Where is the method of least squares applied? The method of least squares in Excel. Regression analysis Least squares regression

Choosing the type of regression function, i.e. the type of the considered model of the dependence of Y on X (or X on Y), for example, a linear model y x = a + bx, it is necessary to determine the specific values ​​of the coefficients of the model.

For different values ​​of a and b, it is possible to construct an infinite number of dependencies of the form y x =a+bx, i.e., there are an infinite number of lines on the coordinate plane, but we need such a dependence that corresponds to the observed values ​​in the best way. Thus, the problem is reduced to the selection of the best coefficients.

We are looking for a linear function a + bx, based only on a certain number of available observations. To find the function with the best fit to the observed values, we use the least squares method.

Denote: Y i - the value calculated by the equation Y i =a+bx i . y i - measured value, ε i =y i -Y i - difference between the measured and calculated values, ε i =y i -a-bx i .

The method of least squares requires that ε i , the difference between the measured y i and the values ​​of Y i calculated from the equation, be minimal. Therefore, we find the coefficients a and b so that the sum of the squared deviations of the observed values ​​from the values ​​on the straight regression line is the smallest:

Investigating this function of arguments a and with the help of derivatives to an extremum, we can prove that the function takes on a minimum value if the coefficients a and b are solutions of the system:

(2)

If we divide both sides of the normal equations by n, we get:

Given that (3)

Get , from here, substituting the value of a in the first equation, we get:

In this case, b is called the regression coefficient; a is called the free member of the regression equation and is calculated by the formula:

The resulting straight line is an estimate for the theoretical regression line. We have:

So, is a linear regression equation.

Regression can be direct (b>0) and inverse (b Example 1. The results of measuring the X and Y values ​​are given in the table:

Assuming that there is a linear relationship between X and Y y=a+bx, determine the coefficients a and b using the least squares method.

Solution. Here n=5
x i =-2+0+1+2+4=5;
x i 2 =4+0+1+4+16=25
x i y i =-2 0.5+0 1+1 1.5+2 2+4 3=16.5
y i =0.5+1+1.5+2+3=8

and normal system (2) has the form

Solving this system, we get: b=0.425, a=1.175. Therefore y=1.175+0.425x.

Example 2. There is a sample of 10 observations of economic indicators (X) and (Y).

It is required to find a sample regression equation Y on X. Construct a sample regression line Y on X.

Solution. 1. Let's sort the data by values ​​x i and y i . We get a new table:

To simplify the calculations, we will compile a calculation table in which we will enter the necessary numerical values.

According to formula (4), we calculate the regression coefficient

and by formula (5)

Thus, the sample regression equation looks like y=-59.34+1.3804x.
Let's plot the points (x i ; y i) on the coordinate plane and mark the regression line.

Fig 4

Figure 4 shows how the observed values ​​are located relative to the regression line. To numerically estimate the deviations of y i from Y i , where y i are observed values, and Y i are values ​​determined by regression, we will make a table:

Y i values ​​are calculated according to the regression equation.

The noticeable deviation of some observed values ​​from the regression line is explained by the small number of observations. When studying the degree of linear dependence of Y on X, the number of observations is taken into account. The strength of the dependence is determined by the value of the correlation coefficient.

The least squares method is one of the most common and most developed due to its simplicity and efficiency of methods for estimating the parameters of linear. At the same time, some caution should be observed when using it, since the models built using it may not meet a number of requirements for the quality of their parameters and, as a result, not “well” reflect the patterns of process development.

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 form can be represented by equation (1.2):

y t = a 0 + a 1 x 1 t +...+ a n x nt + ε t .

The initial data when estimating the parameters a 0 , a 1 ,..., a n is the 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 coefficient of the model .

The method of least squares got its name based on the basic principle that the parameter estimates obtained on its basis should satisfy: the sum of squares of the model error should be minimal.

Examples of solving problems by the least squares method

Example 2.1. The trading enterprise has a network consisting of 12 stores, information on the activities of which is presented in Table. 2.1.

The company's management would like to know how the size of the annual depends on the sales area of ​​the store.

Table 2.1

Least squares solution. Let us designate - the annual turnover of the -th store, million rubles; - selling area of ​​the -th store, thousand m 2.

Fig.2.1. Scatterplot for Example 2.1

To determine the form of the functional relationship between the variables and construct a scatterplot (Fig. 2.1).

Based on the scatter diagram, we can conclude that the annual turnover is positively dependent on the selling area (i.e., y will increase with the growth of ). The most appropriate 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 the linear one-factor econometric model

Table 2.2

In this way,

Therefore, with an increase in the trading area by 1 thousand m 2, other things being equal, the average annual turnover increases by 67.8871 million rubles.

