The distribution of a discrete random variable has a mathematical expectation equal to. Expectation 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.
Purpose of the service. Using the online service are calculated expected value, variance and standard deviation(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 – 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 variance of a constant value is 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 the 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])
- The following computational formula is valid for dispersion:
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.
Solution. 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 properties of dispersion: 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 mathematical expectation
Properties of discrete random variables: all their values can be renumbered by natural numbers; Assign each value a non-zero probability.- We multiply the pairs one by one: x i by p i .
- 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 No. 1.
| x i | 1 | 3 | 4 | 7 | 9 |
| p i | 0.1 | 0.2 | 0.1 | 0.3 | 0.3 |
We find the mathematical expectation using the formula m = ∑x i p i .
Expectation M[X].
M[x] = 1*0.1 + 3*0.2 + 4*0.1 + 7*0.3 + 9*0.3 = 5.9
We find the variance using the formula d = ∑x 2 i p i - M[x] 2 .
Variance 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 No. 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 |
Solution. The value of a is found from the relation: 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 , from where a = 0.08
Example No. 3. Determine the distribution law of a discrete random variable if its variance is known, and x 1
p 1 =0.3; p 2 =0.3; p 3 =0.1; p 4 =0.3
d(x)=12.96
Solution.
Here you need to create 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 the 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, we need to find the roots of the equation, and there will be two of them.
x 3 =8, x 3 =12
Choose the one that satisfies the condition x 1
Distribution law of a discrete random variable
x 1 =6; x 2 =9; x 3 =12; x 4 =15
p 1 =0.3; p 2 =0.3; p 3 =0.1; p 4 =0.3
As is already known, the distribution law completely characterizes a random variable. However, often the distribution law is unknown and one has to limit oneself to less information. Sometimes it is even more profitable to use numbers that describe the random variable in total; such numbers are called numerical characteristics of a random variable.
One of the important numerical characteristics is the mathematical expectation.
The mathematical expectation is approximately equal to the average value of the random variable.
Mathematical expectation of a discrete random variable is the sum of the products of all its possible values and their probabilities.
If a random variable is characterized by a finite distribution series:
| X | x 1 | x 2 | x 3 | … | x n |
| R | p 1 | p 2 | p 3 | … | r p |
then the mathematical expectation M(X) determined by the formula:
The mathematical expectation of a continuous random variable is determined by the equality:
where is the probability density of the random variable X.
Example 4.7. Find the mathematical expectation of the number of points that appear when throwing a dice.
Solution:
Random value X takes the values 1, 2, 3, 4, 5, 6. Let’s create the law of its distribution:
| X | ||||||
| R |
Then the mathematical expectation is:
Properties of mathematical expectation:
1. The mathematical expectation of a constant value is equal to the constant itself:
M (S) = S.
2. The constant factor can be taken out of the mathematical expectation sign:
M (CX) = CM (X).
3. The mathematical expectation of the product of two independent random variables is equal to the product of their mathematical expectations:
M(XY) = M(X)M(Y).
Example 4.8. Independent random variables X And Y are given by the following distribution laws:
| X | Y | ||||||
| R | 0,6 | 0,1 | 0,3 | R | 0,8 | 0,2 |
Find the mathematical expectation of the random variable XY.
Solution.
Let's find the mathematical expectations of each of these quantities:
Random variables X And Y independent, therefore the required mathematical expectation is:
M(XY) = M(X)M(Y)=
Consequence. The mathematical expectation of the product of several mutually independent random variables is equal to the product of their mathematical expectations.
4. The mathematical expectation of the sum of two random variables is equal to the sum of the mathematical expectations of the terms:
M (X + Y) = M (X) + M (Y).
Consequence. The mathematical expectation of the sum of several random variables is equal to the sum of the mathematical expectations of the terms.
Example 4.9. 3 shots are fired with probabilities of hitting the target equal to p 1 = 0,4; p2= 0.3 and p 3= 0.6. Find the mathematical expectation of the total number of hits.
