Calculation of mathematical expectation and variance. Discrete random variables

Random variables, in addition to distribution laws, can also be described numerical characteristics .

Mathematical expectation M (x) of a random variable is called its mean value.

Expected value discrete random variable is calculated by the formula

Where – random variable values, p i- their probabilities.

Let's consider the properties of mathematical expectation:

1. The mathematical expectation of a constant is equal to the constant itself

2. If a random variable is multiplied by a certain number k, then the mathematical expectation will be multiplied by the same number

M (kx) = kM (x)

3. The mathematical expectation of the sum of random variables is equal to the sum of their mathematical expectations

M (x 1 + x 2 + … + x n) = M (x 1) + M (x 2) +…+ M (x n)

4. M (x 1 - x 2) = M (x 1) - M (x 2)

5. For independent random variables x 1, x 2, … x n, the mathematical expectation of the product is equal to the product of their mathematical expectations

M (x 1, x 2, ... x n) = M (x 1) M (x 2) ... M (x n)

6. M (x - M (x)) = M (x) - M (M (x)) = M (x) - M (x) = 0

Let's calculate the mathematical expectation for the random variable from Example 11.

M(x) = = .

Example 12. Let the random variables x 1, x 2 be specified accordingly by the distribution laws:

x 1 Table 2

x 2 Table 3

Let's calculate M (x 1) and M (x 2)

M (x 1) = (- 0.1) 0.1 + (- 0.01) 0.2 + 0 0.4 + 0.01 0.2 + 0.1 0.1 = 0

M (x 2) = (- 20) 0.3 + (- 10) 0.1 + 0 0.2 + 10 0.1 + 20 0.3 = 0

The mathematical expectations of both random variables are the same - they are equal to zero. However, the nature of their distribution is different. If the values ​​of x 1 differ little from their mathematical expectation, then the values ​​of x 2 differ to a large extent from their mathematical expectation, and the probabilities of such deviations are not small. These examples show that it is impossible to determine from the average value what deviations from it occur, both smaller and larger. big side. So, with the same average annual precipitation in two areas, it cannot be said that these areas are equally favorable for agricultural work. Similar to the average wages it is not possible to judge specific gravity high and low paid workers. Therefore, a numerical characteristic is introduced - dispersion D(x) , which characterizes the degree of deviation of a random variable from its average value:

D (x) = M (x - M (x)) 2 . (2)

Dispersion is the mathematical expectation of the squared deviation of a random variable from the mathematical expectation. For a discrete random variable, the variance is calculated using the formula:

D(x)= = (3)

From the definition of dispersion it follows that D (x) 0.

Dispersion properties:

1. The variance of the constant is zero

2. If a random variable is multiplied by a certain number k, then the variance will be multiplied by the square of this number

D (kx) = k 2 D (x)

3. D (x) = M (x 2) – M 2 (x)

4. For pairwise independent random variables x 1 , x 2 , … x n the variance of the sum is equal to the sum of the variances.

D (x 1 + x 2 + … + x n) = D (x 1) + D (x 2) +…+ D (x n)

Let's calculate the variance for the random variable from Example 11.

Mathematical expectation M (x) = 1. Therefore, according to formula (3) we have:

D (x) = (0 – 1) 2 1/4 + (1 – 1) 2 1/2 + (2 – 1) 2 1/4 =1 1/4 +1 1/4= 1/2

Note that it is easier to calculate variance if you use property 3:

D (x) = M (x 2) – M 2 (x).

Let's calculate the variances for the random variables x 1 , x 2 from Example 12 using this formula. The mathematical expectations of both random variables are zero.

D (x 1) = 0.01 0.1 + 0.0001 0.2 + 0.0001 0.2 + 0.01 0.1 = 0.001 + 0.00002 + 0.00002 + 0.001 = 0.00204

D (x 2) = (-20) 2 0.3 + (-10) 2 0.1 + 10 2 0.1 + 20 2 0.3 = 240 +20 = 260

How closer value dispersion to zero, the smaller the spread of the random variable relative to the mean value.

The quantity is called standard deviation. Random variable mode x discrete type Md The value of a random variable that has the highest probability is called.

Random variable mode x continuous type Md, is a real number defined as the point of maximum of the probability distribution density f(x).

Median of a random variable x continuous type Mn is a real number that satisfies the equation

Random variable called variable value, which as a result of each test takes one previously unknown value, depending on random reasons. Random variables are denoted by capital Latin letters: $X,\ Y,\ Z,\ \dots $ According to their type, random variables can be discrete And continuous.

