Probabilistic distribution law. Distribution law of a discrete random variable

So, we come to the problem: how to find the probability that during the next test a random variable will fall into a predetermined interval?

To answer this question, first of all, it is necessary to introduce the concept of the law of distribution of a random variable.

The law of distribution of a random variable (CVD) is a way to calculate the probability that a random variable (CV) will take on a particular value (for discrete random variables) or fall into a particular interval (for continuous random variables) as a result of a test.

For discrete CBs, this is most often a table. For example, for a regular dice, this table would look like this:

1/6	1/6	1/6	1/6	1/6	1/6

Rolling 1, 2, 3, 4, 5, or 6 is equally likely and equals one sixth.

For a continuous random variable, the RPV can be specified either as a graph or as a formula. The most important in mathematical statistics is the normal law of distribution of a random variable or the Gauss law.

This is due to the fact that many SWs are distributed precisely according to this law, including in biology and medicine.

So, to calculate probabilities, we need Gauss's law. Consider this law.

Let's state the problem more precisely. Suppose we have some continuous random variable X and we want to know what is the probability that at the next test this value will take the value x i , lying in a small interval from x to x + dx (here dx is the differential of x). Then the probability P(x i) that this will happen during the next test, according to the Gauss law, will be equal to:

Formula (1) allows us to calculate the probability of the next measurement falling into an infinitely small interval dx. But in practice, we need to learn how to calculate the probability of falling into real intervals, for example, in the interval from x=a to x=b. This can be done using formula (2):

Since we set the interval (a, b) ourselves, therefore, to calculate the probability that the result of the next test will fall into this interval, we need to know only two numbers: μ - expected value and σ- standard deviation.

Thus, the evaluation of these two numbers is one of the main tasks of mathematical statistics.

So, in order to solve the main problem, which, as we know, is to learn how to calculate the probability of a random variable falling into one or another predetermined interval, we need to learn how to calculate these two numbers. This is where we fail, because it turned out to be impossible to accurately calculate these two numbers! It turned out that in order to accurately obtain these two numbers, for example, for the random variable "height", it is necessary to measure the height of all people in the world! It is clear that we cannot do this. What is left for us? And it remains for us to measure the growth of those people whom we can reach, and according to the obtained values, EVALUATE the values ​​of μ and σ. I emphasize: do not get exact values, but only estimate what they are approximately equal to. These estimates, which are called the sample arithmetic mean () and the standard deviation estimate (s), are the very first goal of most statistical studies.

In our consideration, the word “selective” unexpectedly appeared. Let's try to explain what it means. To do this, we introduce the following definition:

The set of objects from which some of its members are selected for study is called the general set, and the part of the general set selected in one way or another is called the sampling set or sample.

In the case of height, the population is the height of all people, while those people for whom we were able to measure the height are called the sample from this population. Obviously, this definition is valid for any random variable.

CALCULATION AND S.

The calculation of these two quantities is very simple and is given by the following two formulas:

(3)

(4)

To explain formulas (3) and (4), imagine that we measured the height of 50 people. This means that n=50. Next, add up all 50 numbers obtained and divide the result by 50. We get the value of the arithmetic mean. These are all calculations according to formula (3). Calculations by formula (4) are somewhat more complicated. First, from all the 50 numbers obtained as a result of measurements, we subtract the previously obtained estimate of the average. We get 50 difference values. Then we square all 50 differences, after which we add them all up. The result is divided by 49 (n-1). From what happened, we extract the square root. Calculations of the arithmetic mean and estimates of the standard deviation are completed.

Now that we have estimates of the mean and standard deviation, we need to return to formula (2). Indeed, we have estimates for μ and σ, we set the interval (a, b) ourselves, it remains to take the integral ... But here we are in for a new trouble! An indefinite integral of this kind is not taken in elementary functions. Fortunately for us, we are not dealing with an indefinite integral, but with a definite integral. As we remember from the previous course, the definite integral is a number and there are quite a few numerical methods obtaining this number with any predetermined accuracy. Applying one of these methods, we will get a number, which will be the probability that the next measurement of a random variable will fall into the interval (a, b). By changing the boundaries of the interval and performing similar calculations, we will get the probability of a random variable falling into this new interval, etc. The task seems to be solved. We have a method for calculating the probability of a random variable falling into any predetermined interval. However, carrying out such calculations is not very convenient, since it requires a lot of calculations. Can you make your life easier? Well, the first thing that comes to mind is to calculate all the integral values ​​for intervals that change with a certain (small step) and put them in a table. Then you can use this table and count nothing. But this table will be correct only for the random variable for which it was calculated. It turns out that we need to create an infinite number of tables for all kinds of random variables. Clearly, something needs to be done here as well. Mankind has figured out how to get by with one table for all cases. To do this, from our random variable X (any one that we study) we need to go to another random variable Z, using the following relation:

What will we get as a result of this operation? We will get a new random variable for which = 0 and s = 1. This random variable is called normalized normally distributed random variable Z. Since this operation can be performed for ANY random variable obeying the Gauss law, we can reduce any random variable to a random variable Z, and, therefore, to calculate the probability of the original random variable falling into a predetermined interval, build ONLY ONE table. Of course, such a table was built a long time ago, it is given in Appendix 3 and is called a table values ​​of the distribution function of a normalized normally distributed random variable

.(6).

Let's learn how to use this table. For example, consider the number at the intersection of the row starting with 0.5 and the column marked with the number 5. This number is equal to 0.7088. It shows that at the next test, the probability that the random variable will take a value LESS than 0.5 5 equals 0.7088. Please note that the column number is the hundredth character of the number we specified. Now let's set the task like this. How, using the table, to find the probability of falling into the interval (z 1, z 2), because this is our main task. If z 2 > z 1, then the desired probability will be equal to the difference Ф(z 2)–Ф(z 1). For example, let's find the probability that during the next test the value of the normalized random variable will fall into the interval (0.95; 1.54). First we find Ф(1.54). To do this, we find in the table a line that starts with 1.5, then we move along this line to the column marked with the number 4. There is the value Ф (1.54) = 0.9382. Similarly, we find Ф(0.95) = 0.8289. Then the desired probability will be equal to: P \u003d 0.9382 - 0.8289 \u003d 0.1093.