Solution.
The number of hits on the first shot is a random variable X 1, which can only take two values: 1 (hit) with probability p 1= 0.4 and 0 (miss) with probability q 1 = 1 – 0,4 = 0,6.
The mathematical expectation of the number of hits on the first shot is equal to the probability of a hit:
Similarly, we find the mathematical expectations of the number of hits for the second and third shots:
M(X 2)= 0.3 and M(X 3)= 0,6.
The total number of hits is also a random variable consisting of the sum of hits in each of the three shots:
X = X 1 + X 2 + X 3.
The required mathematical expectation X We find it using the theorem on the mathematical expectation of the sum.
Probability theory is a special branch of mathematics that is studied only by students of higher educational institutions. Do you like calculations and formulas? Aren't you scared by the prospects of getting acquainted with the normal distribution, ensemble entropy, mathematical expectation and dispersion of a discrete random variable? Then this subject will be very interesting to you. Let's get acquainted with several of the most important basic concepts of this branch 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 point is that without a clear understanding of the basics, you will not be able to work with the formulas discussed below.
So, some random event occurs, some experiment. As a result of the actions we take, we can get several outcomes - some of them occur more often, others less often. 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 can you begin to study the mathematical expectation and dispersion of continuous random variables.
Average
Back in school, during math lessons, you started working with the arithmetic mean. This concept is widely used in probability theory, and therefore 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 dispersion 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 up everything available and divide by the number of elements in the sequence. Let us have numbers from 1 to 9. The sum of the elements will be equal to 45, and we will divide this value by 9. Answer: - 5.
Dispersion
In scientific terms, dispersion is the average square of deviations of the obtained values of a characteristic from the arithmetic mean. It is denoted by one capital Latin letter D. What is needed to calculate it? For each element of the sequence, we calculate the difference between the existing 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 sum up everything received and divide by the number of elements in the sequence. If we have five possible outcomes, then divide by five.
Dispersion also has properties that need to be remembered in order to be used when solving problems. For example, when increasing a random variable by X times, the variance increases by X squared times (i.e. X*X). It is never less than zero and does not depend on shifting values up or down by equal amounts. Additionally, 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 ran 21 experiments and got 7 different outcomes. We observed each of them 1, 2, 2, 3, 4, 4 and 5 times, respectively. What will the variance be equal to?
First, let's calculate the arithmetic mean: the sum of the elements, of course, is 21. Divide it by 7, getting 3. Now subtract 3 from each number in the original sequence, square each value, and add the results together. The result is 12. Now all we have to do is divide the number by the number of elements, and, it would seem, that’s all. But there's a catch! Let's discuss it.
Dependence on the number of experiments
It turns out that when calculating variance, the denominator can contain 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 this 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. Scientists decided to draw the border quite symbolically: today it passes through 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 return to our example of solving the problem of variance and mathematical expectation. We got an intermediate number 12, which needed 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 obtained value, as well as the result of calculating the variance, is obtained only once for the entire problem, no matter how many outcomes are considered in it.
The formula for mathematical expectation 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 not difficult to calculate. For example, the sum of the expected values is equal to the expected value of the sum. The same is true for the work. Not every quantity in probability theory allows you to perform such simple operations. Let's take the problem and calculate the meaning of two concepts we have studied at once. Besides, we were distracted by theory - it's time to practice.
One more example
We ran 50 trials and got 10 types of outcomes - numbers from 0 to 9 - appearing in different percentages. These are, respectively: 2%, 10%, 4%, 14%, 2%,18%, 6%, 16%, 10%, 18%. Recall that to obtain 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 convert the probabilities into the number of outcomes “in pieces” to make it easier to count. We get 1, 5, 2, 7, 1, 9, 3, 8, 5 and 9. From each value obtained, we subtract the arithmetic mean, after which we square each of the results obtained. See how to do this using the first element as an example: 1 - 5 = (-4). Next: (-4) * (-4) = 16. For other values, do these operations yourself. If you did everything correctly, then after adding them all up you will get 90.
