How to determine checkmate expectation. Mathematical expectation formula. Basics of probability theory
Mathematical expectation of a discrete random variable is the sum of the products of all its possible values and their probabilities.
Let a random variable take only probability values which are respectively equal. Then the mathematical expectation of a random variable is determined by the equality
If a discrete random variable takes a countable set of possible values, then
Moreover, the mathematical expectation exists if the series on the right side of the equality converges absolutely.
Comment. From the definition it follows that the mathematical expectation of a discrete random variable is a non-random (constant) quantity.
Definition of mathematical expectation in the general case
Let us determine the mathematical expectation of a random variable whose distribution is not necessarily discrete. Let's start with the case of non-negative random variables. The idea will be to approximate such random variables using discrete ones, for which the mathematical expectation has already been determined, and put the mathematical expectation equal to the limit mathematical expectations of discrete random variables approximating it. By the way, this is a very useful general idea, which is that some characteristic is first determined for simple objects, and then for more complex objects it is determined by approximating them by simpler ones.
Lemma 1. Let there be an arbitrary non-negative random variable. Then there is a sequence of discrete random variables such that
Proof. Let us divide the semi-axis into equal length segments and determine
Then properties 1 and 2 easily follow from the definition of a random variable, and
Lemma 2. Let be a non-negative random variable and and two sequences of discrete random variables possessing properties 1-3 from Lemma 1. Then
Proof. Note that for non-negative random variables we allow
Due to property 3, it is easy to see that there is a sequence positive numbers, such that
It follows that
Using the properties of mathematical expectations for discrete random variables, we obtain
Passing to the limit at we obtain the statement of Lemma 2.
Definition 1. Let be a non-negative random variable, - a sequence of discrete random variables that have properties 1-3 from Lemma 1. The mathematical expectation of a random variable is the number
Lemma 2 guarantees that it does not depend on the choice of approximating sequence.
Let now be an arbitrary random variable. Let's define
From the definition and it easily follows that
Definition 2. The mathematical expectation of an arbitrary random variable is the number
If at least one of the numbers on the right side of this equality is finite.
Properties of mathematical expectation
Property 1. Expected value constant value is equal to the constant itself:
Proof. We will consider a constant as a discrete random variable that has one possible value and takes it with probability, therefore,
Remark 1. Let us define the product of a constant variable by a discrete random variable as a discrete random whose possible values are equal to the products of the constant by the possible values; the probabilities of possible values are equal to the probabilities of the corresponding possible values. For example, if the probability of a possible value is equal then the probability that the value will take the value is also equal
Property 2. The constant factor can be taken out of the sign of the mathematical expectation:
Proof. Let the random variable be given by the probability distribution law:
Taking into account Remark 1, we write the distribution law of the random variable
Remark 2. Before moving on to the next property, we point out that two random variables are called independent if the distribution law of one of them does not depend on what possible values the other variable took. Otherwise, the random variables are dependent. Several random variables are called mutually independent if the laws of distribution of any number of them do not depend on what possible values the remaining variables took.
Remark 3. Let us define the product of independent random variables and as a random variable whose possible values are equal to the products of each possible value by each possible value, the probabilities of the possible values of the product are equal to the products of the probabilities of the possible values of the factors. For example, if the probability of a possible value is, the probability of a possible value is then the probability of a possible value is
Property 3. The mathematical expectation of the product of two independent random variables is equal to the product of their mathematical expectations:
Proof. Let independent random variables be specified by their own probability distribution laws:
Let's compile all the values that a random variable can take. To do this, let's multiply all possible values by each possible value; As a result, we obtain and, taking into account Remark 3, we write the distribution law, assuming for simplicity that all possible values of the product are different (if this is not the case, then the proof is carried out in a similar way):
The mathematical expectation is equal to the sum of the products of all possible values and their probabilities:
Consequence. The mathematical expectation of the product of several mutually independent random variables is equal to the product of their mathematical expectations.