To complete the solution of the problem, it remains to answer only one question: what if the z values ​​turn out to be negative? Indeed, in the table of application 3 there are no negative values. The answer to this question is given by the following formula:

Ф(-z) = 1 - Ф(z) (7).

From formula (7) it follows: if z turned out to be negative, then you need to find the value of Ф (z) according to the table, considering z to be positive, and then subtract the found value from unity, this will be the answer. Now the problem of finding the probability of hitting a random variable distributed according to the Gauss law in any predetermined interval has been completely solved

Let us consider the following example to illustrate the concepts introduced into consideration. Let in maternity hospital 20 children were born per day, the weight of which, with an accuracy of 0.1 kilograms, is given in Table 1.

Table 1

Newborn weight in kilograms

X	1,8	2,1	3,1	4,2	3,6	2,2	3,1	3,9	4,4	2,6
3,3	3,8	3,3	4,8	2,8	3,6	3,4	2,8	3,7	3,2

So, in formula (2) a=2, b=3

Task 2.

The solution of the first problem, although important, is certainly not sufficient for practical purposes. The next most important task of statistics is to get an answer to the question whether it can be considered that some effect really exists or whether it is necessary to recognize that in fact there is no effect, and everything that we observe is a game of chance. Anything can be meant by effect, for example, are Scandinavians really taller than Africans, is one drug really more effective than another, are physiological parameters really changing in the process of adaptation, is performance in one class really higher than performance in another, etc.

Obviously, all these tasks are aimed at comparing two samples. The question is how to do it. Let's say we measured the height of 10,000 Scandinavians and 10,000 Africans. Thus, we have two sets of 10,000 numbers. It is clear that simply by looking at these numbers, we will achieve little. There is a need to describe each of the sets with a small number of parameters derived from them and only then compare not the numbers themselves included in a particular set, but these newly obtained parameters that characterize each of the sets. Since the newly obtained parameters describe the sample made, they are called "descriptive statistics". Descriptive statistics can be divided into several groups. We will consider two of them: measures of central tendency and measures of dispersion.

Measures of central tendency characterize the central value around which the values ​​of a random variable are distributed. These include the arithmetic mean (introduced in the previous section) and the median. The arithmetic mean is well suited for describing distributions that are close to normal. If the distribution differs significantly from the normal one (for example, it has very long and wide tails), then in this case it makes sense to use the median to estimate the "central" value.

The median of the distribution of any random variable X is a number Me for which the probability that the next test will result in a value of the random variable under study greater than Me is equal to 1/2. This means that the probability of getting a value less than or equal to Me is also 1/2. Thus, the median characterizes the center of the distribution in the sense that the occurrence of values ​​greater than the median and less than the median are equally likely.

Now let's consider an algorithm for estimating the median from the sample values. (Note the word "evaluate").

The first thing to do is to rank, i.e. Arrange all values ​​in the sample in ascending order. If we do this procedure with the sample presented in the previous section, we will get the following table:

X	1,8	2,1	2,2	2,6	2,8	2,8	3,1	3,1	3,2	3,3
3,3	3,4	3,6	3,6	3,7	3,8	3,9	4,2	4,4	4,8

Next, you need to determine an even or odd number of values ​​in the sample. If the number of values ​​is odd, then the median is equal to the value at the center of the sample; if the number of values ​​is even, then the median is half the sum of the values ​​at the center of the sample. In our case, the number of values ​​in the sample is 20, i.e. even. In 10th place is the number 3.3, and in 11th place is also the number 3.3. Therefore, the median is: . In our case, it turned out that the median and the arithmetic mean are equal, but this is not always the case.

Scattering measures characterize the spread with which a random variable is distributed around its central value. These measures include variance, standard deviation (introduced in the previous section), standard error of the mean, coefficient of variation.

If we take the arithmetic mean as the central value, then the dispersion estimate can be calculated using the following formula:

(8).

For our case

As can be seen from the comparison of formulas (4) and (8), the standard deviation estimate is related to the dispersion estimate by the following relation:

(9)

In our case .

In medicine, when making calculations, such a measure of dispersion as the standard error of the mean (m) is of great importance, since the results of studies are often presented in the form: . The formula for calculating the estimate of the standard error of the mean is given by the following simple relation:

(10)

For our case

The dispersion measures described above (variance, standard deviation, standard error of the mean) have one drawback: they give an indicator of the variability of a feature in named values, and not in relative ones. For example, for the sample presented in Table 1, the variance will be expressed in kg 2 and the standard deviation and standard error in kilograms. Therefore, it is impossible to compare (or compare) different characteristics according to these parameters. For example, if we measured not only the weight of newborns, but also their height, then using these dispersion measures it would be impossible to answer the question of where the variability is greater: in the case of weight or in the case of height.

To compare the variability of two heterogeneous samples, it is convenient to use the coefficient of variability (variation) of a trait, which is expressed in relative terms, namely in percent, and is calculated by the formula:

(11).

In our case

The more V, the more variable the sign. Values ​​of the coefficient of variation that do not go beyond 10% , are considered normal.

If V>20%, then the sample is non-compact with respect to the given attribute.

Now that we have introduced descriptive statistics into consideration, the problem of determining whether there is an effect or not can be reduced to the question of whether any descriptive statistics of one sample differ from another.

It would seem that the solution to the problem is very simple: calculate the descriptive statistics of one and the second sample and compare them with each other. However, the matter is far from being so simple. Indeed, if we measured the weight of not 20 newborns, but say, for example, only 19, would the value of the mean and all other descriptive statistics be the same? Most likely no! As mentioned above, we are always dealing with a sample, and not with the general population, so we always get ESTIMATES of descriptive statistics, and not their true values. Therefore, to solve the problem, it is impossible to draw conclusions by directly comparing the values ​​themselves. How then to solve the problem?

The concept of confidence interval comes to the rescue. The idea of ​​confidence intervals comes from the question: well, we don't know the exact value of this or that descriptive statistic, but can we at least specify the interval in which it is? The answer to this question is: yes, we can build an interval that contains the exact value of a particular descriptive statistic with a predetermined probability. Thus, we can construct a confidence interval in which the exact value of the descriptive statistic is contained with a probability of, for example, 80% or 90%, or 95% or 99%, etc.

