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5.9. Poisson law

Lecture



In many problems of practice, we have to deal with random variables distributed according to a peculiar law, which is called the Poisson law.

Consider a discontinuous random variable   5.9.  Poisson law which can only accept integer, non-negative values:

  5.9.  Poisson law ,

and the sequence of these values ​​is theoretically unlimited.

It is said that a random variable   5.9.  Poisson law distributed according to the Poisson law, if the probability that it will take a certain value   5.9.  Poisson law expressed by the formula

  5.9.  Poisson law   5.9.  Poisson law , (5.9.1)

where a is some positive quantity, called the Poisson law parameter.

Random distribution series   5.9.  Poisson law distributed according to the Poisson law, has the form:

  5.9.  Poisson law

We first verify that the sequence of probabilities defined by formula (5.9.1) can be a series of distribution, i.e. what is the sum of all probabilities   5.9.  Poisson law equals one. We have:

  5.9.  Poisson law

But

  5.9.  Poisson law ,

from where

  5.9.  Poisson law .

In fig. Figure 5.9.1. Shows polygons of a random variable.   5.9.  Poisson law distributed according to the Poisson law, corresponding to different values ​​of the parameter   5.9.  Poisson law . Table 8 of the appendix shows the values   5.9.  Poisson law for various   5.9.  Poisson law .

  5.9.  Poisson law

Fig. 5.9.1.

We define the main characteristics - the expectation and variance - random variable   5.9.  Poisson law distributed according to the Poisson law. By the definition of expectation

  5.9.  Poisson law .

The first member of the amount (corresponding to   5.9.  Poisson law ) is equal to zero, therefore, the summation can begin with   5.9.  Poisson law :

  5.9.  Poisson law

Denote   5.9.  Poisson law ; then

  5.9.  Poisson law . (5.9.2)

Thus, the parameter   5.9.  Poisson law is nothing more than the expectation of a random variable   5.9.  Poisson law .

To determine the variance, we first find the second initial moment of magnitude   5.9.  Poisson law :

  5.9.  Poisson law

According to previously proven

  5.9.  Poisson law

Besides,

  5.9.  Poisson law

Consequently,

  5.9.  Poisson law

Next, we find the variance of   5.9.  Poisson law :

  5.9.  Poisson law (5.9.3)

Thus, the variance of a random variable distributed according to Poisson’s law is equal to its expectation   5.9.  Poisson law .

This property of the Poisson distribution is often used in practice to decide whether the hypothesis that the random variable   5.9.  Poisson law distributed according to the Poisson law. To do this, determine from the experience of statistical characteristics - the expectation and variance - a random variable. If their values ​​are close, then this may serve as an argument in favor of the Poisson distribution hypothesis; a sharp difference in these characteristics, on the contrary, is against the hypothesis.

Define for a random variable   5.9.  Poisson law distributed according to the Poisson law, the probability that it will take a value not less than the specified   5.9.  Poisson law . Denote this probability   5.9.  Poisson law :

  5.9.  Poisson law .

Obvious likelihood   5.9.  Poisson law can be calculated as the sum

  5.9.  Poisson law

However, it is much easier to determine it from the probability of the opposite event:

  5.9.  Poisson law (5.9.4)

In particular, the probability that the magnitude   5.9.  Poisson law take a positive value, expressed by the formula

  5.9.  Poisson law (5.9.5)

We have already mentioned that many practical tasks lead to the Poisson distribution. Consider one of the typical tasks of this kind.

  5.9.  Poisson law

Fig. 5.9.2.

Suppose that points are randomly distributed on the abscissa axis Ox (Fig. 5.9.2). Suppose that a random distribution of points satisfies the following conditions:

1. The probability of hitting a particular number of points on the segment   5.9.  Poisson law depends only on the length of this segment, but does not depend on its position on the abscissa axis. In other words, the points are distributed on the x-axis with the same average density. Denote this density (ie, the expectation of the number of points per unit length)   5.9.  Poisson law .

2. The points are distributed on the x-axis independently of each other, i.e. the probability of hitting one or another number of points on a given segment does not depend on how many of them fell on any other segment that does not overlap with it.

3. The probability of hitting a small area   5.9.  Poisson law two or more points are negligible compared to the probability of hitting a single point (this condition means the practical impossibility of two or more points coinciding).

Select on the abscissa a certain segment of length   5.9.  Poisson law and consider a discrete random variable   5.9.  Poisson law - the number of points falling on this segment. Possible values ​​will be

  5.9.  Poisson law (5.9.6)

Since the points fall on the segment independently of each other, it is theoretically possible that there will be as many of them there, i.e. the series (5.9.6) is unlimited.

Prove that the random variable   5.9.  Poisson law has a Poisson distribution law. To do this, we calculate the probability   5.9.  Poisson law the fact that the segment   5.9.  Poisson law will fall exactly   5.9.  Poisson law points.