Property 4. The mathematical expectation of the sum of two random variables is equal to the sum of the mathematical expectations of the terms:
Proof. Let random variables and be specified by the following distribution laws:
Let's compile all possible values of a quantity. To do this, we add each possible value to each possible value; we obtain. Let us assume for simplicity that these possible values are different (if this is not the case, then the proof is carried out in a similar way), and we denote their probabilities, respectively, by and
The mathematical expectation of a value is equal to the sum of the products of possible values and their probabilities:
Let us prove that an Event that will take on the value (the probability of this event is equal) entails an event that will take on the value or (the probability of this event by the addition theorem is equal), and vice versa. Hence it follows that the equalities are proved similarly
Substituting the right-hand sides of these equalities into relation (*), we obtain
or finally
Variance and standard deviation
In practice, it is often necessary to estimate the dispersion of possible values of a random variable around its mean value. For example, in artillery it is important to know how closely the shells will fall near the target that is to be hit.
At first glance, it may seem that the easiest way to estimate dispersion is to calculate all possible deviations of a random variable and then find their average value. However, this path will not give anything, since the average value of the deviation, i.e. for any random variable is equal to zero. This property is explained by the fact that some possible deviations are positive, while others are negative; as a result of their mutual cancellation, the average deviation value is zero. These considerations indicate the advisability of replacing possible deviations absolute values or their squares. This is what they do in practice. True, in the case when possible deviations are replaced by absolute values, one has to operate with absolute values, which sometimes leads to serious difficulties. Therefore, most often they take a different path, i.e. calculate the average value of the squared deviation, which is called dispersion.
Mathematical expectation is the definition
Checkmate waiting is one of the most important concepts V mathematical statistics and probability theory, characterizing the distribution of values or probabilities random variable. Typically expressed as a weighted average of all possible parameters of a random variable. Widely used in technical analysis, research number series, the study of continuous and long-term processes. It is important in assessing risks, predicting price indicators when trading on financial markets, and is used in developing strategies and methods of gaming tactics in gambling theories.
Checkmate waiting- This mean value of a random variable, distribution probabilities random variable is considered in probability theory.
Checkmate waiting is a measure of the average value of a random variable in probability theory. Checkmate the expectation of a random variable x denoted by M(x).
Mathematical expectation (Population mean) is
Checkmate waiting is
Checkmate waiting is in probability theory, a weighted average of all possible values that a random variable can take.
Checkmate waiting is the sum of the products of all possible values of a random variable and the probabilities of these values.
Mathematical expectation (Population mean) is
Checkmate waiting is the average benefit from a particular decision, provided that such a decision can be considered within the framework of the theory of large numbers and long distance.
Checkmate waiting is in gambling theory, the amount of winnings that a speculator can earn or lose, on average, on each bet. In the language of gambling speculators this is sometimes called "advantage" speculator" (if it is positive for the speculator) or "house edge" (if it is negative for the speculator).
Mathematical expectation (Population mean) is
There will also be tasks for independent decision, to which you can see the answers.
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 scattering. The expected value is often called simply the average. random variable. Dispersion of a random variable - characteristic of dispersion, spread of a random variable about its mathematical expectation.
In many practical problems, a complete, exhaustive characteristic of a random variable - the distribution law - either cannot be obtained or is not needed at all. In these cases, one is limited to an approximate description of a random variable using numerical characteristics.
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 corresponding mass with a probability of p1 , p 2 , ..., p n. It is required to select one point on the abscissa axis, characterizing the position of the entire system 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, to which the abscissa of each point xi enters with a “weight” equal to the corresponding probability. The average value of the random variable obtained in this way 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 has been organized. There are 1000 winnings, of which 400 are 10 rubles. 300 - 20 rubles each. 200 - 100 rubles each. and 100 - 200 rubles each. What is the average winnings for someone who buys one ticket?
Solution. We will find the average winnings if we divide the total amount of winnings, which is 10*400 + 20*300 + 100*200 + 200*100 = 50,000 rubles, by 1000 (total amount of winnings). Then we get 50000/1000 = 50 rubles. But the expression for calculating the average winnings can be presented in the following form:
On the other hand, in these conditions, the winning size is a random variable, which can take 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 equal to the sum products of the size of winnings and the probability of receiving them.
