It is more convenient to present the result in the form of Table 9.6.

Table 9.6

Let's use the formulas for the initial moments and the results of tables 9.3 and 9.4 and calculate the mathematical expectations of the components X And Y:

We calculate the variances using the second initial moment and the results of the table. 9.3 and 9.4:

To calculate covariance TO(X,Y) we use a similar formula through the initial moment:

The correlation coefficient is determined by the formula:

The required probability is defined as the probability of falling into a region on a plane defined by the corresponding inequality:

9.2. The ship transmits an “SOS” message, which can be received by two radio stations. This signal can be received by one radio station independently of the other. The probability that the signal is received by the first radio station is 0.95; the probability that the signal is received by the second radio station is 0.85. Find the distribution law of a two-dimensional random variable characterizing the reception of a signal by two radio stations. Write the distribution function.

Solution: Let X– an event consisting in the fact that the signal is received by the first radio station. Y– the event is that the signal is received by a second radio station.

Multiple meanings .

X=1 – signal received by the first radio station;

X=0 – the signal was not received by the first radio station.

Multiple meanings .

Y=l – signal received by the second radio station,

Y=0 – the signal is not received by the second radio station.

The probability that the signal is not received by either the first or second radio stations is:

Probability of signal reception by the first radio station:

Probability that the signal is received by the second radio station:

The probability that the signal is received by both the first and second radio stations is equal to: .

Then the distribution law of a two-dimensional random variable is equal to:

X,y) meaning F(X,y) is equal to the sum of the probabilities of those possible values ​​of the random variable ( X,Y), which fall inside the specified rectangle.

Then the distribution function will look like:

9.3. Two companies produce identical products. Each, independently of the other, can decide to modernize production. The probability that the first firm made such a decision is 0.6. The probability of making such a decision by the second firm is 0.65. Write the law of distribution of a two-dimensional random variable that characterizes the decision to modernize the production of two firms. Write the distribution function.

Answer: Distribution law:

For each fixed value of a point with coordinates ( x,y) the value is equal to the sum of the probabilities of those possible values ​​that fall inside the specified rectangle .

9.4. Piston rings for car engines are made on an automatic lathe. The thickness of the ring is measured (random value X) and hole diameter (random value Y). It is known that about 5% of all piston rings are defective. Moreover, 3% of defects are caused by non-standard hole diameters, 1% - by non-standard thickness, and 1% - are rejected on both grounds. Find: joint distribution of a two-dimensional random variable ( X,Y); one-dimensional distributions of components X And Y;mathematical expectations of the components X And Y; correlation moment and correlation coefficient between components X And Y two-dimensional random variable ( X,Y).

Answer: Distribution law:

; ; ; ; ; .

9.5. Factory products are defective due to defects A is 4%, and due to a defect IN– 3.5%. Standard production is 96%. Determine what percentage of all products have both types of defects.

9.6. Random variable ( X,Y)distributed with constant density inside the square R, whose vertices have coordinates (–2;0), (0;2), (2;0), (0;–2). Determine the distribution density of the random variable ( X,Y) and conditional distribution densities r(X\at), p(at\X).

Solution. Let's build on a plane x 0y given square (Fig. 9.5) and determine the equations of the sides of the square ABCD, using the equation of a straight line passing through two given points: Substituting the coordinates of the vertices A And IN we obtain sequentially the equation of the side AB: or .

Similarly, we find the equation of the side Sun: ;sides CD: and sides D.A.: . : .D X , Y) is a hemisphere centered at the origin of the radius R.Find the probability distribution density.

Answer:

9.10. Given a discrete two-dimensional random variable:

Find: a) conditional distribution law X, provided that y= 10;

b) conditional distribution law Y, provided that x =10;

c) mathematical expectation, dispersion, correlation coefficient.

9.11. Continuous two-dimensional random variable ( X,Y)evenly distributed inside a right triangle with vertices ABOUT(0;0), A(0;8), IN(8,0).

Find: a) probability distribution density;

Definition 2.7. is a pair of random numbers (X, Y), or a point on the coordinate plane (Fig. 2.11).