Let's continue calculating the variance and expected value by dividing 90 by N. Why do we choose N rather than N-1? Correct, because the number of experiments performed exceeds 30. So: 90/10 = 9. We got the variance. If you get a different number, don't despair. Most likely, you made a simple mistake in the calculations. Double-check what you wrote, and everything will probably fall into place.
Finally, remember the formula for mathematical expectation. We will not give all the calculations, we will only write an answer that you can check with after completing all the required procedures. The expected value will be 5.48. Let us only recall how to carry out operations, using the first elements as an example: 0*0.02 + 1*0.1... and so on. As you can see, we simply multiply the outcome value by its probability.
Deviation
Another concept closely related to dispersion and mathematical expectation is standard deviation. It is denoted either by the Latin letters sd, or by the Greek lowercase “sigma”. This concept shows how much on average the 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 graph and want to see the squared deviation directly on it, this can be done in several stages. 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 size of the segment between the middle of the distribution and the resulting projection onto the horizontal axis will represent the standard deviation.
Software
As can be seen from the descriptions of the formulas and the examples presented, calculating variance and mathematical expectation is not the simplest procedure from an arithmetic point of view. In order not to waste time, it makes sense to use the program used in higher education institutions - it is called “R”. It has functions that allow you to calculate values for many concepts from statistics and probability theory.
For example, you specify 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 discussed 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 bad grades at the end of the session, which deprives them of scholarships.
Practice for at least one week, half an hour a day, solving tasks similar to those presented in this article. Then, on any test in probability theory, you will be able to cope with the examples without extraneous tips and cheat sheets.
Magnitude
Basic numerical characteristics of random
The density distribution law characterizes a random variable. But often it is unknown, and one has to limit oneself to less information. Sometimes it is even more profitable to use numbers that describe a random variable in total. Such numbers are called numerical characteristics random variable. Let's look at the main ones.
Definition:The mathematical expectation M(X) of a discrete random variable is the sum of the products of all possible values of this quantity and their probabilities:
If a discrete random variable X takes a countably many possible values, then
Moreover, the mathematical expectation exists if this series is absolutely convergent.
From the definition it follows that M(X) a discrete random variable is a non-random (constant) variable.
Example: Let X– number of occurrences of the event A in one test, P(A) = p. We need to find the mathematical expectation X.
Solution: Let's create a tabular distribution law X:
| X | 0 | 1 |
| P | 1 - p | p |
Let's find the mathematical expectation:
Thus, the mathematical expectation of the number of occurrences of an event in one trial is equal to the probability of this event.
Origin of the term expected value associated with the initial period of the emergence of probability theory (XVI-XVII centuries), when the scope of its application was limited to gambling. The player was interested in the average value of the expected win, i.e. mathematical expectation of winning.
Let's consider probabilistic meaning of mathematical expectation.
Let it be produced n tests in which the random variable X accepted m 1 times value x 1, m 2 times value x 2, and so on, and finally she accepted m k times value x k, and m 1 + m 2 +…+ + m k = n.
Then the sum of all values taken by the random variable X, is equal x 1 m 1 +x 2 m 2 +…+x k m k.
Arithmetic mean of all values taken by a random variable X,equals:
since is the relative frequency of a value for any value i = 1, …, k.
As is known, if the number of tests n is sufficiently large, then the relative frequency is approximately equal to the probability of the event occurring, therefore,
Thus, .
Conclusion:The mathematical expectation of a discrete random variable is approximately equal (the more accurately, the greater the number of tests) to the arithmetic mean of the observed values of the random variable.
Let's consider the basic properties of mathematical expectation.
Property 1:The mathematical expectation of a constant value is equal to the constant value itself:
M(C) = C.
Proof: Constant WITH can be considered , which has one possible meaning WITH and accepts it with probability p = 1. Hence, M(C) =C 1= S.
