A continuous random variable has exponential (exponential )distribution law with parameter if its probability density has the form:

(12.1)

Here is a constant positive value. That. exponential distribution is determined by one positive parameter ... Let us find the cumulative function of the exponential distribution:

(12.3)

Rice. 12.1. Differential exponential distribution function ()

Rice. 12.2. Cumulative function of exponential distribution ()

Numerical characteristics of exponential distribution

Let's calculate the mathematical expectation and variance of the exponential distribution:

To calculate the variance, we will use one of its properties:

Because , then it remains to calculate:

Substituting (12.6) into (12.5), we finally get:

(12.7)

For a random variable distributed according to exponential law, the mathematical expectation is equal to the standard deviation.

Example 1. Write the differential and cumulative functions of the exponential distribution, if the parameter.

Solution... a) The distribution density has the form:

b) The corresponding integral function is equal to:

Example 2. Find the probability of hitting a given interval for the SV, distributed according to the exponential law

Solution... Let's find a solution, remembering that:. Now, taking into account (12.3), we get:

Reliability function

Let's call a device an element, regardless of whether it is "simple" or "complex". Let the element begin to work at a moment in time, and after the expiration of the duration, a failure occurs. Let us denote by continuous SV - the duration of the element's uptime. If the element will operate without failure (until the failure occurs) for a time less than, then, consequently, a failure will occur during the duration. In this way, probability of failure over time, the duration is determined by the integral function:

. (12.8)

Then the probability of failure-free operation for the same time duration is equal to the probability of the opposite event, i.e.

Reliability functionis called a function that determines the probability of failure-free operation of an element for a time duration.

Often the duration of the uptime of an element has an exponential distribution, the integral function of which is:

. (12.10)

Then, in the case of an exponential distribution of the element's uptime and taking into account (12.9), the reliability function will be equal to:

. (12.11)

Example 3. The uptime of the element is distributed according to the exponential law at (time in hours). Find the probability that the element will work 100 hours without failure.

Solution... In our example, then we will use (12.11):

The exponential law of reliability is very simple and convenient for solving practical problems. This law has the following important property:

The probability of failure-free operation of an element over a time interval of duration does not depend on the time of the previous operation before the beginning of the considered interval, but depends only on the duration of time(at a given failure rate).

Let us prove this property by introducing the following notation:

failure-free operation of an element on an interval of duration;

Then the event is that the element works flawlessly for an interval of duration. Let us find the probabilities of these events using the formula (12.11), assuming that the uptime of the element is subject to the exponential law:

Let us find the conditional probability that the element will work flawlessly on the time interval, provided that it has already worked flawlessly on the previous time interval:

(12.13)

We see that the resulting formula does not depend on, but only on. Comparing (12.12) and (12.13), we can conclude that the conditional probability of no-failure operation of an element on an interval of duration, calculated on the assumption that the element has worked without failure in the previous interval, is equal to the unconditional probability.

So, in the case of an exponential law of reliability, the failure-free operation of an element "in the past" does not affect the value of the probability of its failure-free operation "in the near future."

Combinatorial elements

The space of elementary events. Random events.

Probability

The modern concept of probability

Classical probabilistic scheme

Geometric probabilities

The law of addition of probabilities

Probability multiplication theorem

Total Probability Formula

Conjecture theorem. Bayes' formula.

Repetition of tests. Bernoulli's scheme.

Local Moivre-Laplace theorem

Integral de Moivre-Laplace theorem

Poisson's Theorem (Law of Rare Events)

Random variables

Distribution functions

Continuous random variable and distribution density

Basic properties of distribution density

Numerical characteristics of a one-dimensional random variable

Mathematical expectation properties

Moments of a random variable

Dispersion properties

Asymmetries and kurtosis

Multidimensional random variables

Properties of a two-dimensional distribution function

Probability density of a two-dimensional random variable

Buffon's problem

Conditional distribution density

Numerical characteristics of a system of random variables

Correlation Coefficient Properties

Normal (Gaussian) distribution law

The probability of hitting the interval

Properties of the normal distribution function

Distribution (chi-square)

Exponential (exponential) distribution law

Numerical characteristics of exponential distribution

Reliability function

A continuous random variable $ X $ obeys an exponential (exponential) probability distribution if its probability density $ f \ left (x \ right) $ has the following form:

$$ f (x) = \ left \ (\ begin (matrix)
0, \ x< 0\\
\ lambda e ^ (- \ lambda x), \ x \ ge 0
\ end (matrix) \ right .. $$

Then the distribution function:

$$ F (x) = \ left \ (\ begin (matrix)
0, \ x< 0\\
1-e ^ (- \ lambda x), \ x \ ge 0
\ end (matrix) \ right. $$

Graphs of density functions $ f \ left (x \ right) $ and distributions $ F \ left (x \ right) $ are shown in the figure:

