Graph of the distribution function of a random variable. Probability distribution function of a random variable and its properties


A universal way of specifying the distribution law, suitable for both discrete and continuous random variables, is the distribution function.

Distribution function random variable X called function F(x), defining for each value x the probability that the random variable X will take a value less than x, that is

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

Basic properties of the distribution function F(x) :

1. Since by definition F(x) is equal to the probability of the event, all possible values ​​of the distribution function belong to the segment:

0 £ F(x) £ 1.

2. If, then, that is F(x) is a non-decreasing function of its argument.

3. The probability that a random variable will take a value belonging to the half-interval [ a, b), is equal to the increment of the distribution function on this interval:

P(a £ X < b) = F(b) - F(a).

4. If all possible values ​​of a random variable belong to the interval [ a, b], That

F(x) = 0, at x £ a; F(x) = 1, with x > b.

The distribution function of discrete random variables can be determined by the formula

. (15)

If the distribution series of a discrete random variable is known, it is easy to calculate and construct its distribution function. Let us demonstrate how this is done using example 23.

Example 25. Calculate and construct the distribution function for a discrete random variable, the distribution law of which has the form:

x i 0,1 1,2 2,3 4,5
p i 0,1 0,2 0,6 0,1

Solution. Let's determine the function values F(x) = P(X < x) for all possible values x:

at xО (- ¥; 0.1] there is not a single value of the random variable X, less than these values x, that is, there is not a single term in the sum (15):

F(x) = 0;

at xО (0.1; 1.2] only one possible value ( X= 0.1) less than the considered values x. That is, when xО (0.1; 1.2] F(x) = P(X = 0,1) = 0,1;

at xО (1.2; 2.3] two values ​​( X= 0.1 and X= 1.2) less than these values x, hence, F(x) = P(X = 0,1) + P(X = 1,2) = 0,1 + 0,2 = 0,3;

at xО (2.3; 4.5] three values ​​( X = 0,1, X= 1.2 and X= 2.3) less than these values x, hence, F(x) = P(X = 0,1) + P(X = 1,2) + P(X = 2,3) = 0,1 + 0,2 + 0,6 = 0,9 ;

at xО (4,5, ¥) all possible values ​​of the random variable X will be less than these values x, And F(x) = P(X = 0,1) + P(X = 1,2) + P(X = 2,3) +

+ P(X = 4,5) = 0,1 + 0,2 + 0,6 + 0,1 = 1.

Thus,

Graph of a function F(x) is shown in Figure 8.

In general, the distribution function F(x) discrete random variable X is a discontinuous step function, continuous on the left, whose jumps occur at points corresponding to possible values X 1 , X 2 , ... random variable X and are equal to the probabilities p 1 , p 2 , ... these values.


Distribution function of continuous random variables. Now we can give a more precise definition of continuous random variables: random variable X called continuous, if its distribution function F(x) for all values x is continuous and, in addition, has a derivative everywhere, with the possible exception of individual points.

From the continuity of the function F(x) follows that the probability of each individual value of a continuous random variable is zero.

Since the probability of each individual value of a continuous random variable is 0, property 3 of the distribution function for a continuous random variable will be of the form

P(a £ X < b) = P(a £ X £ b) = P(a < X £ b) = P(a < X < b) = F(b) - F(a).

Example 26. The probabilities of hitting the target for each of the two shooters are respectively equal to: 0.7; 0.6. Random value X- the number of misses, provided that each shooter fired one shot. Create a series of distributions of a random variable X, build a bar chart and distribution function.

Solution. Possible values ​​of this random variable X: 0, 1, 2. The problem condition can be considered as a series of n= 2 independent trials. IN in this case to calculate the probabilities of possible values ​​of a random variable X you can use the theorems of addition of probabilities incompatible events and multiplying the probabilities of independent events:

Let's denote the events:

A i = ( i-th shooter hit the target) i = 1, 2.

According to the condition, the probability of an event A 1 P(A 1) = 0.7, probability of event A 2 - P(A 2) = 0.6. Then the probabilities of opposite events: , .

Let us determine all the elementary events of a given random experiment and the corresponding probabilities:

Elementary Events Events Probabilities
Total

(Let's check that ).

Distribution series of a given random variable X looks like

x i Total
p i 0,42 0,46 0,12

The bar graph corresponding to this distribution series is shown in Figure 9.

Let's calculate the distribution function of this random variable:

:

at x Î (- ¥, 0] ;

at xО (0, 1] ;

at xО (1, 2] ;

at xО (2, +¥);

So, the distribution function of the random variable under consideration has the form:

Graph of a function F(x) is shown in Figure 10.

Probability density function of a continuous random variable.

Probability distribution density continuous random variable X at the point x the derivative of its distribution function at this point is called:

f(x) = F¢( x).

According to the meaning of the function values f(x) are proportional to the probability that the random variable under study will take a value somewhere in the immediate vicinity of the point x.

Density distribution function f(x), as well as the distribution function F(x), is one of the forms of specifying the distribution law, but it is applicable only for continuous random variables. Probability density function f(x) also called differential distribution function, while the distribution function F(x) are called, respectively, cumulative distribution function.

Density distribution plot f(x) is called distribution curve.

Let us consider the properties that the distribution density function of a continuous random variable has.

Property 1. The probability distribution density is a non-negative function:

f(x) ³ 0

(geometrically: the distribution curve lies not below the x-axis).

Property 2. The probability of a random variable falling into the area from a to b is determined by the formula

;

(geometrically: this probability is equal to the area of ​​the curvilinear trapezoid bounded by the curve f(x), axis Oh and straight x= a and x= b).

Property 3.

(geometrically: the area of ​​the figure bounded by the distribution curve and the x-axis is equal to one).

In particular, if all possible values ​​of a random variable belong to the interval [ a, b], That

Property 4. Distribution function F(x) can be found by known function density distribution as follows:

.

Example 27. A continuous random variable is specified by a distribution function

Determine the differential distribution density function.

Solution. Let us define the differential distribution density function

Example 28. Is each of the following functions the distribution density of some random variable?

Questions for self-control

1. What is called a random variable?

2. What quantities are called discrete? continuous?

3. What is the law of distribution of a random variable called?

4. In what ways can the distribution law of a discrete random variable be specified? continuous?

5. What characterizes the distribution function F(x) random variable?

6. How to determine the probability of a random variable falling into a certain interval using the distribution function?

7. What does the distribution density function of a random variable characterize? Indicate its probabilistic meaning.

8. For what quantities is the distribution density function defined?

9. Can the distribution density function take negative values?

10. How functions are related to each other F(x) And f(x)?

11. What random variables are called continuous?

12. What is the area of ​​the figure bounded by the distribution curve and the x-axis?

13. How to determine the probability of a continuous random variable falling into a certain interval using the distribution density function?

RANDOM VARIABLES

Example 2.1. Random value X given by the distribution function

Find the probability that as a result of the test X will take values ​​contained in the interval (2.5; 3.6).

Solution: X in the interval (2.5; 3.6) can be determined in two ways:

Example 2.2. At what parameter values A And IN function F(x) = A + Be - x can be a distribution function for non-negative values ​​of a random variable X.

Solution: Since all possible values ​​of the random variable X belong to the interval , then in order for the function to be a distribution function for X, the property must be satisfied:

.

Answer: .

Example 2.3. The random variable X is specified by the distribution function

Find the probability that, as a result of four independent tests, the value X exactly 3 times will take a value belonging to the interval (0.25;0.75).

Solution: Probability of hitting a value X in the interval (0.25;0.75) we find using the formula:

Example 2.4. The probability of the ball hitting the basket with one shot is 0.3. Draw up a distribution law for the number of hits with three throws.

Solution: Random value X– the number of hits in the basket with three shots – can take the following values: 0, 1, 2, 3. Probabilities that X

X:

Example 2.5. Two shooters each fire one shot at a target. The probability of the first shooter hitting it is 0.5, the second - 0.4. Draw up a distribution law for the number of hits on a target.

Solution: Let's find the law of distribution of a discrete random variable X– number of hits on the target. Let the event be the first shooter hitting the target, and let the second shooter hit the target, and be their misses, respectively.