Let's define product of a constant variable C and a discrete random variable X as a discrete random variable CX, the possible values of which are equal to the products of the constant WITH to possible values X CX equal to the probabilities of the corresponding possible values X:
| CX | C | C | … | C |
| X | … | |||
| R | … |
Property 2:The constant factor can be taken out of the mathematical expectation sign:
M(CX) = CM(X).
Proof: Let the random variable X is given by the law of probability distribution:
| X | … | |||
| P | … |
Let's write the law of probability distribution of a random variable CX:
| CX | C | C | … | C |
| P | … |
M(CX) = C +C =C + ) = C M(X).
Definition: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.
Definition:Several random variables are said to be mutually independent if the distribution laws of any number of them do not depend on what possible values the remaining variables took.
Let's define product of independent discrete random variables X and Y as a discrete random variable XY, the possible values of which are equal to the products of each possible value X for every possible value Y. Probabilities of possible values XY are equal to the products of the probabilities of possible values of the factors.
Let the distributions of random variables be given X And Y:
| X | … | |||
| P | … |
| Y | … | |||
| G | … |
Then the distribution of the random variable XY has the form:
| XY | … | |||
| P | … |
Some works may be equal. In this case, the probability of a possible value of the product is equal to the sum of the corresponding probabilities. For example, if = , then the probability of the value is
Property 3:The mathematical expectation of the product of two independent random variables is equal to the product of their mathematical expectations:
M(XY) = M(X) M(Y).
Proof: Let independent random variables X And Y are specified by their own probability distribution laws:
| X | ||
| P |
| Y | ||
| G |
To simplify the calculations, we will limit ourselves to a small number of possible values. In the general case the proof is similar.
Let's create a law of distribution of a random variable XY:
| XY | ||||
| P |
M(XY) =
M(X) M(Y).
Consequence:The mathematical expectation of the product of several mutually independent random variables is equal to the product of their mathematical expectations.
Proof: Let us prove for three mutually independent random variables X,Y,Z. Random variables XY And Z independent, then we get:
M(XYZ) = M(XY Z) = M(XY) M(Z) = M(X) M(Y) M(Z).
For an arbitrary number of mutually independent random variables, the proof is carried out by the method of mathematical induction.
Example: Independent random variables X And Y
| X | 5 | 2 | |
| P | 0,6 | 0,1 | 0,3 |
| Y | 7 | 9 |
| G | 0,8 | 0,2 |
Need to find M(XY).
Solution: Since random variables X And Y are independent, then M(XY)=M(X) M(Y)=(5 0,6+2 0,1+4 0,3) (7 0,8+9 0,2)= 4,4 7,4 = =32,56.
Let's define sum of discrete random variables X and Y as a discrete random variable X+Y, the possible values of which are equal to the sums of each possible value X with every possible value Y. Probabilities of possible values X+Y for independent random variables X And Y are equal to the products of the probabilities of the terms, and for dependent random variables - to the products of the probability of one term by the conditional probability of the second.
If = and the probabilities of these values are respectively equal, then the probability (the same as ) is equal to .
Property 4:The mathematical expectation of the sum of two random variables (dependent or independent) is equal to the sum of the mathematical expectations of the terms:
M(X+Y) = M(X) + M(Y).
Proof: Let two random variables X And Y are given by the following distribution laws:
| X | ||
| P |
| Y | ||
| G |
To simplify the conclusion, we will limit ourselves to two possible values of each quantity. In the general case the proof is similar.
Let's compose all possible values of a random variable X+Y(assume, for simplicity, that these values are different; if not, then the proof is similar):
| X+Y | ||||
| P |
Let's find the mathematical expectation of this value.
M(X+Y) = + + + +
Let's prove that + = .
Event X = ( its probability P(X = ) entails the event that the random variable X+Y will take the value or (the probability of this event, according to the addition theorem, is equal to ) and vice versa. Then = .
