Statistics comes to our aid in solving many problems, for example: when it is not possible to build a deterministic model, when there are too many factors, or when we need to assess the likelihood of the constructed model taking into account the available data. The attitude towards statistics is ambiguous. There is an opinion that there are three types of lies: lies, damned lies and statistics. On the other hand, many “users” of statistics believe it too much, without fully understanding how it works: for example, applying a test to any data without checking its normality. Such negligence can generate serious errors and turn test “fans” into statistics haters. Let's try to put currents over i and figure out which models of random variables should be used to describe certain phenomena and what genetic connection exists between them.

First of all, this material will be of interest to students studying probability theory and statistics, although “mature” specialists will be able to use it as a reference. In one of the following works, I will show an example of using statistics to construct a test for assessing the significance of indicators of exchange trading strategies.

The work will consider:

At the end of the article there will be a question for reflection. I will present my thoughts on this matter in the next article.

Some of the above continuous distributions are special cases.

Discrete distributions

A discrete distribution of the probability of the occurrence of each of the possible outcomes of an event is described. As for any distribution (including continuous), the concepts of expectation and dispersion are defined for discrete events. However, it should be understood that the mathematical expectation for a discrete random event is a value in the general case that cannot be realized as the outcome of a single random event, but rather as a value to which the arithmetic mean of the outcomes of events will tend as their number increases.

In modeling discrete random events, combinatorics plays an important role, since the probability of the outcome of an event can be defined as the ratio of the number of combinations that give the required outcome to the total number of combinations. For example: there are 3 white balls and 7 black balls in a basket. When we select 1 ball from the basket, we can do this in 10 different ways (total number of combinations), but only 3 options in which the white ball will be selected (3 combinations that give the required outcome). Thus, the probability of choosing the white ball is: ().

One should also distinguish between samples with and without return. For example, to describe the probability of choosing two white balls, it is important to determine whether the first ball will be returned to the basket. If not, then we are dealing with a sample without return () and the probability will be as follows: - the probability of choosing a white ball from the initial sample multiplied by the probability of again choosing a white ball from those remaining in the basket. If the first ball returns to the basket, then this is a fetch with return(). In this case, the probability of choosing two white balls is .

If we formalize the example with a basket somewhat as follows: let the outcome of an event take one of two values ​​0 or 1 with probabilities and respectively, then the probability distribution of obtaining each of the proposed outcomes will be called the Bernoulli distribution:

According to established tradition, an outcome with a value of 1 is called “success”, and an outcome with a value of 0 is called “failure”. Obviously, obtaining the outcome “success or failure” occurs with probability .

Expectation and variance of the Bernoulli distribution:

The number of successes in trials, the outcome of which is distributed according to the probability of success (the example of returning balls to the basket), is described by a binomial distribution:

In other words, we can say that the binomial distribution describes the sum of independent random variables that can be distributed with a probability of success.
Expectation and variance:

If the quantities and have binomial distributions with parameters and , respectively, then their sum will also be distributed binomially with parameters .

Let's imagine a situation where we pull balls out of the basket and return them back until a white ball is pulled out. The number of such operations is described by a geometric distribution. In other words: the geometric distribution describes the number of trials until the first success with the probability of success in each trial. If the number of the trial in which success occurred is implied, then the geometric distribution will be described by the following formula:

The Pascal distribution is a generalization of the distribution: it describes the distribution of the number of failures in independent trials, the outcome of which is distributed over the probability of success before the total success occurs. When , we obtain a distribution for the quantity .

Expectation and variance of the negative binomial distribution:

So far we have considered examples of samples with reversion, that is, the probability of the outcome did not change from trial to trial.

Now consider the situation without return and describe the probability of the number of successful selections from a population with a pre-known number of successes and failures (a pre-known number of white and black balls in the basket, trump cards in the deck, defective parts in the game, etc.).

Let the total collection contain objects, some of them are marked as “1”, and as “0”. We will consider the selection of an object with a label “1” as a success, and with a label “0” as a failure. We will carry out n tests, and the selected objects will no longer participate in further tests. The probability of success will obey the hypergeometric distribution:

Expectation and variance:

Poisson distribution

The Poisson distribution differs significantly from the distributions discussed above in its “subject” area: now it is not the probability of the occurrence of one or another test outcome that is considered, but the intensity of events, that is, the average number of events per unit of time.

