Unit 12, lesson 76

Poisson Process Basics

How a homogeneous Poisson process counts arrivals in continuous time, why rate times interval length is the expected count, how stationary and independent increments differ, and why overlapping counts are dependent.

  • Video 76
  • Poisson Processes

Messages reach a server at random moments, not at numbered trials. A model for them has to say how many arrive in any stretch of time and how the counts in different stretches relate to each other. The homogeneous Poisson process is the basic model of this kind, with a constant rate, a Poisson count in every interval and independent counts in disjoint intervals. This lesson states those assumptions, computes interval count probabilities, and separates disjoint windows, whose counts are independent, from overlapping windows, whose counts are not.

Key terms

Counting process

Definition A counting process is a random process {N(t):t≥0}\{N(t) : t \ge 0\} in continuous time in which N(t)N(t) is the number of events, called arrivals, in the time interval (0,t](0, t]. Each N(t)N(t) takes values in {0,1,2,… }\{0, 1, 2, \dots\}, the process starts at N(0)=0N(0) = 0, and N(s)≤N(t)N(s) \le N(t) whenever s<ts < t. For s<ts < t the increment N(t)−N(s)N(t) - N(s) is the number of arrivals in the interval (s,t](s, t]. Each sample path is a nondecreasing step function that jumps at the arrival times.

In plain words A counting process is a running tally of arrivals that starts at 00 and never goes down. Its value at time tt says how many arrivals have occurred so far, and the difference of two values says how many occurred in between.

Example Suppose one observed run has arrivals at times 0.40.4, 1.11.1 and 1.31.3 minutes and no others in (0,2](0, 2]. Then N(1)=1N(1) = 1, N(1.2)=2N(1.2) = 2 and N(2)=3N(2) = 3, and the count in the interval (1,2](1, 2] is N(2)−N(1)=2N(2) - N(1) = 2. The path stays at 11 between the arrivals at 0.40.4 and 1.11.1 and rises by one at each arrival.

Homogeneous Poisson process

Definition Let λ>0\lambda > 0. A counting process {N(t):t≥0}\{N(t) : t \ge 0\} is a homogeneous Poisson process with rate λ\lambda if the following hold.

  1. The process starts at N(0)=0N(0) = 0.

  2. For all s≥0s \ge 0 and t>0t > 0, the count in (s,s+t](s, s + t] has the Pois⁡(λt)\operatorname{Pois}(\lambda t) distribution, so for k=0,1,2,…k = 0, 1, 2, \dots

    P(N(s+t)−N(s)=k)=e−λt(λt)kk!.\begin{aligned} &P\bigl(N(s + t) - N(s) = k\bigr) \\[0.4em] &\quad = \frac{e^{-\lambda t} (\lambda t)^{k}}{k!}. \end{aligned}
  3. For every k≥1k \ge 1 and all times 0≤t0<t1<⋯<tk0 \le t_0 < t_1 < \cdots < t_k, the counts N(t1)−N(t0),…,N(tk)−N(tk−1)N(t_1) - N(t_0), \dots, N(t_k) - N(t_{k-1}) are mutually independent.

Homogeneous means that the rate is the same at every time. A Poisson distribution for the count in one interval does not by itself make a process a Poisson process, because the third condition, about how counts in different intervals relate, is part of the definition.

In plain words The model fixes the distribution of the count in every window and says how the windows relate. Even with a constant rate, arrivals come at random times, so the gaps between them vary and arrivals can bunch together. A perfectly regular schedule, or a rate that changes during the day, is a different model.

Example In a constructed model, messages arrive as a homogeneous Poisson process with rate λ=2\lambda = 2 per minute. Over three minutes the parameter is 2⋅3=62 \cdot 3 = 6, so the count N(3)N(3) has mean 66 and variance 66. No arrivals has probability e−6≈0.00248e^{-6} \approx 0.00248, and

P(N(3)=2)=e−6 622!=18e−6,\begin{aligned} P\bigl(N(3) = 2\bigr) &= \frac{e^{-6}\, 6^{2}}{2!} \\[0.4em] &= 18e^{-6}, \end{aligned}

about 0.04460.0446. At most two arrivals includes the counts 00, 11 and 22, so

P(N(3)≤2)=e−6(1+6+18)=25e−6≈0.0620.\begin{aligned} &P\bigl(N(3) \le 2\bigr) \\[0.4em] &\quad = e^{-6}(1 + 6 + 18) \\[0.4em] &\quad = 25e^{-6} \approx 0.0620. \end{aligned}

Rate of arrivals

Definition The rate λ\lambda is the expected number of arrivals per unit time, since E[N(t)]=λtE[N(t)] = \lambda t. It carries a unit, such as arrivals per minute. For an interval of length tt measured in the same time unit, the product λt\lambda t is the Poisson parameter of the count, a pure number equal to the expected count. Changing the time unit changes λ\lambda and tt together and leaves λt\lambda t unchanged. The probability of at least one arrival in an interval of length tt is

P(N(t)≥1)=1−e−λt,P\bigl(N(t) \ge 1\bigr) = 1 - e^{-\lambda t},

which is less than λt\lambda t for every t>0t > 0 and close to λt\lambda t when λt\lambda t is small. The product λt\lambda t is an expected count, not a probability, and it can exceed 11.