For an exponential distribution law, numerical characteristics can be calculated using known formulas. Expected value and standard deviation are equal to each other and equal to $ 1 / \ lambda $, that is:

$$ M \ left (X \ right) = \ sigma \ left (X \ right) = ((1) \ over (\ lambda)). $$

Dispersion:

$$ D \ left (X \ right) = ((1) \ over ((\ lambda) ^ 2)). $$

The distribution parameter $ \ lambda $ in the statistical sense characterizes the average number of events occurring per unit of time. So, if the average device uptime is $ 1 / \ lambda $, then the $ \ lambda $ parameter will characterize the average number of failures per unit time. Examples of random variables subject to an exponential distribution law can be:

• Duration of a telephone conversation;
• Time spent on customer service;
• The period of operation of the device between breakdowns;
• The time intervals between the appearance of cars at the gas station.

Example ... The random variable $ X $ is exponentially distributed $ f \ left (x \ right) = \ left \ (\ begin (matrix)
0, \ x< 0\\
5e ^ (- 5x), \ x \ ge 0
\ end (matrix) \ right. $. Then the expectation $ = $ standard deviation $ \ sigma (X) = 1 / \ lambda = 1/5 = 0.2 $, variance $ D (X) = 1 / (\ lambda) ^ 2 = 1/25 = 0 , 04. $

Example ... The operating time of the device is a random variable $ X $, subject to exponential distribution. It is known that the average operating time of this device is $ 500 $ hours. What is the probability that this device will work for at least $ 600 $ hours?

The mathematical expectation of the random variable $ X $ is $ M \ left (X \ right) = 500 = 1 / \ lambda $, hence the distribution parameter $ \ lambda = 1/500 = 0.002. $ We can write the distribution function:

$$ F (x) = \ left \ (\ begin (matrix)
0, \ x< 0\\
1-e ^ (- \ lambda x) = 1-e ^ (- 0.002x), \ x \ ge 0
\ end (matrix) \ right. $$

Then the probability that the device will work less than $ 600 $ hours is:

$$ P \ left (X \ ge 600 \ right) = 1-P \ left (X< 600\right)=1-F\left(600\right)=1-\left(1-e^{-0,002\cdot 600}\right)\approx 0,301.$$

For baseline factors, we will link factors 1 to 7 to factors from section VI. 3 in the order in which they are written, that is, factor 1 is truncation, factor 2 is symmetry, etc. Then we will link the levels + and - of the factors in table. 4 with two levels of VI factors. 3 randomly. This random order was achieved by using a table of random numbers and comparing those numbers with 1/2. The results of this procedure are shown in table. 5. Combining the table. 4 and 5 gives the plan in the initial factors, given in table. 6, where A1, (i = 1, ..., 4) denote unknown random variables having an exponential distribution with the parameter br - b. As an example, consider combination 1 in table. 6. Factors 1 and 2 are at the + level in the table. 4. Consequently, from table. 5 we have to take a truncated, skewed distribution with tails up. Table 1 we see that this distribution is an exponential distribution of a random variable x. Factor 6 is at the level

In our case, objective reasons for technological products do not allow using these distribution laws. First, the condition for obtaining a normal law is the joint action of many random factors, none of which is dominant. This does not correspond to the operating conditions and rejection of products for technological purposes, where the dominant factors necessarily appear. Secondly, the exponential law requires the conditions of ordinariness, stationarity and aftereffect, which are often not met for these products. In particular, the flow of failures cannot be considered stationary due to its probabilistic regime changing in time.

Such information reflects the prevailing conditions of production processes and therefore is a sample from the general population. Based on the law of large numbers, it can be argued that if the general population obeys a certain distribution law, then the sample from this population, with a sufficiently large volume of it, will obey this law. Most often, this law is unknown, and its definition causes considerable difficulties. In such cases, preference is given to well-known distribution laws, most often exponential and normal.

By the word we will accidentally mean that the probability of one car arriving at the gas station for any small time interval starting at an arbitrary time moment / and having a length m, up to negligible values, is proportional to m with a certain proportionality coefficient X> 0. The value of K can be interpreted as the average number of cars that appear at the station per unit of time, and its inverse value 1L, as the average time of the appearance of one car. The probability that no car will arrive during this period of time is considered to be approximately equal to 1 - m, and the probability of the arrival of two or more cars is negligible in comparison with the value of Yal. The following conclusions can be drawn from the assumptions made. First, the time intervals / between two successive arrivals of cars satisfy the exponential distribution

Losses arising from the operation of automation equipment during this period can be calculated based on the use of reliability theory, according to which sudden failures are defined as system failure due to the occurrence of unforeseen, sudden concentrations of external loads and internal stresses exceeding the calculated ones. If some of the elements and connections are made or repaired poorly, then they will fail at lower loads. Therefore, failures of defective elements are distributed exponentially (the Poisson nature of the distribution of sudden failures is considered), with an average operating time several times less than that of other elements.