Consider the construction of a confidence interval for the mean value. In this case, the following relation is obtained:

-mt< μ < +mt(12)

In formula (12) - average , μ is the mathematical expectation (this is the “true” value, see (2)), m is the standard error of the mean (see (10)). It remains to figure out what t is. The letter t usually denotes the value of the Student's distribution. The calculation of a specific value of the Student's distribution for any particular case is a rather difficult task, therefore this distribution has long been tabulated and presented in the table of Appendix 4.

Consider this table. To find the value we need, we must first of all answer the question for ourselves: with what probability are we going to build a confidence interval? Appendix 4 contains a table that allows you to build confidence intervals with probabilities of 0.95, 0.99 and 0.999. If we are given, for example, a probability of 0.95, then we will use the first column of the table. In order to find the number we need in this column, we need to find a line that starts with a number equal to n-1, where n is the number of dimensions. In our case, n=20, which means we are looking for a line starting with 19. At the intersection of the selected column and the desired row, the value we need is located. In our case, this number is 2.093. Therefore, the confidence interval will be () or, after calculations (2.965; 3.635). So, the true mean (expectation) with a probability of 0.95 lies SOMEWHERE between these two numbers. We wrote the word "somewhere" to illustrate one of the properties of confidence intervals: any value within the interval can be the expected value with the same probability. The second property is that we built the interval with a probability of 0.95, which means that with this probability the true mean lies inside the interval, but it also means that with a probability of 0.05 it is not in this interval. Here, for the first time, we encounter a fundamental property of any statistical inference: there is always the possibility that it is wrong. Statistical inference is the calculation of the probability of the validity of two hypotheses: null and alternative. The null hypothesis always says no. There are no differences in descriptive statistics between the two samples, no relationship between the two samples, and so on. Obviously, the alternative says "yes" accordingly. The question arises, when can the null hypothesis be considered refuted and the alternative one accepted? To do this, you need to set the level of significance. The significance level is the maximum acceptable probability for the researcher to erroneously reject the null hypothesis when in fact it is true. In medicine, a minimum significance level of 0.05 is accepted. What does it mean? If, as a result of calculations, we get that the probability of the validity of the null hypothesis is less than 0.05, we have the right to refute it and accept an alternative hypothesis, thereby considering it proved that there are differences (and, consequently, the effect).

Now we have everything necessary concepts, to solve the problem "is there an effect or not". Suppose we have a group of men of 20 patients with hypertension of the same age range and the same severity of the disease. Let them take new drug to lower blood pressure. It is necessary to answer the question: is this drug really effective. Conducted background (before treatment) daily monitoring of systolic blood pressure and obtained average daily values ​​for each of 20 people. After applying the treatment regimen, daily monitoring of systolic blood pressure was again carried out and average daily values ​​were also obtained for each patient. As a result, the values ​​presented in Table 2 were obtained.

table 2

Mean daily values ​​of systolic blood pressure before and after treatment

patient number	Average daily systolic pressure (before treatment), mm Hg	Average daily systolic pressure (after treatment) mm Hg	The difference in systolic pressure before treatment and after treatment, mm Hg
			+10
			+10
			-2
			+11
			+8
			+1
			+4
			+9
			+8
			+17
			+9
			+17
			+11
			+11
			+27
			+11
			+22
			-1
			+15
			+21
	177,1	166.2	11,0
s	6,8	8,4	7,5
m	1,5	1,9	1,7

Algorithm for solving the problem using confidence intervals.

Thus, Problem 2 can be solved by constructing confidence intervals. However, a different approach is more commonly used to solve this problem. It is based on calculating the experimental value of the Student's distribution and comparing it with the table.

To construct this algorithm for solving Problem 2, we need to introduce two more concepts. Let's ask ourselves whether it is possible to rearrange the experimental data in the columns in Table 2 in an arbitrary order? Answer: of course not, because in this case, the data obtained on one patient will get to another! Such samples are called linked samples. In our case, they are linked by the patient number. For such samples, the experimental value of the Student's distribution is calculated by the formula:

(13)

In formula (13) - arithmetic mean of the difference, - standard deviation for the difference, - error of the mean for the difference. Using the values ​​in Table 2, we calculate .

As we already know, the tabular value () for the level of 0.95 and the number of degrees of freedom 19 (20-1) is 2.086, therefore, in our case . Therefore, the observed differences in blood pressure do exist. We currently draw this conclusion at a significance level of 0.05. But now that we have the experimental value of the Student's distribution, we can compare it with tabular values ​​for other confidence probabilities. Let's see, for example, what is the tabular value of Student's distribution for a confidence level of 0.99 (significance level 0.01). As follows from the table in Appendix 4, this value is 2.861, and for a confidence level of 0.999 (significance level 0.001) - 3.883. Since 6.47 > 3.883, we can conclude that the studied drug is effective not only at the level of 0.05, i.e. assuming that the error probability is not more than 5%, but also at the level of 0.001, i.e. the probability that our conclusion is not correct does not exceed 0.1% !!!

The above calculations are valid for linked samples. Now we will solve the same problem (whether there really is an effect or the obtained differences are nothing more than a game of chance) for unrelated samples.

Consider how the hypothesis of inequality of means for unrelated samples is tested. In this case, the experimental value of the Student's distribution can be calculated using the formula:

(14)

In formula (14) and, respectively, the arithmetic mean for the first sample and the arithmetic mean for the second sample. Similarly, - the volume of the first sample, - the volume of the second sample, s - the combined estimate of the standard deviation of the two groups, which is calculated by the formula:
(15)

In formula (15) - an estimate of the standard deviation for the first group, and - for the second. - the value of the Student's distribution, calculated from the experimental data.

Table 3 shows the values ​​of the performance averaged over all grades for two groups of students in the first semester. It is necessary to determine whether it can be considered that one group learned better than the other.

Obviously, in this case we are dealing with unrelated samples.

Table 3

The average performance of students of the two groups for the first semester.

