The distribution function contains complete information about the random variable. In practice, the allocation function cannot always be established; sometimes such exhaustive knowledge is not required. Partial information about a random variable is given by numerical characteristics, which, depending on the type of information, are divided into the following groups.
1. Characteristics of the position of a random variable on the numerical axis (mode Mo, median Me, expected value M(X)).
2. Characteristics of the spread of a random variable around the mean value (dispersion D(X), standard deviation σ( X)).
3. Characteristics of the curve shape y = φ( x) (asymmetry As, kurtosis Ex).
Let's take a closer look at each of these characteristics.
Expected value
random variable X indicates some average value around which all possible values are grouped X. For a discrete random variable that can take only a finite number of possible values, the mathematical expectation is the sum of the products of all possible values of the random variable and the probability of these values:
. (2.4)
For a continuous random variable X, which has a given distribution density φ( x) the mathematical expectation is the following integral: 
. (2.5)
Here it is assumed that the improper integral converges absolutely, i.e. exist.
Properties of mathematical expectation:
1. M(S) = C, where With = const;
2. M(C∙X) = C∙M(X);
3. M(X ± Y) = M(X) ± M(Y), where X and Y– any random variables;
4. M(X∙Y)=M(X)∙M(Y), where X and Y are independent random variables.
Two random variables are called independent
, if the distribution law of one of them does not depend on what possible values the other value has taken.
Fashion
discrete random variable, denoted Mo, its most probable value is called (Fig. 2.3), and the mode of a continuous random variable is the value at which the probability density is maximum (Fig. 2.4). 
Rice. 2.3 Fig. 2.4
Median
continuous random variable X its value Me is called such, for which it is equally probable whether the random variable will turn out to be less or more Me, i.e.
P(X < Me) = P(X > Me)
From the definition of the median, it follows that P(X<Me) = 0.5, i.e. F (Me) = 0.5. Geometrically, the median can be interpreted as the abscissa, in which the ordinate φ( x) bisects the area bounded by the distribution curve (Fig. 2.5). In the case of a symmetrical distribution, the median coincides with the mode and the mathematical expectation (Fig. 2.6). 
Rice. 2.5 Fig. 2.6
Dispersion.
Variance of a random variable- a measure of the spread of a given random variable, that is, its deviation from the mathematical expectation. Denoted D[X] in Russian literature and (eng. variance) in foreign countries. In statistics, the designation or is often used. The square root of the variance, equal to , is called the standard deviation, standard deviation, or standard spread. The standard deviation is measured in the same units as the random variable itself, and the variance is measured in the squares of that unit.
It follows from Chebyshev's inequality that a random variable moves away from its mathematical expectation by more than k standard deviations with probability less than 1/ k². So, for example, in at least 75% of cases, a random variable is removed from its mean by no more than two standard deviations, and in about 89% - by no more than three.
dispersion
random variable is called the mathematical expectation of the square of its deviation from the mathematical expectation
D(X) = M(X –M(X)) 2 .
Variance of a random variable X it is convenient to calculate by the formula:
a) for a discrete quantity
; (2.6)
b) for a continuous random variable
j( X)d x – 2 . (2.7)
The dispersion has the following properties:
1. D(C) = 0, where With = const;
2. D(C× X) = C 2 ∙ D(X);
3. D(X± Y) = D(X) + D(Y), if X and Y independent random variables.
Standard deviation
random variable X is called the arithmetic root of the variance, i.e.
σ( X) = .
Note that the dimension σ( X) coincides with the dimension of the random variable itself X, so the standard deviation is more convenient for scattering characterization.
A generalization of the main numerical characteristics of random variables is the concept of moments of a random variable.
The initial moment of the kth order
α k random variable X is called the mathematical expectation of the quantity X k, i.e. α k = M(X k).
The initial moment of the first order is the mathematical expectation of the random variable.
The central moment of the kth order
μ k random variable X is called the mathematical expectation of the quantity ( X–M(X))k, i.e. μ k = M(X–M(X))k.
The central moment of the second order is the variance of the random variable.
For a discrete random variable, the initial moment is expressed by the sum α k= , and the central one is the sum μ k = 
where p i = p(X=x i). For the initial and central moments of a continuous random variable, the following equalities can be obtained:
α k = , μ k = 
,
where φ( x) is the distribution density of the random variable X.