In plain words Multiply the rate by the length of the window, in matching units, to get the expected count. A rate of two per minute does not mean one arrival every thirty seconds, and it does not guarantee two arrivals in each minute. For a very short window the chance of an arrival is roughly the rate times the length, but for a longer window that product need not be a probability at all.

Example With λ=2\lambda = 2 per minute, a half-minute window has parameter 2⋅12=12 \cdot \tfrac{1}{2} = 1, so exactly two arrivals there has probability e−1/2!=e−1/2≈0.1839e^{-1}/2! = e^{-1}/2 \approx 0.1839. In seconds the rate is 2/60=1/302/60 = 1/30 per second and the window lasts 3030 seconds, which again gives parameter 11. For a whole minute λt=2\lambda t = 2, which cannot be a probability, and the chance of at least one arrival is 1−e−2≈0.86471 - e^{-2} \approx 0.8647. For a window of 0.010.01 minutes, λt=0.02\lambda t = 0.02 and 1−e−0.02≈0.01981 - e^{-0.02} \approx 0.0198, close to 0.020.02.

Stationary increments

Definition A counting process has stationary increments if for every t>0t > 0 the distribution of N(s+t)−N(s)N(s + t) - N(s) is the same for all s≥0s \ge 0, so that it depends only on the length tt of the interval (s,s+t](s, s + t]. In a homogeneous Poisson process with rate λ\lambda this common distribution is Pois⁡(λt)\operatorname{Pois}(\lambda t). Stationarity is a statement about distributions only. It does not say that the counts in two windows of equal length are equal, and it says nothing about whether they are independent.

In plain words Sliding a window of fixed length along the time axis does not change the distribution of its count. The constant rate is what makes this true, so a stream of arrivals whose expected count per hour is higher in the morning than at night does not have stationary increments.

Example With λ=2\lambda = 2 per minute, the counts in (0,1](0, 1] and in (10,11](10, 11] both have the Pois⁡(2)\operatorname{Pois}(2) distribution, so each is zero with probability e−2≈0.1353e^{-2} \approx 0.1353. The counts in (0,1](0, 1] and in (0.5,1.5](0.5, 1.5] also share the Pois⁡(2)\operatorname{Pois}(2) distribution, but those windows overlap, and their counts are dependent. Equal distributions say nothing about independence.

Independent increments

Definition A counting process has independent increments if for every k≥1k \ge 1 and all times 0≤t0<t1<⋯<tk0 \le t_0 < t_1 < \cdots < t_k the counts N(t1)−N(t0),…,N(tk)−N(tk−1)N(t_1) - N(t_0), \dots, N(t_k) - N(t_{k-1}) are mutually independent. Equivalently, the counts in any finite collection of disjoint intervals of the form (a,b](a, b] are mutually independent. Half-open intervals that share only an endpoint, such as (0,1](0, 1] and (1,2](1, 2], are disjoint.

In plain words What happens in one window carries no information about the count in a window that does not overlap it. A quiet minute does not make the next minute busier or quieter. Independence is a separate assumption from stationarity. Equal distributions do not imply independence, and a process can have independent increments without stationary ones. An example is a model with independent counts in disjoint windows that expects more arrivals per hour in the morning than at night.

Example With λ=2\lambda = 2 per minute, the counts in (0,1](0, 1] and (1,2](1, 2] are independent, so no arrivals in either minute has probability e−2⋅e−2=e−4≈0.0183e^{-2} \cdot e^{-2} = e^{-4} \approx 0.0183. This agrees with P(N(2)=0)=e−4P(N(2) = 0) = e^{-4} computed from the two-minute count. Exactly one arrival in each of the two minutes has probability (2e−2)2=4e−4≈0.0733(2e^{-2})^2 = 4e^{-4} \approx 0.0733.