Example 2. The publisher decided to publish new book. He plans to sell the book for 280 rubles, of which he himself will receive 200, 50 - the bookstore and 30 - the author. The table provides information about the costs of publishing a book and the probability 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 sales and the cost of expenses. 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 publication 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. Probability of hitting with one shot p= 0.2. Determine the consumption of projectiles that provide a mathematical expectation of the number of hits equal to 5.
Solution. From the same mathematical expectation formula that we have used so far, we express x- shell consumption:
.
Example 4. Determine the mathematical expectation of a random variable x number of hits with three shots, if the probability of a hit with each shot p = 0,4 .
Hint: find the probability of random variable values by Bernoulli's formula .
Properties of mathematical expectation
Let's 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 mathematical 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 a product of random variables is equal to the product of their mathematical expectations:
Property 5. If all values of a random variable X decrease (increase) by the same number WITH, then its mathematical expectation will decrease (increase) by the same number:
When you can’t limit yourself only to mathematical expectation
In most cases, only the mathematical expectation cannot sufficiently characterize a random variable.
Let the 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 patterns are different. Random value X can only take values that differ little 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 share of high- and low-paid workers. In other words, one cannot judge from the mathematical expectation what deviations from it, at least on average, are possible. To do this, you need to find the variance of the random variable.
Variance of a discrete random variable
Variance 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 the arithmetic value of the square root of its variance is called:
.
Example 5. Calculate variances and standard deviations of random variables X And Y, the distribution laws of which 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 at E(X)=E(y)=0 we get:
Then the standard deviations of random variables X And Y make up
.
Thus, with the same mathematical expectations, the variance of the random variable X very small, but a random variable Y- significant. This is a consequence of differences in their distribution.
Example 6. The investor has 4 alternative investment projects. The table summarizes 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 the mathematical expectation, variance and standard deviation for each alternative.
Solution. Let us show how these values are calculated for the 3rd alternative:
The table summarizes the found values for all alternatives.
All alternatives have the same mathematical expectations. This means that in the long run everyone has the same income. Standard deviation can be interpreted as a measure of risk - the higher it is, the greater the risk of the investment. An investor who does not want much risk will choose project 1 since 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 dispersion.
Property 1. The variance 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. Let us denote by p the probability with which a random variable takes a value x1 = −3 . Then the probability of the value x2 = 7 will be 1 − p. Let us derive the equation for the 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 .
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 dispersion:
D(X) = 2,7 + 34,3 − 16 = 21 .
Find the mathematical expectation of a random variable yourself, and then look at the solution
Example 8. Discrete random variable X takes only two values. It accepts the greater of the values 3 with probability 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. There are 6 white and 4 black balls in an urn. 3 balls are drawn 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 values 0, 1, 2, 3. The corresponding probabilities can be calculated from probability multiplication rule. 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 .
Expectation and variance 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). Unlike a discrete random variable, whose 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 average value.
To find the mathematical expectation and variance of a continuous random variable, you need to find definite integrals . If the density function of a continuous random variable is given, then it directly enters 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 .
2. Basics of probability theory
Expected value
Consider a random variable with numerical values. It is often useful to associate a number with this function - its "average value" or, as they say, " average value", "index of central tendency". For a number of reasons, some of which will become clear later, the mathematical expectation is usually used as the “average value”.
Definition 3. Mathematical expectation of a random variable X called number
those. the mathematical expectation of a random variable is a weighted sum of the values of a random variable with weights equal to the probabilities of the corresponding elementary events.
Example 6. Let's calculate the mathematical expectation of the number that appears on the top face of the die. It follows directly from Definition 3 that
Statement 2. Let the random variable X takes values x 1, x 2,…, xm. Then the equality is true
(5)
those. The mathematical expectation of a random variable is a weighted sum of the values of the random variable with weights equal to the probabilities that the random variable takes certain values.
Unlike (4), where the summation is carried out directly over elementary events, a random event can consist of several elementary events.