Let's compose the law of probability distribution of SV X:

Example 2.6. Three elements are tested, operating independently of each other. The duration of time (in hours) of failure-free operation of elements has a distribution density function: for the first: F 1 (t) =1-e- 0,1 t, for the second: F 2 (t) = 1-e- 0,2 t, for the third: F 3 (t) =1-e- 0,3 t. Find the probability that in the time interval from 0 to 5 hours: only one element will fail; only two elements will fail; all three elements will fail.

Solution: Let's use the definition of the probability generating function:

The probability that in independent trials, in the first of which the probability of an event occurring A equal to , in the second, etc., event A appears exactly once, equal to the coefficient in the expansion of the generating function in powers of . Let us find the probabilities of failure and non-failure, respectively, of the first, second and third element in the time interval from 0 to 5 hours:

Let's create a generating function:

The coefficient at is equal to the probability that the event A will appear exactly three times, that is, the probability of failure of all three elements; the coefficient at is equal to the probability that exactly two elements will fail; the coefficient at is equal to the probability that only one element will fail.

Example 2.7. Given the probability density f(x)random variable X:

Find the distribution function F(x).

Solution: We use the formula:

.

Thus, the distribution function looks like:

Example 2.8. The device consists of three independently operating elements. The probability of failure of each element in one experiment is 0.1. Draw up a distribution law for the number of failed elements in one experiment.

Solution: Random value X– the number of elements that failed in one experiment – ​​can take the following values: 0, 1, 2, 3. Probabilities that X takes these values, we find using Bernoulli’s formula:

Thus, we obtain the following law of probability distribution of a random variable X:

Example 2.9. In a batch of 6 parts there are 4 standard ones. 3 parts were selected at random. Draw up a distribution law for the number of standard parts among the selected ones.

Solution: Random value X– the number of standard parts among the selected ones – can take the following values: 1, 2, 3 and has a hypergeometric distribution. Probabilities that X

Where -- number of parts in the batch;

-- number of standard parts in a batch;

number of selected parts;

-- number of standard parts among those selected.

.

.

.

Example 2.10. The random variable has a distribution density

and are not known, but , a and . Find and.

Solution: In this case, the random variable X has a triangular distribution (Simpson distribution) on the interval [ a, b]. Numerical characteristics X:

Hence, . Solving this system, we obtain two pairs of values: . Since according to the conditions of the problem, we finally have: .

Answer: .

Example 2.11. On average, the insurance company pays out 10% of contracts insurance amounts in connection with the occurrence of an insured event. Calculate the mathematical expectation and dispersion of the number of such contracts among four randomly selected ones.

Solution: The mathematical expectation and variance can be found using the formulas:

.

Possible values ​​of SV (number of contracts (out of four) with the occurrence of an insured event): 0, 1, 2, 3, 4.

We use Bernoulli's formula to calculate the probabilities various numbers contracts (out of four) for which the insured amounts were paid:

.

The IC distribution series (the number of contracts with the occurrence of an insured event) has the form:

0,6561 0,2916 0,0486 0,0036 0,0001

Answer: , .

Example 2.12. Of the five roses, two are white. Draw up a law of distribution of a random variable expressing the number of white roses among two simultaneously taken.

Solution: In a selection of two roses, there may either be no white rose, or there may be one or two white roses. Therefore, the random variable X can take values: 0, 1, 2. Probabilities that X takes these values, we find it using the formula:

Where -- number of roses;

-- number of white roses;

number of roses taken at the same time;

-- the number of white roses among those taken.

.

.

.

Then the distribution law of the random variable will be as follows:

Example 2.13. Among the 15 assembled units, 6 require additional lubrication. Draw up a distribution law for the number of units that need additional lubrication among five randomly selected from the total number.

Solution: Random value X– the number of units that require additional lubrication among the five selected – can take the following values: 0, 1, 2, 3, 4, 5 and has a hypergeometric distribution. Probabilities that X takes these values, we find it using the formula:

Where -- number of assembled units;

-- the number of units that require additional lubrication;

number of selected units;

-- the number of units that require additional lubrication among those selected.

.

.

.

.

.

.

Then the distribution law of the random variable will be as follows:

Example 2.14. Of the 10 watches received for repair, 7 require general cleaning of the mechanism. The watches are not sorted by type of repair. The master, wanting to find watches that need cleaning, examines them one by one and, having found such watches, stops further viewing. Find the mathematical expectation and variance of the number of hours watched.

Solution: Random value X– the number of units that need additional lubrication among the five selected – can take the following values: 1, 2, 3, 4. Probabilities that X takes these values, we find it using the formula:

.

.

.

.

Then the distribution law of the random variable will be as follows:

Now let's calculate the numerical characteristics of the quantity:

Answer: , .

Example 2.15. The subscriber has forgotten the last digit of the phone number he needs, but remembers that it is odd. Find the mathematical expectation and variance of the number of times he dials a phone number before reaching the desired number, if he dials the last digit at random and does not subsequently dial the dialed digit.

Solution: The random variable can take the following values: . Since the subscriber does not dial the dialed digit in the future, the probabilities of these values ​​are equal.

Let's compile a distribution series of a random variable:

0,2

Let's calculate the mathematical expectation and variance of the number of dialing attempts:

Answer: , .

Example 2.16. The probability of failure during reliability tests for each device in the series is equal to p. Determine the mathematical expectation of the number of devices that failed if they were tested N devices.

Solution: Discrete random variable X is the number of failed devices in N independent tests, in each of which the probability of failure is equal p, distributed according to the binomial law. The mathematical expectation of a binomial distribution is equal to the number of trials multiplied by the probability of an event occurring in one trial:

Example 2.17. Discrete random variable X takes 3 possible values: with probability ; with probability and with probability. Find and , knowing that M( X) = 8.

Solution: Using definitions mathematical expectation and the law of distribution of a discrete random variable:

We find: .

Example 2.18. Department technical control checks products for standardness. The probability that the product is standard is 0.9. Each batch contains 5 products. Find the mathematical expectation of a random variable X– the number of batches, each of which contains exactly 4 standard products, if 50 batches are subject to inspection.

Solution: In this case, all experiments conducted are independent, and the probabilities that each batch contains exactly 4 standard products are the same, therefore, the mathematical expectation can be determined by the formula:

,

where is the number of parties;

The probability that a batch contains exactly 4 standard products.

We find the probability using Bernoulli's formula:

Answer: .

Example 2.19. Find the variance of a random variable X– number of occurrences of the event A in two independent trials, if the probabilities of the occurrence of an event in these trials are the same and it is known that M(X) = 0,9.

Solution: The problem can be solved in two ways.

1) Possible values ​​of SV X: 0, 1, 2. Using the Bernoulli formula, we determine the probabilities of these events:

, , .

Then the distribution law X has the form:

From the definition of mathematical expectation, we determine the probability:

Let's find the dispersion of SV X:

.

2) You can use the formula:

.

Answer: .

Example 2.20. Mathematical expectation and average standard deviation normally distributed random variable X respectively equal to 20 and 5. Find the probability that as a result of the test X will take the value contained in the interval (15; 25).

Solution: Probability of hitting a normal random variable X on the section from to is expressed through the Laplace function:

Example 2.21. Given function:

At what parameter value C this function is the distribution density of some continuous random variable X? Find the mathematical expectation and variance of a random variable X.

Solution: In order for a function to be the distribution density of some random variable, it must be non-negative, and it must satisfy the property:

.

Hence:

Let's calculate the mathematical expectation using the formula:

.

Let's calculate the variance using the formula:

T is equal p. It is necessary to find the mathematical expectation and variance of this random variable.

Solution: The distribution law of a discrete random variable X - the number of occurrences of an event in independent trials, in each of which the probability of the event occurring is equal to , is called binomial. The mathematical expectation of the binomial distribution is equal to the product of the number of trials and the probability of occurrence of event A in one trial:

.

Example 2.25. Three independent shots are fired at the target. The probability of hitting each shot is 0.25. Determine the standard deviation of the number of hits with three shots.