Overlapping counts

Definition Counts over intervals that overlap share the arrivals in the overlap. For a homogeneous Poisson process with rate λ>0\lambda > 0 and times 0<s<t0 < s < t, the count N(t)N(t) splits as

N(t)=N(s)+(N(t)−N(s))\begin{aligned} N(t) &= N(s) \\[0.4em] &\quad + \bigl(N(t) - N(s)\bigr) \end{aligned}

into independent counts over (0,s](0, s] and (s,t](s, t]. The cumulative counts N(s)N(s) and N(t)N(t) themselves are dependent, since

Cov⁡(N(s),N(t))=Var⁡(N(s))=λs>0.\begin{aligned} &\operatorname{Cov}\bigl(N(s), N(t)\bigr) \\[0.4em] &\quad = \operatorname{Var}\bigl(N(s)\bigr) = \lambda s > 0. \end{aligned}

More generally, the covariance of the counts in two intervals is λ\lambda times the length of their overlap. Given N(s)=jN(s) = j, the later count is jj plus an independent Pois⁡(λ(t−s))\operatorname{Pois}\bigl(\lambda(t - s)\bigr) count, so it cannot be less than jj.

In plain words A longer cumulative count contains the shorter one, so knowing the shorter count changes what is likely, and sometimes what is possible, for the longer one. Independent increments apply to the first count and the additional count, not to two cumulative counts. Before multiplying probabilities for several windows, check whether the windows overlap.

Example With λ=2\lambda = 2 per minute, five arrivals in the first minute has probability e−2 25/5!≈0.0361>0e^{-2}\, 2^5/5! \approx 0.0361 > 0, and given that event, the event N(2)≤4N(2) \le 4 is impossible. Without that information N(2)∼Pois⁡(4)N(2) \sim \operatorname{Pois}(4), and P(N(2)≤4)≈0.6288P(N(2) \le 4) \approx 0.6288. The covariance is Cov⁡(N(1),N(2))=2\operatorname{Cov}(N(1), N(2)) = 2, and the correlation is 2/2⋅4≈0.7072/\sqrt{2 \cdot 4} \approx 0.707.

Check yourself

  1. A homogeneous Poisson process has rate 33 arrivals per hour. For a two-hour window, find the mean and the variance of the count, the probability of no arrivals and the probability of at least one arrival.
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The parameter is λt=3⋅2=6\lambda t = 3 \cdot 2 = 6, so the count N(2)N(2) is Pois⁡(6)\operatorname{Pois}(6) with mean 66 and variance 66. No arrivals has probability e−6≈0.00248e^{-6} \approx 0.00248, and at least one arrival has probability 1−e−6≈0.997521 - e^{-6} \approx 0.99752.

  1. For the same process, one classmate says that a twenty-minute window has expected count 11, so an arrival in it is certain. Another uses t=20t = 20 and gets the parameter 6060. Find the correct parameter and the probability of at least one arrival.
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The rate is per hour, so the window must be measured in hours. Twenty minutes is 13\tfrac{1}{3} of an hour, and the parameter is 3⋅13=13 \cdot \tfrac{1}{3} = 1. The value 6060 mixes hours with minutes and is the expected count for twenty hours. The expected count 11 is not a probability, and

P(N(1/3)≥1)=1−e−1≈0.632.\begin{aligned} P\bigl(N(1/3) \ge 1\bigr) &= 1 - e^{-1} \\[0.4em] &\approx 0.632. \end{aligned}

So the window has no arrivals with probability e−1≈0.368e^{-1} \approx 0.368, and an arrival is far from certain.

  1. For the same process, let N(1)N(1) be the count in the first hour and N(2)N(2) the count in the first two hours. Are N(1)N(1) and N(2)−N(1)N(2) - N(1) independent? Are N(1)N(1) and N(2)N(2) independent? Check both answers with the events of no arrivals.
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The counts N(1)N(1) and N(2)−N(1)N(2) - N(1) cover the disjoint intervals (0,1](0, 1] and (1,2](1, 2], so they are independent, and no arrivals in both hours has probability e−3⋅e−3=e−6e^{-3} \cdot e^{-3} = e^{-6}. This matches P(N(2)=0)=e−6P(N(2) = 0) = e^{-6} from the two-hour count. For the cumulative counts, N(2)=0N(2) = 0 forces N(1)=0N(1) = 0, so

P(N(1)=0, N(2)=0)=P(N(2)=0)=e−6,\begin{aligned} &P\bigl(N(1) = 0,\ N(2) = 0\bigr) \\[0.4em] &\quad = P\bigl(N(2) = 0\bigr) = e^{-6}, \end{aligned}

about 0.002480.00248. The product of the separate probabilities is e−3⋅e−6=e−9≈0.000123e^{-3} \cdot e^{-6} = e^{-9} \approx 0.000123. The two numbers differ, so N(1)N(1) and N(2)N(2) are dependent, as they must be, since the two-hour count contains the first-hour count.

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