Least squares solution with explanation. Linear pairwise regression analysis. Examples of solving problems using the least squares method
If a certain physical quantity depends on another quantity, then this dependence can be studied by measuring y at different values of x. As a result of measurements, a number of values are obtained:
x 1, x 2, ..., x i, ..., x n;
y 1 , y 2 , ..., y i , ... , y n .
Based on the data of such an experiment, it is possible to construct a graph of the dependence y = ƒ(x). The resulting curve makes it possible to judge the form of the function ƒ(x). However constant odds, which are included in this function remain unknown. The method allows us to determine them least squares. Experimental points, as a rule, do not lie exactly on the curve. The least squares method requires that the sum of the squares of the deviations of the experimental points from the curve, i.e. 2 was the smallest.
In practice, this method is most often (and most simply) used in the case of a linear relationship, i.e. When
y = kx or y = a + bx.
Linear dependence is very widespread in physics. And even when the relationship is nonlinear, they usually try to construct a graph so as to get a straight line. For example, if it is assumed that the refractive index of glass n is related to the light wavelength λ by the relation n = a + b/λ 2, then the dependence of n on λ -2 is plotted on the graph.
Consider the dependency y = kx(a straight line passing through the origin). Let's compose the value φ the sum of the squares of the deviations of our points from the straight line
The value of φ is always positive and turns out to be smaller the closer our points are to the straight line. The least squares method states that the value for k should be chosen such that φ has a minimum
or
(19)
The calculation shows that the root-mean-square error in determining the value of k is equal to
, (20)
where n is the number of measurements.
Let us now consider a slightly more difficult case, when the points must satisfy the formula y = a + bx(a straight line that does not pass through the origin).
The task is to find the best values of a and b from the available set of values x i, y i.
Let us again compose the quadratic form φ, equal to the sum of the squared deviations of points x i, y i from the straight line
and find the values of a and b for which φ has a minimum
;
.
The joint solution of these equations gives
(21)
The root mean square errors of determination of a and b are equal
(23)
. (24)
When processing measurement results using this method, it is convenient to summarize all the data in a table in which all the amounts included in formulas (19)(24) are preliminarily calculated. The forms of these tables are given in the examples below.
Example 1. The basic equation of dynamics was studied rotational movementε = M/J (line passing through the origin). At different values of the moment M, the angular acceleration ε of a certain body was measured. It is required to determine the moment of inertia of this body. The results of measurements of the moment of force and angular acceleration are listed in the second and third columns table 5.
Table 5
| n | M, N m | ε, s -1 | M 2 | M ε | ε - kM | (ε - kM) 2 |
| 1 | 1.44 | 0.52 | 2.0736 | 0.7488 | 0.039432 | 0.001555 |
| 2 | 3.12 | 1.06 | 9.7344 | 3.3072 | 0.018768 | 0.000352 |
| 3 | 4.59 | 1.45 | 21.0681 | 6.6555 | -0.08181 | 0.006693 |
| 4 | 5.90 | 1.92 | 34.81 | 11.328 | -0.049 | 0.002401 |
| 5 | 7.45 | 2.56 | 55.5025 | 19.072 | 0.073725 | 0.005435 |
| ∑ | | | 123.1886 | 41.1115 | | 0.016436 |
Using formula (19) we determine:
.
To determine the root mean square error, we use formula (20)
0.005775kg-1 · m -2 .
According to formula (18) we have
S J = (2.996 0.005775)/0.3337 = 0.05185 kg m2.
Having set the reliability P = 0.95, using the table of Student coefficients for n = 5, we find t = 2.78 and determine the absolute error ΔJ = 2.78 0.05185 = 0.1441 ≈ 0.2 kg m2.
Let's write the results in the form:
J = (3.0 ± 0.2) kg m2;
Example 2. Let's calculate the temperature coefficient of metal resistance using the least squares method. Resistance depends linearly on temperature
R t = R 0 (1 + α t°) = R 0 + R 0 α t°.
The free term determines the resistance R 0 at a temperature of 0 ° C, and the slope coefficient is the product of the temperature coefficient α and the resistance R 0 .
The results of measurements and calculations are given in the table ( see table 6).