Rice. 2.11.

A two-dimensional random variable is a special case of a multivariate random variable, or random vector.

Definition 2.8. Random vector - is it a random function?,(/) with a finite set of possible argument values t, whose value for any value t is a random variable.

A two-dimensional random variable is called continuous if its coordinates are continuous, and discrete if its coordinates are discrete.

To set the law of distribution of two-dimensional random variables means to establish a correspondence between its possible values ​​and the probability of these values. According to the methods of specifying, random variables are divided into continuous and discrete, although there are general ways to specify the distribution law of any random variable.

Discrete two-dimensional random variable

A discrete two-dimensional random variable is specified using a distribution table (Table 2.1).

Table 2.1

Distribution table (joint distribution) SV ( X, U)

The elements of the table are determined by the formula

Properties of distribution table elements:

The distribution over each coordinate is called one-dimensional or marginal:

r 1> = P(X =.g,) - marginal distribution of SV X;

p^2) = P(Y= y,)- marginal distribution of SV U.

Relationship between the joint distribution of CBs X and Y, specified by a set of probabilities [p () ), i = 1,..., n,j = 1,..., T(distribution table), and marginal distribution.

Similarly for SV U p- 2)= X r, g

Problem 2.14. Given:

Continuous two-dimensional random variable

/(X, y)dxdy- element of probability for a two-dimensional random variable (X, Y) - probability of a random variable (X, Y) falling into a rectangle with sides cbc,dy at dx,dy -* 0:

f(x, y) - distribution density two-dimensional random variable (X, Y). The task /(x, y) we give complete information about the distribution of a two-dimensional random variable.

Marginal distributions are specified as follows: for X - by the distribution density of SV X/,(x); By Y- distribution density of SV U f>(y).

Specifying the distribution law of a two-dimensional random variable by the distribution function

A universal way to specify the distribution law for a discrete or continuous two-dimensional random variable is the distribution function F(x, y).

Definition 2.9. Distribution function F(x, y)- probability of joint occurrence of events (Xy), i.e. F(x 0 ,y n) = = P(X y), thrown onto the coordinate plane, fall into an infinite quadrant with the vertex at the point M(x 0, y i)(in the shaded area in Fig. 2.12).

Rice. 2.12. Illustration of the distribution function F( x, y)

Function Properties F(x, y)

• 1) 0 1;
• 2) F(-oo,-oo) = F(x,-oo) = F(-oo, y) = 0; F( oo, oo) = 1;
• 3) F(x, y)- non-decreasing for each argument;
• 4) F(x, y) - continuous on the left and below;
• 5) consistency of distributions:

F(x, X: F(x, oo) = F,(x); F(y, oo) - marginal distribution over Y F( oh, y) = F 2 (y). Connection /(x, y) With F(x, y):

Relationship between joint density and marginal density. Dana f(x, y). Let's get the marginal distribution densities f(x),f 2 (y)".

The case of independent coordinates of a two-dimensional random variable

Definition 2.10. NE X And Yindependent(nz), if any events associated with each of these SVs are independent. From the definition of NZ SV it follows:

• 1 )Pij = p X) pf
• 2 )F(x,y) = F l (x)F 2 (y).

It turns out that for independent SVs X And Y completed and

3 )f(x,y) = J(x)f,(y).

Let us prove that for independent SVs X And Y 2) 3). Proof, a) Let 2 be satisfied, i.e.

at the same time F(x,y) = f J f(u,v)dudv, hence follows 3);

b) let now 3) be fulfilled, then

those. true 2).

Let's consider the tasks.

Problem 2.15. The distribution is given by the following table:

We construct marginal distributions:

We get P(X = 3, U = 4) = 0,17 * P(X = 3)P(U = 4) = 0.1485 => => SV X and Dependent.

Distribution function:

Problem 2.16. The distribution is given by the following table:

We get P tl = 0.2 0.3 = 0.06; R 12 = 0.2? 0.7 = 0.14; P2l = 0,8 ? 0,3 = = 0,24; R 22 - 0.8 0.7 = 0.56 => NE X And Y nz.

