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Mathematical expectation and variance are the most commonly used numerical characteristics of a random variable. They characterize the most important features of the distribution: its position and degree of dispersion. The mathematical expectation is often referred to simply as the mean. random variable. Dispersion of a random variable - a characteristic of dispersion, dispersion of a random variable around its mathematical expectation.

In many problems of practice, a complete, exhaustive description of a random variable - the law of distribution - either cannot be obtained, or is not needed at all. In these cases, they are limited to an approximate description of a random variable using numerical characteristics.

Mathematical expectation of a discrete random variable

Let's come to the concept of mathematical expectation. Let the mass of some substance be distributed between the points of the x-axis x1 , x 2 , ..., x n. Moreover, each material point has a mass corresponding to it with a probability of p1 , p 2 , ..., p n. It is required to choose one point on the x-axis, which characterizes the position of the entire system of material points, taking into account their masses. It is natural to take the center of mass of the system of material points as such a point. This is the weighted average of the random variable X, in which the abscissa of each point xi enters with a "weight" equal to the corresponding probability. The mean value of the random variable thus obtained X is called its mathematical expectation.

The mathematical expectation of a discrete random variable is the sum of the products of all its possible values ​​and the probabilities of these values:

Example 1 A win-win lottery was organized. There are 1000 winnings, 400 of which are 10 rubles each. 300 - 20 rubles each 200 - 100 rubles each. and 100 - 200 rubles each. What is the average winnings for a person who buys one ticket?

Solution. We will find the average win if the total amount of winnings, which is equal to 10*400 + 20*300 + 100*200 + 200*100 = 50,000 rubles, is divided by 1000 (the total amount of winnings). Then we get 50000/1000 = 50 rubles. But the expression for calculating the average gain can also be represented in the following form:

On the other hand, under these conditions, the amount of winnings is a random variable that can take on the values ​​of 10, 20, 100 and 200 rubles. with probabilities equal to 0.4, respectively; 0.3; 0.2; 0.1. Therefore, the expected average payoff is equal to the sum of the products of the size of the payoffs and the probability of receiving them.

Example 2 The publisher decided to publish a new book. He is going to sell the book for 280 rubles, of which 200 will be given to him, 50 to the bookstore, and 30 to the author. The table gives information about the cost of publishing a book and the likelihood of selling a certain number of copies of the book.

Find the publisher's expected profit.

Solution. The random variable "profit" is equal to the difference between the income from the sale and the cost of the costs. For example, if 500 copies of a book are sold, then the income from the sale is 200 * 500 = 100,000, and the cost of publishing is 225,000 rubles. Thus, the publisher faces a loss of 125,000 rubles. The following table summarizes the expected values ​​of the random variable - profit:

NumberProfit xi Probability pi xi p i
500 -125000 0,20 -25000
1000 -50000 0,40 -20000
2000 100000 0,25 25000
3000 250000 0,10 25000
4000 400000 0,05 20000
Total: 1,00 25000

Thus, we obtain the mathematical expectation of the publisher's profit:

.

Example 3 Chance to hit with one shot p= 0.2. Determine the consumption of shells that provide the mathematical expectation of the number of hits equal to 5.

Solution. From the same expectation formula that we have used so far, we express x- consumption of shells:

.

Example 4 Determine the mathematical expectation of a random variable x number of hits with three shots, if the probability of hitting with each shot p = 0,4 .

Hint: find the probability of the values ​​of a random variable by Bernoulli formula .

Expectation Properties

Consider the properties of mathematical expectation.

Property 1. The mathematical expectation of a constant value is equal to this constant:

Property 2. The constant factor can be taken out of the expectation sign:

Property 3. The mathematical expectation of the sum (difference) of random variables is equal to the sum (difference) of their mathematical expectations:

Property 4. The mathematical expectation of the product of random variables is equal to the product of their mathematical expectations:

Property 5. If all values ​​of the random variable X decrease (increase) by the same number WITH, then its mathematical expectation will decrease (increase) by the same number:

When you can not be limited only to mathematical expectation

In most cases, only the mathematical expectation cannot adequately characterize a random variable.

Let random variables X And Y are given by the following distribution laws:

Meaning X Probability
-0,1 0,1
-0,01 0,2
0 0,4
0,01 0,2
0,1 0,1
Meaning Y Probability
-20 0,3
-10 0,1
0 0,2
10 0,1
20 0,3

The mathematical expectations of these quantities are the same - equal to zero:

However, their distribution is different. Random value X can only take values ​​that are little different from the mathematical expectation, and the random variable Y can take values ​​that deviate significantly from the mathematical expectation. A similar example: the average wage does not make it possible to judge the proportion of high- and low-paid workers. In other words, by mathematical expectation one cannot judge what deviations from it, at least on average, are possible. To do this, you need to find the variance of a random variable.

Dispersion of a discrete random variable

dispersion discrete random variable X is called the mathematical expectation of the square of its deviation from the mathematical expectation:

The standard deviation of a random variable X is the arithmetic value of the square root of its variance:

.

Example 5 Calculate variances and standard deviations of random variables X And Y, whose distribution laws are given in the tables above.

Solution. Mathematical expectations of random variables X And Y, as found above, are equal to zero. According to the dispersion formula for E(X)=E(y)=0 we get:

Then the standard deviations of random variables X And Y constitute

.

Thus, with the same mathematical expectations, the variance of the random variable X very small and random Y- significant. This is a consequence of the difference in their distribution.