Exponential distribution. This distribution, as a rule, obeys the operating time of sudden failures (i.e., failures due to latent technology defects) and the distribution of time between two successive failures, if the products operate in a steady state.

Let us consider the case when the investigated parameter is distributed exponentially.

Ya.B. Shor gives the following formula for determining the confidence interval for the general average in the case of an exponential distribution of a random variable

Despite the seeming ease of the conditions under which the last expression was obtained, in theoretical terms, for a number of interesting cases, they turn out to be impracticable. This happens when the derivative g (x) at the point x = v becomes infinity. In particular, this is the case with the two-sided exponential distribution, which we have already met in examples 2 and 3 from. In one version of constructing the optimal

In this chapter we will consider the most common laws of distribution of random variables, as well as the main parameters of these laws. Methods will be given for finding the probability distribution function of a random variable in the case of a non-integrable probability density, as well as algorithms for obtaining sequences of random variables with an arbitrary distribution law, which is necessary when simulating random processes. Special attention will be given a generalized exponential distribution, which is most suitable for studying asset pricing.

One of the most important distributions found in statistics is the normal distribution (Gaussian distribution), which belongs to the class of exponential. The probability density of this distribution is

Another type of exponential distribution, along with the normal one, is the Laplace distribution, the density of which is expressed by the formula

Generalized exponential distribution.

Earlier in this chapter, two types of exponential distributions of Gauss and Laplace were considered. They have a lot in common, they are symmetric, they depend on two parameters (//, s),

In VI. 2, we briefly describe the MMD and the purpose of the experiment, i.e., the study of the MMD sensitivity to violation of its premises. In VI.3 we will discuss in detail various factors that can affect this sensitivity. We define the distribution abnormality as factor 1. This factor describes the possibility or impossibility for random variables to become less than a given constant (the so-called truncated distribution factor), the asymmetry and tails of the distribution, we will take factor 2. Combining factors 1 and 2, we choose four types of distributions (exponential , Erlang, weighted difference of two random variables with exponential distribution and the sum of differences of random variables with exponential distribution). Variance heterogeneity will be denoted as factor 3. This means that the variance of the best population (afki) can be either greater or less than the variance of the competing worst population (in the least favorable situation). Factor 4 measures whether the two variances differ greatly or not at all. Factor 5 indicates whether the variances of the worst populations (in the least favorable situation) are equal or are all different. Factor 6 determines the number of populations (three or seven), factor 7 determines the distance 8 = 6 between the best and the next populations in the least favorable situation. Factor P, guaranteeing the minimum value of the probability the right choice considered

Such information is a sample from the general population, which has a certain distribution law. More often than not, this law is unknown and its definition causes constructive difficulties. In such cases, preference is given to x> oso known distribution laws, most often - exponential and normal.

distribution laws. In particular, for b = 1 it turns into an exponential law, for b = 2 - into Rayleigh's law, with b = 3.25 - it is close to normal. This circumstance makes it possible to use one and the same mathematical apparatus in the study of the most diverse flows of product failures. Moreover, this

In a number of studies, it is argued that for failures of technical products due to wear, fatigue, corrosion and aging, a normal or logarithmically normal distribution law will be quite satisfactory, while in the case of sudden failures arising from random overloads, accidents, etc., an exponential law is suitable. distribution law.

The universality of this law is explained by the fact that for different values ​​of the parameter b it approaches a number of distribution laws. In particular, when b = it turns into an exponential law, when 6 = 2 - into Rayleigh's law, when b = 3.25 - it is close to normal.

In this example, we considered the simplest case of Poisson input stream, exponential service time, one server installation. In fact, in reality, distributions are much more complicated, and gas stations include a larger number of gas stations. In order to streamline the classification of queuing systems, the American mathematician D. Kendall proposed a convenient notation system that has become widespread by now. Kendall designated the type of queuing system using three symbols, the first of which describes the type of input stream, the second - the type of probabilistic description of the service system, and the third - the number of serving devices. The symbol M denoted the Poisson distribution of the input stream (with an exponential distribution of intervals between claims), the same symbol was used for the exponential distribution of the service duration. Thus, the queuing system described and studied in this section has the designation M / M / 1. M / G / 3 system, for example, stands for a system with Poisson input stream, general (in English - general) service time distribution function and three servers. There are also other notations: D is a deterministic distribution of intervals between claims or service durations, E is an Erlang distribution of order n, etc.

Based on the methods outlined here for constructing sequences of random numbers with different distributions, it is possible to construct the procedures randl and rand2, which were used in the Algol program for calculations on the gas station model. If the used random intervals between cars and the duration of service have an exponential distribution, then it is better to use the method of inverse functions, and if there is some empirical distribution, then a method based on storing discrete values ​​in the computer's RAM.