No. in order	Progress in the first group (average score per semester)	Performance in the second group (average score per semester)
	4,1	3,1
	3,8	3,7
	4,1	3,8
	3,5	3,2
	3,2	4,0
	2,9	3,4
	3,7	3,6
	4,2	4,1
	5,0	3,3
	2,8	4,2
	3,6	2,7
	4,9	3,2
		2,7
		3,9
N
	3,82	3,49
s by groups	0,69	0,49
s combined standard deviation	0,59

Calculated by formula (3), and calculated by formula (4). Using formula (15) we calculate s:

Now, using formula (14), we calculate the experimental value of the Student's distribution:

Next, we find the theoretical value of the Student's distribution for a confidence probability of 0.95 and the number of degrees of freedom. That is, we are looking for a number that is at the intersection of the first column of the Appendix 4 table and line 24. From the table it follows that this number is equal to .

Therefore, in our case: , and we have no right to say that one group learns better (or worse) than another. We are forced to admit that the differences observed in the performance of groups are of a random nature, and, in general, the performance of the groups is the same.

This completes the solution of Problem 2. It remains to make only two remarks.

Remark 1 is that the above calculation schemes are valid if both samples are made from populations distributed according to the Gaussian law.

Remark 2. We are aware that at present no one will count manually in real calculations. However, to consolidate the material, it is very useful to carry out calculations using a calculator. For these purposes, the complete solution of the model problem is given below.

Task The content of free heparin in the blood in two different age groups took the following values:

X 1 (mg%)	5,7	5,9	6,3	5,6	4,1	4,0	4,5	5,0	5,1	6,7
X 2 (mg%)	5,1	3,2	6,0	5,1	4,9	3,8	6,2	4,5	5,6	5,8

1. Calculate the sample arithmetic mean, standard deviation, standard error of the mean, median, coefficient of variation for each series, and confidence intervals for the means. Compare mean heparin values ​​for two age groups.

Solution:

The number of measurements in each row n=10.

The sample mean is determined by the formula:

Therefore, for the first row it is equal to:

Let's find the dispersion by the formula:

Therefore, for the first row, the sample variance is:

Calculate the standard deviation
.

Calculate the standard error of the mean

To determine the median ( Me 1) by given values x 1 i we build a variational series:

4,0 4,1 4,5 5,0 5,1 5,6 5,7 5,9 6,3 6,7

With an even number, the median option is defined as the arithmetic mean of the two central options:

(mg,%)

Calculate the coefficient of variation .

Calculate the 95% confidence interval for the mean. In our case, the number of measurements is 10, and the confidence level is 0.95. We enter the table of application 4. At the intersection of column 0.95 and the ninth row is the number t = 2.262.

Therefore, in our case, , and hence the confidence interval will be or finally .

Having carried out similar calculations for the second row, we get:

Comparing the confidence interval for the mean first series with the confidence interval for the second series, it is easy to see that they overlap a lot. Therefore, the observed differences between the means are random and we must conclude that there are no differences between them.

2. Compare the means using the calculation of the experimental value of the Student's distribution.

In this case, we are dealing with unrelated samples, so to calculate the experimental value, we will use the following formula:
.

Compute the combined estimate of the standard deviation of the two groups:

Then
. The number of degrees of freedom in our case is n=10+10-2=18. So, we enter the tables of Appendix 4 on the eighteenth line and the first column. At the intersection is the number 2.103. This number is much larger than the resulting 0.71. Therefore, we come to the same conclusion that the means of the two samples do not differ.

So, the answer in this case will look like this:.

Random a quantity is called, which, as a result of the test, can take one or another possible value, unknown in advance. There are discrete and continuous random variables.
If the set of possible values ​​of a random variable is finite or form an infinite numerical sequence, then such a random variable is called discrete (examples 3.1, 3.3, 3.4).
A random variable whose set of values ​​completely fills a certain numerical gap is called continuous (example 3.2). Note that discrete and continuous variables do not exhaust all types of random variables.
If a random variable is neither a discrete nor a continuous random variable, then it is called mixed .
Obviously, to fully characterize a discrete random variable, it is not enough to know its value. They need to be matched with probabilities.
The correspondence between all possible values ​​of a discrete random variable and their probabilities is called distribution law given random variable.
The simplest form of specifying the distribution law of a discrete random variable is a table that lists the possible values ​​of a random variable (usually in ascending order) and their corresponding probabilities:

Such a table is called a distribution row. Assume that the number of possible values ​​of a random variable is finite: X 1 , X 2 , …, x n. In one trial, the random variable takes on one and only one constant value. Therefore, events X = x i (i = 1, 2, … , n) form full group pairwise independent events. Hence, R 1 + R 2 + … + p n = 1.
You can also depict the distribution law graphically, plotting the possible values ​​​​of a random variable on the abscissa axis, and the corresponding probabilities on the ordinate axis. For greater expressiveness, the obtained points are connected by straight line segments. The resulting figure is called a distribution polygon (polygon).
There are a number of distribution laws:

Binomial

Poisson

Normal(Gauss)

exponential (exponential)

Uniform

Binomial distribution of a random variable

n is the number of trials

Poisson distribution.
A situation where the probability of occurrence of an event in each trial is close to 0 (such events are called rare events), and the number of trials is large. The probability that an event will occur m times in n independent trials is approximately equal to:

n is the number of trials
m - the expected occurrence of the desired event
p is the probability of an event occurring in one trial
Example: It has been established that when transporting more than 5000 products in a wagon, on average one product deteriorates. Find the probability that three items will fail. (0.06).

– mathematical expectation

Dispersion

The exponential (exponential) distribution

- intensity (average number of events per unit of time)

A continuous random variable X, whose density function is given by this expression, is called a random variable having an exponential, or exponential, distribution.

The value of the service life of various devices and the time of failure-free operation of individual elements of these devices, under certain conditions, usually obeys an exponential distribution. In other words, the amount of time interval between the occurrence of two successive rare events often obeys an exponential distribution.

As can be seen from the formula, the exponential distribution is determined by only one parameter . This feature of the exponential distribution indicates its advantage over distributions that depend on a larger number of parameters.

The graph of exponential distribution functions looks like:

The probability of hitting a random variable X in the interval :

,expected value

, dispersion

standard deviation

Thus, for an exponential distribution, it is characteristic that the standard deviation is numerically equal to the mathematical expectation.

Uniform distribution
The uniform probability distribution is the simplest and can be either discrete or continuous. A discrete uniform distribution is such a distribution for which the probability of each of the values ​​of CB is the same, that is:

where N is the number of possible SW values.