Table 6
| n | t°, s | r, Ohm | t-¯t | (t-¯t) 2 | (t-¯t)r | r - bt - a | (r - bt - a) 2 .10 -6 |
| 1 | 23 | 1.242 | -62.8333 | 3948.028 | -78.039 | 0.007673 | 58.8722 |
| 2 | 59 | 1.326 | -26.8333 | 720.0278 | -35.581 | -0.00353 | 12.4959 |
| 3 | 84 | 1.386 | -1.83333 | 3.361111 | -2.541 | -0.00965 | 93.1506 |
| 4 | 96 | 1.417 | 10.16667 | 103.3611 | 14.40617 | -0.01039 | 107.898 |
| 5 | 120 | 1.512 | 34.16667 | 1167.361 | 51.66 | 0.021141 | 446.932 |
| 6 | 133 | 1.520 | 47.16667 | 2224.694 | 71.69333 | -0.00524 | 27.4556 |
| ∑ | 515 | 8.403 | | 8166.833 | 21.5985 | | 746.804 |
| ∑/n | 85.83333 | 1.4005 | | | | | |
Using formulas (21), (22) we determine
R 0 = ¯ R- α R 0 ¯ t = 1.4005 - 0.002645 85.83333 = 1.1735 Ohm.
Let's find an error in the definition of α. Since , then according to formula (18) we have:
.
Using formulas (23), (24) we have
;
0.014126 Ohm.
Having set the reliability to P = 0.95, using the table of Student coefficients for n = 6, we find t = 2.57 and determine the absolute error Δα = 2.57 0.000132 = 0.000338 deg -1.
α = (23 ± 4) 10 -4 hail-1 at P = 0.95.
Example 3. It is required to determine the radius of curvature of the lens using Newton's rings. The radii of Newton's rings r m were measured and the numbers of these rings m were determined. The radii of Newton's rings are related to the radius of curvature of the lens R and the ring number by the equation
r 2 m = mλR - 2d 0 R,
where d 0 the thickness of the gap between the lens and the plane-parallel plate (or the deformation of the lens),
λ wavelength of incident light.
λ = (600 ± 6) nm;
r 2 m = y;
m = x;
λR = b;
-2d 0 R = a,
then the equation will take the form y = a + bx.
.The results of measurements and calculations are entered into table 7.
Table 7
| n | x = m | y = r 2, 10 -2 mm 2 | m -¯ m | (m -¯m) 2 | (m -¯ m)y | y - bx - a, 10 -4 | (y - bx - a) 2 , 10 -6 |
| 1 | 1 | 6.101 | -2.5 | 6.25 | -0.152525 | 12.01 | 1.44229 |
| 2 | 2 | 11.834 | -1.5 | 2.25 | -0.17751 | -9.6 | 0.930766 |
| 3 | 3 | 17.808 | -0.5 | 0.25 | -0.08904 | -7.2 | 0.519086 |
| 4 | 4 | 23.814 | 0.5 | 0.25 | 0.11907 | -1.6 | 0.0243955 |
| 5 | 5 | 29.812 | 1.5 | 2.25 | 0.44718 | 3.28 | 0.107646 |
| 6 | 6 | 35.760 | 2.5 | 6.25 | 0.894 | 3.12 | 0.0975819 |
| ∑ | 21 | 125.129 | | 17.5 | 1.041175 | | 3.12176 |
| ∑/n | 3.5 | 20.8548333 | | | | | |
Least square method
In the final lesson of the topic, we will get acquainted with the most famous application FNP, which finds the widest application in various fields of science and practical activity. This could be physics, chemistry, biology, economics, sociology, psychology, and so on and so forth. By the will of fate, I often have to deal with the economy, and therefore today I will arrange for you a trip to an amazing country called Econometrics=) ...How can you not want it?! It’s very good there – you just need to make up your mind! ...But what you probably definitely want is to learn how to solve problems least squares method. And especially diligent readers will learn to solve them not only accurately, but also VERY QUICKLY ;-) But first general statement of the problem+ accompanying example:
Let us study indicators in a certain subject area that have a quantitative expression. At the same time, there is every reason to believe that the indicator depends on the indicator. This assumption can be either a scientific hypothesis or based on basic common sense. Let's leave science aside, however, and explore more appetizing areas - namely, grocery stores. Let's denote by:
– retail area of a grocery store, sq.m.,
– annual turnover of a grocery store, million rubles.