Let's move on to the description of the car service time. Since drivers take different amounts of gasoline and differ in skill, the service time can hardly be considered constant. Let the probability that the service of a car, which is at a gas station at any moment t, will be completed in a small interval U, f + rJ, is approximately equal to JLIT, where u> 0. The probability that the service will not end during this period of time is considered to be approximately equal to 1 - ct, and the probability that the service will be completed is. bath of two or more cars - negligible value. Then

where λ Is a constant positive value.

From expression (3.1), it follows that the exponential distribution is determined by one parameter λ.

This feature of the exponential distribution points to his advantage over distributions , depending on a larger number of parameters. Parameters are usually unknown and we have to find their estimates (approximate values), of course, it's easier to estimate one parameter than two or three, etc. . An example of a continuous random variable distributed according to the exponential law , the time between occurrences of two consecutive events of the simplest flow can serve.

Let us find the distribution function of the exponential law .

so

The graphs of the density and the distribution function of the exponential law are shown in Fig. 3.1.

Considering that we get:

Function values ​​can be found from the table.

Numerical characteristics of exponential distribution

Let a continuous random variableΧ distributed according to the exponential law

Find the expected value , using the formula for its calculation for a continuous random variable:

Hence:

Find the standard deviation , for which we extract the square root of the variance:

Comparing (3.4), (3.5), and (3.6), we see that

i.e.the mathematical expectation and the standard deviation of the exponential distribution are equal to each other.

The exponential distribution is widely used in various applications of financial and technical problems, for example, in the theory of reliability.

4. Chi-square distribution and Student's t-distribution.

4.1 Chi-square distribution (- distribution)

Let Χ i (ί = 1, 2, ..., n) be normal independent random variables , and the mathematical expectation of each of them is equal to zero , a standard deviation - unit .

Then the sum of the squares of these quantities

distributed by lawWithdegrees of freedom , if these quantities are related by one linear relationship, for example, then the number of degrees of freedom

The chi-square distribution is widely used in mathematical statistics.

The density of this distribution

where is the gamma function, in particular.

This shows that the chi-square distribution is determined by one parameter - number of degrees of freedomk.

As the number of degrees of freedom increases, the chi-square distribution slowly approaches normal.

The chi-square distribution is obtained if we take in the Erlang distribution law λ = ½ and k = n /2 – 1.

The mathematical expectation and variance of a random variable with a chi-squared distribution, are determined by simple formulas, which we present without derivation:

It follows from the formula that atthe chi-squared distribution coincides with the exponential distribution atλ = ½ .

The cumulative distribution function for the chi-square distribution is determined through special incomplete tabulated gamma functions

Figure 4.1. are given plots of the probability density and the distribution function of a random variable having a chi-square distribution for n = 4, 6, 10.

Figure 4.1. a ) Probability density plots for the chi-squared distribution

Figure 4.1. b) Plots of the distribution function for the chi-squared distribution

4.2 Student distribution

Let Z be a normal random variable, and

a V Is a quantity independent of Z, which is distributed according to the chi-square law withk degrees of freedom. Then magnitude:

has a distribution calledt -distribution or Student's distribution (pseudonym of the English statistician V. Gosset),

Withk = n - 1 degrees of freedom (n - the size of the statistical sample when solving statistical problems).

so , the ratio of the normalized normal value to the square root of an independent random variable distributed according to the chi-square law with k degrees of freedom , divided by k, distributed according to Student's law with k degrees of freedom.

Distribution density of Student:

We note here the basic concepts and formulas related to the exponential distribution of a continuous random variable $ X $ without going into the details of their derivation.

Definition 1

An exponential or exponential distribution of a continuous random variable $ X $ is a distribution whose density has the form:

Picture 1.

The exponential distribution density graph has the form (Fig. 1):

Figure 2. Plot of exponential distribution density.

Exponential distribution function

As it is easy to check, the exponential distribution function has the form:

Figure 3.

where $ \ gamma $ is a positive constant.

The graph of the exponential distribution function is as follows:

Figure 4. Graph of the exponential distribution function.

The probability of hitting a random variable with an exponential distribution

The probability of a continuous random variable falling into the interval $ (\ alpha, \ beta) $ with an exponential distribution is calculated by the following formula:

Expectation: $ M \ left (X \ right) = \ frac (1) (\ gamma). $

Variance: $ D \ left (X \ right) = \ frac (1) ((\ gamma) ^ 2). $

Standard deviation: $ \ sigma \ left (X \ right) = \ frac (1) (\ gamma) $.

An example of an exponential distribution problem

Example 1

The random variable $ X $ obeys an exponential distribution law. In the region of definition $ \ left \