The Poisson distribution describes the probability of occurrence of independent events over time at an average intensity of events:

The Poisson distribution, in combination with , which describes the time intervals between the occurrences of independent events, constitutes the mathematical basis of the reliability theory.

The probability density of the product of random variables x and y () with distributions and can be calculated as follows:

Some of the distributions below are special cases of the Pearson distribution, which in turn is a solution to the equation:

The distributions that will be discussed in this section have close relationships with each other. These connections are expressed in the fact that some distributions are special cases of other distributions, or describe transformations of random variables that have other distributions.

The diagram below shows the relationships between some of the continuous distributions that will be considered in this paper. In the diagram, solid arrows show the transformation of random variables (the beginning of the arrow indicates the initial distribution, the end of the arrow indicates the resulting one), and the dotted arrows indicate the generalization relation (the beginning of the arrow indicates the distribution, which is a special case of the one to which the end of the arrow points). For special cases of the Pearson distribution, the corresponding type of Pearson distribution is indicated above the dotted arrows.

Normal distribution (Gaussian distribution)

The probability density of a normal distribution with parameters and is described by the Gaussian function:

If and , then such a distribution is called standard.

Expectation and variance of normal distribution:

The normal distribution is a type VI distribution.

The sum of the squares of independent normal quantities has , and the ratio of independent Gaussian quantities is distributed over .

The normal distribution is infinitely divisible: the sum of normally distributed quantities and with parameters and, accordingly, also has a normal distribution with parameters , where and .

The normal distribution well models quantities that describe natural phenomena, noise of a thermodynamic nature, and measurement errors.

In addition, according to the central limit theorem, the sum of a large number of independent terms of the same order converges to a normal distribution, regardless of the distributions of the terms. Due to this property, the normal distribution is popular in statistical analysis; many statistical tests are designed for normally distributed data.

The z-test is based on the infinite divisibility of the normal distribution. This test is used to check whether the expected value of a sample of normally distributed values ​​is equal to a certain value. The variance value should be known. If the variance value is unknown and is calculated based on the analyzed sample, then a t-test based on .

Let us assume that we have a sample of n independent normally distributed values ​​from the general population with standard deviation, let us hypothesize that . Then the value will have a standard normal distribution. By comparing the obtained z value with the quantiles of the standard distribution, you can accept or reject the hypothesis with the required level of significance.

Due to the widespread use of the Gaussian distribution, many researchers who are not very familiar with statistics forget to check the data for normality, or evaluate the distribution density graph “by eye”, blindly believing that they are dealing with Gaussian data. Accordingly, you can safely use tests designed for normal distribution and get completely incorrect results. This is probably where the rumor about statistics as the most terrible type of lie came from.

Let's consider an example: we need to measure the resistance of a set of resistors of a certain value. Resistance has a physical nature; it is logical to assume that the distribution of resistance deviations from the nominal value will be normal. We measure and obtain a bell-shaped probability density function for the measured values ​​with a mode in the vicinity of the resistor value. Is this a normal distribution? If yes, then we will look for defective resistors using , or the z-test, if we know the dispersion of the distribution in advance. I think that many will do just that.

But let's take a closer look at resistance measurement technology: Resistance is defined as the ratio of applied voltage to current flow. We measured current and voltage with instruments, which, in turn, have normally distributed errors. That is, the measured values ​​of current and voltage are normally distributed random variables with mathematical expectations corresponding to the true values ​​of the measured quantities. This means that the obtained resistance values ​​are distributed according to , and not according to Gaussian.

The distribution describes the sum of squares of random variables, each of which is distributed according to the standard normal law:

Where is the number of degrees of freedom, .

Expectation and dispersion of distribution:

It is a special case (and therefore a Type III distribution) and a generalization. The ratio of quantities distributed over distributed over .

The Pearson goodness-of-fit test is based on the distribution. Using this criterion, you can check the reliability of a sample of a random variable belonging to a certain theoretical distribution.

Let's assume that we have a sample of some random variable. Based on this sample, we calculate the probabilities of values ​​falling into the intervals (). Let there also be an assumption about the analytical expression of the distribution, according to which the probabilities of falling into the selected intervals should be . Then the quantities will be distributed according to the normal law.