Problem 2.17. Dana /(x, y) = 1st exp| -0.5(d" + 2xy + 5g/ 2)]. Find Oh) And /Ay)-

Solution

(count it yourself).

two-dimensional discrete distribution random

Often the result of an experiment is described by several random variables: . For example, the weather in a given place at a certain time of day can be characterized by the following random variables: X 1 - temperature, X 2 - pressure, X 3 - air humidity, X 4 - wind speed.

In this case, we speak of a multidimensional random variable or a system of random variables.

Consider a two-dimensional random variable whose possible values ​​are pairs of numbers. Geometrically, a two-dimensional random variable can be interpreted as a random point on a plane.

If the components X And Y are discrete random variables, then - a discrete two-dimensional random variable, and if X And Y are continuous, then is a continuous two-dimensional random variable.

The law of probability distribution of a two-dimensional random variable is the correspondence between possible values ​​and their probabilities.

The distribution law of a two-dimensional discrete random variable can be specified in the form of a table with a double input (see Table 6.1), where is the probability that the component X took on the meaning x i, and the component Y- meaning y j .

Table 6.1.1.

Since the events constitute a complete group of pairwise incompatible events, the sum of the probabilities is equal to 1, i.e.

From Table 6.1 you can find the laws of distribution of one-dimensional components X And Y.

Example 6.1.1 . Find the laws of distribution of components X And Y, if the distribution of a two-dimensional random variable is given in the form of table 6.1.2.

Table 6.1.2.

If we fix the value of one of the arguments, for example, then the resulting distribution of the value X called conditional distribution. The conditional distribution is defined similarly Y.

Example 6.1.2 . According to the distribution of a two-dimensional random variable given in Table. 6.1.2, find: a) the conditional distribution law of the component X given that; b) conditional distribution law Y provided that.

Solution. Conditional probabilities of components X And Y calculated using formulas

Conditional distribution law X provided it has the form

Control: .

The distribution law of a two-dimensional random variable can be specified in the form distribution functions, which determines for each pair of numbers the probability that X will take a value less than X, and at the same time Y will take a value less than y:

Geometrically, the function means the probability of a random point falling into an infinite square with its vertex at the point (Fig. 6.1.1).

Let's note the properties.

• 1. The range of values ​​of the function is , i.e. .
• 2. Function - a non-decreasing function for each argument.
• 3. There are limiting relations:

When the distribution function of the system becomes equal to the distribution function of the component X, i.e. .

Likewise, .

Knowing this, you can find the probability of a random point falling within the rectangle ABCD.

Namely,

Example 6.1.3. A two-dimensional discrete random variable is specified by a distribution table

Find the distribution function.

Solution. Value in case of discrete components X And Y is found by summing all probabilities with indices i And j, for which, . Then, if and, then (the events and are impossible). Similarly we get:

if and, then;

if and, then;

if and, then;

if and, then;

if and, then;

if and, then;

if and, then;

if and, then;

if and, then.

Let us present the results obtained in the form of a table (6.1.3) of values:

For two-dimensional continuous random variable, the concept of probability density is introduced

Geometric probability density is a distribution surface in space

The two-dimensional probability density has the following properties:

3. The distribution function can be expressed through the formula

4. The probability of a continuous random variable falling into the region is equal to

5. In accordance with property (4) of the function, the following formulas hold:

Example 6.1.4. The distribution function of a two-dimensional random variable is given

An ordered pair (X, Y) of random variables X and Y is called a two-dimensional random variable, or a random vector in two-dimensional space. A two-dimensional random variable (X,Y) is also called a system of random variables X and Y. The set of all possible values ​​of a discrete random variable with their probabilities is called the distribution law of this random variable. A discrete two-dimensional random variable (X, Y) is considered given if its distribution law is known:

P(X=x i , Y=y j) = p ij , i=1,2...,n, j=1,2...,m

Purpose of the service. Using the service, according to a given distribution law, you can find:

• distribution series X and Y, mathematical expectation M[X], M[Y], variance D[X], D[Y];
• covariance cov(x,y), correlation coefficient r x,y, conditional distribution series X, conditional expectation M;

Instructions. Specify the dimension of the probability distribution matrix (number of rows and columns) and its type. The resulting solution is saved in a Word file.