Discrete random variable- this is a random variable whose values ​​can be no more than countable, that is, either finite or countable. By countability we mean that the values ​​of a random variable can be numbered.

Example 1 . Here are examples of discrete random variables:

a) the number of hits on the target with $n$ shots, here the possible values ​​are $0,\ 1,\ \dots ,\ n$.

b) the number of emblems dropped when tossing a coin, here the possible values ​​are $0,\ 1,\ \dots ,\ n$.

c) the number of ships arriving on board (a countable set of values).

d) the number of calls arriving at the PBX (countable set of values).

1. Law of probability distribution of a discrete random variable.

A discrete random variable $X$ can take values ​​$x_1,\dots ,\ x_n$ with probabilities $p\left(x_1\right),\ \dots ,\ p\left(x_n\right)$. The correspondence between these values ​​and their probabilities is called law of distribution of a discrete random variable. As a rule, this correspondence is specified using a table, the first line of which indicates the values ​​$x_1,\dots ,\ x_n$, and the second line contains the probabilities $p_1,\dots ,\ p_n$ corresponding to these values.

$\begin(array)(|c|c|)
\hline
X_i & x_1 & x_2 & \dots & x_n \\
\hline
p_i & p_1 & p_2 & \dots & p_n \\
\hline
\end(array)$

Example 2 . Let the random variable $X$ be the number of points rolled when tossing a die. Such a random variable $X$ can take the following values: $1,\ 2,\ 3,\ 4,\ 5,\ 6$. The probabilities of all these values ​​are equal to $1/6$. Then the law of probability distribution of the random variable $X$:

$\begin(array)(|c|c|)
\hline
1 & 2 & 3 & 4 & 5 & 6 \\
\hline

\hline
\end(array)$

Comment. Since in the distribution law of a discrete random variable $X$ the events $1,\ 2,\ \dots ,\ 6$ form full group events, then the sum of the probabilities must be equal to one, that is, $\sum(p_i)=1$.

2. Mathematical expectation of a discrete random variable.

Expectation of a random variable sets its “central” meaning. For a discrete random variable, the mathematical expectation is calculated as the sum of the products of the values ​​$x_1,\dots ,\ x_n$ and the probabilities $p_1,\dots ,\ p_n$ corresponding to these values, that is: $M\left(X\right)=\sum ^n_(i=1)(p_ix_i)$. In English-language literature, another notation $E\left(X\right)$ is used.

Properties of mathematical expectation$M\left(X\right)$:

1. $M\left(X\right)$ is contained between the smallest and highest values random variable $X$.
2. The mathematical expectation of a constant is equal to the constant itself, i.e. $M\left(C\right)=C$.
3. The constant factor can be taken out of the sign of the mathematical expectation: $M\left(CX\right)=CM\left(X\right)$.
4. The mathematical expectation of the sum of random variables is equal to the sum of their mathematical expectations: $M\left(X+Y\right)=M\left(X\right)+M\left(Y\right)$.
5. The mathematical expectation of the product of independent random variables is equal to the product of their mathematical expectations: $M\left(XY\right)=M\left(X\right)M\left(Y\right)$.

Example 3 . Let's find the mathematical expectation of the random variable $X$ from example $2$.

$$M\left(X\right)=\sum^n_(i=1)(p_ix_i)=1\cdot ((1)\over (6))+2\cdot ((1)\over (6) )+3\cdot ((1)\over (6))+4\cdot ((1)\over (6))+5\cdot ((1)\over (6))+6\cdot ((1 )\over (6))=3.5.$$

We can notice that $M\left(X\right)$ lies between the smallest ($1$) and largest ($6$) values ​​of the random variable $X$.

Example 4 . It is known that the mathematical expectation of the random variable $X$ is equal to $M\left(X\right)=2$. Find the mathematical expectation of the random variable $3X+5$.

Using the above properties, we get $M\left(3X+5\right)=M\left(3X\right)+M\left(5\right)=3M\left(X\right)+5=3\cdot 2 +5=$11.

Example 5 . It is known that the mathematical expectation of the random variable $X$ is equal to $M\left(X\right)=4$. Find the mathematical expectation of the random variable $2X-9$.

Using the above properties, we get $M\left(2X-9\right)=M\left(2X\right)-M\left(9\right)=2M\left(X\right)-9=2\cdot 4 -9=-1$.

3. Dispersion of a discrete random variable.

Possible values ​​of random variables with equal mathematical expectations can disperse differently around their average values. For example, in two student groups GPA for the exam in probability theory it turned out to be equal to 4, but in one group everyone turned out to be good students, and in the other group there were only C students and excellent students. Therefore, there is a need for a numerical characteristic of a random variable that would show the spread of the values ​​of the random variable around its mathematical expectation. This characteristic is dispersion.

