The Poisson probability law has become increasingly important in recent years as more and more random phenomena to which the law applies have been studied. In physics the random emission of electrons from the filament of a vacuum tube, or from a photosensitive substance under the influence of light, and the spontaneous decomposition of radioactive atomic nuclei lead to phenomena obeying a Poisson probability law. This law arises frequently in the fields of operations research and management science, since demands for service , whether upon the cashiers or salesmen of a department store, the stock clerk of a factory, the runways of an airport, the cargo-handling facilities of a port, the maintenance man of a machine shop, and the trunk lines of a telephone exchange, and also the rate at which service is rendered , often lead to random phenomena either exactly or approximately obeying a Poisson probability law. Such random phenomena also arise in connection with the occurrence of accidents, errors, breakdowns, and other similar calamities.
The kinds of random phenomena that lead to a Poisson probability law can best be understood by considering the kinds of phenomena that lead to a binomial probability law. The usual situation to which the binomial probability law applies is one in which \(n\) independent occurrences of some experiment are observed. One may then determine (i) the number of trials on which a certain event occurred and (ii) the number of trials on which the event did not occur. There are random events, however, that do not occur as the outcomes of definite trials of an experiment but rather at random points in time or space. For such events one may count the number of occurrences of the event in a period of time (or space). However, it makes no sense to speak of the number of non-occurrences of such an event in a period of time (or space). For example, suppose one observes the number of airplanes arriving at a certain airport in an hour. One may report how many airplanes arrived at the airport; however, it makes no sense to inquire how many airplanes did not arrive at the airport. Similarly, if one is observing the number of organisms in a unit volume of some fluid, one may count the number of organisms present, but it makes no sense to speak of counting the number of organisms not present.
We next indicate some conditions under which one may expect that the number of occurrences of a random event occurring in time or space (such as the presence of an organism at a certain point in 3-dimensional space, or the arrival of an airplane at a certain point in time) obeys a Poisson probability law. We make the basic assumption that there exists a positive quantity \(\mu\) such that, for any small positive number \(h\) and any time interval of length \(h\) ,
(i) the probability that exactly one event will occur in the interval is approximately equal to \(\mu h\) , in the sense that it is equal to \(\mu h+r_{1}(h)\) , and \(r_{1}(h) / h\) tends to 0 as \(h\) tends to 0;
(ii) the probability that exactly zero events occur in the interval is approximately equal to \(1-\mu h\) , in the sense that it is equal to \(1-\mu h+\) \(r_{2}(h)\) , and \(r_{2}(h) / h\) tends to 0 as \(h\) tends to 0; and,
(iii) the probability that two or more events occur in the interval is equal to a quantity \(r_{3}(h)\) such that the quotient \(r_{3}(h) / h\) tends to 0 as the length \(h\) of the interval tends to 0.
The parameter \(\mu\) may be interpreted as the mean rate at which events occur per unit time (or space); consequently, we refer to \(\mu\) as the mean rate of occurrence (of events).
Example 3A. Suppose one is observing the times at which automobiles arrive at a toll collector’s booth on a toll bridge. Let us suppose that we are informed that the mean rate \(\mu\) of arrival of automobiles is given by \(\mu=1.5\) automobiles per minute. The foregoing assumption then states that in a time period of length \(h=1\) second \(=\left(\frac{1}{60}\right)\) minute, exactly one car will arrive with approximate probability \(\mu h=(1.5)\left(\frac{1}{60}\right)=\frac{1}{40}\) , whereas exactly zero cars will arrive with approximate probability \(1-\mu h=\frac{39}{40}\) .
In addition to the assumption concerning the existence of the parameter \(\mu\) with the properties stated, we also make the assumption that if an interval of time is divided into \(n\) sub-intervals and, for \(i=1, \ldots, n, A_{i}\) denotes the event that at least one event of the kind we are observing occurs in the \(i\) th sub-interval then, for any integer \(n, A_{1}, \ldots, A_{n}\) are independent events.
We now show, under these assumptions, that the number of occurrences of the event in a period of time (or space) of length (or area or volume) \(t\) obeys a Poisson probability law with parameter \(\mu t\) ; more precisely, the probability that exactly \(k\) events occur in a time period of length \(t\) is equal to
\[e^{-\mu t} \frac{(\mu t)^{k}}{k !}. \tag{3.1}\]
Consequently, we may describe briefly a sequence of events occurring in time (or space), and which satisfy the foregoing assumptions, by saying that the events obey a Poisson probability law at the rate of \(\mu\) events per unit time (or unit space) .