Example No. 1. A two-dimensional discrete random variable has a distribution table:

Solution. We find the value of q from the condition Σp ij = 1
Σp ij = 0.02 + 0.03 + 0.11 + … + 0.03 + 0.02 + 0.01 + q = 1
0.91+q = 1. Where does q = 0.09 come from?

Using the formula ∑P(x i,y j) = p i(j=1..n), we find the distribution series X.

Covariance cov(X,Y) = M - M[X] M[Y] = 2 10 0.11 + 3 10 0.12 + 4 10 0.03 + 2 20 0.13 + 3 20 0.09 + 4 ·20·0.02 + 1·30·0.02 + 2·30·0.11 + 3·30·0.08 + 4·30·0.01 + 1·40·0.03 + 2·40·0.11 + 3·40·0.05 + 4·40 ·0.09 - 25.2 · 2.59 = -0.068
Correlation coefficient r xy = cov(x,y)/σ(x)&sigma(y) = -0.068/(11.531*0.801) = -0.00736

Example 2. Data from statistical processing of information regarding two indicators X and Y are reflected in the correlation table. Required:

1. write distribution series for X and Y and calculate sample means and sample standard deviations for them;
2. write conditional distribution series Y/x and calculate conditional averages Y/x;
3. graphically depict the dependence of conditional averages Y/x on X values;
4. calculate the sample correlation coefficient Y on X;
5. write a sample forward regression equation;
6. depict geometrically the data of the correlation table and construct a regression line.