Value As= μ 3 / σ 3 is called asymmetry coefficient
.
If the asymmetry coefficient is negative, then this indicates a large influence on the value of m 3 negative deviations. In this case, the distribution curve (Fig. 2.7) is more flat to the left of M(X). If the coefficient As is positive, which means that the influence of positive deviations prevails, then the distribution curve (Fig. 2.7) is flatter on the right. In practice, the sign of the asymmetry is determined by the location of the distribution curve relative to the mode (maximum point of the differential function). 
Rice. 2.7
kurtosis
Ek is called the quantity
Ek\u003d μ 4 / σ 4 - 3.
Question 24: Correlation
Correlation (correlation dependence) - statistical relationship of two or more random variables (or variables that can be considered as such with some acceptable degree of accuracy). In this case, changes in the values of one or more of these quantities are accompanied by a systematic change in the values of another or other quantities. The mathematical measure of the correlation of two random variables is correlation relation, or correlation coefficient (or ) . If a change in one random variable does not lead to a regular change in another random variable, but leads to a change in another statistical characteristic of this random variable, then such a relationship is not considered a correlation, although it is statistical.
For the first time, the term “correlation” was introduced into scientific circulation by the French paleontologist Georges Cuvier in the 18th century. He developed the "law of correlation" of parts and organs of living beings, with the help of which it is possible to restore the appearance of a fossil animal, having at its disposal only a part of its remains. In statistics, the word "correlation" was first used by the English biologist and statistician Francis Galton at the end of the 19th century.
Some types of correlation coefficients can be positive or negative (it is also possible that there is no statistical relationship - for example, for independent random variables). If it is assumed that a strict order relation is given on the values of the variables, then negative correlation- correlation, in which an increase in one variable is associated with a decrease in another variable, while the correlation coefficient can be negative; positive correlation in such conditions, a correlation in which an increase in one variable is associated with an increase in another variable, while the correlation coefficient can be positive.
Probability theory is a special branch of mathematics that is studied only by students of higher educational institutions. Do you love calculations and formulas? Are you not afraid of the prospects of acquaintance with the normal distribution, the entropy of the ensemble, the mathematical expectation and the variance of a discrete random variable? Then this subject will be of great interest to you. Let's get acquainted with some of the most important basic concepts of this section of science.
Let's remember the basics
Even if you remember the simplest concepts of probability theory, do not neglect the first paragraphs of the article. The fact is that without a clear understanding of the basics, you will not be able to work with the formulas discussed below.
So, there is some random event, some experiment. As a result of the actions performed, we can get several outcomes - some of them are more common, others less common. The probability of an event is the ratio of the number of actually obtained outcomes of one type to the total number of possible ones. Only knowing the classical definition of this concept, you can begin to study the mathematical expectation and dispersion of continuous random variables.
Average
Back in school, in mathematics lessons, you started working with the arithmetic mean. This concept is widely used in probability theory, and therefore it cannot be ignored. The main thing for us at the moment is that we will encounter it in the formulas for the mathematical expectation and variance of a random variable.
We have a sequence of numbers and want to find the arithmetic mean. All that is required of us is to sum everything available and divide by the number of elements in the sequence. Let we have numbers from 1 to 9. The sum of the elements will be 45, and we will divide this value by 9. Answer: - 5.
Dispersion
In scientific terms, variance is the average square of the deviations of the obtained feature values from the arithmetic mean. One is denoted by a capital Latin letter D. What is needed to calculate it? For each element of the sequence, we calculate the difference between the available number and the arithmetic mean and square it. There will be exactly as many values as there can be outcomes for the event we are considering. Next, we summarize everything received and divide by the number of elements in the sequence. If we have five possible outcomes, then divide by five.
The variance also has properties that you need to remember in order to apply it when solving problems. For example, if the random variable is increased by X times, the variance increases by X times the square (i.e., X*X). It is never less than zero and does not depend on shifting values by an equal value up or down. Also, for independent trials, the variance of the sum is equal to the sum of the variances.
Now we definitely need to consider examples of the variance of a discrete random variable and the mathematical expectation.