Note that if \(X\) is the number of events occurring in a time interval of length \(t\) , then \(X\) obeys a Poisson probability law with mean \(\mu t\) . Consequently, \(\mu\) is the mean rate of occurrence of events per unit time, in the sense that the number of events occurring in a time interval of length 1 obeys a Poisson probability law with mean \(\mu\) .
To prove (3.1), we divide the time period of length \(t\) into \(n\) time periods of length \(h=t / n\) . Then the probability that \(k\) events will occur in the time \(t\) is approximately equal to the probability that exactly one event has occurred in exactly \(k\) of the \(n\) sub-intervals of time into which the original interval was divided. By the foregoing assumptions, this is equal to the probability of scoring exactly \(k\) successes in \(n\) independent repeated Bernoulli trials in which the probability of success at each trial is \(p=\) \(h \mu=(\mu t) / n\) ; this is equal to
\[\left(\begin{array}{l} n \tag{3.2} \\ k \end{array}\right)\left(\frac{\mu t}{n}\right)^{k}\left(1-\frac{\mu t}{n}\right)^{n-k}.\]
Now (3.2) is only an approximation to the probability that \(k\) events will occur in time \(t\) . To get an exact evaluation, we must let the number of sub-intervals increase to infinity. Then (3.2) tends to (3.1) since rewriting (3.2)
\[\frac{1}{k !}(\mu t)^{k}\left(1-\frac{\mu t}{n}\right)^{n-k} \frac{(n)_{k}}{n^{k}} \rightarrow \frac{1}{k !}(\mu t)^{k} e^{-\mu t}\]
as \(n \rightarrow \infty\) .
It should be noted that the foregoing derivation of (3.1) is not completely rigorous. To give a rigorous proof of (3.1), one must treat the random phenomenon under consideration as a stochastic process. A sketch of such proof, using differential equations, is given in section 5.
Example 3B. It is known that bacteria of a certain kind occur in water at the rate of two bacteria per cubic centimeter of water. Assuming that this phenomenon obeys a Poisson probability law, what is the probability that a sample of two cubic centimeters of water will contain (i) no bacteria, (ii) at least two bacteria?
Solution
Under the assumptions made, it follows that the number of bacteria in a two-cubic-centimeter sample of water obeys a Poisson probability law with parameter \(\mu t=(2)(2)=4\) , in which \(\mu\) denotes the rate at which bacteria occur in a unit volume and \(t\) represents the volume of the sample of water under consideration. Consequently, the probability that there will be no bacteria in the sample is equal to \(e^{-4}\) , and the probability that there will be two or more bacteria in the sample is equal to \(1-5 e^{-4}\) .
Example 3C. Misprints. In a certain published book of 520 pages 390 typographical errors occur. What is the probability that four pages, selected randomly by the printer as examples of his work, will be free from errors?
Solution
The problem as stated is incapable of mathematical solution. However, let us recast the problem as follows. Assume that typographical errors occur in the work of a certain printer in accordance with the Poisson probability law at the rate of \(390 / 520=\frac{3}{4}\) errors per page. The number of errors in four pages then obeys a Poisson probability law with parameter (3) \(4=3\) ; consequently, the probability is \(e^{-3}\) that there will be no errors in the four pages.
Example 3D. Shot noise in electron tubes. The sensitivity attainable with electronic amplifiers and apparatus is inherently limited by the spontaneous current fluctuations present in such devices, usually called noise. One source of noise in vacuum tubes is shot noise, which is due to the random emission of electrons from the heated cathode. Suppose that the potential difference between the cathode and the anode is so great that all electrons emitted by the cathode have such high velocities that there is no accumulation of electrons between the cathode and the anode (and thus no space charge). If we consider an emission of an electron from the cathode as an event, then the assumptions preceding (3.1) may be shown as satisfied (see W. B. Davenport, Jr. and W. L. Root, An Introduction to the Theory of Random Signals and Noise , McGraw-Hill, New York, 1958, pp. 112–119). Consequently, the number of electrons emitted from the cathode in a time interval of length \(t\) obeys a Poisson probability law with parameter \(\lambda t\) , in which \(\lambda\) is the mean rate of emission of electrons from the cathode.