It is absolutely clear that the larger the store area, the greater in most cases its turnover will be.
Suppose that after carrying out observations/experiments/calculations/dances with a tambourine we have numerical data at our disposal: 
With grocery stores, I think everything is clear: - this is the area of the 1st store, - its annual turnover, - the area of the 2nd store, - its annual turnover, etc. By the way, it is not at all necessary to have access to classified materials– a fairly accurate assessment of trade turnover can be obtained by means of mathematical statistics
. However, let’s not get distracted, the commercial espionage course is already paid =)
Tabular data can also be written in the form of points and depicted in the familiar form Cartesian system .
Let's answer an important question: How many points are needed for a qualitative study?
The bigger, the better. The minimum acceptable set consists of 5-6 points. In addition, when the amount of data is small, “anomalous” results cannot be included in the sample. So, for example, a small elite store can earn orders of magnitude more than “its colleagues,” thereby distorting general pattern, which is what you need to find!
To put it very simply, we need to select a function, schedule which passes as close as possible to the points 
. This function is called approximating
(approximation - approximation) or theoretical function
. Generally speaking, an obvious “contender” immediately appears here - the polynomial high degree, whose graph passes through ALL points. But this option is complicated and often simply incorrect. (since the graph will “loop” all the time and poorly reflect the main trend).
Thus, the sought function must be quite simple and at the same time adequately reflect the dependence. As you might guess, one of the methods for finding such functions is called least squares method. First, let's look at its essence in general terms. Let some function approximate experimental data: 
How to evaluate the accuracy of this approximation? Let us also calculate the differences (deviations) between the experimental and functional values (we study the drawing). The first thought that comes to mind is to estimate how large the sum is, but the problem is that the differences can be negative (For example, 
)
and deviations as a result of such summation will cancel each other out. Therefore, as an estimate of the accuracy of the approximation, it begs to take the sum modules deviations:
or collapsed: (in case anyone doesn't know:
is the sum icon, and
– an auxiliary “counter” variable, which takes values from 1 to
)
.
By approximating experimental points with different functions, we will obtain different values, and obviously, where this sum is smaller, that function is more accurate.
Such a method exists and it is called least modulus method. However, in practice it has become much more widespread least square method, in which possible negative values are eliminated not by the module, but by squaring the deviations:
, after which efforts are aimed at selecting a function such that the sum of squared deviations 
was as small as possible. Actually, this is where the name of the method comes from.
And now we return to another important point: as noted above, the selected function should be quite simple - but there are also many such functions: linear , hyperbolic , exponential , logarithmic , quadratic etc. And, of course, here I would immediately like to “reduce the field of activity.” Which class of functions should I choose for research? A primitive but effective technique:
– The easiest way is to depict points on the drawing and analyze their location. If they tend to run in a straight line, then you should look for equation of a line 
with optimal values and . In other words, the task is to find SUCH coefficients so that the sum of squared deviations is the smallest.
If the points are located, for example, along hyperbole, then it is obviously clear that the linear function will give a poor approximation. In this case, we are looking for the most “favorable” coefficients for the hyperbola equation 
– those that give the minimum sum of squares 
.
Now note that in both cases we are talking about functions of two variables, whose arguments are searched dependency parameters:
And essentially we need to solve a standard problem - find minimum function of two variables.