The probability distribution of a continuous CB X, taking all its values ​​from the segment [a; b] is called uniform if its probability density on this segment is constant, and outside it is equal to zero:

11. Distribution function and its properties.

distribution function random variable X is the probability that it will take a value less than the function argument x:

F(x)=P( X<x}.

Geometrically, the distribution function is interpreted as the probability that a random point X will fall to the left of a given point X. From a geometric interpretation, one can clearly deduce basic properties of the distribution function.

1. F(-¥) = 0.

2. F(+¥) = 1.

3. F(x) is a non-decreasing function of its argument, i.e. at x 1 < x 2

F(x 1) £ F(x 2).

4. P(α£ X < β) = F(β) - F(α), for "[α,β[нR. (5.4)

The probability that a random variable X as a result of the experiment will fall on the segment from α to β (including α) equal to the increment of the distribution function I am in this area.

Thus, the distribution function F(x) of any random variable is a non-decreasing function of its argument, the values ​​of which are between 0 and 1: 0≤F(x)≤1, and F(-∞)=0, F(+∞)= 1.

12. Distribution function of a discrete and continuous random variable.

Distribution function of a discrete random variable

If x is a discrete random variable taking the values x 1 <x 2 < … <x i < … с вероятностями p 1 <p 2 < … <pi < …, то таблица вида

called distribution of a discrete random variable.

The distribution function of a random variable with such a distribution has the form

A discrete random variable has a stepwise distribution function.

The set of values ​​of a continuous random variable is uncountable and usually represents some finite or infinite interval.

A random variable x(w) given in a probability space (W, S, P) is called continuous(absolutely continuous) W if there exists a non-negative function such that, for any x, the distribution function Fx(x) can be represented as an integral

13. Density of distribution of a continuous random variable.

The function is called a function probability distribution density.

The properties of the distribution density function follow from the definition:

1. The distribution density is non-negative: .

2. The integral over the entire number line of the probability distribution density is equal to one:

3. At points of continuity, the distribution density is equal to the derivative of the distribution function: .

4. The distribution density determines the distribution law of a random variable, since it determines the probability of a random variable falling into the interval:

5. The probability that a continuous random variable will take a specific value is zero: . Therefore, the following equalities are true:

The plot of the distribution density function is called distribution curve, and the area bounded by the distribution curve and the x-axis is equal to one. Then, geometrically, the value of the distribution function Fx(x) at the point x0 is the area bounded by the distribution curve and the x-axis and lying to the left of the point x0.

14. Relationship between distribution function and distribution density. Integral formula of total probability.

Knowing the distribution density F(X), we can find the distribution function F(X) according to the formula

.

Really, F(X) = P(X< X ) = P(-∞< X < X) .

Hence,

.

.

Thus, Knowing the distribution density, you can find the distribution function. Of course, from the known distribution function, one can find the distribution density, namely:

F(X) = F"(X).
15. Numerical characteristics of random variables.

The distribution law completely describes a random variable with

probabilistic point of view. But often it is enough to indicate only a separate

nye numerical parameters that allow you to express in a compressed form

the most significant features of the distribution. Such parameters are called

are numerical characteristics of a random variable.

Among the numerical characteristics, one can single out the characteristics of

positions, i.e., some average, approximate values ​​of a random variable

ranks around which its possible meanings are grouped.

Numerical features include:

· Expected value

Dispersion

median

Moments

quantile

asymmetry

eccentricity

16.Mathematical expectation of a random variable and its properties.

Expected value - the number around which the values ​​of the random variable are concentrated. The mathematical expectation of a random variable x is denoted M x.

Mathematical expectation of a discrete random variable x with distribution

is called a quantity if the number of values ​​of the random variable is finite.

If the number of values ​​of a random variable is countable, then . Moreover, if the series on the right side of the equality diverges, then the random variable x is said to have no mathematical expectation.

Mathematical expectation of a continuous random variable with probability density px(x) is calculated by the formula . Moreover, if the integral on the right side of the equality diverges, then the random variable x is said to have no mathematical expectation.

If the random variable h is a function of the random variable x, h = f(x), That

.

Similar formulas are valid for functions of a discrete random variable:

, .

The main properties of mathematical expectation:

the mathematical expectation of a constant is equal to this constant, M c=c ;

· mathematical expectation - a linear functional on the space of random variables, i.e. for any two random variables x, h and arbitrary constants a And b fair: M (ax+ bh) = a M (x)+ b M (h);

the mathematical expectation of the product of two independent random variables is equal to the product of their mathematical expectations, i.e. M (x h) = M (x) M (h).

17. Dispersion of a random variable and its properties.

The dispersion of a random variable characterizes the measure of the spread of a random variable around its mathematical expectation.

If the random variable x has the mathematical expectation M x, That dispersion random variable x is called the quantity D x=M (x - M x) 2 .

It is easy to show that D x= M (x - M x) 2 =M x 2 - M (x) 2 .

This universal formula is equally well applicable for both discrete random variables and continuous ones. Value M x 2 >for discrete and continuous random variables, respectively, is calculated by the formulas

, .

To determine the measure of the spread of values ​​of a random variable, it is often used standard deviation, related to the dispersion by the relation .

The main properties of the dispersion:

the variance of any random variable is non-negative, D x 0;

the variance of the constant is zero, D c=0;

for an arbitrary constant D (cx) = c 2 D (x);

the variance of the sum of two independent random variables is equal to the sum of their variances: D (x ± h) = D (x) + D (h).

18. Moment of order k of a random variable, absolute and central moments.

The initial moment of the kth order random variable x is called the mathematical expectation k th power of the random variable x, i.e. a k = M x k.

The central moment of the kth order random variable x is called the variable m k, defined by the formula m k = M (x - M x)k.

Note that the mathematical expectation of a random variable is the initial moment of the first order, a 1 = M x, and the variance is the central moment second order,

a 2 = M x 2 = M (x - M x) 2 =D x.

There are formulas that allow expressing the central moments of a random variable in terms of its initial moments, for example:

m2 =a 2 -a 1 2 , m 3 = a 3 - 3a 2 a 1 + 2a 1 3 .