Variance of a discrete random variable$X$ is equal to:

$$D\left(X\right)=\sum^n_(i=1)(p_i(\left(x_i-M\left(X\right)\right))^2).\ $$

In English literature the notation $V\left(X\right),\ Var\left(X\right)$ is used. Very often the variance $D\left(X\right)$ is calculated using the formula $D\left(X\right)=\sum^n_(i=1)(p_ix^2_i)-(\left(M\left(X \right)\right))^2$.

Dispersion properties$D\left(X\right)$:

1. The variance is always greater than or equal to zero, i.e. $D\left(X\right)\ge 0$.
2. The variance of the constant is zero, i.e. $D\left(C\right)=0$.
3. The constant factor can be taken out of the sign of the dispersion provided that it is squared, i.e. $D\left(CX\right)=C^2D\left(X\right)$.
4. The variance of the sum of independent random variables is equal to the sum of their variances, i.e. $D\left(X+Y\right)=D\left(X\right)+D\left(Y\right)$.
5. The variance of the difference between independent random variables is equal to the sum of their variances, i.e. $D\left(X-Y\right)=D\left(X\right)+D\left(Y\right)$.

Example 6 . Let's calculate the variance of the random variable $X$ from example $2$.

$$D\left(X\right)=\sum^n_(i=1)(p_i(\left(x_i-M\left(X\right)\right))^2)=((1)\over (6))\cdot (\left(1-3.5\right))^2+((1)\over (6))\cdot (\left(2-3.5\right))^2+ \dots +((1)\over (6))\cdot (\left(6-3.5\right))^2=((35)\over (12))\approx 2.92.$$

Example 7 . It is known that the variance of the random variable $X$ is equal to $D\left(X\right)=2$. Find the variance of the random variable $4X+1$.

Using the above properties, we find $D\left(4X+1\right)=D\left(4X\right)+D\left(1\right)=4^2D\left(X\right)+0=16D\ left(X\right)=16\cdot 2=32$.

Example 8 . It is known that the variance of the random variable $X$ is equal to $D\left(X\right)=3$. Find the variance of the random variable $3-2X$.

Using the above properties, we find $D\left(3-2X\right)=D\left(3\right)+D\left(2X\right)=0+2^2D\left(X\right)=4D\ left(X\right)=4\cdot 3=12$.

4. Distribution function of a discrete random variable.

The method of representing a discrete random variable in the form of a distribution series is not the only one, and most importantly, it is not universal, since a continuous random variable cannot be specified using a distribution series. There is another way to represent a random variable - the distribution function.

Distribution function random variable $X$ is called a function $F\left(x\right)$, which determines the probability that the random variable $X$ will take a value less than some fixed value $x$, that is, $F\left(x\right )=P\left(X< x\right)$

Properties of the distribution function:

1. $0\le F\left(x\right)\le 1$.
2. The probability that the random variable $X$ will take values ​​from the interval $\left(\alpha ;\ \beta \right)$ is equal to the difference between the values ​​of the distribution function at the ends of this interval: $P\left(\alpha< X < \beta \right)=F\left(\beta \right)-F\left(\alpha \right)$
3. $F\left(x\right)$ - non-decreasing.
4. $(\mathop(lim)_(x\to -\infty ) F\left(x\right)=0\ ),\ (\mathop(lim)_(x\to +\infty ) F\left(x \right)=1\ )$.

Example 9 . Let us find the distribution function $F\left(x\right)$ for the distribution law of the discrete random variable $X$ from example $2$.

$\begin(array)(|c|c|)
\hline
1 & 2 & 3 & 4 & 5 & 6 \\
\hline
1/6 & 1/6 & 1/6 & 1/6 & 1/6 & 1/6 \\
\hline
\end(array)$

If $x\le 1$, then, obviously, $F\left(x\right)=0$ (including for $x=1$ $F\left(1\right)=P\left(X< 1\right)=0$).

If $1< x\le 2$, то $F\left(x\right)=P\left(X=1\right)=1/6$.

If $2< x\le 3$, то $F\left(x\right)=P\left(X=1\right)+P\left(X=2\right)=1/6+1/6=1/3$.

If $3< x\le 4$, то $F\left(x\right)=P\left(X=1\right)+P\left(X=2\right)+P\left(X=3\right)=1/6+1/6+1/6=1/2$.