Sometimes relation (5) is taken as the definition of mathematical expectation. However, using Definition 3, as shown below, it is easier to establish the properties of the mathematical expectation necessary for constructing probabilistic models of real phenomena than using relation (5).
To prove relation (5), we group into (4) terms with identical values of the random variable:
Since the constant factor can be taken out of the sign of the sum, then
By determining the probability of an event
Using the last two relations we obtain the required:
The concept of mathematical expectation in probabilistic-statistical theory corresponds to the concept of the center of gravity in mechanics. Let's put it in points x 1, x 2,…, xm on the mass number axis P(X= x 1 ), P(X= x 2 ),…, P(X= x m) respectively. Then equality (5) shows that the center of gravity of this system of material points coincides with the mathematical expectation, which shows the naturalness of Definition 3.
Statement 3. Let X- random value, M(X)– its mathematical expectation, A– a certain number. Then
1) M(a)=a; 2) M(X-M(X))=0; 3M[(X- a) 2 ]= M[(X- M(X)) 2 ]+(a- M(X)) 2 .
To prove this, let us first consider a random variable that is constant, i.e. the function maps the space of elementary events to a single point A. Since the constant multiplier can be taken beyond the sign of the sum, then
If each member of a sum is divided into two terms, then the whole sum is divided into two sums, of which the first is made up of the first terms, and the second is made up of the second. Therefore, the mathematical expectation of the sum of two random variables X+Y, defined on the same space of elementary events, is equal to the sum of mathematical expectations M(X) And M(U) these random variables:
M(X+Y) = M(X) + M(Y).
And therefore M(X-M(X)) = M(X) - M(M(X)). As shown above, M(M(X)) = M(X). Hence, M(X-M(X)) = M(X) - M(X) = 0.
Because the (X - a) 2 = ((X – M(X)) + (M(X) - a)} 2 = (X - M(X)) 2 + 2(X - M(X))(M(X) - a) + (M(X) – a) 2 , That M[(X - a) 2 ] =M(X - M(X)) 2 + M{2(X - M(X))(M(X) - a)} + M[(M(X) – a) 2 ]. Let's simplify the last equality. As shown at the beginning of the proof of Statement 3, the mathematical expectation of a constant is the constant itself, and therefore M[(M(X) – a) 2 ] = (M(X) – a) 2 . Since the constant multiplier can be taken beyond the sign of the sum, then M{2(X - M(X))(M(X) - a)} = 2(M(X) - a)M(X - M(X)). The right side of the last equality is 0 because, as shown above, M(X-M(X))=0. Hence, M[(X- a) 2 ]= M[(X- M(X)) 2 ]+(a- M(X)) 2 , which was what needed to be proven.
From the above it follows that M[(X- a) 2 ] reaches a minimum A, equal M[(X- M(X)) 2 ], at a = M(X), since the second term in equality 3) is always non-negative and equals 0 only for the specified value A.
Statement 4. Let the random variable X takes values x 1, x 2,…, xm, and f is some function of the numerical argument. Then
To prove this, let’s group on the right side of equality (4), which defines the mathematical expectation, terms with the same values:
Using the fact that the constant factor can be taken out of the sign of the sum, and the definition of the probability of a random event (2), we obtain
Q.E.D.
Statement 5. Let X And U– random variables defined on the same space of elementary events, A And b- some numbers. Then M(aX+ bY)= aM(X)+ bM(Y).
Using the definition of the mathematical expectation and the properties of the summation symbol, we obtain a chain of equalities:
The required has been proven.
The above shows how the mathematical expectation depends on the transition to another reference point and to another unit of measurement (transition Y=aX+b), as well as to functions of random variables. The results obtained are constantly used in technical and economic analysis, in assessing the financial and economic activities of an enterprise, during the transition from one currency to another in foreign economic calculations, in regulatory and technical documentation, etc. The results under consideration allow the use of the same calculation formulas for various parameters scale and shift.
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– the number of boys among 10 newborns.
It is absolutely clear that this number is not known in advance, and the next ten children born may include:
Or boys - one and only one from the listed options.
And, in order to keep in shape, a little physical education:
– long jump distance (in some units).