The Poisson probability law was first published in 1837 by Poisson in his book Recherches sur la probabilité des jugements en matière criminelle et en matière civile . In 1898, in a work entitled Das Gesetz der kleinen Zahlen , Bortkewitz described various applications of the Poisson distribution. However until 1907 the Poisson distribution was regarded as more of a curiosity than a useful scientific tool, since the applications made of it were to such phenomena as the suicides of women and children and deaths from the kick of a horse in the Prussian army. Because of its derivation as a limit of the binomial law, the Poisson law was usually described as the probability law of the number of successes in a very large number of independent repeated trials, each with a very small probability of success.
In 1907 the celebrated statistician W. S. Gosset (writing, as was his wont, under the pseudonym “Student”) deduced the Poisson law as the probability law of the number of minute corpuscles to be found in sample drops of a liquid, under the assumption that the corpuscles are distributed at random throughout the liquid; see “Student”, “On the error of counting with a Haemocytometer”, Biometrika , Vol. 5, p. 351. In 1910 the Poisson law was shown to fit the number of “ \(\alpha\) -particles discharged per \(\frac{1}{8}\) -minute or \(\frac{1}{4}\) -minute interval from a film of polonium”; see Rutherford and Geiger, “The probability variations in the distribution of \(\alpha\) -particles”, Philosophical Magazine , Vol. 20, p. 700.
Although one is able to state assumptions under which a random phenomenon will obey a Poisson probability law with some parameter \(\lambda\) , the value of the constant \(\lambda\) cannot be deduced theoretically but must be determined empirically. The determination of \(\lambda\) is a statistical problem. The following procedure for the determination of \(\lambda\) can be justified on various grounds. Given events occurring in time, choose an interval of length \(t\) . Observe a large number \(N\) of time intervals of length \(t\) . For each integer \(k=0,1,2, \ldots\) let \(N_{k}\) be the number of intervals in which exactly \(k\) events have occurred. Let
\[T=0 \cdot N_{0}+1 \cdot N_{1}+2 \cdot N_{2}+\cdots+k \cdot N_{k}+\cdots \tag{3.3}\]
be the total number of events observed in the \(N\) intervals of length \(t\) . Then the ratio \(T / N\) represents the observed average number of events happening per time interval of length \(t\) . As an estimate \(\hat{\lambda}\) of the value of the parameter \(\lambda\) , we take
\[\hat{\lambda}=\frac{T}{N}=\frac{1}{N} \sum_{k=0}^{\infty} k N_{k}. \tag{3.4}\]
If we believe that the random phenomenon under observation obeys a Poisson probability law with parameter \(\hat{\lambda}\) , then we may compute the probability \(p(k ; \hat{\lambda})\) that in a time interval of length \(t\) exactly \(k\) successes will occur.
Example 3E. Vacancies in the United States Supreme Court. W. A. Wallis, writing on “The Poisson Distribution and the Supreme Court”, Journal of the American Statistical Association , Vol. 31 (1936), pp. 376–380, reports that vacancies in the United States Supreme Court, either by death or resignation of members, occurred as follows during the 96 years, 1837 to 1932:
| \(k=\) number of vacancies during the year | \(N_{k}=\) number of years with \(k\) vacancies |
|---|---|
| 0 | 59 |
| 1 | 27 |
| 2 | 9 |
| over 3 | 1 |
Since \(T=27+2 \cdot 9+1 \cdot 3=48\) and \(N=96\) , it follows from (3.4) that \(\hat{\lambda}=0.5\) . If it is believed that vacancies in the Supreme Court occur in accord with a Poisson probability law at a mean rate of 0.5 a year, then it follows that the probability is equal to \(e^{-2}\) that during his four-year term of office the next president will make no appointments to the Supreme Court.
The foregoing data also provide a method of testing the hypothesis that vacancies in the Supreme Court obey a Poisson probability at the rate of 0.5 vacancies per year. If this is the case, then the probability that in a year there will be \(k\) vacancies is given by \[p(k ; 0.5)=e^{-0.5} \frac{(0.5)^{k}}{k !}, \quad k=0,1,2, \cdots.\]
The expected number of years in \(N\) years in which \(k\) vacancies occur, which is equal to \(N p(k ; 0.5)\) , may be computed and compared with the observed number of years in which \(k\) vacancies have occurred; refer to Table 3A.
| Number of Years out of 96 in which \(k\) Vacancies Occur | |||
|---|---|---|---|
| 3-4 Number of Vacancies \(k\) | Probability \(p(k; 0.5)\) of \(k\) Vacancies | Expected Number \((96)p(k ; 0.5)\) | Observed Number \(N_{k}\) |
| 0 | 0.6065 | 58.224 | 59 |
| 1 | 0.3033 | 29.117 | 27 |
| 2 | 0.0758 | 7.277 | 9 |
| 3 | 0.0126 | 1.210 | 1 |
| over 3 | 0.0018 | 0.173 | 0 |
The observed and expected numbers may then be compared by various statistical criteria (such as the \(\chi^{2}\) -test for goodness of fit) to determine whether the observations are compatible with the hypothesis that the number of vacancies obeys a Poisson probability law at a mean rate of 0.5.