Solution: Since three independent trials are performed, and the probability of the occurrence of event A (a hit) in each trial is the same, we will assume that the discrete random variable X - the number of hits on the target - is distributed according to the binomial law.

The variance of the binomial distribution is equal to the product of the number of trials and the probability of the occurrence and non-occurrence of an event in one trial:

Example 2.26. Average number of clients visiting insurance company in 10 minutes, equals three. Find the probability that at least one client will arrive in the next 5 minutes.

Average number of clients arriving in 5 minutes: . .

Example 2.29. The waiting time for an application in the processor queue obeys an exponential distribution law with an average value of 20 seconds. Find the probability that the next (random) request will wait on the processor for more than 35 seconds.

Solution: In this example, the mathematical expectation , and the failure rate is equal to .

Then the desired probability:

Example 2.30. A group of 15 students holds a meeting in a hall with 20 rows of 10 seats each. Each student takes a place in the hall randomly. What is the probability that no more than three people will be in the seventh place of the row?

Solution:

Example 2.31.

Then, according to the classical definition of probability:

Where -- number of parts in the batch;

-- number of non-standard parts in the batch;

number of selected parts;

-- number of non-standard parts among those selected.

Then the distribution law of the random variable will be as follows.

Definition of a function of random variables. Function of discrete random argument and its numerical characteristics. Function of continuous random argument and its numerical characteristics. Functions of two random arguments. Determination of the probability distribution function and density for a function of two random arguments.

Law of probability distribution of a function of one random variable

When solving problems related to assessing the accuracy of operation of various automatic systems, the accuracy of production of individual elements of systems, etc., it is often necessary to consider functions of one or more random variables. Such functions are also random variables. Therefore, when solving problems, it is necessary to know the distribution laws of the random variables appearing in the problem. In this case, the distribution law of the system of random arguments and the functional dependence are usually known.

Thus, a problem arises that can be formulated as follows.

Given a system of random variables (X_1,X_2,\ldots,X_n), the distribution law of which is known. Some random variable Y is considered as a function of these random variables:

Y=\varphi(X_1,X_2,\ldots,X_n).

It is required to determine the law of distribution of the random variable Y, knowing the form of functions (6.1) and the law of joint distribution of its arguments.

Let us consider the problem of the distribution law of a function of one random argument

Y=\varphi(X).

\begin(array)(|c|c|c|c|c|)\hline(X)&x_1&x_2&\cdots&x_n\\\hline(P)&p_1&p_2&\cdots&p_n\\\hline\end(array)

Then Y=\varphi(X) is also a discrete random variable with possible values ​​. If all values y_1,y_2,\ldots,y_n are different, then for each k=1,2,\ldots,n the events \(X=x_k\) and \(Y=y_k=\varphi(x_k)\) are identical. Hence,

P\(Y=y_k\)=P\(X=x_k\)=p_k


and the required distribution series has the form

\begin(array)(|c|c|c|c|c|)\hline(Y)&y_1=\varphi(x_1)&y_2=\varphi(x_2)&\cdots&y_n=\varphi(x_n)\\\hline (P)&p_1&p_2&\cdots&p_n\\\hline\end(array)

If among the numbers y_1=\varphi(x_1),y_2=\varphi(x_2),\ldots,y_n=\varphi(x_n) there are identical ones, then each group identical values y_k=\varphi(x_k) you need to allocate one column in the table and add the corresponding probabilities.

For continuous random variables, the problem is posed as follows: knowing the distribution density f(x) of the random variable X, find the distribution density g(y) of the random variable Y=\varphi(X). When solving the problem, we consider two cases.

Let us first assume that the function y=\varphi(x) is monotonically increasing, continuous and differentiable on the interval (a;b) on which all possible values ​​of X lie. Then the inverse function x=\psi(y) exists, while also being monotonically increasing, continuous and differentiable. In this case we get

G(y)=f\bigl(\psi(y)\bigr)\cdot |\psi"(y)|.

Example 1. Random variable X distributed with density

F(x)=\frac(1)(\sqrt(2\pi))e^(-x^2/2)

Find the law of distribution of the random variable Y associated with the value X by the dependence Y=X^3.

Solution. Since the function y=x^3 is monotonic on the interval (-\infty;+\infty), we can apply formula (6.2). The inverse function with respect to the function \varphi(x)=x^3 is \psi(y)=\sqrt(y) , its derivative \psi"(y)=\frac(1)(3\sqrt(y^2)). Hence,

G(y)=\frac(1)(3\sqrt(2\pi))e^(-\sqrt(y^2)/2)\frac(1)(\sqrt(y^2))

Let us consider the case of a nonmonotonic function. Let the function y=\varphi(x) be such that the inverse function x=\psi(y) is ambiguous, i.e. one value of y corresponds to several values ​​of the argument x, which we denote x_1=\psi_1(y),x_2=\psi_2(y),\ldots,x_n=\psi_n(y), where n is the number of sections in which the function y=\varphi(x) changes monotonically. Then

G(y)=\sum\limits_(k=1)^(n)f\bigl(\psi_k(y)\bigr)\cdot |\psi"_k(y)|.

Example 2. Under the conditions of example 1, find the distribution of the random variable Y=X^2.

Solution. The inverse function x=\psi(y) is ambiguous. One value of the argument y corresponds to two values ​​of the function x


Applying formula (6.3), we obtain:

\begin(gathered)g(y)=f(\psi_1(y))|\psi"_1(y)|+f(\psi_2(y))|\psi"_2(y)|=\\\\ =\frac(1)(\sqrt(2\pi))\,e^(-\left(-\sqrt(y^2)\right)^2/2)\!\left|-\frac(1 )(2\sqrt(y))\right|+\frac(1)(\sqrt(2\pi))\,e^(-\left(\sqrt(y^2)\right)^2/2 )\!\left|\frac(1)(2\sqrt(y))\right|=\frac(1)(\sqrt(2\pi(y)))\,e^(-y/2) .\end(gathered)

Distribution law of a function of two random variables

Let the random variable Y be a function of two random variables forming the system (X_1;X_2), i.e. Y=\varphi(X_1;X_2). The task is to find the distribution of the random variable Y using the known distribution of the system (X_1;X_2).

Let f(x_1;x_2) be the distribution density of the system of random variables (X_1;X_2) . Let us introduce into consideration a new quantity Y_1 equal to X_1 and consider the system of equations

We will assume that this system is uniquely solvable with respect to x_1,x_2


and satisfies the differentiability conditions.

Distribution density of random variable Y

G_1(y)=\int\limits_(-\infty)^(+\infty)f(x_1;\psi(y;x_1))\!\left|\frac(\partial\psi(y;x_1)) (\partial(y))\right|dx_1.

Note that the reasoning does not change if the introduced new value Y_1 is set equal to X_2.

Mathematical expectation of a function of random variables

In practice, there are often cases when there is no particular need to completely determine the distribution law of a function of random variables, but it is enough only to indicate its numerical characteristics. Thus, the problem arises of determining the numerical characteristics of functions of random variables in addition to the laws of distribution of these functions.

Let the random variable Y be a function of the random argument X with a given distribution law

Y=\varphi(X).

It is required, without finding the law of distribution of the quantity Y, to determine its mathematical expectation

M(Y)=M[\varphi(X)].

Let X be a discrete random variable having a distribution series

\begin(array)(|c|c|c|c|c|)\hline(x_i)&x_1&x_2&\cdots&x_n\\\hline(p_i)&p_1&p_2&\cdots&p_n\\\hline\end(array)

Let's make a table of the values ​​of the value Y and the probabilities of these values:

\begin(array)(|c|c|c|c|c|)\hline(y_i=\varphi(x_i))&y_1=\varphi(x_1)&y_2=\varphi(x_2)&\cdots&y_n=\varphi( x_n)\\\hline(p_i)&p_1&p_2&\cdots&p_n\\\hline\end(array)

This table is not a series of the distribution of the random variable Y, since in the general case some of the values ​​may coincide with each other and the values ​​in the top row are not necessarily in ascending order. However, the mathematical expectation of the random variable Y can be determined by the formula

M[\varphi(X)]=\sum\limits_(i=1)^(n)\varphi(x_i)p_i,


since the value determined by formula (6.4) cannot change due to the fact that under the sum sign some terms will be combined in advance, and the order of the terms will be changed.