The equalities = = = are proved in a similar way
Substituting the right-hand sides of these equalities into the resulting formula for the mathematical expectation, we obtain:
M(X + Y) = + ) = M(X) + M(Y).
Consequence:The mathematical expectation of the sum of several random variables is equal to the sum of the mathematical expectations of the terms.
Proof: Let us prove for three random variables X,Y,Z. Let's find the mathematical expectation of random variables X+Y And Z:
M(X+Y+Z)=M((X+Y Z)=M(X+Y) M(Z)=M(X)+M(Y)+M(Z)
For an arbitrary number of random variables, the proof is carried out by the method of mathematical induction.
Example: Find the average of the sum of the number of points that can be obtained when throwing two dice.
Solution: Let X– the number of points that can appear on the first die, Y- On the second. It is obvious that random variables X And Y have the same distributions. Let's write down the distribution data X And Y into one table:
| X | 1 | 2 | 3 | 4 | 5 | 6 |
| Y | 1 | 2 | 3 | 4 | 5 | 6 |
| P | 1/6 | 1/6 | 1/6 | 1/6 | 1/6 | 1/6 |
M(X) = M(Y) (1+2+3+4+5+6) = =
M(X + Y) = 7.
So, the average value of the sum of the number of points that can appear when throwing two dice is 7 .
Theorem:The mathematical expectation M(X) of the number of occurrences of event A in n independent trials is equal to the product of the number of trials and the probability of the occurrence of the event in each trial: M(X) = np.
Proof: Let X– number of occurrences of the event A V n independent tests. Obviously the total number X occurrences of the event A in these trials is the sum of the number of occurrences of the event in individual trials. Then, if the number of occurrences of an event in the first trial, in the second, and so on, finally, is the number of occurrences of the event in n-th test, then the total number of occurrences of the event is calculated by the formula:
By property 4 of mathematical expectation we have:
M(X) = M( ) + … + M( ).
Since the mathematical expectation of the number of occurrences of an event in one trial is equal to the probability of the event, then
M( ) = M( )= … = M( ) = p.
Hence, M(X) = np.
Example: The probability of hitting the target when firing from a gun is p = 0.6. Find the average number of hits if made 10 shots.
Solution: The hit for each shot does not depend on the outcomes of other shots, therefore the events under consideration are independent and, therefore, the required mathematical expectation is equal to:
M(X) = np = 10 0,6 = 6.
So the average number of hits is 6.
Now consider the mathematical expectation of a continuous random variable.
Definition:The mathematical expectation of a continuous random variable X, the possible values of which belong to the interval,called the definite integral:
where f(x) is the probability distribution density.
If possible values of a continuous random variable X belong to the entire Ox axis, then
It is assumed that this improper integral converges absolutely, i.e. the integral converges If this requirement were not met, then the value of the integral would depend on the rate at which (separately) the lower limit tends to -∞, and the upper limit tends to +∞.
It can be proven that all properties of the mathematical expectation of a discrete random variable are preserved for a continuous random variable. The proof is based on the properties of definite and improper integrals.
It is obvious that the mathematical expectation M(X) greater than the smallest and less than the largest possible value of the random variable X. Those. on the number axis, possible values of a random variable are located to the left and to the right of its mathematical expectation. In this sense, the mathematical expectation M(X) characterizes the location of the distribution and is therefore often called distribution center.
Chapter 6.
Numerical characteristics of random variables
Mathematical expectation and its properties
To solve many practical problems, knowledge of all possible values of a random variable and their probabilities is not always required. Moreover, sometimes the distribution law of the random variable under study is simply unknown. However, it is necessary to highlight some features of this random variable, in other words, numerical characteristics.
Numerical characteristics– these are some numbers that characterize certain properties, distinctive features of a random variable.