If the probability distribution density of a continuous random variable is symmetric with respect to the straight line x= M x, then all its central moments of odd order are equal to zero.

ABSOLUTE MOMENT

random variable s X- mathematical expectation The usual designation of A. m. thus,

Number r called. order of A. m. If F(x) is the distribution function x, That

and, e.g., if the distribution X has a density p(x) , That

19. Fashion and Fashion

random variable X is called its most probable value, i.e., then,

for which the probability pi

or the distribution density f (x) reaches

go to the maximum. Mode is usually denoted by Mx

If the probability polygon or distribution density reaches a maximum at

several points, then such distributions are called polymodal median random variables.

The median of a continuous random variable X is called

its value xm is given for which

20. Level x quantile of the distribution of a random variable.

-quantile random variable with distribution function is any number that satisfies two conditions:

2)

Note that these conditions are equivalent to the following:

If - strictly continuous monotonic function, then there is a single quantile of any order that is uniquely determined from the equation and, therefore, is expressed in terms of the function inverse to the distribution function:

In addition to the indicated situation, when the equation has a unique solution (which gives the corresponding quantile), two others are also possible:

§ if the specified equation has no solutions, then this means that there is a single point at which the distribution function has a discontinuity that satisfies this definition and is a quantile of order . For this point, the following relations are satisfied: and (the first inequality is strict, and the second can be both strict and turn into equality).

§ if the equation has more than one solution, then all its solutions form an interval on which the distribution function is constant. Any point of this interval can be taken as the order quantile. Meaningful conclusions, in which the quantile is involved, will not change significantly because of this, since the probability of a random variable falling into this interval is zero.

21. Asymmetry and eccentricity of the distribution of a random variable.

Asymmetry

In probability theory and in mathematical statistics, the measure of distribution skewness is the skewness coefficient, which is determined by the formula ,

where m 3 is the central moment of the third order, is the standard deviation.

The normal distribution is most often used in probability theory and in mathematical statistics, so the probability density plot of the normal distribution has become a kind of benchmark against which other distributions are compared. One of the parameters that determine the difference between the distribution of a random variable x from the normal distribution is kurtosis.

Kurtosis g random variable x is defined by the equality .

For a normal distribution, of course, g = 0. If g (x) > 0, then this means that the probability density graph px(x) is more “pointed” than the normal distribution, if g (x)< 0, то “заостренность” графикаpx(x) is smaller than the normal distribution.

22. Binomial distribution law.

P is the probability of an event occurring in one trial.
q - probability of not occurrence of an event in one test q = (1-p)
n is the number of trials
k is the estimated number of occurrences of the desired event
The Bernoulli formula allows you to calculate the probability that an event will appear in n trials exactly k times.

23. Normal law of distribution of a random variable. Theory of Laplace-Lyapunov.
Normal (Gaussian) distribution
This is the basic law of probability theory. In the limit, all laws tend to normal distribution laws. The sum of an infinite number of random variables distributed according to any laws eventually acquires a normal distribution law.

A continuous random variable X is called normally distributed if its distribution density is:

– mathematical expectation discrete random variable is called the sum of products of possible values ​​of a random variable by the probability of their occurrence

Dispersion- to assess the degree of dispersion (deviation) of an indicator from its average value, the concepts of dispersion are used.

Sample variance or sample variance is a measure of the variability of a variable. The dispersion is calculated by the formula:

where x is the sample mean, N is the number of observations in the sample. The dispersion varies from zero to infinity. An extreme value of 0 means no variability when the values ​​of the variable are constant. - standard deviation of a random variable (square root of the variance.)

The graph of the normal distribution function, as can be seen from the figure, has the form of a domed curve called
Gaussian, the maximum point has coordinates This means that this ordinate decreases with increasing value (the curve “shrinks” towards the Ox axis) and increases with decreasing value (the curve “stretches” in the positive direction of the Oy axis). Changing the values ​​of the parameter u (with a constant value of ) does not affect the shape of the curve, but only moves the curve along the Ox axis. Normal distribution with parameters =0 and =1 is called normalized. The distribution function of a random variable in this case will look like:

For =0, =1 the graph takes the form:

This curve at =0, =1 received the status of a standard, it is called a unit normal curve, that is, any collected data is sought to be transformed so that their distribution curve is as close as possible to this standard curve.

The normalized curve was invented to solve problems in probability theory, but it turned out in practice that it perfectly approximates the frequency distribution with a large number of observations for many variables

Let x 1 , x 2 , …, x n , … be an unlimited sequence of independent random variables with mathematical expectations m 1 , m 2 , …, m n , … and dispersions s 1 2 , s 2 2 , …, s n 2 … . Let's denote , and .

In practice, most random variables, which are affected by a large number of random factors, obey the normal law of probability distribution. Therefore, in various applications of probability theory, this law is of particular importance.

A random variable $X$ obeys the normal probability distribution law if its probability distribution density has the following form

$$f\left(x\right)=((1)\over (\sigma \sqrt(2\pi )))e^(-(((\left(x-a\right))^2)\over ( 2(\sigma )^2)))$$

Schematically, the graph of the function $f\left(x\right)$ is shown in the figure and has the name "Gaussian curve". To the right of this graphic is the German 10 Mark banknote, which was in use even before the introduction of the euro. If you look closely, then on this banknote you can see the Gaussian curve and its discoverer, the greatest mathematician Carl Friedrich Gauss.

Let's go back to our density function $f\left(x\right)$ and give some explanation about the distribution parameters $a,\ (\sigma )^2$. The parameter $a$ characterizes the center of dispersion of the values ​​of the random variable, that is, it has the meaning of the mathematical expectation. When the parameter $a$ changes and the parameter $(\sigma )^2$ remains unchanged, we can observe the shift of the graph of the function $f\left(x\right)$ along the abscissa axis, while the density graph itself does not change its shape.

The parameter $(\sigma )^2$ is the variance and characterizes the shape of the density curve $f\left(x\right)$. When changing the parameter $(\sigma )^2$ with the parameter $a$ unchanged, we can observe how the density graph changes its shape, shrinking or stretching, while not shifting along the abscissa.