Formula (6.4) does not explicitly contain the distribution law of the function \varphi(X) itself, but contains only the distribution law of the argument X. Thus, to determine the mathematical expectation of the function Y=\varphi(X), it is not at all necessary to know the distribution law of the function \varphi(X), but rather to know the distribution law of the argument X.

For a continuous random variable, the mathematical expectation is calculated using the formula

M[\varphi(X)]=\int\limits_(-\infty)^(+\infty)\varphi(x)f(x)\,dx,


where f(x) is the probability distribution density of the random variable X.

Let us consider cases when, in order to find the mathematical expectation of a function of random arguments, knowledge of even the laws of distribution of arguments is not required, but it is enough to know only some of their numerical characteristics. Let us formulate these cases in the form of theorems.

Theorem 6.1. The mathematical expectation of the sum of both dependent and independent two random variables is equal to the sum of the mathematical expectations of these variables:

M(X+Y)=M(X)+M(Y).

Theorem 6.2. The mathematical expectation of the product of two random variables is equal to the product of their mathematical expectations plus the correlation moment:

M(XY)=M(X)M(Y)+\mu_(xy).

Corollary 6.1. The mathematical expectation of the product of two uncorrelated random variables is equal to the product of their mathematical expectations.

Corollary 6.2. The mathematical expectation of the product of two independent random variables is equal to the product of their mathematical expectations.

Variance of a function of random variables

By definition of dispersion we have D[Y]=M[(Y-M(Y))^2].. Hence,

D[\varphi(x)]=M[(\varphi(x)-M(\varphi(x)))^2], Where .

We present the calculation formulas only for the case of continuous random arguments. For a function of one random argument Y=\varphi(X), the variance is expressed by the formula

D[\varphi(x)]=\int\limits_(-\infty)^(+\infty)(\varphi(x)-M(\varphi(x)))^2f(x)\,dx,

Where M(\varphi(x))=M[\varphi(X)]- mathematical expectation of the function \varphi(X) ; f(x) - distribution density of the value X.

Formula (6.5) can be replaced with the following:

D[\varphi(x)]=\int\limits_(-\infty)^(+\infty)\varphi^2(x)f(x)\,dx-M^2(X)

Let's consider dispersion theorems, which play an important role in probability theory and its applications.

Theorem 6.3. The variance of the sum of random variables is equal to the sum of the variances of these quantities plus the doubled sum of the correlation moments of each of the summands with all subsequent ones:

D\!\left[\sum\limits_(i=1)^(n)X_i\right]=\sum\limits_(i=1)^(n)D+2\sum\limits_(i

Corollary 6.3. The variance of the sum of uncorrelated random variables is equal to the sum of the variances of the terms:

D\!\left[\sum\limits_(i=1)^(n)X_i\right]=\sum\limits_(i=1)^(n)D\mu_(y_1y_2)= M(Y_1Y_2)-M(Y_1)M(Y_2).

\mu_(y_1y_2)=M(\varphi_1(X)\varphi_2(X))-M(\varphi_1(X))M(\varphi_2(X)).


that is, the correlation moment of two functions of random variables is equal to the mathematical expectation of the product of these functions minus the product of the mathematical expectations.

Let's look at the main properties of the correlation moment and correlation coefficient.

Property 1. Adding constants to random variables does not change the correlation moment and the correlation coefficient.

Property 2. For any random variables X and Y, the absolute value of the correlation moment does not exceed the geometric mean of the variances of these values:

|\mu_(xy)|\leqslant\sqrt(D[X]\cdot D[Y])=\sigma_x\cdot \sigma_y,

3. The distribution function is non-decreasing: if, then

4. Distribution function left continuous: for anyone .

Note. The last property indicates what values ​​the distribution function takes at the break points. Sometimes the definition of the distribution function is formulated using a loose inequality: . In this case, continuity on the left is replaced by continuity on the right: when . This does not change any meaningful properties of the distribution function, so this question is only terminological.

Properties 1-4 are characteristic, i.e. any function satisfying these properties is a distribution function of some random variable.

The distribution function uniquely defines the probability distribution of a random variable. In fact, it is a universal and most visual way of describing this distribution.

The more the distribution function grows on a given interval of the number line, the higher the probability of a random variable falling into this interval. If the probability of falling into an interval is zero, then the distribution function on it is constant.

In particular, the probability that a random variable will take a given value is equal to the jump in the distribution function at a given point:

.

If the distribution function is continuous at point , then the probability of taking this value for a random variable is zero. In particular, if the distribution function is continuous on the entire numerical axis (in this case, the corresponding distribution is called continuous), then the probability of accepting any given value is zero.

From the definition of the distribution function it follows that the probability of a random variable falling into an interval closed on the left and open on the right is equal to:

Using this formula and the above method of finding the probability of hitting any given point, the probabilities of a random variable getting into intervals of other types are easily determined: , and . Further, by the measure extension theorem, we can uniquely extend the measure to all Borel sets of the number line. In order to apply this theorem, it is necessary to show that the measure thus defined on intervals is sigma-additive on them; when proving this, properties 1-4 are used exactly (in particular, the property of left continuity 4, so it cannot be discarded).

Generating a random variable with a given distribution

Let's consider a random variable that has a distribution function. Let's pretend that continuous. Consider the random variable

.

It is easy to show that then will have a uniform distribution on the segment .

1.2.4. Random variables and their distributions

Distributions of random variables and distribution functions. The distribution of a numerical random variable is a function that uniquely determines the probability that the random variable takes a given value or belongs to some given interval.

The first is if the random variable takes a finite number of values. Then the distribution is given by the function P(X = x), assigning to each possible value X random variable X the probability that X = x.

The second is if the random variable takes on infinitely many values. This is possible only when the probabilistic space on which the random variable is defined consists of an infinite number of elementary events. Then the distribution is given by a set of probabilities P(a < X for all pairs of numbers a, b such that a . The distribution can be specified using the so-called. distribution function F(x) = P(X defining for all real X the probability that the random variable X takes values ​​less than X. It's clear that

P(a < X

This relationship shows that both the distribution can be calculated from the distribution function, and, conversely, the distribution function can be calculated from the distribution.

Used in probabilistic-statistical methods of decision making and others applied research Distribution functions are either discrete or continuous, or combinations thereof.

Discrete distribution functions correspond to discrete random variables that take a finite number of values ​​or values ​​from a set whose elements can be numbered by natural numbers (such sets are called countable in mathematics). Their graph looks like a stepped ladder (Fig. 1).

Example 1. Number X defective items in a batch takes on a value of 0 with a probability of 0.3, a value of 1 with a probability of 0.4, a value of 2 with a probability of 0.2, and a value of 3 with a probability of 0.1. Graph of the distribution function of a random variable X shown in Fig. 1.

Fig.1. Graph of the distribution function of the number of defective products.

Continuous distribution functions do not have jumps. They increase monotonically as the argument increases - from 0 at to 1 at . Random variables that have continuous distribution functions are called continuous.

Continuous distribution functions used in probabilistic statistical methods decision making, have derivatives. First derivative f(x) distribution functions F(x) is called the probability density,

Using the probability density, you can determine the distribution function:

For any distribution function

The listed properties of distribution functions are constantly used in probabilistic and statistical methods of decision making. In particular, the last equality implies a specific form of constants in the formulas for probability densities considered below.

Example 2. The following distribution function is often used:

(1)

Where a And b– some numbers, a . Let's find the probability density of this distribution function:

(at points x = a And x = b derivative of a function F(x) does not exist).

A random variable with distribution function (1) is called “uniformly distributed on the interval [ a; b]».

Mixed distribution functions occur, in particular, when observations stop at some point. For example, when analyzing statistical data obtained from the use of reliability test plans that provide for termination of testing after a certain period. Or when analyzing data on technical products that required warranty repairs.