Let us reduce to the standard normal distribution: ,
where and .

The resulting values ​​have a normal distribution with parameters (0, 1), and therefore the sum of their squares is distributed over a degree of freedom. The decrease in the degree of freedom is associated with an additional restriction on the sum of the probabilities of values ​​falling into the intervals: it must be equal to 1.

By comparing the value with the quantiles of the distribution, you can accept or reject the hypothesis about the theoretical distribution of the data with the required level of significance.

The Student distribution is used to conduct a t-test: a test for the equality of the expected value of a sample of distributed random variables to a certain value, or the equality of the expected value of two samples with the same variance (the equality of variances must be checked). The Student distribution describes the ratio of a distributed random variable to a variable distributed over .

Let and be independent random variables having degrees of freedom and respectively. Then the quantity will have a Fisher distribution with degrees of freedom, and the quantity will have a Fisher distribution with degrees of freedom.
The Fisher distribution is defined for real non-negative arguments and has a probability density:


Expectation and variance of the Fisher distribution:

A number of statistical tests are based on the Fisher distribution, such as assessing the significance of regression parameters, a test for heteroskedasticity and a test for equality of sample variances (f-test, should be distinguished from accurate Fisher test).

F-test: let there be two independent samples and distributed data volumes and respectively. Let us put forward a hypothesis about the equality of sample variances and test it statistically.

Let's calculate the value. It will have a Fisher distribution with degrees of freedom.

By comparing the value with the quantiles of the corresponding Fisher distribution, we can accept or reject the hypothesis of equality of sample variances with the required level of significance.

Exponential (exponential) distribution and Laplace distribution (double exponential, double exponential)

The exponential distribution describes the time intervals between independent events occurring at an average intensity. The number of occurrences of such an event over a certain period of time is described as discrete. The exponential distribution together with form the mathematical basis of the theory of reliability.

In addition to the theory of reliability, exponential distribution is used in the description of social phenomena, in economics, in the theory of queuing, in transport logistics - wherever it is necessary to model the flow of events.

The exponential distribution is a special case (for n=2), and therefore . Since an exponentially distributed quantity is a chi-square quantity with 2 degrees of freedom, it can be interpreted as the sum of the squares of two independent normally distributed quantities.

Also, exponential distribution is a fair case

LECTURE 8

Probability distributions of discrete random variables.Binomial distribution. Poisson distribution. Geometric distribution. Generating function.

6. PROBABILITY DISTRIBUTIONS
DISCRETE RANDOM VARIABLES

Binomial distribution

Let it be produced n independent trials, in each of which the event A It may or may not appear. Probability p occurrence of an event A in all tests is constant and does not change from test to test. Consider as a random variable X the number of occurrences of the event A in these tests. Formula to find the probability of an event occurring A
smooth k once every n tests, as is known, are described Bernoulli's formula

The probability distribution defined by Bernoulli's formula is called binomial .

This law is called "binomial" because the right-hand side can be considered as a general term in the expansion of Newton's binomial

Let's write the binomial law in the form of a table

Let us find the numerical characteristics of this distribution.

.

Let us write down the equality, which is a binary of Newton

.

and differentiate it with respect to p. As a result we get

.

Multiply the left and right sides by p:

.

Considering that p+q=1, we have

(6.2)

So, the mathematical expectation of the number of occurrences of events in n independent trials is equal to the product of the number of trials n by the probability p of the occurrence of an event in each trial.

Let's calculate the variance using the formula

For this we will find

.

Let us first differentiate Newton's binomial formula twice with respect to p:

and multiply both sides of the equality by p 2:

Hence,

So, the variance of the binomial distribution is

. (6.3)

These results can also be obtained from purely qualitative reasoning. The total number X of occurrences of event A across all trials is the sum of the number of occurrences of the event in individual trials. Therefore, if X 1 is the number of occurrences of the event in the first trial, X 2 – in the second, etc., then the total number of occurrences of event A in all trials is equal to X = X 1 +X 2 +…+X n. According to the property of mathematical expectation:

Each of the terms on the right side of the equality is the mathematical expectation of the number of events in one trial, which is equal to the probability of the event. Thus,

According to the dispersion property:

Since , and the mathematical expectation of a random variable, which can take only two values, namely 1 2 with probability p and 0 2 with probability q, That . Thus, As a result, we get

Using the concept of initial and central moments, we can obtain formulas for asymmetry and kurtosis:

. (6.4)

The polygon of the binomial distribution has the following form (see Fig. 6.1). Probability P n(k) first increases with increasing k, reaches its highest value and then begins to decrease. The binomial distribution is skewed except for the case p=0.5. Note that with a large number of tests n The binomial distribution is very close to normal. (The rationale for this proposal is related to the local theorem of Moivre-Laplace.)