Let's remember our example: suppose that “store” points tend to be located in a straight line and there is every reason to believe that linear dependence turnover from retail space. Let's find SUCH coefficients “a” and “be” such that the sum of squared deviations 
was the smallest. Everything is as usual - first 1st order partial derivatives. According to linearity rule You can differentiate right under the sum icon: 
If you want to use this information for an essay or coursework - I will be very grateful for the link in the list of sources; you will find such detailed calculations in few places: 
Let's create a standard system: 
We reduce each equation by “two” and, in addition, “break up” the sums: 
Note
: independently analyze why “a” and “be” can be taken out beyond the sum icon. By the way, formally this can be done with the sum 
Let's rewrite the system in “applied” form: 
after which the algorithm for solving our problem begins to emerge:
Do we know the coordinates of the points? We know. Amounts 
can we find it? Easily. Let's make the simplest system of two linear equations with two unknowns(“a” and “be”). We solve the system, for example, Cramer's method, as a result of which we obtain a stationary point. Checking sufficient condition extremum, we can verify that at this point the function 
reaches exactly minimum. The check involves additional calculations and therefore we will leave it behind the scenes (if necessary, the missing frame can be viewedHere
)
. We draw the final conclusion:
Function 
the best way (at least compared to any other linear function)
brings experimental points closer . Roughly speaking, its graph passes as close as possible to these points. In tradition econometrics the resulting approximating function is also called paired linear regression equation
.
The problem under consideration has a large practical significance. In our example situation, Eq. allows you to predict what trade turnover ("Igrek") the store will have at one or another value of the sales area (one or another meaning of “x”). Yes, the resulting forecast will only be a forecast, but in many cases it will turn out to be quite accurate.
I will analyze just one problem with “real” numbers, since there are no difficulties in it - all calculations are at the level school curriculum 7-8 grades. In 95 percent of cases, you will be asked to find just a linear function, but at the very end of the article I will show that it is no more difficult to find the equations of the optimal hyperbola, exponential and some other functions.
In fact, all that remains is to distribute the promised goodies - so that you can learn to solve such examples not only accurately, but also quickly. We carefully study the standard:
Task
As a result of studying the relationship between two indicators, the following pairs of numbers were obtained: 
Using the least squares method, find the linear function that best approximates the empirical (experienced) data. Make a drawing on which to construct experimental points and a graph of the approximating function in a Cartesian rectangular coordinate system 
. Find the sum of squared deviations between the empirical and theoretical values. Find out if the feature would be better (from the point of view of the least squares method) bring experimental points closer.
Please note that the “x” meanings are natural, and this has a characteristic meaningful meaning, which I will talk about a little later; but they, of course, can also be fractional. In addition, depending on the content of a particular task, both “X” and “game” values can be completely or partially negative. Well, we have been given a “faceless” task, and we begin it solution:
We find the coefficients of the optimal function as a solution to the system: 
For the purpose of more compact recording, the “counter” variable can be omitted, since it is already clear that the summation is carried out from 1 to .
It is more convenient to calculate the required amounts in tabular form: 
Calculations can be carried out on a microcalculator, but it is much better to use Excel - both faster and without errors; watch a short video:
Thus, we get the following system:
Here you can multiply the second equation by 3 and subtract the 2nd from the 1st equation term by term. But this is luck - in practice, systems are often not a gift, and in such cases it saves Cramer's method:
, which means the system has a unique solution.
Let's check. I understand that you don’t want to, but why skip errors where they can absolutely not be missed? Let us substitute the found solution into the left side of each equation of the system:
The right-hand sides of the corresponding equations are obtained, which means that the system is solved correctly.
Thus, the desired approximating function: – from all linear functions It is she who best approximates the experimental data.
Unlike straight
dependence of the store's turnover on its area, the found dependence is reverse
(principle “the more, the less”), and this fact is immediately revealed by the negative slope. Function 
tells us that with an increase in a certain indicator by 1 unit, the value of the dependent indicator decreases average by 0.65 units. As they say, the higher the price of buckwheat, the less it is sold.
To plot the graph of the approximating function, we find its two values:
and execute the drawing: 
The constructed straight line is called trend line
(namely, a linear trend line, i.e. in the general case, a trend is not necessarily a straight line). Everyone is familiar with the expression “to be in trend,” and I think that this term does not need additional comments.
Let's calculate the sum of squared deviations 
between empirical and theoretical values. Geometrically, this is the sum of the squares of the lengths of the “raspberry” segments (two of which are so small that they are not even visible).
