of several independent exponential lifetimes ends be computed without any integration?
2. How does splitting a Poisson process differ from merging two Poisson processes, and what happens to the rate in each case?
3. Why is the average gap observed by a rider arriving at a random bus stop time larger than the average interarrival time between buses?
4. In the 5-and-10-minute gap example, why does a randomly arriving observer see a longer average gap than 7.5 minutes?
5. How does the "used is as good as new" property simplify the fishing problem's calculations?