The number m 0 of occurrences of an event is called most likely, if the probability of an event occurring a given number of times in this series of tests is greatest (maximum in the distribution polygon). For binomial distribution

. (6.5)

Comment. This inequality can be proven using the recurrent formula for binomial probabilities:

(6.6)

Example 6.1. The share of premium products at this enterprise is 31%. What are the mathematical expectation and variance, as well as the most probable number of premium products in a randomly selected batch of 75 products?

Solution. Because the p=0,31, q=0,69, n=75, then

M[ X] = n.p.= 75×0.31 = 23.25; D[ X] = npq= 75×0.31×0.69 = 16.04.

To find the most probable number m 0, let's create a double inequality

It follows that m 0 = 23.

Poisson distribution

As already noted, the binomial distribution approaches normal when n®¥. However, this does not take place if, along with an increase n one of the quantities p or q tends to zero. In this case, the asymptotic Poisson formula holds, i.e. at n®¥, p®0

, (6.7)

where l= n.p.. This formula determines Poisson distribution law , which has independent meaning, and not just as a special case of the binomial distribution. Unlike the binomial distribution, here the random variable k can take on an infinite number of values: k=0,1,2,…

Poisson's law describes the number of events k occurring over equal periods of time, provided that the events occur independently of each other with a constant average intensity, which is characterized by the parameter l. The Poisson distribution polygon is shown in Fig. 6.2. Note that for large l races
Poisson's distribution approaches normal. Therefore, the Poisson distribution is used, as a rule, in cases where l is of the order of unity, and the number of trials n must be large, and the probability of the event occurring p in each test is small. In this regard, Poisson's law is often also called law of distribution of rare phenomena.

Examples of situations in which the Poisson distribution occurs are the distributions of: 1) the number of certain microbes per unit volume; 2) the number of electrons emitted from the heated cathode per unit time; 3) the number of a-particles emitted by a radioactive source over a certain period of time; 4) the number of calls arriving at the telephone exchange at a certain time of day, etc.

Let's write Poisson's law in the form of a table

Let's check that the sum of all probabilities is equal to one:

Let us find the numerical characteristics of this distribution. By definition of the mathematical expectation for the DSV, we have

Note that in the last sum, the summation begins with k=1, because the first term of the sum corresponding to k=0, equal to zero.

To find the variance, we first find the mathematical expectation of the square of the random:

Thus, the mathematical expectation and variance of a random variable distributed according to Poisson’s law coincide and are equal to the parameter of this distribution

. (6.8)

This is the distinctive feature of the Poisson distribution. Thus, if, based on experimental data, it was found that the mathematical expectation and the variance of a certain value are close to each other, then there is reason to assume that this random variable is distributed in accordance with Poisson’s law.

Using the concept of initial and central moments, we can show that for the Poisson distribution the skewness coefficient and kurtosis are equal:

. (6.9)

Since the parameter l is always positive, the Poisson distribution always has positive skewness and kurtosis.

Let us now show that Poisson's formula can be considered as a mathematical model of the simplest flow of events.

The flow of events call a sequence of events that occur at random times. The stream is called the simplest, if it has the properties stationarity, no aftereffect And ordinariness.

Flow intensity l is the average number of events that occur per unit time.

If the flow intensity constant l is known, then the probability of occurrence k events of the simplest flow over time t is determined by the Poisson formula:

. (6.10)

This formula reflects all the properties of the simplest flow. Moreover, any simplest flow is described by the Poisson formula, therefore the simplest flows are often called Poisson.

Stationarity property k events in any period of time depends only on the number k and on duration t period of time and does not depend on the beginning of its counting. In other words, if the flow has the property of stationarity, then the probability of occurrence k events over a period of time t there is a function that depends only on k and from t.