Let's summarize the calculations in a table: 
Again, they can be done manually; just in case, I’ll give an example for the 1st point: 
but it is much more effective to do it in the already known way:
We repeat once again: What is the meaning of the result obtained? From all linear functions y function the indicator is the smallest, that is, in its family it is the best approximation. And here, by the way, the final question of the problem is not accidental: what if the proposed exponential function 
would it be better to bring the experimental points closer?
Let's find the corresponding sum of squared deviations - to distinguish, I will denote them by the letter “epsilon”. The technique is exactly the same: 
And again, just in case, the calculations for the 1st point: 
In Excel we use the standard function EXP (syntax can be found in Excel Help).
Conclusion: , which means that the exponential function approximates the experimental points worse than a straight line .
But here it should be noted that “worse” is doesn't mean yet, what is wrong. Now I have built a graph of this exponential function - and it also passes close to the points 
- yes, so without analytical research and it is difficult to say which function is more accurate.
This concludes the solution, and I return to the question of the natural values of the argument. IN various studies, as a rule, economic or sociological, natural “X’s” are used to number months, years or other equal time intervals. Consider, for example, the following problem:
The following data is available on the store’s retail turnover for the first half of the year:
Using analytical straight line alignment, determine the volume of turnover for July.
Yes, no problem: we number the months 1, 2, 3, 4, 5, 6 and use the usual algorithm, as a result of which we get an equation - the only thing is that when it comes to time, they usually use the letter “te” (although this is not critical). The resulting equation shows that in the first half of the year trade turnover increased by an average of 27.74 units. per month. Let's get the forecast for July (month no. 7): d.e.
And there are countless tasks like this. Those who wish can use an additional service, namely my Excel calculator (demo version), which solves the analyzed problem almost instantly! Working version of the program is available in exchange or for symbolic fee.
At the end of the lesson, brief information about finding dependencies of some other types. Actually, there’s not much to tell, since the fundamental approach and solution algorithm remain the same.
Let us assume that the arrangement of the experimental points resembles a hyperbola. Then, to find the coefficients of the best hyperbola, you need to find the minimum of the function - anyone can carry out detailed calculations and come to similar system:
From a formal technical point of view, it is obtained from a “linear” system 
(let's denote it with an asterisk) replacing "x" with . Well, what about the amounts? 
calculate, after which to the optimal coefficients “a” and “be” close at hand.
If there is every reason to believe that the points are located along a logarithmic curve, then to find the optimal values we find the minimum of the function 
. Formally, in the system (*) needs to be replaced with: 
When performing calculations in Excel, use the function LN. I confess that it would not be particularly difficult for me to create calculators for each of the cases under consideration, but it would still be better if you “programmed” the calculations yourself. Lesson videos to help.
With exponential dependence the situation is a little more complicated. To reduce the matter to the linear case, we take the function logarithm and use properties of the logarithm:
Now, comparing the resulting function with the linear function, we come to the conclusion that in the system (*) must be replaced by , and – by . For convenience, let's denote: 
Please note that the system is resolved with respect to and, and therefore, after finding the roots, you must not forget to find the coefficient itself.
To bring experimental points closer optimal parabola 
, should be found minimum function of three variables 
. After performing standard actions, we get the following “working” system:
Yes, of course, there are more amounts here, but there are no difficulties at all when using your favorite application. And finally, I’ll tell you how to quickly perform a check using Excel and build the desired trend line: create a scatter plot, select any of the points with the mouse and right click select the option "Add trend line". Next, select the chart type and on the tab "Options" activate the option "Show equation on diagram". OK
As always, I would like to end the article with some in a beautiful phrase, and I almost typed “Be trendy!” But he changed his mind in time. And not because it is stereotyped. I don’t know how it is for anyone, but I don’t really want to follow the promoted American and especially European trend =) Therefore, I wish each of you to stick to your own line!
http://www.grandars.ru/student/vysshaya-matematika/metod-naimenshih-kvadratov.html
The least squares method is one of the most common and most developed due to its simplicity and efficiency of methods for estimating parameters of linear econometric models. At the same time, when using it, some caution should be observed, since models constructed using it may not satisfy a number of requirements for the quality of their parameters and, as a result, do not reflect the patterns of process development “well” enough.