Conditional mathematical expectation M = 11*0.33 + 16*0.67 + 21*0 + 26*0 + 31*0 + 36*0 = 14.33
Conditional variance D = 11 2 *0.33 + 16 2 *0.67 + 21 2 *0 + 26 2 *0 + 31 2 *0 + 36 2 *0 - 14.33 2 = 5.56
Conditional distribution law X(Y=30).
P(X=11/Y=30) = 0/9 = 0
P(X=16/Y=30) = 6/9 = 0.67
P(X=21/Y=30) = 3/9 = 0.33
P(X=26/Y=30) = 0/9 = 0
P(X=31/Y=30) = 0/9 = 0
P(X=36/Y=30) = 0/9 = 0
Conditional mathematical expectation M = 11*0 + 16*0.67 + 21*0.33 + 26*0 + 31*0 + 36*0 = 17.67
Conditional variance D = 11 2 *0 + 16 2 *0.67 + 21 2 *0.33 + 26 2 *0 + 31 2 *0 + 36 2 *0 - 17.67 2 = 5.56
Conditional distribution law X(Y=40).
P(X=11/Y=40) = 0/55 = 0
P(X=16/Y=40) = 0/55 = 0
P(X=21/Y=40) = 6/55 = 0.11
P(X=26/Y=40) = 45/55 = 0.82
P(X=31/Y=40) = 4/55 = 0.0727
P(X=36/Y=40) = 0/55 = 0
Conditional mathematical expectation M = 11*0 + 16*0 + 21*0.11 + 26*0.82 + 31*0.0727 + 36*0 = 25.82
Conditional variance D = 11 2 *0 + 16 2 *0 + 21 2 *0.11 + 26 2 *0.82 + 31 2 *0.0727 + 36 2 *0 - 25.82 2 = 4.51
Conditional distribution law X(Y=50).
P(X=11/Y=50) = 0/16 = 0
P(X=16/Y=50) = 0/16 = 0
P(X=21/Y=50) = 2/16 = 0.13
P(X=26/Y=50) = 8/16 = 0.5
P(X=31/Y=50) = 6/16 = 0.38
P(X=36/Y=50) = 0/16 = 0
Conditional mathematical expectation M = 11*0 + 16*0 + 21*0.13 + 26*0.5 + 31*0.38 + 36*0 = 27.25
Conditional variance D = 11 2 *0 + 16 2 *0 + 21 2 *0.13 + 26 2 *0.5 + 31 2 *0.38 + 36 2 *0 - 27.25 2 = 10.94
Conditional distribution law X(Y=60).
P(X=11/Y=60) = 0/14 = 0
P(X=16/Y=60) = 0/14 = 0
P(X=21/Y=60) = 0/14 = 0
P(X=26/Y=60) = 4/14 = 0.29
P(X=31/Y=60) = 7/14 = 0.5
P(X=36/Y=60) = 3/14 = 0.21
Conditional mathematical expectation M = 11*0 + 16*0 + 21*0 + 26*0.29 + 31*0.5 + 36*0.21 = 30.64
Conditional variance D = 11 2 *0 + 16 2 *0 + 21 2 *0 + 26 2 *0.29 + 31 2 *0.5 + 36 2 *0.21 - 30.64 2 = 12.37
3. Conditional distribution law Y.
Conditional distribution law Y(X=11).
P(Y=20/X=11) = 2/2 = 1
P(Y=30/X=11) = 0/2 = 0
P(Y=40/X=11) = 0/2 = 0
P(Y=50/X=11) = 0/2 = 0
P(Y=60/X=11) = 0/2 = 0
Conditional mathematical expectation M = 20*1 + 30*0 + 40*0 + 50*0 + 60*0 = 20
Conditional variance D = 20 2 *1 + 30 2 *0 + 40 2 *0 + 50 2 *0 + 60 2 *0 - 20 2 = 0
Conditional distribution law Y(X=16).
P(Y=20/X=16) = 4/10 = 0.4
P(Y=30/X=16) = 6/10 = 0.6
P(Y=40/X=16) = 0/10 = 0
P(Y=50/X=16) = 0/10 = 0
P(Y=60/X=16) = 0/10 = 0
Conditional mathematical expectation M = 20*0.4 + 30*0.6 + 40*0 + 50*0 + 60*0 = 26
Conditional variance D = 20 2 *0.4 + 30 2 *0.6 + 40 2 *0 + 50 2 *0 + 60 2 *0 - 26 2 = 24
Conditional distribution law Y(X=21).
P(Y=20/X=21) = 0/11 = 0
P(Y=30/X=21) = 3/11 = 0.27
P(Y=40/X=21) = 6/11 = 0.55
P(Y=50/X=21) = 2/11 = 0.18
P(Y=60/X=21) = 0/11 = 0
Conditional mathematical expectation M = 20*0 + 30*0.27 + 40*0.55 + 50*0.18 + 60*0 = 39.09
Conditional variance D = 20 2 *0 + 30 2 *0.27 + 40 2 *0.55 + 50 2 *0.18 + 60 2 *0 - 39.09 2 = 44.63
Conditional distribution law Y(X=26).
P(Y=20/X=26) = 0/57 = 0
P(Y=30/X=26) = 0/57 = 0
P(Y=40/X=26) = 45/57 = 0.79
P(Y=50/X=26) = 8/57 = 0.14
P(Y=60/X=26) = 4/57 = 0.0702
Conditional mathematical expectation M = 20*0 + 30*0 + 40*0.79 + 50*0.14 + 60*0.0702 = 42.81
Conditional variance D = 20 2 *0 + 30 2 *0 + 40 2 *0.79 + 50 2 *0.14 + 60 2 *0.0702 - 42.81 2 = 34.23
Conditional distribution law Y(X=31).
P(Y=20/X=31) = 0/17 = 0
P(Y=30/X=31) = 0/17 = 0
P(Y=40/X=31) = 4/17 = 0.24
P(Y=50/X=31) = 6/17 = 0.35
P(Y=60/X=31) = 7/17 = 0.41
Conditional mathematical expectation M = 20*0 + 30*0 + 40*0.24 + 50*0.35 + 60*0.41 = 51.76
Conditional variance D = 20 2 *0 + 30 2 *0 + 40 2 *0.24 + 50 2 *0.35 + 60 2 *0.41 - 51.76 2 = 61.59
Conditional distribution law Y(X=36).
P(Y=20/X=36) = 0/3 = 0
P(Y=30/X=36) = 0/3 = 0
P(Y=40/X=36) = 0/3 = 0
P(Y=50/X=36) = 0/3 = 0
P(Y=60/X=36) = 3/3 = 1
Conditional mathematical expectation M = 20*0 + 30*0 + 40*0 + 50*0 + 60*1 = 60
Conditional variance D = 20 2 *0 + 30 2 *0 + 40 2 *0 + 50 2 *0 + 60 2 *1 - 60 2 = 0
Covariance.
cov(X,Y) = M - M[X]·M[Y]
cov(X,Y) = (20 11 2 + 20 16 4 + 30 16 6 + 30 21 3 + 40 21 6 + 50 21 2 + 40 26 45 + 50 26 8 + 60 26 4 + 40 31 4 + 50 31 6 + 60 31 7 + 60 36 3)/100 - 25.3 42.3 = 38.11
If random variables are independent, then their covariance is zero. In our case, cov(X,Y) ≠ 0.
Correlation coefficient.