Let's say we run 21 experiments and get 7 different outcomes. We observed each of them, respectively, 1,2,2,3,4,4 and 5 times. What will be the variance?
First, we calculate the arithmetic mean: the sum of the elements, of course, is 21. We divide it by 7, getting 3. Now we subtract 3 from each number in the original sequence, square each value, and add the results together. It turns out 12. Now it remains for us to divide the number by the number of elements, and, it would seem, that's all. But there is a catch! Let's discuss it.
Dependence on the number of experiments
It turns out that when calculating the variance, the denominator can be one of two numbers: either N or N-1. Here N is the number of experiments performed or the number of elements in the sequence (which is essentially the same thing). What does it depend on?
If the number of tests is measured in hundreds, then we must put N in the denominator. If in units, then N-1. The scientists decided to draw the border quite symbolically: today it runs along the number 30. If we conducted less than 30 experiments, then we will divide the amount by N-1, and if more, then by N.
Task
Let's go back to our example of solving the variance and expectation problem. We got an intermediate number of 12, which had to be divided by N or N-1. Since we conducted 21 experiments, which is less than 30, we will choose the second option. So the answer is: the variance is 12 / 2 = 2.
Expected value
Let's move on to the second concept, which we must consider in this article. The mathematical expectation is the result of adding all possible outcomes multiplied by the corresponding probabilities. It is important to understand that the resulting value, as well as the result of calculating the variance, is obtained only once for the whole task, no matter how many outcomes it considers.
The mathematical expectation formula is quite simple: we take the outcome, multiply it by its probability, add the same for the second, third result, etc. Everything related to this concept is easy to calculate. For example, the sum of mathematical expectations is equal to the mathematical expectation of the sum. The same is true for the work. Not every quantity in probability theory allows such simple operations to be performed. Let's take a task and calculate the value of two concepts we have studied at once. In addition, we were distracted by theory - it's time to practice.
One more example
We ran 50 trials and got 10 kinds of outcomes - numbers 0 to 9 - appearing in varying percentages. These are, respectively: 2%, 10%, 4%, 14%, 2%, 18%, 6%, 16%, 10%, 18%. Recall that to get the probabilities, you need to divide the percentage values by 100. Thus, we get 0.02; 0.1 etc. Let us present an example of solving the problem for the variance of a random variable and the mathematical expectation.
We calculate the arithmetic mean using the formula that we remember from elementary school: 50/10 = 5.
Now let's translate the probabilities into the number of outcomes "in pieces" to make it more convenient to count. We get 1, 5, 2, 7, 1, 9, 3, 8, 5 and 9. Subtract the arithmetic mean from each value obtained, after which we square each of the results obtained. See how to do this with the first element as an example: 1 - 5 = (-4). Further: (-4) * (-4) = 16. For other values, do these operations yourself. If you did everything right, then after adding everything you get 90.
Let's continue calculating the variance and mean by dividing 90 by N. Why do we choose N and not N-1? That's right, because the number of experiments performed exceeds 30. So: 90/10 = 9. We got the dispersion. If you get a different number, don't despair. Most likely, you made a banal error in the calculations. Double-check what you wrote, and for sure everything will fall into place.
Finally, let's recall the mathematical expectation formula. We will not give all the calculations, we will only write the answer with which you can check after completing all the required procedures. The expected value will be 5.48. We only recall how to carry out operations, using the example of the first elements: 0 * 0.02 + 1 * 0.1 ... and so on. As you can see, we simply multiply the value of the outcome by its probability.
Deviation
Another concept closely related to dispersion and mathematical expectation is the standard deviation. It is denoted either by the Latin letters sd, or by the Greek lowercase "sigma". This concept shows how, on average, values deviate from the central feature. To find its value, you need to calculate the square root of the variance.
If you plot a normal distribution and want to see the squared deviation directly on it, this can be done in several steps. Take half of the image to the left or right of the mode (central value), draw a perpendicular to the horizontal axis so that the areas of the resulting figures are equal. The value of the segment between the middle of the distribution and the resulting projection on the horizontal axis will be the standard deviation.
Software
As can be seen from the descriptions of the formulas and the examples presented, calculating the variance and mathematical expectation is not the easiest procedure from an arithmetic point of view. In order not to waste time, it makes sense to use the program used in higher education - it is called "R". It has functions that allow you to calculate values for many concepts from statistics and probability theory.