For example, the average value of a random variable, the average spread of all values of a random variable around its average, etc. The main purpose of numerical characteristics is to express in a concise form the most important features of the distribution of the random variable under study. Numerical characteristics play a huge role in probability theory. They help solve, even without knowledge of the laws of distribution, many important practical problems.
Among all the numerical characteristics, we first highlight position characteristics. These are characteristics that fix the position of a random variable on the numerical axis, i.e. a certain average value around which the remaining values of the random variable are grouped.
Of the characteristics of a position, the greatest role in probability theory is played by the mathematical expectation.
Expected value sometimes called simply the mean of a random variable. It is a kind of distribution center.
Expectation of a discrete random variable
Let us first consider the concept of mathematical expectation for a discrete random variable.
Before introducing a formal definition, let us solve the following simple problem.
Example 6.1. Let a certain shooter fire 100 shots at a target. As a result, the following picture was obtained: 50 shots - hitting the "eight", 20 shots - hitting the "nine" and 30 - hitting the "ten". What is the average score for one shot?
Solution This problem is obvious and boils down to finding the average value of 100 numbers, namely, points.
We transform the fraction by dividing the numerator by the denominator term by term, and present the average value in the form of the following formula:
Let us now assume that the number of points in one shot are the values of some discrete random variable X. From the problem statement it is clear that X 1 =8; X 2 =9; X 3 =10. The relative frequencies of occurrence of these values are known, which, as is known, with a large number of tests are approximately equal to the probabilities of the corresponding values, i.e. R 1 ≈0,5;R 2 ≈0,2; R 3 ≈0.3. So, . The value on the right side is the mathematical expectation of a discrete random variable.
Mathematical expectation of a discrete random variable X is the sum of the products of all its possible values and the probabilities of these values.
Let the discrete random variable X is given by its distribution series:
| X | X 1 | X 2 | … | X n |
| R | R 1 | R 2 | … | R n |
Then the mathematical expectation M(X) of a discrete random variable is determined by the following formula:
If a discrete random variable takes on an infinite countable set of values, then the mathematical expectation is expressed by the formula:
,
Moreover, the mathematical expectation exists if the series on the right side of the equality converges absolutely.
Example 6.2 . Find the mathematical expectation of winning X under the conditions of example 5.1.
Solution . Recall that the distribution series X has the following form:
| X | |||
| R | 0,7 | 0,2 | 0,1 |
We get M(X)=0∙0.7+10∙0.2+50∙0.1=7. Obviously, 7 rubles is a fair price for a ticket in this lottery, without various costs, for example, associated with the distribution or production of tickets. ■
Example 6.3 . Let the random variable X is the number of occurrences of some event A in one test. The probability of this event is R. Find M(X).
Solution. Obviously, the possible values of the random variable are: X 1 =0 – event A didn't show up and X 2 =1 – event A appeared. The distribution series looks like:
| X | ||
| R | 1−R | R |
Then M(X) = 0∙(1−R)+1∙R= R. ■
So, the mathematical expectation of the number of occurrences of an event in one trial is equal to the probability of this event.
At the beginning of the paragraph, a specific problem was given, where the connection between the mathematical expectation and the average value of a random variable was indicated. Let us explain this in general terms.
Let it be produced k tests in which the random variable X accepted k 1 time value X 1 ; k 2 times the value X 2, etc. and finally k n times value xn. It's obvious that k 1 +k 2 +…+k n = k. Let us find the arithmetic mean of all these values, we have
Note that a fraction is the relative frequency of occurrence of a value x i V k tests. With a large number of tests, the relative frequency is approximately equal to the probability, i.e. . It follows that
.
Thus, the mathematical expectation is approximately equal to the arithmetic mean of the observed values of the random variable, and the more accurate the greater the number of tests - this is probabilistic meaning of mathematical expectation.
The expected value is sometimes called center distribution of a random variable, since it is obvious that the possible values of the random variable are located on the numerical axis to the left and to the right of its mathematical expectation.
Let us now move on to the concept of mathematical expectation for a continuous random variable.