Example 6 The investor has 4 alternative investment projects. The table summarizes the data on the expected profit in these projects with the corresponding probability.

Project 1Project 2Project 3Project 4
500, P=1 1000, P=0,5 500, P=0,5 500, P=0,5
0, P=0,5 1000, P=0,25 10500, P=0,25
0, P=0,25 9500, P=0,25

Find for each alternative the mathematical expectation, variance and standard deviation.

Solution. Let us show how these quantities are calculated for the 3rd alternative:

The table summarizes the found values ​​for all alternatives.

All alternatives have the same mathematical expectation. This means that in the long run everyone has the same income. The standard deviation can be interpreted as a measure of risk - the larger it is, the greater the risk of the investment. An investor who doesn't want much risk will choose project 1 because it has the smallest standard deviation (0). If the investor prefers risk and high returns in a short period, then he will choose the project with the largest standard deviation - project 4.

Dispersion Properties

Let us present the properties of the dispersion.

Property 1. The dispersion of a constant value is zero:

Property 2. The constant factor can be taken out of the dispersion sign by squaring it:

.

Property 3. The variance of a random variable is equal to the mathematical expectation of the square of this value, from which the square of the mathematical expectation of the value itself is subtracted:

,

Where .

Property 4. The variance of the sum (difference) of random variables is equal to the sum (difference) of their variances:

Example 7 It is known that a discrete random variable X takes only two values: −3 and 7. In addition, the mathematical expectation is known: E(X) = 4 . Find the variance of a discrete random variable.

Solution. Denote by p the probability with which a random variable takes on a value x1 = −3 . Then the probability of the value x2 = 7 will be 1 − p. Let's derive the equation for mathematical expectation:

E(X) = x 1 p + x 2 (1 − p) = −3p + 7(1 − p) = 4 ,

where we get the probabilities: p= 0.3 and 1 − p = 0,7 .

The law of distribution of a random variable:

X −3 7
p 0,3 0,7

We calculate the variance of this random variable using the formula from property 3 of the variance:

D(X) = 2,7 + 34,3 − 16 = 21 .

Find the mathematical expectation of a random variable yourself, and then see the solution

Example 8 Discrete random variable X takes only two values. It takes the larger value of 3 with a probability of 0.4. In addition, the variance of the random variable is known D(X) = 6 . Find the mathematical expectation of a random variable.

Example 9 An urn contains 6 white and 4 black balls. 3 balls are taken from the urn. The number of white balls among the drawn balls is a discrete random variable X. Find the mathematical expectation and variance of this random variable.

Solution. Random value X can take the values ​​0, 1, 2, 3. The corresponding probabilities can be calculated from rule of multiplication of probabilities. The law of distribution of a random variable:

X 0 1 2 3
p 1/30 3/10 1/2 1/6

Hence the mathematical expectation of this random variable:

M(X) = 3/10 + 1 + 1/2 = 1,8 .

The variance of a given random variable is:

D(X) = 0,3 + 2 + 1,5 − 3,24 = 0,56 .

Mathematical expectation and dispersion of a continuous random variable

For a continuous random variable, the mechanical interpretation of the mathematical expectation will retain the same meaning: the center of mass for a unit mass distributed continuously on the x-axis with density f(x). In contrast to a discrete random variable, for which the function argument xi changes abruptly, for a continuous random variable, the argument changes continuously. But the mathematical expectation of a continuous random variable is also related to its mean value.

To find the mathematical expectation and variance of a continuous random variable, you need to find definite integrals . If a density function of a continuous random variable is given, then it enters directly into the integrand. If a probability distribution function is given, then by differentiating it, you need to find the density function.

The arithmetic average of all possible values ​​of a continuous random variable is called its mathematical expectation, denoted by or .

Expected value

Dispersion continuous random variable X, the possible values ​​of which belong to the entire axis Ox, is determined by the equality:

Service assignment. The online calculator is designed to solve problems in which either distribution density f(x) , or distribution function F(x) (see example). Usually in such tasks it is required to find mathematical expectation, standard deviation, plot the functions f(x) and F(x).

Instruction. Select the type of input data: distribution density f(x) or distribution function F(x) .

The distribution density f(x) is given:

The distribution function F(x) is given:

A continuous random variable is defined by a probability density
(Rayleigh distribution law - used in radio engineering). Find M(x) , D(x) .

The random variable X is called continuous , if its distribution function F(X)=P(X< x) непрерывна и имеет производную.
The distribution function of a continuous random variable is used to calculate the probabilities of a random variable falling into a given interval:
P(α< X < β)=F(β) - F(α)
moreover, for a continuous random variable, it does not matter whether its boundaries are included in this interval or not:
P(α< X < β) = P(α ≤ X < β) = P(α ≤ X ≤ β)
Distribution density continuous random variable is called the function
f(x)=F'(x) , derivative of the distribution function.

Distribution Density Properties

1. The distribution density of a random variable is non-negative (f(x) ≥ 0) for all values ​​of x.
2. Normalization condition:

The geometric meaning of the normalization condition: the area under the distribution density curve is equal to one.
3. The probability of hitting a random variable X in the interval from α to β can be calculated by the formula

Geometrically, the probability that a continuous random variable X falls into the interval (α, β) is equal to the area of ​​the curvilinear trapezoid under the distribution density curve based on this interval.
4. The distribution function is expressed in terms of density as follows:

The distribution density value at the point x is not equal to the probability of taking this value; for a continuous random variable, we can only talk about the probability of falling into a given interval. Let )

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