In the case of the simplest flow, it follows from Poisson’s formula (6.10) that the probability k events during t, at a given intensity, is a function of only two arguments: k And t, which characterizes the property of stationarity.

No aftereffect property is that the probability of occurrence k events in any period of time depends on whether events appeared or did not appear at points in time preceding the beginning of the period in question. In other words, the flow's history does not affect the probabilities of events occurring in the near future.

In the case of the simplest flow, the Poisson formula (6.10) does not use information about the occurrence of events before the beginning of the time period under consideration, which characterizes the property of the absence of aftereffects.

Ordinariness property is that the occurrence of two or more events in a short period of time is practically impossible. In other words, the probability of more than one event occurring in a short period of time is negligible compared to the probability of only one event occurring.

Let us show that the Poisson formula (6.10) reflects the property of ordinaryness. Putting k=0 and k=1, we find, respectively, the probabilities of no events occurring and the occurrence of one event:

Therefore, the probability of more than one event occurring is

Using the expansion of the function in the Maclaurin series, after elementary transformations we obtain

.

Comparing P t(1) and P t(k>1), we conclude that for small values t the probability of the occurrence of more than one event is negligible compared to the probability of the occurrence of one event, which characterizes the property of ordinaryness.

Example 6.2. In the observations of Rutherford and Geiger, a radioactive substance over a period of time of 7.5 sec emitted an average of 3.87 a-particles. Find the probability that for 1 sec this substance will emit at least one particle.

Solution. As we have already noted, the distribution of the number of a-particles emitted by a radioactive source over a certain period of time is described by the Poisson formula, i.e. forms the simplest flow of events. Since the intensity of emission of a-particles for 1 sec equals

,

then the Poisson formula (6.10) takes the form

Thus, the probability that t=1 sec the substance will emit at least one particle will be equal

Geometric distribution

Let shooting be carried out at a given target until the first hit, and the probability p hitting the target in each shot is the same and does not depend on the results of previous shots. In other words, in the experiment under consideration, the Bernoulli scheme is implemented. As a random variable X we will consider the number of shots fired. Obviously, the possible values ​​of the random variable X are natural numbers: x 1 =1, x 2 =2, ... then the probability that it will be needed k shots will be equal

. (6.11)

Assuming in this formula k=1,2, ... we get a geometric progression with the first term p and a multiplier q:

For this reason, the distribution defined by formula (6.11) is called geometric .

Using the formula for the sum of an infinitely decreasing geometric progression, it is easy to verify that

.

Let's find the numerical characteristics of the geometric distribution.

By definition of the mathematical expectation for the DSV, we have

.

Let's calculate the variance using the formula

.

For this we will find

.

Hence,

.

So, the mathematical expectation and variance of the geometric distribution are equal to

. (6.12)

6.4.* Generating function

When solving problems related to DSV, combinatorics methods are often used. One of the most developed theoretical methods of combinatorial analysis is the method of generating functions, which is one of the most powerful methods in applications. Let's get to know him briefly.

If the random variable x takes only non-negative integer values, i.e.

,

That generating function the probability distribution of a random variable x is called a function

, (6.13)

Where z– real or complex variable. Note that between multiple generating functions j x ( x)and many distributions(P(x= k)} there is a one-to-one correspondence.

Let the random variable x have binomial distribution

.

Then, using Newton's binomial formula, we get

,

those. binomial distribution generating function looks like

. (6.14)

Addition. Poisson generating function

looks like

. (6.15)

Generating function of geometric distribution

looks like

. (6.16)

Using generating functions, it is convenient to find the main numerical characteristics of the DSV. For example, the first and second initial moments are related to the generating function by the following equalities:

, (6.17)

. (6.18)

The method of generating functions is often convenient because in some cases the distribution function of the DSV is very difficult to determine, while the generating function is sometimes easy to find. For example, consider Bernoulli's sequential independent test design, but make one change to it. Let the probability of an event occur A varies from trial to trial. This means that Bernoulli's formula becomes inapplicable for such a scheme. The task of finding the distribution function in this case presents significant difficulties. However, for this scheme, the generating function is easy to find, and, therefore, the corresponding numerical characteristics are easy to find.