First we solve a simpler problem. Consider on the axis oh a small section   5.9.  Poisson law and calculate the probability that at least one point falls on this area. We will argue as follows. The mathematical expectation of the number of points falling on this area is obviously equal to   5.9.  Poisson law (since per unit length falls on average   5.9.  Poisson law points). According to condition 3 for a small segment   5.9.  Poisson law You can neglect the possibility of hitting him with two or more points. Therefore, the expectation   5.9.  Poisson law the number of points falling on the plot   5.9.  Poisson law , will be approximately equal to the probability of hitting one point on it (or, which in our conditions is equivalent, at least one).

Thus, up to infinitesimal higher order, with   5.9.  Poisson law can be considered the probability that the plot   5.9.  Poisson law there will be one (at least one) point equal to   5.9.  Poisson law , and the probability that no one will be equal to   5.9.  Poisson law .

We use this to calculate the probability   5.9.  Poisson law hits on a segment   5.9.  Poisson law smooth   5.9.  Poisson law points. Divide the segment   5.9.  Poisson law on   5.9.  Poisson law equal parts long   5.9.  Poisson law . We agree to call an elementary segment   5.9.  Poisson law “Empty” if not a single point has fallen into it, and “busy” if at least one has fallen into it. According to the above, the probability that the segment   5.9.  Poisson law will be "busy", approximately equal to   5.9.  Poisson law ; the probability that it will be "empty" is equal to   5.9.  Poisson law . Since, according to condition 2, the points falling into non-overlapping segments are independent, our n segments can be considered as   5.9.  Poisson law independent “experiments”, in each of which a segment can be “occupied” with probability   5.9.  Poisson law . Find the probability that among   5.9.  Poisson law segments will be exactly   5.9.  Poisson law "Busy". By the repetition theorem, this probability is

  5.9.  Poisson law

or denoting   5.9.  Poisson law ,

  5.9.  Poisson law (5.9.7)

When large enough   5.9.  Poisson law this probability is approximately equal to the probability of hitting the segment   5.9.  Poisson law smooth   5.9.  Poisson law points, since getting two or more points per segment   5.9.  Poisson law has a negligible probability. In order to find the exact value   5.9.  Poisson law , it is necessary in expression (5.9.7) to go to the limit at   5.9.  Poisson law :

  5.9.  Poisson law (5.9.8)

Convert the expression under the limit sign:

  5.9.  Poisson law (5.9.9)

The first fraction and the denominator of the last fraction in the expression (5.9.9) with   5.9.  Poisson law obviously aiming for a unit. Expression   5.9.  Poisson law from   5.9.  Poisson law does not depend. The numerator of the last fraction can be converted as follows:

  5.9.  Poisson law (5.9.10)

With   5.9.  Poisson law   5.9.  Poisson law and expression (5.9.10) tends to   5.9.  Poisson law . Thus, it is proved that the probability of hitting exactly   5.9.  Poisson law points per segment   5.9.  Poisson law expressed by the formula

  5.9.  Poisson law ,

Where   5.9.  Poisson law i.e. the value of X is distributed according to the Poisson law with the parameter   5.9.  Poisson law .

Note that the value   5.9.  Poisson law meaning is the average number of points per segment   5.9.  Poisson law .

Magnitude   5.9.  Poisson law (the probability that the magnitude of X will take a positive value) in this case expresses the probability that   5.9.  Poisson law at least one point will fall:

  5.9.  Poisson law . (5.9.11)

Thus, we have seen that the Poisson distribution occurs where some points (or other elements) occupy a random position independently of each other, and the number of these points that fall into some area is calculated. In our case, this “area” was a segment   5.9.  Poisson law on the x-axis. However, our conclusion can be easily extended to the case of distribution of points on a plane (random flat field of points) and in space (random spatial field of points). It is easy to prove that if the conditions are met:

1) points are distributed in a field statistically uniform with average density   5.9.  Poisson law ;

2) the points fall into non-overlapping regions in an independent manner;

3) dots appear singly, not in pairs, triples, etc., then the number of dots   5.9.  Poisson law falling into any area   5.9.  Poisson law (flat or spatial) are distributed according to the Poisson law:

  5.9.  Poisson law   5.9.  Poisson law ,

Where   5.9.  Poisson law - the average number of points falling into the area   5.9.  Poisson law .

For a flat case

  5.9.  Poisson law ,

Where   5.9.  Poisson law - area of ​​the region   5.9.  Poisson law ; for spatial

  5.9.  Poisson law ,

Where   5.9.  Poisson law - area volume   5.9.  Poisson law .