If $4< x\le 5$, то $F\left(X\right)=P\left(X=1\right)+P\left(X=2\right)+P\left(X=3\right)+P\left(X=4\right)=1/6+1/6+1/6+1/6=2/3$.

If $5< x\le 6$, то $F\left(x\right)=P\left(X=1\right)+P\left(X=2\right)+P\left(X=3\right)+P\left(X=4\right)+P\left(X=5\right)=1/6+1/6+1/6+1/6+1/6=5/6$.

If $x > 6$, then $F\left(x\right)=P\left(X=1\right)+P\left(X=2\right)+P\left(X=3\right) +P\left(X=4\right)+P\left(X=5\right)+P\left(X=6\right)=1/6+1/6+1/6+1/6+ 1/6+1/6=1$.

So $F(x)=\left\(\begin(matrix)
0,\ at\ x\le 1,\\
1/6,at\ 1< x\le 2,\\
1/3,\ at\ 2< x\le 3,\\
1/2,at\ 3< x\le 4,\\
2/3,\ at\ 4< x\le 5,\\
5/6,\ at\ 4< x\le 5,\\
1,\ for\ x > 6.
\end(matrix)\right.$

Mathematical expectation is the definition

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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.

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Mathematical expectation (Population mean) is

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Mathematical expectation (Population mean) is

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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 short form.

These quantities include primarily expected value And dispersion .

Expected value— the average value of a random variable in probability theory. Denoted as .

The most in a simple way mathematical expectation of a random variable X(w), find how integralLebesgue in relation 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 found through distribution R X. For example, If X- a random variable with values ​​in and f(x)- unambiguous Borel'sfunction X , That:

If F(x)- distribution function X, then the mathematical expectation is representable integralLebesgue - Stieltjes (or Riemann - Stieltjes):

in this case integrability X In terms of ( * ) 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 probability density p(x), That

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 randomly taken variables = the product of their mathematical expectations:

M=M[X]+M[Y]

If X And Y independent.

if the series converges:

Algorithm for calculating mathematical expectation.

Properties of discrete random variables: all their values ​​can be renumbered natural numbers; assign each value a non-zero probability.

1. Multiply the pairs one by one: x i on p i.

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 using the formula.

The mathematical expectation of a random variable X is the mean value.

1. M(C) = C

2. M(CX) = CM(X), Where C= const

3. M(X ± Y) = M(X) ± M(Y)

4. If random variables X And Y are independent, then M(XY) = M(X) M(Y)

Dispersion

The variance of a random variable X is called

D(X) = S(x – M(X)) 2 p = M(X 2 ) – M 2 (X).

Dispersion is a measure of the deviation of the values ​​of a random variable from its mean value.

1. D(C) = 0

2. D(X + C) = D(X)

3. D(CX) = C 2 D(X), Where C= const

4. For independent random variables

D(X ± Y) = D(X) + D(Y)

5. D(X ± Y) = D(X) + D(Y) ± 2Cov(x, y)

Square root from the variance of the random variable X is called the standard deviation .

@Task 3: Let the random variable X take only two values ​​(0 or 1) with probabilities q, p, Where p + q = 1. Find the mathematical expectation and variance.

Solution:

M(X) = 1 p + 0 q = p; D(X) = (1 – p) 2 p + (0 – p) 2 q = pq.

@Task 4: Expectation and variance of a random variable X are equal to 8. Find the mathematical expectation and variance of random variables: a) X – 4; b) 3X – 4.

Solution: M(X – 4) = M(X) – 4 = 8 – 4 = 4; D(X – 4) = D(X) = 8; M(3X – 4) = 3M(X) – 4 = 20; D(3X – 4) = 9D(X) = 72.

@Task 5: The totality of families has the following distribution by number of children:

x i	x 1	x 2
p i	0,1	p 2	0,4	0,35

Define x 1, x 2 And p 2, if it is known that M(X) = 2; D(X) = 0.9.

Solution: Probability p 2 is equal to p 2 = 1 – 0.1 – 0.4 – 0.35 = 0.15. The unknown x are found from the equations: M(X) = x 1 ·0.1 + x 2 ·0.15 + 2·0.4 + 3·0.35 = 2; D(X) = ·0.1 + ·0.15 + 4·0.4 + 9·0.35 – 4 = 0.9. x 1 = 0; x 2 = 1.

Population and sample. Parameter estimates

Selective observation

Statistical observation can be organized continuous or not continuous. Continuous observation involves examining all units of the population being studied (general population). Population is a set of physical or legal entities, which the researcher studies according to his task. This is often not economically viable and sometimes impossible. In this regard, only part of the general population is studied - sample population .