The linear regression equation from y to x is:

The linear regression equation from x to y is:

Let's find the necessary numerical characteristics.
Sample averages:
x = (20(2 + 4) + 30(6 + 3) + 40(6 + 45 + 4) + 50(2 + 8 + 6) + 60(4 + 7 + 3))/100 = 42.3
y = (20(2 + 4) + 30(6 + 3) + 40(6 + 45 + 4) + 50(2 + 8 + 6) + 60(4 + 7 + 3))/100 = 25.3
Variances:
σ 2 x = (20 2 (2 + 4) + 30 2 (6 + 3) + 40 2 (6 + 45 + 4) + 50 2 (2 + 8 + 6) + 60 2 (4 + 7 + 3) )/100 - 42.3 2 = 99.71
σ 2 y = (11 2 (2) + 16 2 (4 + 6) + 21 2 (3 + 6 + 2) + 26 2 (45 + 8 + 4) + 31 2 (4 + 6 + 7) + 36 2 (3))/100 - 25.3 2 = 24.01
Where do we get standard deviations from:
σ x = 9.99 and σ y = 4.9
and covariance:
Cov(x,y) = (20 11 2 + 20 16 4 + 30 16 6 + 30 21 3 + 40 21 6 + 50 21 2 + 40 26 45 + 50 26 8 + 60 26 4 + 40 31 4 + 50 31 6 + 60 31 7 + 60 36 3)/100 - 42.3 25.3 = 38.11
Let's determine the correlation coefficient:


Let us write down the equations of the regression lines y(x):

and calculating, we get:
y x = 0.38 x + 9.14
Let us write down the equations of the regression lines x(y):

and calculating, we get:
x y = 1.59 y + 2.15
If we plot the points determined by the table and the regression lines, we will see that both lines pass through the point with coordinates (42.3; 25.3) and the points are located close to the regression lines.
Significance of the correlation coefficient.

Using the Student's table with significance level α=0.05 and degrees of freedom k=100-m-1 = 98, we find t crit:
t crit (n-m-1;α/2) = (98;0.025) = 1.984
where m = 1 is the number of explanatory variables.
If t observed > t critical, then the resulting value of the correlation coefficient is considered significant (the null hypothesis stating that the correlation coefficient is equal to zero is rejected).
Since t obs > t crit, we reject the hypothesis that the correlation coefficient is equal to 0. In other words, the correlation coefficient is statistically significant.

Exercise. The number of hits of pairs of values ​​of random variables X and Y in the corresponding intervals is given in the table. Using these data, find the sample correlation coefficient and sample equations of straight regression lines of Y on X and X on Y.
Solution

Example. The probability distribution of a two-dimensional random variable (X, Y) is given by a table. Find the laws of distribution of the component quantities X, Y and the correlation coefficient p(X, Y).
Download solution

Exercise. A two-dimensional discrete quantity (X, Y) is given by a distribution law. Find the laws of distribution of components X and Y, covariance and correlation coefficient.