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 in continuous time in which is the number of events, called arrivals, in the time interval . Each takes values in , the process starts at , and whenever . For the increment is the number of arrivals in the interval . 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 and never goes down. Its value at time 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 , and minutes and no others in . Then , and , and the count in the interval is . The path stays at between the arrivals at and and rises by one at each arrival.
Homogeneous Poisson process
Definition Let . A counting process is a homogeneous Poisson process with rate if the following hold.
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The process starts at .
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For all and , the count in has the distribution, so for
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For every and all times , the counts 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 per minute. Over three minutes the parameter is , so the count has mean and variance . No arrivals has probability , and
about . At most two arrivals includes the counts , and , so
Rate of arrivals
Definition The rate is the expected number of arrivals per unit time, since . It carries a unit, such as arrivals per minute. For an interval of length measured in the same time unit, the product is the Poisson parameter of the count, a pure number equal to the expected count. Changing the time unit changes and together and leaves unchanged. The probability of at least one arrival in an interval of length is
which is less than for every and close to when is small. The product is an expected count, not a probability, and it can exceed .
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 per minute, a half-minute window has parameter , so exactly two arrivals there has probability . In seconds the rate is per second and the window lasts seconds, which again gives parameter . For a whole minute , which cannot be a probability, and the chance of at least one arrival is . For a window of minutes, and , close to .
Stationary increments
Definition A counting process has stationary increments if for every the distribution of is the same for all , so that it depends only on the length of the interval . In a homogeneous Poisson process with rate this common distribution is . 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 per minute, the counts in and in both have the distribution, so each is zero with probability . The counts in and in also share the 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 and all times the counts are mutually independent. Equivalently, the counts in any finite collection of disjoint intervals of the form are mutually independent. Half-open intervals that share only an endpoint, such as and , 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 per minute, the counts in and are independent, so no arrivals in either minute has probability . This agrees with computed from the two-minute count. Exactly one arrival in each of the two minutes has probability .
Overlapping counts
Definition Counts over intervals that overlap share the arrivals in the overlap. For a homogeneous Poisson process with rate and times , the count splits as
into independent counts over and . The cumulative counts and themselves are dependent, since
More generally, the covariance of the counts in two intervals is times the length of their overlap. Given , the later count is plus an independent count, so it cannot be less than .
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 per minute, five arrivals in the first minute has probability , and given that event, the event is impossible. Without that information , and . The covariance is , and the correlation is .
Check yourself
- A homogeneous Poisson process has rate 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.
Show the answer
The parameter is , so the count is with mean and variance . No arrivals has probability , and at least one arrival has probability .
- For the same process, one classmate says that a twenty-minute window has expected count , so an arrival in it is certain. Another uses and gets the parameter . Find the correct parameter and the probability of at least one arrival.
Show the answer
The rate is per hour, so the window must be measured in hours. Twenty minutes is of an hour, and the parameter is . The value mixes hours with minutes and is the expected count for twenty hours. The expected count is not a probability, and
So the window has no arrivals with probability , and an arrival is far from certain.
- For the same process, let be the count in the first hour and the count in the first two hours. Are and independent? Are and independent? Check both answers with the events of no arrivals.
Show the answer
The counts and cover the disjoint intervals and , so they are independent, and no arrivals in both hours has probability . This matches from the two-hour count. For the cumulative counts, forces , so
about . The product of the separate probabilities is . The two numbers differ, so and are dependent, as they must be, since the two-hour count contains the first-hour count.