Example 3. Let, for example, the service life of an electric light bulb be a random variable with a distribution function F(t), and the test is carried out until the light bulb fails, if this occurs in less than 100 hours from the start of the test, or until t 0= 100 hours. Let G(t)– distribution function of the operating time of the light bulb in good condition during this test. Then

Function G(t) has a jump at a point t 0, since the corresponding random variable takes the value t 0 with probability 1- F(t 0)> 0.

Characteristics of random variables. In probabilistic-statistical methods of decision-making, a number of characteristics of random variables are used, expressed through distribution functions and probability densities.

When describing income differentiation, when finding confidence limits for the parameters of distributions of random variables, and in many other cases, such a concept as “order quantile” is used. R", where 0< p < 1 (обозначается x p). Order quantile R– the value of a random variable for which the distribution function takes the value R or there is a “jump” from a value less R to a value greater R(Fig. 2). It may happen that this condition is satisfied for all values ​​of x belonging to this interval (i.e. the distribution function is constant on this interval and is equal to R). Then each such value is called a “quantile of order” R" For continuous distribution functions, as a rule, there is a single quantile x p order R(Fig. 2), and

F(x p) = p. (2)

Fig.2. Definition of quantile x p order R.

Example 4. Let's find the quantile x p order R for the distribution function F(x) from (1).

At 0< p < 1 квантиль x p is found from the equation

those. x p = a + p(b – a) = a( 1- p) +bp. At p= 0 any x < a is a quantile of order p= 0. Order quantile p= 1 is any number x > b.

For discrete distributions, as a rule, there is no x p, satisfying equation (2). More precisely, if the distribution of a random variable is given in Table 1, where x 1< x 2 < … < x k , then equality (2), considered as an equation with respect to x p, has solutions only for k values p, namely,

p = p 1 ,

p = p 1 + p 2 ,

p = p 1 + p 2 + p 3 ,

p = p 1 + p 2 + …+ p m, 3 < m < k,

p = p 1 + p 2 + … + p k.

Table 1.

Distribution of a discrete random variable

For those listed k probability values p solution x p equation (2) is not unique, namely,

F(x) = p 1 + p 2 + … + p m

for all X such that x m< x < x m+1 . Those. x p – any number from the interval (x m; x m+1 ]. For everyone else R from the interval (0;1), not included in the list (3), there is a “jump” from a value less R to a value greater R. Namely, if

p 1 + p 2 + … + p m

That x p = x m+1.

The considered property of discrete distributions creates significant difficulties when tabulating and using such distributions, since it is impossible to accurately maintain typical numerical values ​​of the distribution characteristics. In particular, this is true for the critical values ​​and significance levels of nonparametric statistical tests (see below), since the distributions of the statistics of these tests are discrete.

Quantile order is of great importance in statistics R= ½. It is called the median (random variable X or its distribution function F(x)) and is designated Me(X). In geometry there is the concept of “median” - a straight line passing through the vertex of a triangle and dividing its opposite side in half. In mathematical statistics, the median divides in half not the side of the triangle, but the distribution of a random variable: equality F(x 0.5)= 0.5 means that the probability of getting to the left x 0.5 and the probability of getting to the right x 0.5(or directly to x 0.5) are equal to each other and equal to ½, i.e.

P(X < x 0,5) = P(X > x 0.5) = ½.

The median indicates the "center" of the distribution. From the point of view of one of the modern concepts - the theory of stable statistical procedures - the median is a better characteristic of a random variable than the mathematical expectation. When processing measurement results on an ordinal scale (see the chapter on measurement theory), the median can be used, but the mathematical expectation cannot.

A characteristic of a random variable such as mode has a clear meaning - the value (or values) of a random variable corresponding to the local maximum of the probability density for a continuous random variable or the local maximum of the probability for a discrete random variable.

If x 0– mode of a random variable with density f(x), then, as is known from differential calculus, .

A random variable can have many modes. So, for uniform distribution (1) each point X such that a< x < b , is fashion. However, this is an exception. Most random variables used in probabilistic statistical methods of decision making and other applied research have one mode. Random variables, densities, distributions that have one mode are called unimodal.

The mathematical expectation for discrete random variables with a finite number of values ​​is discussed in the chapter “Events and Probabilities”. For a continuous random variable X expected value M(X) satisfies the equality

which is an analogue of formula (5) from statement 2 of chapter “Events and Probabilities”.

Example 5. Expectation for a uniformly distributed random variable X equals

For the random variables considered in this chapter, all those properties of mathematical expectations and variances that were considered earlier for discrete random variables with a finite number of values ​​are true. However, we do not provide proof of these properties, since they require deepening into mathematical subtleties, which is not necessary for understanding and qualified application of probabilistic-statistical methods of decision-making.

Comment. This textbook consciously avoids mathematical subtleties associated, in particular, with the concepts of measurable sets and measurable functions, algebra of events, etc. Those wishing to master these concepts should turn to specialized literature, in particular, the encyclopedia.

Each of the three characteristics – mathematical expectation, median, mode – describes the “center” of the probability distribution. The concept of "center" can be defined in different ways - hence three different characteristics. However, for an important class of distributions—symmetric unimodal—all three characteristics coincide.

Distribution density f(x)– density of symmetric distribution, if there is a number x 0 such that

. (3)

Equality (3) means that the graph of the function y = f(x) symmetrical about a vertical line passing through the center of symmetry X = X 0 . From (3) it follows that the symmetric distribution function satisfies the relation

(4)

For a symmetric distribution with one mode, the mathematical expectation, median and mode coincide and are equal x 0.

The most important case is symmetry about 0, i.e. x 0= 0. Then (3) and (4) become equalities

(6)

respectively. The above relations show that there is no need to tabulate symmetric distributions for all X, it is enough to have tables at x > x 0.

Let us note one more property of symmetric distributions, which is constantly used in probabilistic-statistical methods of decision-making and other applied research. For a continuous distribution function

P(|X| < a) = P(-a < X < a) = F(a) – F(-a),

Where F– distribution function of a random variable X. If the distribution function F is symmetrical about 0, i.e. formula (6) is valid for it, then

P(|X| < a) = 2F(a) – 1.

Another formulation of the statement in question is often used: if

.

If and are quantiles of order and, respectively (see (2)) of a distribution function symmetric about 0, then from (6) it follows that

From the characteristics of the position - mathematical expectation, median, mode - let's move on to the characteristics of the spread of the random variable X: variance, standard deviation and coefficient of variation v. The definition and properties of dispersion for discrete random variables were discussed in the previous chapter. For continuous random variables

The standard deviation is the non-negative value of the square root of the variance:

The coefficient of variation is the ratio of the standard deviation to the mathematical expectation:

The coefficient of variation is applied when M(X)> 0. It measures the spread in relative units, while the standard deviation is in absolute units.

Example 6. For a uniformly distributed random variable X Let's find the dispersion, standard deviation and coefficient of variation. The variance is:

Changing the variable makes it possible to write:

Where c = (ba)/ 2. Therefore, the standard deviation is equal to and the coefficient of variation is:

For each random variable X determine three more quantities - centered Y, normalized V and given U. Centered random variable Y is the difference between a given random variable X and its mathematical expectation M(X), those. Y = X – M(X). Expectation of a centered random variable Y equals 0, and the variance is the variance of a given random variable: M(Y) = 0, D(Y) = D(X). Distribution function F Y(x) centered random variable Y related to the distribution function F(x) original random variable X ratio:

F Y(x) = F(x + M(X)).

The densities of these random variables satisfy the equality

f Y(x) = f(x + M(X)).

Normalized random variable V is the ratio of a given random variable X to its standard deviation, i.e. . Expectation and variance of a normalized random variable V expressed through characteristics X So:

,

Where v– coefficient of variation of the original random variable X. For the distribution function F V(x) and density f V(x) normalized random variable V we have:

Where F(x) – distribution function of the original random variable X, A f(x) – its probability density.

Reduced random variable U is a centered and normalized random variable:

.

For the given random variable

Normalized, centered and reduced random variables are constantly used both in theoretical studies and in algorithms, software products, regulatory, technical and instructional documentation. In particular, because the equalities make it possible to simplify the justification of methods, the formulation of theorems and calculation formulas.