The widespread use of generating functions is based on the fact that the study of sums of random variables can be replaced by the study of products of the corresponding generating functions. So, if x 1, x 2, …, x n are independent, then

Let p k=Pk(A) – probability of “success” in k-th test in the Bernoulli circuit (respectively, q k=1–p k– probability of “failure” in k th test). Then, in accordance with formula (6.19), the generating function will have the form

. (6.20)

Using this generating function, we can write

.

It is taken into account here that p k +q k=1. Now, using formula (6.1), we find the second initial moment. To do this, let us first calculate

And .

In a special case p 1 =p 2 =…=p n=p(i.e. in the case of binomial distribution) from the obtained formulas it follows that Mx= n.p., Dx= npq.

Those. discrete random the value of X has a geom. distributor with parameter R and denominator q, if it takes values ​​1,2,3,… k, ... with probabilities

P(X) = pq k-1 , where q=1-R.

The distribution is called geom., because. verity p 1, p 2, ... form a geometric progression, the first member of which is R, and the denominator is q.

If the number of tests is not limited, i.e. if a random variable can take values ​​1, 2, ..., ∞, then the expected value and variance are geometric. distributions can be found using the formulas Mх = 1/p, Dх = q/p 2

Example. The gun is fired at the target until the first hit is made. The probability of hitting the target is p = 0.6 with each shot. S.v. X is the number of possible shots before the first hit.

A) Compile a distribution series, find the distribution function, construct its graph and find all the numerical characteristics. b) Find the mathematical expectation and variance for the case if the shooter intends to fire no more than three shots.

A) The random variable can take values ​​1, 2, 3, 4,..., ∞
P(1) = p = 0.6
P(2) = qp = 0.4 0.6 = 0.24
P(3) = q 2 p = 0.4 2 0.6 = 0.096 ...
P(k) = q k-1 p = 0.4 k-1 0.6 ...
Distribution Range:

Control: Σp i = 0.6/(1-0.4) = 1 (sum of geometric progression)

The distribution function is the probability that r.v. X will take on a value less than the specific numeric value of x. The distribution function values ​​are found by summing the probabilities.

If x ≤ 1, then F(x) = 0

If 1< x ≤ 2, то F(x) = 0,6
If 2< x ≤ 3, то F(x) = 0,6 + 0,24 = 0,84
If 3< x ≤ 4, то F(x) = 0,84 + 0,096 = 0,936 ...
If k-1< x ≤ k, то F(x) = 0,6(1-0,4 k-1)/(1-0,4) = 1-0,4 k-1 (частичная сумма геом-ой прогрессии) ...

Mx = 1/p = 1/0.6 ≈ 1.667
Dх = q/p 2 = 0.4/0.36 ≈ 1.111
σ = √Dх ≈ 1.054

b) The random variable can take values ​​1, 2, 3.
P(1) = p = 0.6
P(2) = qp = 0.4 0.6 = 0.24
P(3) = q 2 p + q 3 = 0.4 2 0.6 + 0.4 3 = 0.16
Distribution Range:

Control: Σp i = 0.6 + 0.24 + 0.16 = 1
Distribution function.

If x ≤ 1, then F(x) = 0
If 1< x ≤ 2, то F(x) = 0,6
If 2< x ≤ 3, то F(x) = 0,6 + 0,24 = 0,84
If x > 3, then F(x) = 0.84 + 0.16 = 1
M(X) = 1 0.6 + 2 0.24 + 3 0.16 = 1.56
D(X) = 1 2 0.6 + 2 2 0.24 + 3 2 0.16 - 1.56 2 = 0.5664
σ(X) ≈ 0.752

Skewness and kurtosis

Asymmetry is a property of the sample distribution that characterizes the asymmetry of the distribution of a random variable. In practice, symmetric distributions are rare, and in order to identify and evaluate the degree of asymmetry, the concept of asymmetry is introduced. In the case of a negative asymmetry coefficient, a gentler “descent” is observed on the left, otherwise – on the right. In the first case, the asymmetry is called left-sided, and in the second - right-sided.