The results obtained from a sample population can be extended to the general population if the following principles are followed:

1. The sample population must be determined randomly.

2. The number of units in the sample population must be sufficient.

3. Must be provided representativeness ( representativeness) of the sample. A representative sample is a smaller but accurate model of the population it is intended to reflect.

Sample types

In practice they are used following types samples:

a) strictly random, b) mechanical, c) typical, d) serial, e) combined.

Proper random sampling

At actual random sample the selection of units in the sample population is carried out randomly, for example, by drawing lots or using a random number generator.

Samples can be repeated or non-repeated. When resampling, the unit that was included in the sample is returned and stored equal opportunity be included in the sample again. In non-repetitive sampling, a population unit that is included in the sample does not participate in the sample in the future.

Errors inherent in sampling observation, arising due to the fact that the sample population does not completely reproduce the general population, are called standard errors . They represent the mean square difference between the values ​​of the indicators obtained from the sample and the corresponding values ​​of the indicators of the general population.

Calculation formulas standard error with random repeated selection the following: , and with random non-repetitive selection the following: , where S 2 is the variance of the sample population, n/N – sample share, n, N- the number of units in the sample and general population. At n = N standard error m = 0.

Mechanical sampling

At mechanical sampling The population is divided into equal intervals and one unit is randomly selected from each interval.

For example, with a 2% sampling rate, every 50th unit is selected from the population list.

The standard error of mechanical sampling is defined as the error of a truly random non-repetitive sampling.

Typical sample

At typical sample the general population is divided into homogeneous typical groups, then units are randomly selected from each group.

A typical sample is used in the case of a heterogeneous population. A typical sample provides more accurate results because it ensures representativeness.

For example, teachers, as a general population, are divided into groups according to the following criteria: gender, experience, qualifications, education, urban and rural schools etc.

Standard errors of a typical sample are defined as errors of a truly random sample, with the only difference that S 2 is replaced by the average of the within-group variances.

Serial sampling

At serial sampling the general population is divided into separate groups (series), then randomly selected groups are subjected to continuous observation.

The standard errors of a serial sample are defined as the errors of a truly random sample, with the only difference being that S 2 is replaced by the average of the between-group variances.

Combined sample

Combined sample is a combination of two or more sample types.

Point estimate

The ultimate goal sample observation is to find the characteristics of the population. Since this cannot be done directly, the characteristics of the sample population are extended to the general population.

The fundamental possibility of determining the arithmetic mean of the population from the data of the average sample is proven Chebyshev's theorem. With unlimited magnification n the probability that the difference between the sample mean and the general mean will be arbitrarily small tends to 1.

This means that the characteristics of the population with an accuracy of . This assessment is called point .

Interval estimation

The basis of interval estimation is central limit theorem.

Interval estimation allows us to answer the question: within what interval and with what probability is the unknown, desired value of the population parameter located?

Usually we talk about confidence probability p = 1 – a, with which it will be in the interval – D< < + D, где D = t cr m > 0 marginal error samples, a - significance level (probability that the inequality will be false), t cr - critical value, which depends on the values n and a. For a small sample n< 30 t cr is specified using the critical value of the Student t-distribution for a two-sided test with n– 1 degrees of freedom with significance level a ( t cr(n – 1, a) is found from the table “Critical values ​​of Student’s t-distribution”, Appendix 2). For n > 30, t cr is a quantile of the normal distribution law ( t cr is found from the table of values ​​of the Laplace function F(t) = (1 – a)/2 as argument). At p = 0.954 the critical value t cr= 2 at p = 0.997 critical value t cr= 3. This means that the marginal error is usually 2-3 times larger than the standard error.

Thus, the essence of the sampling method is that, based on the statistical data of a certain small part of the population, it is possible to find an interval in which, with a confidence probability p the desired characteristic of the general population is found (average number of workers, average score, average yield, standard deviation, etc.).

@Task 1. To determine the speed of settlements with creditors of corporation enterprises in commercial bank A random sample of 100 payment documents was carried out, according to which average term transfer and receipt of money turned out to be 22 days (= 22) with a standard deviation of 6 days (S = 6). With probability p= 0.954 determine the maximum error of the sample mean and confidence interval average duration settlements of enterprises of this corporation.

Solution: Marginal error of sample average according to(1)equal to D= 2· 0.6 = 1.2, and the confidence interval is defined as (22 – 1.2; 22 + 1.2), i.e. (20.8; 23.2).

§6.5 Correlation and regression