Transformations of random variables and more general ones are used. So, if Y = aX + b, Where a And b– some numbers, then

Example 7. If then Y is the reduced random variable, and formulas (8) transform into formulas (7).

With each random variable X you can associate many random variables Y, given by the formula Y = aX + b at different a> 0 and b. This set is called scale-shift family, generated by the random variable X. Distribution functions F Y(x) constitute a scale-shift family of distributions generated by the distribution function F(x). Instead of Y = aX + b often use recording

Number With is called the shift parameter, and the number d- scale parameter. Formula (9) shows that X– the result of measuring a certain quantity – goes into U– the result of measuring the same quantity if the beginning of the measurement is moved to the point With, and then use the new unit of measurement, in d times larger than the old one.

For the scale-shift family (9), the distribution of X is called standard. In probabilistic statistical methods of decision making and other applied research, the standard normal distribution, standard Weibull-Gnedenko distribution, standard gamma distribution, etc. are used (see below).

Other transformations of random variables are also used. For example, for a positive random variable X are considering Y= log X, where lg X– decimal logarithm of a number X. Chain of equalities

F Y (x) = P( lg X< x) = P(X < 10x) = F( 10x)

connects distribution functions X And Y.

When processing data, the following characteristics of a random variable are used X as moments of order q, i.e. mathematical expectations of a random variable Xq, q= 1, 2, ... Thus, the mathematical expectation itself is a moment of order 1. For a discrete random variable, the moment of order q can be calculated as

For a continuous random variable

Moments of order q also called initial moments of order q, in contrast to related characteristics - central moments of order q, given by the formula

So, dispersion is a central moment of order 2.

Normal distribution and the central limit theorem. In probabilistic-statistical methods of decision-making we often talk about normal distribution. Sometimes they try to use it to model the distribution of initial data (these attempts are not always justified - see below). More importantly, many data processing methods are based on the fact that the calculated values ​​have distributions close to normal.

Let X 1 , X 2 ,…, Xn M(X i) = m and variances D(X i) = , i = 1, 2,…, n,... As follows from the results of the previous chapter,

Consider the reduced random variable U n for the amount , namely,

As follows from formulas (7), M(U n) = 0, D(U n) = 1.

(for identically distributed terms). Let X 1 , X 2 ,…, Xn, … – independent identically distributed random variables with mathematical expectations M(X i) = m and variances D(X i) = , i = 1, 2,…, n,... Then for any x there is a limit

Where F(x)– standard function normal distribution.

More about the function F(x) – below (read “fi from x”, because F– Greek uppercase letter"fi")

The Central Limit Theorem (CLT) gets its name because it is the central, most commonly used mathematical result probability theory and mathematical statistics. The history of the CLT takes about 200 years - from 1730, when the English mathematician A. Moivre (1667-1754) published the first result related to the CLT (see below about the Moivre-Laplace theorem), until the twenties and thirties of the twentieth century, when Finn J.W. Lindeberg, Frenchman Paul Levy (1886-1971), Yugoslav V. Feller (1906-1970), Russian A.Ya. Khinchin (1894-1959) and other scientists obtained necessary and sufficient conditions for the validity of the classical central limit theorem.

The development of the topic under consideration did not stop there - they studied random variables that do not have dispersion, i.e. those for whom

(academician B.V. Gnedenko and others), a situation when random variables (more precisely, random elements) of a more complex nature than numbers are summed up (academicians Yu.V. Prokhorov, A.A. Borovkov and their associates), etc. .d.

Distribution function F(x) is given by the equality

,

where is the density of the standard normal distribution, which has a rather complex expression:

.

Here =3.1415925… is a number known in geometry, equal to the ratio of the circumference to the diameter, e = 2.718281828... - the base of natural logarithms (to remember this number, please note that 1828 is the year of birth of the writer L.N. Tolstoy). As is known from mathematical analysis,

When processing observation results, the normal distribution function is not calculated using the given formulas, but is found using special tables or computer programs. The best “Tables of mathematical statistics” in Russian were compiled by corresponding members of the USSR Academy of Sciences L.N. Bolshev and N.V. Smirnov.

The form of the density of the standard normal distribution follows from mathematical theory, which we have no opportunity to consider here, as well as the proof of the CLT.

For illustration, we provide small tables of the distribution function F(x)(Table 2) and its quantiles (Table 3). Function F(x) symmetrical about 0, which is reflected in Table 2-3.

Table 2.

Standard normal distribution function.

If the random variable X has a distribution function F(x), That M(X) = 0, D(X) = 1. This statement is proven in probability theory based on the form of the probability density. It is consistent with a similar statement for the characteristics of the reduced random variable U n, which is quite natural, since the CLT states that with an unlimited increase in the number of terms, the distribution function U n tends to the standard normal distribution function F(x), and for any X.

Table 3.

Quantiles of the standard normal distribution.

Order quantile R

Order quantile R

Let us introduce the concept of a family of normal distributions. By definition, a normal distribution is the distribution of a random variable X, for which the distribution of the reduced random variable is F(x). As follows from the general properties of scale-shift families of distributions (see above), a normal distribution is a distribution of a random variable

Where X– random variable with distribution F(X), and m = M(Y), = D(Y). Normal distribution with shift parameters m and scale is usually indicated N(m, ) (sometimes the notation is used N(m, ) ).

As follows from (8), the probability density of the normal distribution N(m, ) There is

Normal distributions form a scale-shift family. In this case, the scale parameter is d= 1/ , and the shift parameter c = - m/ .

For the central moments of the third and fourth order of the normal distribution, the following equalities are valid:

These equalities underlie classical methods checking that observation results follow a normal distribution. Nowadays it is usually recommended to test normality using the criterion W Shapiro - Wilka. The problem of normality testing is discussed below.

If random variables X 1 And X 2 have distribution functions N(m 1 , 1) And N(m 2 , 2) accordingly, then X 1+ X 2 has a distribution Therefore, if random variables X 1 , X 2 ,…, Xn N(m, ) , then their arithmetic mean

has a distribution N(m, ) . These properties of the normal distribution are constantly used in various probabilistic and statistical methods of decision-making, in particular, in the statistical regulation of technological processes and in statistical acceptance control based on quantitative criteria.

Using the normal distribution, three distributions are defined that are now often used in statistical data processing.

Distribution (chi - square) – distribution of a random variable

where are the random variables X 1 , X 2 ,…, Xn independent and have the same distribution N(0,1). In this case, the number of terms, i.e. n, is called the “number of degrees of freedom” of the chi-square distribution.

Distribution t Student's t is the distribution of a random variable

where are the random variables U And X independent, U has a standard normal distribution N(0.1), and X– chi distribution – square c n degrees of freedom. Wherein n is called the “number of degrees of freedom” of the Student distribution. This distribution was introduced in 1908 by the English statistician W. Gosset, who worked at a beer factory. Probabilistic statistical methods have been used to make economic and technical solutions at this factory, therefore its management forbade V. Gosset to publish scientific articles under his own name. In this way, trade secrets and “know-how” in the form of probabilistic and statistical methods developed by V. Gosset were protected. However, he had the opportunity to publish under the pseudonym "Student". The history of Gosset-Student shows that for another hundred years, managers in Great Britain were aware of the greater economic efficiency of probabilistic-statistical methods of decision-making.

The Fisher distribution is the distribution of a random variable

where are the random variables X 1 And X 2 are independent and have chi-square distributions with the number of degrees of freedom k 1 And k 2 respectively. At the same time, the couple (k 1 , k 2 ) – a pair of “degrees of freedom” of the Fisher distribution, namely, k 1 is the number of degrees of freedom of the numerator, and k 2 – number of degrees of freedom of the denominator. The distribution of the random variable F is named after the great English statistician R. Fisher (1890-1962), who actively used it in his works.

Expressions for the chi-square, Student and Fisher distribution functions, their densities and characteristics, as well as tables can be found in the specialized literature (see, for example,).