For example, you define a vector of values. This is done as follows: vector<-c(1,5,2…). Теперь, когда вам потребуется посчитать какие-либо значения для этого вектора, вы пишете функцию и задаете его в качестве аргумента. Для нахождения дисперсии вам нужно будет использовать функцию var. Пример её использования: var(vector). Далее вы просто нажимаете «ввод» и получаете результат.
Finally
Dispersion and mathematical expectation are without which it is difficult to calculate anything in the future. In the main course of lectures at universities, they are considered already in the first months of studying the subject. It is precisely because of the lack of understanding of these simple concepts and the inability to calculate them that many students immediately begin to fall behind in the program and later receive poor marks in the session, which deprives them of scholarships.
Practice at least one week for half an hour a day, solving tasks similar to those presented in this article. Then, on any probability theory test, you will cope with examples without extraneous tips and cheat sheets.
Each individual value is completely determined by its distribution function. Also, to solve practical problems, it is enough to know several numerical characteristics, thanks to which it becomes possible to present the main features of a random variable in a concise form.
These quantities are primarily expected value and dispersion .
Expected value- the average value of a random variable in probability theory. Designated as .
In the simplest way, the mathematical expectation of a random variable X(w), are found as integralLebesgue with respect to the probability measure R original probability space
You can also find the mathematical expectation of a value as Lebesgue integral from X by probability distribution R X quantities X:
where is the set of all possible values X.
Mathematical expectation of functions from a random variable X is through distribution R X. for example, if X- random variable with values in and f(x)- unambiguous Borelfunction X , then:
If a F(x)- distribution function X, then the mathematical expectation is representable integralLebesgue - Stieltjes (or Riemann - Stieltjes):
while the integrability X in what sense ( * ) corresponds to the finiteness of the integral
In specific cases, if X has a discrete distribution with probable values x k, k=1, 2, . , and probabilities , then
if X has an absolutely continuous distribution with a probability density p(x), then
in this case, the existence of a mathematical expectation is equivalent to the absolute convergence of the corresponding series or integral.
Properties of the mathematical expectation of a random variable.
- The mathematical expectation of a constant value is equal to this value:
C- constant;
- M=C.M[X]
- The mathematical expectation of the sum of randomly taken values is equal to the sum of their mathematical expectations:
- The mathematical expectation of the product of independent random variables = the product of their mathematical expectations:
M=M[X]+M[Y]
if X and Y independent.
if the series converges:
Algorithm for calculating the mathematical expectation.
Properties of discrete random variables: all their values can be renumbered by natural numbers; equate each value with a non-zero probability.
1. Multiply the pairs in turn: x i on the pi.
2. Add the product of each pair x i p i.
For example, for n = 4 :
Distribution function of a discrete random variable stepwise, it increases abruptly at those points whose probabilities have a positive sign.
Example: Find the mathematical expectation by the formula.
The mathematical expectation (average value) of a random variable X , given on a discrete probability space, is the number m =M[X]=∑x i p i , if the series converges absolutely.
Service assignment. With an online service the mathematical expectation, variance and standard deviation are calculated(see example). In addition, a graph of the distribution function F(X) is plotted.
Properties of the mathematical expectation of a random variable
- The mathematical expectation of a constant value is equal to itself: M[C]=C , C is a constant;
- M=C M[X]
- The mathematical expectation of the sum of random variables is equal to the sum of their mathematical expectations: M=M[X]+M[Y]
- The mathematical expectation of the product of independent random variables is equal to the product of their mathematical expectations: M=M[X] M[Y] if X and Y are independent.
Dispersion Properties
- The dispersion of a constant value is equal to zero: D(c)=0.
- The constant factor can be taken out from under the dispersion sign by squaring it: D(k*X)= k 2 D(X).
- If random variables X and Y are independent, then the variance of the sum is equal to the sum of the variances: D(X+Y)=D(X)+D(Y).
- If random variables X and Y are dependent: D(X+Y)=DX+DY+2(X-M[X])(Y-M[Y])
- For the variance, the computational formula is valid:
D(X)=M(X 2)-(M(X)) 2
Example. The mathematical expectations and variances of two independent random variables X and Y are known: M(x)=8 , M(Y)=7 , D(X)=9 , D(Y)=6 . Find the mathematical expectation and variance of the random variable Z=9X-8Y+7 .