The Poisson, and related, probability laws arise in a variety of ways in the mathematical theory of queues (waiting lines) and the mathematical theory of inventory and production control. We give a very simple example of an inventory problem. It should be noted that to make the following example more realistic one must take into account the costs of the various actions available.
Example 3F. An inventory problem. Suppose a retailer discovers that the number of items of a certain kind demanded by customers in a given time period obeys a Poisson probability law with known parameter \(\lambda\) . What stock \(K\) of this item should the retailer have on hand at the beginning of the time period in order to have a probability 0.99 that he will be able to supply immediately all customers who demand the item during the time period under consideration?
Solution
The problem is to find the number \(K\) , such that the probability is 0.99 that there will be \(K\) or less occurrences during the time period of the event when the item is demanded. Since the number of occurrences of this event obeys a Poisson probability law with parameter \(\lambda\) , we seek the integer \(K\) such that
\[\sum_{k=0}^{K} e^{-\lambda} \frac{\lambda^{k}}{k !} \geq 0.99, \quad \sum_{k=\tilde{K}+1}^{\infty} e^{-\lambda} \frac{\lambda^{k}}{k !} \leq 0.01. \tag{3.5}\]
The solution \(K\) of the second inequality in (3.5) can be read from Molina’s tables (E. C. Molina, Poisson’s Exponential Binomial Limit, Van Nostrand , New York, 1942). If \(\lambda\) is so large that the normal approximation to the Poisson law may be used, then (3.5) may be solved explicitly for \(K\) . Since the first sum in (3.5) is approximately equal to \[\Phi\left(\frac{K-\lambda+\frac{1}{2}}{\sqrt{\lambda}}\right),\] \(K\) should be chosen so that \(\left(K-\lambda+\frac{1}{2}\right) / \sqrt{\lambda}=2.326\) or \[K=2.326 \sqrt{\lambda}+\lambda-\frac{1}{2}. \tag{3.6}\]
Theoretical Exercises
3.1. A problem of aerial search. State conditions for the validity of the following assertion: if \(N\) ships are distributed at random over a region of the ocean of area \(A\) , and if a plane can search over \(Q\) square miles of ocean per hour of flight, then the number of ships sighted by a plane in a flight of \(T\) hours obeys a Poisson probability law with parameter \(\lambda=N Q T / A\) .
3.2. The number of matches approximately obeys a Poisson probability law. Consider the number of matches obtained by distributing \(M\) balls, numbered 1 to \(M\) , among \(M\) urns in such a way that each urn contains exactly 1 ball. Show that the probability of exactly \(m\) matches tends to \(e^{-1}(1 / m !)\) , as \(M\) tends to infinity, so that for large \(M\) the number of matches approximately obeys a Poisson probability law with parameter 1.
Exercises
State carefully the probabilistic assumptions under which you solve the following problems. Keep in mind the empirically observed fact that the occurrence of accidents, errors, breakdowns, and so on, in many instances appear to obey Poisson probability laws.
3.1. The incidence of polio during the years 1949–1954 was approximately 25 per 100,000 population. In a city of 40,000 what is the probability of having 5 or fewer cases? In a city of \(1,000,000\) what is the probability of having 5 or fewer cases? State your assumptions.
Answer
\(0.0671\) , \(0.000\) .
3.2. A manufacturer of wool blankets inspects the blankets by counting the number of defects. (A defect may be a tear, an oil spot, etc.) From past records it is known that the mean number of defects per blanket is 5. Calculate the probability that a blanket will contain 2 or more defects.
3.3. Bank tellers in a certain bank make errors in entering figures in their ledgers at the rate of 0.75 error per page of entries. What is the probability that in 4 pages there will be 2 or more errors?
Answer
0.8008.
3.4. Workers in a certain factory incur accidents at the rate of 2 accidents per week. Calculate the probability that there will be 2 or fewer accidents during (i) 1 week, (ii) 2 weeks; (iii) calculate the probability that there will be 2 or fewer accidents in each of 2 weeks.