As already noted, normal distributions are now often used in probabilistic models in various applied areas. What is the reason for this two-parameter family of distributions being so widespread? It is clarified by the following theorem.

Central limit theorem(for differently distributed terms). Let X 1 , X 2 ,…, Xn,… - independent random variables with mathematical expectations M(X 1 ), M(X 2 ),…, M(X n), ... and variances D(X 1 ), D(X 2 ),…, D(X n), ... respectively. Let

Then, if certain conditions are true that ensure the small contribution of any of the terms in U n,

for anyone X.

We will not formulate the conditions in question here. They can be found in specialized literature (see, for example,). “The clarification of the conditions under which the CPT operates is the merit of the outstanding Russian scientists A.A. Markov (1857-1922) and, in particular, A.M. Lyapunov (1857-1918).”

The central limit theorem shows that in the case when the result of a measurement (observation) is formed under the influence of many causes, each of them making only a small contribution, and the total result is determined additively, i.e. by addition, then the distribution of the measurement (observation) result is close to normal.

It is sometimes believed that for the distribution to be normal, it is sufficient that the result of the measurement (observation) X is formed under the influence of many reasons, each of which has a small impact. This is wrong. What matters is how these causes operate. If additive, then X has an approximately normal distribution. If multiplicatively(i.e. the actions of individual causes are multiplied and not added), then the distribution X close not to normal, but to the so-called. logarithmically normal, i.e. Not X, and log X has an approximately normal distribution. If there is no reason to believe that one of these two mechanisms for the formation of the final result is at work (or some other well-defined mechanism), then about distribution X nothing definite can be said.

It follows from the above that in a specific applied problem, the normality of measurement results (observations), as a rule, cannot be established from general considerations; it should be checked using statistical criteria. Or use nonparametric statistical methods that are not based on assumptions about the membership of the distribution functions of measurement results (observations) to one or another parametric family.

Continuous distributions used in probabilistic and statistical methods of decision making. In addition to the scale-shift family of normal distributions, a number of other families of distributions are widely used - lognormal, exponential, Weibull-Gnedenko, gamma distributions. Let's look at these families.

Random value X has a lognormal distribution if the random variable Y= log X has a normal distribution. Then Z= log X = 2,3026…Y also has a normal distribution N(a 1 ,σ 1), where ln X- natural logarithm X. The density of the lognormal distribution is:

From the central limit theorem it follows that the product X = X 1 X 2 Xn independent positive random variables X i, i = 1, 2,…, n, at large n can be approximated by a lognormal distribution. In particular, the multiplicative model of formation wages or income leads to the recommendation to approximate the distributions of wages and income by lognormal laws. For Russia, this recommendation turned out to be justified - statistical data confirms it.

There are other probabilistic models that lead to the lognormal law. Classic example such a model was given by A.N. Kolmogorov, who, from a physically based system of postulates, concluded that the particle sizes when crushing pieces of ore, coal, etc. in ball mills have a lognormal distribution.

Let's move on to another family of distributions, widely used in various probabilistic-statistical methods of decision-making and other applied research - the family of exponential distributions. Let's start with a probabilistic model that leads to such distributions. To do this, consider the “flow of events”, i.e. a sequence of events occurring one after another at certain points in time. Examples include: call flow at a telephone exchange; flow of equipment failures in the technological chain; flow of product failures during product testing; flow of customer requests to the bank branch; flow of buyers applying for goods and services, etc. In the theory of streams of events, a theorem similar to the central limit theorem is valid, but in it we're talking about not about the summation of random variables, but about the summation of streams of events. We consider a total flow composed of a large number of independent flows, none of which has a predominant influence on the total flow. For example, a call flow entering a telephone exchange is composed of a large number of independent call flows originating from individual subscribers. It has been proven that in the case when the characteristics of flows do not depend on time, the total flow is completely described by one number - the intensity of the flow. For the total flow, consider the random variable X- the length of the time interval between successive events. Its distribution function has the form

(10)

This distribution is called exponential distribution because participates in formula (10) exponential function ex. The value 1/λ is a scale parameter. Sometimes a shift parameter is also introduced With, the distribution of a random variable is called exponential X + s, where the distribution X is given by formula (10).

Exponential distributions - special case so-called Weibull - Gnedenko distributions. They are named after the names of the engineer V. Weibull, who introduced these distributions into the practice of analyzing the results of fatigue tests, and the mathematician B.V. Gnedenko (1912-1995), who received such distributions as limits when studying the maximum of the test results. Let X- a random variable characterizing the duration of operation of the product, complex system, element (i.e. resource, operating time to the limiting state, etc.), duration of operation of an enterprise or life of a living being, etc. Failure intensity plays an important role

(11)

Where F(x) And f(x) - distribution function and density of a random variable X.

Let us describe the typical behavior of the failure rate. The entire time interval can be divided into three periods. On the first of them the function λ(x) has high values ​​and a clear tendency to decrease (most often it decreases monotonically). This can be explained by the presence in the batch of product units in question with obvious and hidden defects, which lead to a relatively rapid failure of these product units. The first period is called the “break-in period” (or “break-in”). This is what the warranty period usually covers.

Then comes a period of normal operation, characterized by an approximately constant and relatively low failure rate. The nature of failures during this period is sudden (accidents, errors of operating personnel, etc.) and does not depend on the duration of operation of the product unit.

Finally, last period operation - a period of aging and wear. The nature of failures during this period is irreversible physical-mechanical and chemical changes materials leading to a progressive deterioration in the quality of a product unit and its final failure.

Each period has its own type of function λ(x). Let us consider the class of power dependences

λ(x) = λ 0bx b -1 , (12)

Where λ 0 > 0 and b> 0 - some numeric parameters. Values b < 1, b= 0 and b> 1 correspond to the type of failure rate during the periods of running-in, normal operation and aging, respectively.

Relationship (11) at a given failure rate λ(x)- differential equation for a function F(x). From theory differential equations follows that

(13)

Substituting (12) into (13), we obtain that

(14)

The distribution given by formula (14) is called the Weibull - Gnedenko distribution. Because the

then from formula (14) it follows that the quantity A, given by formula (15), is a scale parameter. Sometimes a shift parameter is also introduced, i.e. Weibull-Gnedenko distribution functions are called F(x - c), Where F(x) is given by formula (14) for some λ 0 and b.

The Weibull-Gnedenko distribution density has the form

(16)

Where a> 0 - scale parameter, b> 0 - form parameter, With- shift parameter. In this case, the parameter A from formula (16) is associated with the parameter λ 0 from formula (14) by the relationship specified in formula (15).

The exponential distribution is a very special case of the Weibull-Gnedenko distribution, corresponding to the value of the shape parameter b = 1.

The Weibull-Gnedenko distribution is also used in constructing probabilistic models of situations in which the behavior of an object is determined by the “weakest link”. There is an analogy with a chain, the safety of which is determined by the link that has the least strength. In other words, let X 1 , X 2 ,…, Xn- independent identically distributed random variables,

X(1)=min( X 1, X 2,…, X n), X(n)=max( X 1, X 2,…, X n).

In a number of applied problems, they play an important role X(1) And X(n) , in particular, when studying the maximum possible values ​​("records") of certain values, for example, insurance payments or losses due to commercial risks, when studying the elasticity and endurance limits of steel, a number of reliability characteristics, etc. It is shown that for large n the distributions X(1) And X(n) , as a rule, are well described by Weibull-Gnedenko distributions. Fundamental contribution to the study of distributions X(1) And X(n) contributed by the Soviet mathematician B.V. Gnedenko. The works of V. Weibull, E. Gumbel, V.B. are devoted to the use of the results obtained in economics, management, technology and other fields. Nevzorova, E.M. Kudlaev and many other specialists.

Let's move on to the family of gamma distributions. They are widely used in economics and management, theory and practice of reliability and testing, in various fields of technology, meteorology, etc. In particular, in many situations, the gamma distribution is subject to such quantities as the total service life of the product, the length of the chain of conductive dust particles, the time the product reaches the limiting state during corrosion, the operating time to k-th refusal, k= 1, 2, …, etc. The life expectancy of patients with chronic diseases and the time to achieve a certain effect during treatment in some cases have a gamma distribution. This distribution is most adequate for describing demand in economic and mathematical models of inventory management (logistics).