Even a master of sports cannot predict it :)
However, your hypotheses?
2) Continuous random variable – accepts All numeric values from some finite or infinite interval.
Note : V educational literature popular abbreviations DSV and NSV
First, let's analyze the discrete random variable, then - continuous.
Distribution law of a discrete random variable
- This correspondence between possible values of this quantity and their probabilities. Most often, the law is written in a table: 
The term appears quite often row
distribution, but in some situations it sounds ambiguous, and so I will stick to the "law".
And now Very important point
: since the random variable Necessarily will accept one of the values, then the corresponding events form full group and the sum of the probabilities of their occurrence is equal to one:
or, if written condensed:
So, for example, the law of probability distribution of points rolled on a die has the following form: 
No comments.
You may be under the impression that a discrete random variable can only take on “good” integer values. Let's dispel the illusion - they can be anything:
Example 1
Some game has the following winning distribution law: 
...you've probably dreamed of such tasks for a long time :) I'll tell you a secret - me too. Especially after finishing work on field theory.
Solution: since a random variable can take only one of three values, the corresponding events form full group, which means the sum of their probabilities is equal to one: 
Exposing the “partisan”: 
– thus, the probability of winning conventional units is 0.4.
Control: that’s what we needed to make sure of.
Answer:
It is not uncommon when you need to draw up a distribution law yourself. For this they use classical definition of probability, multiplication/addition theorems for event probabilities and other chips tervera:
Example 2
The box contains 50 lottery tickets, among which 12 are winning, and 2 of them win 1000 rubles each, and the rest - 100 rubles each. Draw up a law for the distribution of a random variable - the size of the winnings, if one ticket is drawn at random from the box.
Solution: as you noticed, the values of a random variable are usually placed in in ascending order. Therefore, we start with the smallest winnings, namely rubles.
There are 50 such tickets in total - 12 = 38, and according to classical definition:
– the probability that a randomly drawn ticket will be a loser.
In other cases everything is simple. The probability of winning rubles is:
Check: – and this is a particularly pleasant moment of such tasks!
Answer: the desired law of distribution of winnings: 
The following task is for you to solve on your own:
Example 3
The probability that the shooter will hit the target is . Draw up a distribution law for a random variable - the number of hits after 2 shots.
...I knew that you missed him :) Let's remember multiplication and addition theorems. The solution and answer are at the end of the lesson.
The distribution law completely describes a random variable, but in practice it can be useful (and sometimes more useful) to know only some of it numerical characteristics .
Expectation of a discrete random variable
Speaking in simple language, This average expected value when testing is repeated many times. Let the random variable take values with probabilities 
respectively. Then the mathematical expectation of this random variable is equal to sum of products all its values to the corresponding probabilities:
or collapsed: 
Let us calculate, for example, the mathematical expectation of a random variable - the number of points rolled on a die:
Now let's remember our hypothetical game: 
The question arises: is it profitable to play this game at all? ...who has any impressions? So you can’t say it “offhand”! But this question can be easily answered by calculating the mathematical expectation, essentially - weighted average by probability of winning:
Thus, the mathematical expectation of this game losing.
Don't trust your impressions - trust the numbers!
Yes, here you can win 10 or even 20-30 times in a row, but in the long run, inevitable ruin awaits us. And I wouldn't advise you to play such games :) Well, maybe only for fun.
From all of the above it follows that the mathematical expectation is no longer a RANDOM value.
Creative task for independent research:
Example 4
Mr. X plays European roulette using the following system: he constantly bets 100 rubles on “red”. Draw up a law of distribution of a random variable - its winnings. Calculate the mathematical expectation of winnings and round it to the nearest kopeck. How many average Does the player lose for every hundred he bet?
Reference : European roulette contains 18 red, 18 black and 1 green sector (“zero”). If a “red” appears, the player is paid double the bet, otherwise it goes to the casino’s income
There are many other roulette systems for which you can create your own probability tables. But this is the case when we do not need any distribution laws or tables, because it has been established for certain that the player’s mathematical expectation will be exactly the same. The only thing that changes from system to system is
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