Let us consider the procedure for estimating the parameters of a linear econometric model using the least squares method in more detail. Such a model in general can be represented by equation (1.2):
y t = a 0 + a 1 x 1t +...+ a n x nt + ε t.
The initial data when estimating the parameters a 0 , a 1 ,..., a n is a vector of values of the dependent variable y= (y 1 , y 2 , ... , y T)" and the matrix of values of independent variables
in which the first column, consisting of ones, corresponds to the model coefficient.
The least squares method received its name based on the basic principle that the parameter estimates obtained on its basis must satisfy: the sum of squares of the model error should be minimal.
Examples of solving problems using the least squares method
Example 2.1. The trading enterprise has a network of 12 stores, information on the activities of which is presented in table. 2.1.
The management of the enterprise would like to know how the size of the annual turnover depends on the retail space of the store.
Table 2.1
| Store number | Annual turnover, million rubles. | Retail area, thousand m2 |
| 19,76 | 0,24 | |
| 38,09 | 0,31 | |
| 40,95 | 0,55 | |
| 41,08 | 0,48 | |
| 56,29 | 0,78 | |
| 68,51 | 0,98 | |
| 75,01 | 0,94 | |
| 89,05 | 1,21 | |
| 91,13 | 1,29 | |
| 91,26 | 1,12 | |
| 99,84 | 1,29 | |
| 108,55 | 1,49 |
Least squares solution. Let us denote the annual turnover of the th store, million rubles; - retail area of the th store, thousand m2.
Fig.2.1. Scatterplot for Example 2.1
To determine the form of the functional relationship between the variables and we will construct a scatter diagram (Fig. 2.1).
Based on the scatter diagram, we can conclude that annual turnover is positively dependent on retail space (i.e., y will increase with increasing ). The most suitable form of functional connection is linear.
Information for further calculations is presented in table. 2.2. Using the least squares method, we estimate the parameters of a linear one-factor econometric model
Table 2.2
| t | y t | x 1t | y t 2 | x 1t 2 | x 1t y t |
| 19,76 | 0,24 | 390,4576 | 0,0576 | 4,7424 | |
| 38,09 | 0,31 | 1450,8481 | 0,0961 | 11,8079 | |
| 40,95 | 0,55 | 1676,9025 | 0,3025 | 22,5225 | |
| 41,08 | 0,48 | 1687,5664 | 0,2304 | 19,7184 | |
| 56,29 | 0,78 | 3168,5641 | 0,6084 | 43,9062 | |
| 68,51 | 0,98 | 4693,6201 | 0,9604 | 67,1398 | |
| 75,01 | 0,94 | 5626,5001 | 0,8836 | 70,5094 | |
| 89,05 | 1,21 | 7929,9025 | 1,4641 | 107,7505 | |
| 91,13 | 1,29 | 8304,6769 | 1,6641 | 117,5577 | |
| 91,26 | 1,12 | 8328,3876 | 1,2544 | 102,2112 | |
| 99,84 | 1,29 | 9968,0256 | 1,6641 | 128,7936 | |
| 108,55 | 1,49 | 11783,1025 | 2,2201 | 161,7395 | |
| S | 819,52 | 10,68 | 65008,554 | 11,4058 | 858,3991 |
| Average | 68,29 | 0,89 |
Thus,
Therefore, with an increase in retail space by 1 thousand m2, other things being equal, the average annual turnover increases by 67.8871 million rubles.
Example 2.2. The company's management noticed that the annual turnover depends not only on the store's sales area (see example 2.1), but also on the average number of visitors. The relevant information is presented in table. 2.3.
Table 2.3
Solution. Let us denote - the average number of visitors to the th store per day, thousand people.
To determine the form of the functional relationship between the variables and we will construct a scatter diagram (Fig. 2.2).
Based on the scatterplot, we can conclude that annual turnover is positively dependent on the average number of visitors per day (i.e., y will increase with increasing ). The form of functional dependence is linear.