Note that for the Poisson distribution of the number of points falling in a segment or region, the condition of constant density (   5.9.  Poisson law ) irrelevant. If two other conditions are fulfilled, then the Poisson law still holds, only the parameter and in it acquires another expression: it is obtained not by a simple density multiplication   5.9.  Poisson law by length, area, or volume of a region, and by integrating a variable density over a segment, area, or volume. (For details on this, see n ° 19.4)

The presence of random points scattered on a line, on a plane or volume is not the only condition under which the Poisson distribution occurs. You can, for example, prove that the Poisson law is the limit for the binomial distribution:

  5.9.  Poisson law , (5.9.12)

if simultaneously push the number of experiences   5.9.  Poisson law to infinity and the probability   5.9.  Poisson law - to zero, and their product   5.9.  Poisson law keeps constant value:

  5.9.  Poisson law . (5.9.13)

Indeed, this limiting property of the binomial distribution can be written as:

  5.9.  Poisson law . (5.9.14)

But from the condition (5.9.13) it follows that

  5.9.  Poisson law . (5.9.15)

Substituting (5.9.15) into (5.9.14), we obtain the equality

  5.9.  Poisson law , (5.9.16)

which has just been proven by us on another occasion.

This limiting property of the binomial law often finds application in practice. Assume that a large number of independent experiments are made.   5.9.  Poisson law in each of which event   5.9.  Poisson law has a very low probability   5.9.  Poisson law . Then to calculate the probability   5.9.  Poisson law that event   5.9.  Poisson law will appear exactly   5.9.  Poisson law Once, you can use the approximate formula:

  5.9.  Poisson law , (5.9.17)

Where   5.9.  Poisson law - the parameter of the Poisson law, which is approximately replaced by the binomial distribution.

From this property of the Poisson law — to express the binomial distribution with a large number of experiments and a low probability of an event — its name, often used in statistical textbooks: the law of rare phenomena, occurs.

Consider a few examples related to the Poisson distribution from different areas of practice.

Example 1. An average telephone density call arrives at an automatic telephone exchange.   5.9.  Poisson law calls per hour. Considering that the number of calls in any part of the time is distributed according to the Poisson law, find the probability that in two minutes exactly three calls will arrive at the station.

Decision. The average number of calls in two minutes is:

  5.9.  Poisson law .

According to the formula (5.9.1) the probability of receipt of exactly three calls is equal to:

  5.9.  Poisson law

Example 2. In the conditions of the previous example, find the probability that in two minutes at least one call will arrive.

Decision. By the formula (5.9.4) we have:

  5.9.  Poisson law .

Example 3. In the same conditions, find the probability that in two minutes at least three calls will arrive.

Decision. By the formula (5.9.4) we have:

  5.9.  Poisson law

Example 4. On a loom the thread breaks on average 0.375 times during the hour of the machine. Find the probability that for a shift (8 hours) the number of thread breaks will be contained in the boundaries 2 and 4 (at least 2 and no more than 4 breaks).

Decision. Obviously

  5.9.  Poisson law

we have:

  5.9.  Poisson law

According to table 8 of the application when   5.9.  Poisson law

  5.9.  Poisson law

Example 5. On average, a cathode is heated from a hot cathode   5.9.  Poisson law electrons where   5.9.  Poisson law - time elapsed since the beginning of the experience. Find the probability that over a period of time   5.9.  Poisson law starting at the moment   5.9.  Poisson law , exactly m electrons will fly from the cathode.

Decision. Find the average number of electrons a, departing from the cathode for a given period of time. We have:

  5.9.  Poisson law .

According to the calculated   5.9.  Poisson law determine the desired probability:

  5.9.  Poisson law .

Example 6. The number of fragments that fall into a small-sized target at a given position of the break point is distributed according to the Poisson law. The average density of the fragmentation field in which the target finds itself at a given position of the break point is 3 sc / sq. The target area is equal to   5.9.  Poisson law sq.m. To hit a target, it is enough to hit at least one fragment in it. Find the probability of hitting the target at a given position of the break point.

Decision.   5.9.  Poisson law . By the formula (5.9.4) we find the probability of hitting at least one fragment:

  5.9.  Poisson law .

(To calculate the value of the exponential function   5.9.  Poisson law use table 2 of the annex).

Example 7. The average density of pathogenic microbes in one cubic meter of air is 100. A sample of 2 cubic meters is taken per sample. dm air. Find the probability that at least one microbe will be detected in it.

Decision. Taking the hypothesis of the Poisson distribution of the number of microbes in the volume, we find:

  5.9.  Poisson law

Example 8. For some goal 50 independent shots are fired. The probability of hitting the target with one shot is 0.04. Using the limit property of the binomial distribution (formula (5.9.17)), we can find the approximate probability that the target will hit: not a single projectile, one projectile, two projectiles.

Decision. We have   5.9.  Poisson law . In table 8 of the application we find the probabilities:

  5.9.  Poisson law


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Probability theory. Mathematical Statistics and Stochastic Analysis

Terms: Probability theory. Mathematical Statistics and Stochastic Analysis