3.5. A radioactive source is observed during 4 time intervals of 6 seconds each. The number of particles emitted during each period are counted. If the particles emitted obey a Poisson probability law, at a rate of 0.5 particles emitted per second, find the probability that (i) in each of the 4 time intervals 3 or more particles will be emitted, (ii) in at least 1 of the 4 time intervals 3 or more particles will be emitted.
Answer
(i) 0.111; (ii) 0.968.
3.6. Suppose that the suicide rate in a certain state is 1 suicide per 250,000 inhabitants per week.
(i) Find the probability that in a certain town of population 500,000 there will be 6 or more suicides in a week.
(ii) What is the expected number of weeks in a year in which 6 or more suicides will be reported in this town.
(iii) Would you find it surprising that during 1 year there were at least 2 weeks in which 6 or more suicides were reported?
3.7. Suppose that customers enter a certain shop at the rate of 30 persons an hour.
(i) What is the probability that during a 2-minute interval either no one will enter the shop or at least 2 persons will enter the shop.
(ii) If you observed the number of persons entering the shop during each of 302 -minute intervals, would you find it surprising that 20 or more of these intervals had the property that either no one or at least 2 persons entered the shop during that time?
Answer
(i) 0.632; (ii) not surprising, since the number of 2 minute intervals in an hour in which either no one enters or 2 or more enter obeys a binomial probability law with mean 19.0 and variance 6.975.
3.8. Suppose that the telephone calls coming into a certain switchboard obey a Poisson probability law at a rate of 16 calls per minute. If the switchboard can handle at most 24 calls per minute, what is the probability, using a normal approximation, that in 1 minute the switchboard will receive more calls than it can handle (assume all lines are clear).
3.9. In a large fleet of delivery trucks the average number inoperative on any day because of repairs is 2. Two standby trucks are available. What is the probability that on any day (i) no standby trucks will be needed, (ii) the number of standby trucks is inadequate.
Answer
(i) 0.1353; (ii) 0.3233.
3.10. Major motor failures occur among the buses of a large bus company at the rate of 2 a day. Assuming that each motor failure requires the services of 1 mechanic for a whole day, how many mechanics should the bus company employ to insure that the probability is at least 0.95 that a mechanic will be available to repair each motor as it fails? (More precisely, find the smallest integer \(K\) such that the probability is greater than or equal to 0.95 that \(K\) or fewer motor failures will occur in a day.)
3.11. Consider a restaurant located in the business section of a city. How many seats should it have available if it wishes to serve at least \(95 \%\) of all those who desire its services in a given hour, assuming that potential customers (each of whom takes at least an hour to eat) arrive in accord with the following schemes:
(i) 1000 persons pass by the restaurant in a given hour, each of whom has probability \(1 / 100\) of desiring to eat in the restaurant (that is, each person passing by the restaurant enters the restaurant once in every 100 times);
(ii) persons, each of whom has probability \(1 / 100\) of desiring to eat in the restaurant, pass by the restaurant at the rate of 1000 an hour;
(iii) persons, desiring to be patrons of the restaurant, arrive at the restaurant at the rate of 10 an hour.
Answer
15.
3.12. Flying-bomb hits on London. The following data (R. D. Clarke, “An application of the Poisson distribution”, Journal of the Institute of Actuaries , Vol. 72 (1946), p. 48) give the number of flying-bomb hits recorded in each of 576 small areas of \(t=\frac{1}{4}\) square kilometers each in the south of London during World War II.
\[\begin{array}{cc} \begin{array}{c} k=\text { number of flying- } \\ \text { bomb hits per area } \end{array} & \begin{array}{c} N_k=\text { number of areas } \\ \text { with } k \text { hits } \end{array} \\ \hline 0 & 229 \\ 1 & 211 \\ 2 & 93 \\ 3 & 35 \\ 4 & 7 \\ 5 \text { or over } & 1 \\ \hline \end{array}\]
Using the procedure in example \(3 \mathrm{E}\) , show that these observations are well fitted by a Poisson probability law.
3.13. For each of the following numerical valued random phenomena state conditions under which it may be expected to obey, either exactly or approximately, a Poisson probability law: (i) the number of telephone calls received at a given switchboard per minute; (ii) the number of automobiles passing a given point on a highway per minute; (iii) the number of bacterial colonies in a given culture per 0.01 square millimeter on a microscope slide; (iv) the number of times one receives 4 aces per 75 hands of bridge; (v) the number of defective screws per box of 100.