Rice. 2.2. Scatterplot for Example 2.2
Table 2.4
| t | x 2t | x 2t 2 | y t x 2t | x 1t x 2t |
| 8,25 | 68,0625 | 163,02 | 1,98 | |
| 10,24 | 104,8575 | 390,0416 | 3,1744 | |
| 9,31 | 86,6761 | 381,2445 | 5,1205 | |
| 11,01 | 121,2201 | 452,2908 | 5,2848 | |
| 8,54 | 72,9316 | 480,7166 | 6,6612 | |
| 7,51 | 56,4001 | 514,5101 | 7,3598 | |
| 12,36 | 152,7696 | 927,1236 | 11,6184 | |
| 10,81 | 116,8561 | 962,6305 | 13,0801 | |
| 9,89 | 97,8121 | 901,2757 | 12,7581 | |
| 13,72 | 188,2384 | 1252,0872 | 15,3664 | |
| 12,27 | 150,5529 | 1225,0368 | 15,8283 | |
| 13,92 | 193,7664 | 1511,016 | 20,7408 | |
| S | 127,83 | 1410,44 | 9160,9934 | 118,9728 |
| Average | 10,65 |
In general, it is necessary to determine the parameters of a two-factor econometric model
y t = a 0 + a 1 x 1t + a 2 x 2t + ε t
The information required for further calculations is presented in table. 2.4.
Let us estimate the parameters of a linear two-factor econometric model using the least squares method.
Thus,
Estimation of the coefficient =61.6583 shows that, other things being equal, with an increase in retail space by 1 thousand m 2, the annual turnover will increase by an average of 61.6583 million rubles.
The coefficient estimate = 2.2748 shows that, other things being equal, with an increase in the average number of visitors per 1 thousand people. per day, annual turnover will increase by an average of 2.2748 million rubles.
Example 2.3. Using the information presented in table. 2.2 and 2.4, estimate the parameter of the one-factor econometric model
where is the centered value of the annual turnover of the th store, million rubles; - centered value of the average daily number of visitors to the t-th store, thousand people. (see examples 2.1-2.2).
Solution. Additional information required for calculations is presented in table. 2.5.
Table 2.5
| -48,53 | -2,40 | 5,7720 | 116,6013 | |
| -30,20 | -0,41 | 0,1702 | 12,4589 | |
| -27,34 | -1,34 | 1,8023 | 36,7084 | |
| -27,21 | 0,36 | 0,1278 | -9,7288 | |
| -12,00 | -2,11 | 4,4627 | 25,3570 | |
| 0,22 | -3,14 | 9,8753 | -0,6809 | |
| 6,72 | 1,71 | 2,9156 | 11,4687 | |
| 20,76 | 0,16 | 0,0348 | 3,2992 | |
| 22,84 | -0,76 | 0,5814 | -17,413 | |
| 22,97 | 3,07 | 9,4096 | 70,4503 | |
| 31,55 | 1,62 | 2,6163 | 51,0267 | |
| 40,26 | 3,27 | 10,6766 | 131,5387 | |
| Amount | 48,4344 | 431,0566 |
Using formula (2.35), we obtain
Thus,
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Example.
Experimental data on the values of variables X And at are given in the table. 
As a result of their alignment, the function is obtained 
Using least square method, approximate these data by a linear dependence y=ax+b(find parameters A And b). Find out which of the two lines better (in the sense of the least squares method) aligns the experimental data. Make a drawing.
Solution.
In our example n=5. We fill out the table for the convenience of calculating the amounts that are included in the formulas of the required coefficients. 
The values in the fourth row of the table are obtained by multiplying the values of the 2nd row by the values of the 3rd row for each number i.
The values in the fifth row of the table are obtained by squaring the values in the 2nd row for each number i.
The values in the last column of the table are the sums of the values across the rows.
We use the formulas of the least squares method to find the coefficients A And b. We substitute the corresponding values from the last column of the table into them: 
Hence, y = 0.165x+2.184- the desired approximating straight line.
It remains to find out which of the lines y = 0.165x+2.184 or better approximates the original data, that is, makes an estimate using the least squares method.
Proof.
So that when found A And b function took smallest value, it is necessary that at this point the matrix of the quadratic form of the second-order differential for the function 
was positive definite. Let's show it.
The second order differential has the form: 
That is
Therefore, the matrix of quadratic form has the form 
and the values of the elements do not depend on A And b.