Decision. Based on the properties of mathematical expectation: M(Z) = M(9X-8Y+7) = 9*M(X) - 8*M(Y) + M(7) = 9*8 - 8*7 + 7 = 23 .
Based on the dispersion properties: D(Z) = D(9X-8Y+7) = D(9X) - D(8Y) + D(7) = 9^2D(X) - 8^2D(Y) + 0 = 81*9 - 64*6 = 345
Algorithm for calculating the mathematical expectation
Properties of discrete random variables: all their values can be renumbered by natural numbers; Assign each value a non-zero probability.- Multiply the pairs one by one: x i by p i .
- We add the product of each pair x i p i .
For example, for n = 4: m = ∑x i p i = x 1 p 1 + x 2 p 2 + x 3 p 3 + x 4 p 4
Example #1.
| x i | 1 | 3 | 4 | 7 | 9 |
| pi | 0.1 | 0.2 | 0.1 | 0.3 | 0.3 |
The mathematical expectation is found by the formula m = ∑x i p i .
Mathematical expectation M[X].
M[x] = 1*0.1 + 3*0.2 + 4*0.1 + 7*0.3 + 9*0.3 = 5.9
The dispersion is found by the formula d = ∑x 2 i p i - M[x] 2 .
Dispersion D[X].
D[X] = 1 2 *0.1 + 3 2 *0.2 + 4 2 *0.1 + 7 2 *0.3 + 9 2 *0.3 - 5.9 2 = 7.69
Standard deviation σ(x).
σ = sqrt(D[X]) = sqrt(7.69) = 2.78
Example #2. A discrete random variable has the following distribution series:
| X | -10 | -5 | 0 | 5 | 10 |
| R | a | 0,32 | 2a | 0,41 | 0,03 |
Decision. The value a is found from the relationship: Σp i = 1
Σp i = a + 0.32 + 2 a + 0.41 + 0.03 = 0.76 + 3 a = 1
0.76 + 3 a = 1 or 0.24=3 a , whence a = 0.08
Example #3. Determine the distribution law of a discrete random variable if its variance is known, and x 1
p 1 =0.3; p2=0.3; p3=0.1; p 4 \u003d 0.3
d(x)=12.96
Decision.
Here you need to make a formula for finding the variance d (x) :
d(x) = x 1 2 p 1 +x 2 2 p 2 +x 3 2 p 3 +x 4 2 p 4 -m(x) 2
where expectation m(x)=x 1 p 1 +x 2 p 2 +x 3 p 3 +x 4 p 4
For our data
m(x)=6*0.3+9*0.3+x 3 *0.1+15*0.3=9+0.1x 3
12.96 = 6 2 0.3+9 2 0.3+x 3 2 0.1+15 2 0.3-(9+0.1x 3) 2
or -9/100 (x 2 -20x+96)=0
Accordingly, it is necessary to find the roots of the equation, and there will be two of them.
x 3 \u003d 8, x 3 \u003d 12
We choose the one that satisfies the condition x 1
Distribution law of a discrete random variable
x 1 =6; x2=9; x 3 \u003d 12; x4=15
p 1 =0.3; p2=0.3; p3=0.1; p 4 \u003d 0.3
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. In many problems of practice, a complete, exhaustive description of a random variable - the law of distribution - either cannot be obtained at all, or is not needed at all. In these cases, they are limited to an approximate description of a random variable using numerical characteristics.
The mathematical expectation is often referred to simply as the average value of a random variable. Dispersion of a random variable is a characteristic of dispersion, dispersion of a random variable around its mathematical expectation.
Mathematical expectation of a discrete random variable
Let's approach the concept of mathematical expectation, first proceeding from the mechanical interpretation of the distribution of a discrete random variable. Let the unit mass be distributed between the points of the x-axis x1 , x 2 , ..., x n, and each material point has a mass corresponding to it from 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?
Decision. 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.
Decision. 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:
| Number | Profit 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.
Decision. 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.
Decision. 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 1 | Project 2 | Project 3 | Project 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.
Decision. 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.
Decision. 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.
Decision. 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 .