Probability of a normally distributed random variable falling into a given interval

As is known, the probability that a random variable $X$ falls into the interval $\left(\alpha ;\ \beta \right)$ can be calculated $P\left(\alpha< X < \beta \right)=\int^{\beta }_{\alpha }{f\left(x\right)dx}$. Для нормального распределения случайной величины $X$ с параметрами $a,\ \sigma $ справедлива следующая формула:

$$P\left(\alpha< X < \beta \right)=\Phi \left({{\beta -a}\over {\sigma }}\right)-\Phi \left({{\alpha -a}\over {\sigma }}\right)$$

Here the function $\Phi \left(x\right)=((1)\over (\sqrt(2\pi )))\int^x_0(e^(-t^2/2)dt)$ is the Laplace function . The values ​​of this function are taken from . The following properties of the function $\Phi \left(x\right)$ can be noted.

1 . $\Phi \left(-x\right)=-\Phi \left(x\right)$, i.e. the function $\Phi \left(x\right)$ is odd.

2 . $\Phi \left(x\right)$ is a monotonically increasing function.

3 . $(\mathop(lim)_(x\to +\infty ) \Phi \left(x\right)\ )=0.5$, $(\mathop(lim)_(x\to -\infty ) \ Phi \left(x\right)\ )=-0.5$.

To calculate the values ​​of the $\Phi \left(x\right)$ function, you can also use the $f_x$ function wizard of the Excel package: $\Phi \left(x\right)=NORMDIST\left(x;0;1;1\right )-0.5$. For example, let's calculate the values ​​of the function $\Phi \left(x\right)$ for $x=2$.

The probability that a normally distributed random variable $X\in N\left(a;\ (\sigma )^2\right)$ falls into an interval symmetric with respect to the expectation $a$ can be calculated by the formula

$$P\left(\left|X-a\right|< \delta \right)=2\Phi \left({{\delta }\over {\sigma }}\right).$$

Three sigma rule. It is practically certain that a normally distributed random variable $X$ falls into the interval $\left(a-3\sigma ;a+3\sigma \right)$.

Example 1 . The random variable $X$ is subject to the normal probability distribution law with parameters $a=2,\ \sigma =3$. Find the probability that $X$ falls into the interval $\left(0,5;1\right)$ and the probability that the inequality $\left|X-a\right|< 0,2$.

Using the formula

$$P\left(\alpha< X < \beta \right)=\Phi \left({{\beta -a}\over {\sigma }}\right)-\Phi \left({{\alpha -a}\over {\sigma }}\right),$$

find $P\left(0,5;1\right)=\Phi \left(((1-2)\over (3))\right)-\Phi \left(((0,5-2)\ over (3))\right)=\Phi \left(-0.33\right)-\Phi \left(-0.5\right)=\Phi \left(0.5\right)-\Phi \ left(0.33\right)=0.191-0.129=$0.062.

$$P\left(\left|X-a\right|< 0,2\right)=2\Phi \left({{\delta }\over {\sigma }}\right)=2\Phi \left({{0,2}\over {3}}\right)=2\Phi \left(0,07\right)=2\cdot 0,028=0,056.$$

Example 2 . Suppose that during the year the price of shares of a certain company is a random variable distributed according to the normal law with a mathematical expectation equal to 50 conventional monetary units and a standard deviation equal to 10. What is the probability that on a randomly chosen day of the period under discussion, the price for the share will be:

a) more than 70 conventional monetary units?

b) below 50 per share?

c) between 45 and 58 conventional monetary units per share?

Let the random variable $X$ be the price of shares of some company. By condition $X$ is subject to normal distribution with parameters $a=50$ - mathematical expectation, $\sigma =10$ - standard deviation. Probability $P\left(\alpha< X < \beta \right)$ попадания $X$ в интервал $\left(\alpha ,\ \beta \right)$ будем находить по формуле:

$$P\left(\alpha< X < \beta \right)=\Phi \left({{\beta -a}\over {\sigma }}\right)-\Phi \left({{\alpha -a}\over {\sigma }}\right).$$

$$a)\ P\left(X>70\right)=\Phi \left(((\infty -50)\over (10))\right)-\Phi \left(((70-50)\ over (10))\right)=0.5-\Phi \left(2\right)=0.5-0.4772=0.0228.$$

$$b)\ P\left(X< 50\right)=\Phi \left({{50-50}\over {10}}\right)-\Phi \left({{-\infty -50}\over {10}}\right)=\Phi \left(0\right)+0,5=0+0,5=0,5.$$

$$c)\ P\left(45< X < 58\right)=\Phi \left({{58-50}\over {10}}\right)-\Phi \left({{45-50}\over {10}}\right)=\Phi \left(0,8\right)-\Phi \left(-0,5\right)=\Phi \left(0,8\right)+\Phi \left(0,5\right)=$$

The density of the normal distribution has the following form:

Where a- the center of the probability distribution or the mathematical expectation of a given random variable, i.e.

the standard deviation of a given random variable.

In practice, the corresponding statistical estimates are calculated. So, the estimate for the mathematical expectation will be the average value:

where is the amount of data in the considered statistical array.

The mathematical expectation is that theoretical value of a given random variable, to which the average value tends with an unlimited increase in the amount of data.

Standard deviation:

In logistics, one or another value of a quantity is estimated by the value

in this case, the coefficient of variation is estimated:

Figure 4 shows a graph of the normal probability distribution.

Figure 4 - Normal law of probability distribution

The density of the exponential probability distribution law has the following form:

where is the base of the natural logarithm.

The exponential law describes the time parameters of random logistic processes. The following random variables fall under the exponential law:

1) customer service time;

2) the time of loading and unloading vehicles;

3) time spent on other logistics operations

4) the interval between applications arriving for service.

A feature of the exponential law is that it is determined by one parameter. Wherein

where is the average value of the studied time parameter.

For quantities obeying an exponential law, the mathematical expectation M and the root mean square value are equal to each other:

Figure 5 shows a graph of the exponential law.

Figure 5 - Exponential probability distribution law

Binomial probability distribution law

The binomial law of probability distribution is expressed by the formula:

This law determines the probabilities of occurrence of events from the total number of events

where is the probability of occurrence of one event from a given group of events;

the probability that the specified event will not occur,

The value of the number of combinations of By , is determined by the formula:

To calculate the number of combinations, the equality is used:

With a binomial distribution, the most likely number of events is:

Comparison of probability distribution laws. Goodness Criteria

In probability theory, there are methods that allow one to estimate the degree of correspondence between actual probability distributions and their theoretical values. For this purpose, so-called goodness-of-fit criteria are used, the best known of which is the criterion. This criterion makes it possible to compare empirical distribution laws obtained from the same actual data.