The gamma distribution density has the form

(17)

The probability density in formula (17) is determined by three parameters a, b, c, Where a>0, b>0. Wherein a is a form parameter, b- scale parameter and With- shift parameter. Factor 1/Γ(а) is normalizing, it was introduced to

Here Γ(a)- one of the special functions used in mathematics, the so-called “gamma function”, after which the distribution given by formula (17) is named,

At fixed A formula (17) specifies a scale-shift family of distributions generated by a distribution with density

(18)

A distribution of the form (18) is called the standard gamma distribution. It is obtained from formula (17) at b= 1 and With= 0.

A special case of gamma distributions for A= 1 are exponential distributions (with λ = 1/b). With natural A And With=0 gamma distributions are called Erlang distributions. From the works of the Danish scientist K.A. Erlang (1878-1929), an employee of the Copenhagen Telephone Company, who studied in 1908-1922. the functioning of telephone networks, the development of queuing theory began. This theory deals with probabilistic and statistical modeling of systems in which a flow of requests is serviced in order to make optimal decisions. Erlang distributions are used in the same application areas in which exponential distributions are used. This is based on the following mathematical fact: the sum of k independent random variables exponentially distributed with the same parameters λ and With, has a gamma distribution with a shape parameter a =k, scale parameter b= 1/λ and shift parameter kc. At With= 0 we obtain the Erlang distribution.

If the random variable X has a gamma distribution with a shape parameter A such that d = 2 a- integer, b= 1 and With= 0, then 2 X has a chi-square distribution with d degrees of freedom.

Random value X with gvmma distribution has the following characteristics:

Expected value M(X) =ab + c,

Variance D(X) = σ 2 = ab 2 ,

The coefficient of variation

Asymmetry

Excess

The normal distribution is an extreme case of the gamma distribution. More precisely, let Z be a random variable having a standard gamma distribution given by formula (18). Then

for any real number X, Where F(x)- standard normal distribution function N(0,1).

In applied research, other parametric families of distributions are also used, of which the most famous are the system of Pearson curves, the Edgeworth and Charlier series. They are not considered here.

Discrete distributions used in probabilistic and statistical methods of decision making. The most commonly used are three families of discrete distributions - binomial, hypergeometric and Poisson, as well as some other families - geometric, negative binomial, multinomial, negative hypergeometric, etc.

As already mentioned, the binomial distribution occurs in independent trials, in each of which with probability R event appears A. If the total number of trials n given, then the number of tests Y, in which the event appeared A, has a binomial distribution. For a binomial distribution, the probability of being accepted as a random variable is Y values y is determined by the formula

Number of combinations of n elements by y, known from combinatorics. For all y, except 0, 1, 2, …, n, we have P(Y= y)= 0. Binomial distribution with fixed sample size n is specified by the parameter p, i.e. binomial distributions form a one-parameter family. They are used in the analysis of data from sample studies, in particular, in the study of consumer preferences, selective control of product quality according to single-stage control plans, when testing populations of individuals in demography, sociology, medicine, biology, etc.

If Y 1 And Y 2 - independent binomial random variables with the same parameter p 0 , determined from samples with volumes n 1 And n 2 accordingly, then Y 1 + Y 2 - binomial random variable having distribution (19) with R = p 0 And n = n 1 + n 2 . This remark extends the applicability of the binomial distribution by allowing the results of several groups of tests to be combined when there is reason to believe that the same parameter corresponds to all these groups.

The characteristics of the binomial distribution were calculated earlier:

M(Y) = n.p., D(Y) = n.p.( 1- p).

In the section "Events and Probabilities" the law of large numbers is proven for a binomial random variable:

for anyone . Using the central limit theorem, the law of large numbers can be refined by indicating how much Y/ n differs from R.

De Moivre-Laplace theorem. For any numbers a and b, a< b, we have

Where F(X) is a function of standard normal distribution with mathematical expectation 0 and variance 1.

To prove it, it suffices to use the representation Y in the form of a sum of independent random variables corresponding to the outcomes of individual tests, formulas for M(Y) And D(Y) and the central limit theorem.

This theorem is for the case R= ½ was proven by the English mathematician A. Moivre (1667-1754) in 1730. In the above formulation, it was proven in 1810 by the French mathematician Pierre Simon Laplace (1749 - 1827).

Hypergeometric distribution occurs during selective control of a finite set of objects of volume N according to an alternative criterion. Each controlled object is classified either as having the attribute A, or as not having this characteristic. The hypergeometric distribution has a random variable Y, equal to the number of objects that have the attribute A in a random sample of volume n, Where n< N. For example, number Y defective units of product in a random sample of volume n from batch volume N has a hypergeometric distribution if n< N. Another example is the lottery. Let the sign A ticket is a sign of “being a winner”. Let the total number of tickets N, and some person acquired n of them. Then the number of winning tickets for this person has a hypergeometric distribution.

For a hypergeometric distribution, the probability of a random variable Y accepting the value y has the form

(20)

Where D– the number of objects that have the attribute A, in the considered set of volume N. Wherein y takes values ​​from max(0, n - (N - D)) to min( n, D), other things y the probability in formula (20) is equal to 0. Thus, the hypergeometric distribution is determined by three parameters - volume population N, number of objects D in it, possessing the characteristic in question A, and sample size n.

Simple random volume sampling n from the total volume N is a sample obtained as a result of random selection in which any of the sets of n objects have the same probability of being selected. Methods for randomly selecting samples of respondents (interviewees) or units of piece goods are discussed in the instructional, methodological, and regulatory documents. One of the selection methods is this: objects are selected one from another, and at each step, each of the remaining objects in the set has the same chance of being selected. In the literature, for the type of samples under consideration, the terms “ random sample", "random sampling without return".

Since the volumes of the population (batch) N and samples n are usually known, then the parameter of the hypergeometric distribution to be estimated is D. In statistical methods of product quality management D– usually the number of defective units in a batch. The distribution characteristic is also of interest D/ N– level of defects.

For hypergeometric distribution

The last factor in the expression for variance is close to 1 if N>10 n. If you make a replacement p = D/ N, then the expressions for the mathematical expectation and variance of the hypergeometric distribution will turn into expressions for the mathematical expectation and variance of the binomial distribution. This is no coincidence. It can be shown that

at N>10 n, Where p = D/ N. The limiting ratio is valid

and this limiting relation can be used when N>10 n.

The third widely used discrete distribution is the Poisson distribution. The random variable Y has a Poisson distribution if

,

where λ is the Poisson distribution parameter, and P(Y= y)= 0 for all others y(for y=0 it is designated 0! =1). For Poisson distribution

M(Y) = λ, D(Y) = λ.

This distribution is named after the French mathematician S. D. Poisson (1781-1840), who first obtained it in 1837. The Poisson distribution is the limiting case of the binomial distribution, when the probability R the implementation of the event is small, but the number of tests n great, and n.p.= λ. More precisely, the limit relation is valid

Therefore, the Poisson distribution (in the old terminology “distribution law”) is often also called the “law of rare events.”

The Poisson distribution originates in event flow theory (see above). It has been proven that for the simplest flow with constant intensity Λ, the number of events (calls) that occurred during the time t, has a Poisson distribution with parameter λ = Λ t. Therefore, the probability that during the time t no event will occur, equal to e - Λ t, i.e. the distribution function of the length of the interval between events is exponential.

The Poisson distribution is used when analyzing the results of sample marketing surveys of consumers, calculating the operational characteristics of statistical acceptance control plans in the case of small values ​​of the acceptance level of defects, to describe the number of breakdowns of a statistically controlled technological process per unit of time, the number of “service requests” entering the queuing system per unit of time, statistical patterns of accidents and rare diseases, etc.

Descriptions of other parametric families of discrete distributions and the possibilities of their practical use are considered in the literature.


In some cases, for example, when studying prices, output volumes or total time between failures in reliability problems, the distribution functions are constant over certain intervals in which the values ​​of the studied random variables cannot fall.

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