Asymmetry coefficient discrete random variable is calculated using the formula:
As(X) = (x 1-M X) 3 p 1 + (x 2 - M X) 3 p 2 + ... + ( x n-M X) 3 p n

Coeff. asymmetry continuous sl.vel. calculated by the formula:

Excess is a measure of the steepness of the distribution curve. The kurtosis coefficient of a discrete random variable is calculated using the formula:

Ex(X) = [(x 1 - M X) 4 p 1 + (x 2 - M X) 4 p 2 + ... + (x n - M X) 4 p n ] / σ 4 - 3

The kurtosis coefficient of a continuous random variable is calculated using the formula:

Example.

The distribution law of a discrete random variable X is a list of all possible values ​​of the next variable. X that it can accept, and the corresponding probabilities. The sum of all beliefs must be equal to 1. Check: 0.1 + 0.2 + 0.5 + 0.1 + 0.1 = 1.

1. Expected value: M(X) = -2 0.1 - 1 0.2 + 0 0.5 + 1 0.1 + 2 0.1 = -0.1
2. Dispersion is the mathematical expectation of the squared deviation of the values ​​of the next vel. X from her mat.ozh.: D(X) = (-2 + 0.1) 2 0.1 + (- 1 + 0.1) 2 0.2 + (0 + 0.1) 2 0.5 + (1 + 0.1) 2 0.1 + (2 + 0.1) 2 0.1 = 1.09
   or D(X) = (-2) 2 0.1 + (-1) 2 0.2 + 0 2 0.5 + 1 2 0.1 + 2 2 0.1 - (-0 ,1) 2 = 1.1 - 0.01 = 1.09
3. Wed. sq. off is the square root of the variance: σ = √1.09 ≈ 1.044
4. Coef. asymmetry As(X) = [(-2 + 0.1) 3 0.1 + (- 1 + 0.1) 3 0.2 + (0 + 0.1) 3 0.5 + (1 + 0.1) 3 0.1 + (2 + 0.1) 3 0.1] / 1.044 3 = 0.200353
5. Coef. excess E x(X) = [(-2 + 0.1) 4 0.1 + (- 1 + 0.1) 4 0.2 + (0 + 0.1) 4 0.5 + (1 + 0 ,1) 4 ·0.1 + (2 + 0.1) 4 ·0.1]/1.044 4 - 3 = 0.200353
6. The distribution function is the probability that the random variable X will take a value less than some numerical value x: F(X) = P(X< x). The distribution function is a non-decreasing function. It takes values ​​in the range from 0 to 1.

P(X< -0,1) = F(-0,1) = 0,3 P(X >-0.05) = P(0) + P(1) + P(2) = 0.5 + 0.1 + 0.1 = 0.7

2) Continuous random variables. Normal distribution.

Continuous a random variable does not take any specific numerical values, but any values ​​on a numerical interval. The description of the distribution law in the continuous case is much more complicated than in the discrete case.

Continuous called a random variable that can take any value from a certain given interval, for example, the waiting time for transport, the air temperature in any month, the deviation of the actual size of a part from the nominal one, etc. The interval at which it is set can be infinite in one or both directions.

The main difference in the problems of calculating probabilities for discrete and continuous cases is as follows. In a discrete case for events like x = c(the random variable takes a certain value) the probability is sought R(With). In the continuous case probabilities of this type are equal to zero, therefore, the probabilities of events of the type “a random variable takes values ​​from a certain segment” are of interest, i.e. A≤ X≤ b. Or for events like X≤ With looking for probability R(X≤ With). We obtained a graph of the distribution function F( X≤ With).

So, the variety of random variables is very large. The number of values ​​they accept can be finite, countable, or uncountable; values ​​can be located discretely or fill intervals completely. In order to specify the probabilities of the values ​​of random variables that are so different in nature, and, moreover, to specify them in the same way, the concept of distribution function of a random variable.

Let be a random variable and X- an arbitrary real number. The probability that it will take a value less than X, called probability distribution function random variable: F(x)= P(<х}.

Let's summarize what has been said: random variable is a quantity whose values ​​depend on the case and for which the probability distribution function is defined.

For continuous random variables (when the set of possible values ​​of a random variable is uncountable), the distribution law is specified using a function. Most often this distribution function :F( x) = P(X<X) .

Function F( x) has the following properties:

1. 0 ≤ F( x) ≤ 1 ;

2.F( x) does not decrease;

3.F( x) left continuous;

4.F(- ∞ ) = 0, F( ∞ ) = 1.