The smaller the value, the better this empirical law agrees with the theoretical one. To compare the empirical laws of probability distribution, the values ​​are calculated according to the following formula:

Where, respectively, are the actual and theoretical values ​​of the frequencies of the distribution laws under study.

The value is also random, and therefore obeys its own distribution law. The approach to comparing empirical distribution laws can be shown by an example.

Let us establish which probability distribution law - normal or exponential - better reflects the distribution of a given value, i.e. Let's test the hypotheses. As the investigated value, we take the volume of sales of a certain product. The initial data are presented in table 3:

Table 3. Information about the sale of goods

	Sales (thousand rubles)

The task is formulated as follows: to build the probability distribution of the demand for a given product, if as a result of the study, the results of the implementation were obtained, in thousand rubles. in a day.

The solution to the problem is presented in Appendix 4.

In the general case, a number of logistics processes, namely: sales, shipment of products from wholesale and trade enterprises, movement of stocks, provision of services for the supply of products, spending material resources and so on. is described by the normal law of probability distribution. A distinctive feature of this distribution is the presence of a pronounced symmetry of random variables relative to their average value. For these processes, the normal law applies to all products, certain product groups or individual items.

In the ABC analysis of the structure of logistics processes, the resulting characteristics in value or physical terms are subject to exponential distribution.

The fact that the sale of products complies with the normal law is important for logistics, since it allows you to determine the amount of inventory, for which the following formula is recommended:

where is the required amount of inventory for an indefinite period,

average sales per unit of time (day, week, month),

standard deviation.

For this example, inventory is:

This model shows that any requirement of the buyer for a particular quality of goods must be satisfied with a probability close to 1. This model uses the "three sigma" rule. In the normal law corresponds to a probability of 0.99.

In modern conditions, computer technologies make it possible to track the average sales and standard deviations in the current time mode, as well as adjust the value of the inventory.

The inventory model provided can be used for both retail and wholesale.

LAW OF DISTRIBUTION AND CHARACTERISTICS

RANDOM VALUES

Random variables, their classification and methods of description.

A random value is a quantity that, as a result of an experiment, can take on one or another value, but which one is not known in advance. For a random variable, therefore, only values ​​can be specified, one of which it will necessarily take as a result of the experiment. These values ​​will be referred to as possible values ​​of the random variable. Since a random variable quantitatively characterizes the random result of an experiment, it can be considered as quantitative characteristic random event.

Random variables are usually denoted by capital letters of the Latin alphabet, for example, X..Y..Z, and their possible values ​​by the corresponding small letters.

There are three types of random variables:

discrete; Continuous; Mixed.

Discrete such a random variable is called, the number of possible values ​​of which forms a countable set. In turn, a countable set is a set whose elements can be numbered. The word "discrete" comes from the Latin discretus, which means "discontinuous, consisting of separate parts."

Example 1. A discrete random variable is the number of defective parts X in a batch of nfl. Indeed, the possible values ​​of this random variable are a series of integers from 0 to n.

Example 2. A discrete random variable is the number of shots before the first hit on the target. Here, as in Example 1, the possible values ​​can be numbered, although in the limiting case the possible value is an infinitely large number.

continuous is a random variable whose possible values ​​continuously fill a certain interval numerical axis, sometimes called the interval of existence of this random variable. Thus, on any finite interval of existence, the number of possible values ​​of a continuous random variable is infinitely large.

Example 3. A continuous random variable is the electricity consumption at the enterprise for a month.

Example 4. A continuous random variable is the error in the height measurement using an altimeter. Let it be known from the principle of operation of the altimeter that the error lies in the range from 0 to 2 m. Therefore, the interval of existence of this random variable is the interval from 0 to 2 m.

Law of distribution of random variables.

A random variable is considered to be completely specified if its possible values ​​are indicated on the numerical axis and the distribution law is established.

The law of distribution of a random variable is called a relation that establishes a relationship between the possible values ​​of a random variable and the corresponding probabilities.

A random variable is said to be distributed according to a given law, or subject to a given distribution law. A number of probabilities, a distribution function, a probability density, a characteristic function are used as distribution laws.

The distribution law gives a complete probable description of a random variable. According to the distribution law, it is possible to judge before experience which possible values ​​of a random variable will appear more often, and which ones less often.

For a discrete random variable, the distribution law can be given in the form of a table, analytically (in the form of a formula) and graphically.

The simplest form setting the distribution law of a discrete random variable is a table (matrix), which lists in ascending order all possible values ​​of a random variable and their corresponding probabilities, i.e.

Such a table is called a series of distribution of a discrete random variable. 1

The events X 1 , X 2 ,..., X n , consisting in the fact that, as a result of the test, the random variable X will take the values ​​x 1 , x 2 ,... x n, respectively, are inconsistent and the only possible ones (because the table lists all possible values ​​of a random variable), i.e. form a complete group. Therefore, the sum of their probabilities is equal to 1. Thus, for any discrete random variable

(This unit is somehow distributed among the values ​​of the random variable, hence the term "distribution").

A distribution series can be displayed graphically if the values ​​of a random variable are plotted along the abscissa axis, and their corresponding probabilities along the ordinate axis. The connection of the obtained points forms a broken line, called a polygon or polygon of the probability distribution (Fig. 1).

Example The lottery is played: a car worth 5000 den. units, 4 TVs worth 250 den. unit, 5 VCRs worth 200 den. units In total, 1000 tickets are sold for 7 den. units Draw up the law of distribution of the net winnings received by the lottery participant who bought one ticket.

Solution. Possible values ​​of the random variable X - net winnings per ticket - are 0-7 = -7 den. units (if the ticket did not win), 200-7 = 193, 250-7 = 243, 5000-7 = 4993 den. units (if the ticket won the VCR, TV or car, respectively). Given that out of 1000 tickets the number of non-winners is 990, and the indicated winnings are 5, 4 and 1, respectively, and using the classical definition of probability, we get.