Let us show that the matrix is positive definite. To do this, the angular minors must be positive.
Angular minor of the first order 
. The inequality is strict, since the points
The essence of the least squares method is in finding the parameters of a trend model that best describes the tendency of development of any random phenomenon in time or space (a trend is a line that characterizes the tendency of this development). The task of the least squares method (LSM) comes down to finding not just some trend model, but to finding the best or optimal model. This model will be optimal if the sum of square deviations between the observed actual values and the corresponding calculated trend values is minimal (smallest):
where is the square deviation between the observed actual value
and the corresponding calculated trend value,
The actual (observed) value of the phenomenon being studied,
The calculated value of the trend model,
The number of observations of the phenomenon being studied.
MNC is used quite rarely on its own. As a rule, most often it is used only as a necessary technical technique in correlation studies. It should be remembered that the information basis of OLS can only be a reliable statistical series, and the number of observations should not be less than 4, otherwise the smoothing procedures of OLS may lose common sense.
The MNC toolkit boils down to the following procedures:
First procedure. It turns out whether there is any tendency at all to change the resultant attribute when the selected factor-argument changes, or in other words, is there a connection between “ at " And " X ».
Second procedure. It is determined which line (trajectory) can best describe or characterize this trend.
Third procedure.
Example. Let's say we have information about the average sunflower yield for the farm under study (Table 9.1).
Table 9.1
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Observation number |
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Productivity, c/ha |
Since the level of technology in sunflower production in our country has remained virtually unchanged over the past 10 years, it means that, apparently, fluctuations in yield during the analyzed period were very much dependent on fluctuations in weather and climatic conditions. Is this really true?
First OLS procedure. The hypothesis about the existence of a trend in sunflower yield changes depending on changes in weather and climatic conditions over the analyzed 10 years is tested.
In this example, for " y " it is advisable to take the sunflower yield, and for " x » – number of the observed year in the analyzed period. Testing the hypothesis about the existence of any relationship between " x " And " y "can be done in two ways: manually and using computer programs. Of course, with the availability of computer technology, this problem can be solved by itself. But in order to better understand the MNC tools, it is advisable to test the hypothesis about the existence of a relationship between “ x " And " y » manually, when only a pen and an ordinary calculator are at hand. In such cases, the hypothesis about the existence of a trend is best checked visually by location graphic image analyzed series of dynamics - correlation field:
The correlation field in our example is located around a slowly increasing line. This in itself indicates the existence of a certain trend in changes in sunflower yields. It is impossible to talk about the presence of any tendency only when the correlation field looks like a circle, a circle, a strictly vertical or strictly horizontal cloud, or consists of chaotically scattered points. In all other cases, the hypothesis about the existence of a relationship between “ x " And " y ", and continue research.
Second OLS procedure. It is determined which line (trajectory) can best describe or characterize the trend of changes in sunflower yield over the analyzed period.
If you have computer technology, the selection of the optimal trend occurs automatically. In “manual” processing, the selection of the optimal function is carried out, as a rule, visually - by the location of the correlation field. That is, based on the type of graph, the equation of the line that best fits the empirical trend (the actual trajectory) is selected.
As you know, there is a huge diversity in nature functional dependencies, therefore, visually analyzing even a small part of them is extremely difficult. Fortunately, in real economic practice, most relationships can be described quite accurately either by a parabola, or a hyperbola, or a straight line. In this regard, with the “manual” option of selecting the best function, you can limit yourself to only these three models.
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Hyperbola: |
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Second order parabola: 
:
It is easy to see that in our example, the trend in sunflower yield changes over the analyzed 10 years is best characterized by a straight line, so the regression equation will be the equation of a straight line.
Third procedure. The parameters of the regression equation characterizing this line are calculated, or in other words, an analytical formula is determined that describes the best trend model.
Finding the values of the parameters of the regression equation, in our case the parameters and , is the core of the OLS. This process comes down to solving a system of normal equations.
(9.2)
This system of equations can be solved quite easily by the Gauss method. Let us recall that as a result of the solution, in our example, the values of the parameters and are found. Thus, the found regression equation will have the following form: 