Example 2.2. The management of the enterprise noticed that the annual turnover depends not only on the sales area of ​​the store (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. 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 construct a scatterplot (Fig. 2.2).

Based on the scatter diagram, we can conclude that the annual turnover is positively related to the average number of visitors per day (i.e., y will increase with the growth of ). The form of functional dependence is linear.

Rice. 2.2. Scatterplot for example 2.2

Table 2.4

In general, it is necessary to determine the parameters of the two-factor econometric model

y t \u003d a 0 + a 1 x 1 t + a 2 x 2 t + 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.

In this way,

Evaluation of the coefficient = 61.6583 shows that, all other things being equal, with an increase in sales area by 1 thousand m 2, the annual turnover will increase by an average of 61.6583 million rubles.

Which finds the widest application in various fields of science and practice. It can 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 ticket to an amazing country called Econometrics=) … How do you not want that?! It's very good there - you just have to decide! …But what you probably definitely want is to learn how to solve problems least squares. And especially diligent readers will learn to solve them not only accurately, but also VERY FAST ;-) But first general statement of the problem+ related example:

Let indicators be studied in some 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 both a scientific hypothesis and based on elementary common sense. Let's leave science aside, however, and explore more appetizing areas - namely, grocery stores. Denote by:

– retail space of a grocery store, sq.m.,
- annual turnover of a grocery store, million rubles.

It is quite clear that the larger the area of ​​the store, the greater its turnover in most cases.

Suppose that after conducting observations / experiments / calculations / dancing with a tambourine, we have at our disposal numerical data:

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 the turnover can be obtained using mathematical statistics. However, do not be distracted, the course of commercial espionage is already paid =)

Tabular data can also be written in the form of points and depicted in the usual way for us. Cartesian system .

Let's answer an important question: how many points are needed for a qualitative study?

The bigger, the better. The minimum admissible set consists of 5-6 points. In addition, with a small amount of data, “abnormal” results should not be included in the sample. So, for example, a small elite store can help out orders of magnitude more than “their colleagues”, thereby distorting the general pattern that needs to be found!

If it’s quite simple, we need to choose a function , schedule which passes as close as possible to the points . Such a function is called approximating (approximation - approximation) or theoretical function . Generally speaking, here immediately appears an obvious "pretender" - a polynomial of high degree, the graph of which passes through ALL points. But this option is complicated, and often simply incorrect. (because the chart will “wind” all the time and poorly reflect the main trend).

Thus, the desired function must be sufficiently simple and at the same time reflect the dependence adequately. As you might guess, one of the methods for finding such functions is called least squares. First, let's analyze its essence in a general way. Let some function approximate the 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 big 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 suggests itself to take the sum modules deviations:

or in folded form: (suddenly, who doesn’t know: is the sum icon, and is an auxiliary variable-“counter”, which takes values ​​from 1 to ).

By approximating the experimental points with different functions, we will obtain different values ​​of , and it is obvious that where this sum is smaller, that function is more accurate.

Such a method exists and 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 modulus, but by squaring the deviations:

, after which efforts are directed to the selection of such a function that the sum of the squared deviations was as small as possible. Actually, hence the name of the method.

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." What class of functions to choose for research? Primitive but effective technique:

- The easiest way to draw points on the drawing and analyze their location. If they tend to be in a straight line, then you should look for straight line equation with optimal values ​​and . In other words, the task is to find SUCH coefficients - so that the sum of the squared deviations is the smallest.

If the points are located, for example, along hyperbole, then it is 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 notice that in both cases we are talking about functions of two variables, whose arguments are searched dependency options:

And in essence, we need to solve a standard problem - to find minimum of a function of two variables.

Recall our example: suppose that the "shop" points tend to be located in a straight line and there is every reason to believe the presence linear dependence turnover from the trading area. Let's find SUCH coefficients "a" and "be" so that the sum of squared deviations was the smallest. Everything as usual - first partial derivatives of the 1st order. 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 not find such detailed calculations anywhere:

Let's make a standard system:

We reduce each equation by a “two” and, in addition, “break apart” the sums:

Note : independently analyze why "a" and "be" can be taken out of the sum icon. By the way, formally this can be done with the sum

Let's rewrite the system in an "applied" form:

after which the algorithm for solving our problem begins to be drawn:

Do we know the coordinates of the points? We know. Sums can we find? Easily. We compose the simplest system of two linear equations with two unknowns("a" and "beh"). We solve the system, for example, Cramer's method, resulting in a stationary point . Checking sufficient condition for an extremum, we can verify that at this point the function reaches precisely minimum. Verification is associated with additional calculations and therefore we will leave it behind the scenes. (if necessary, the missing frame can be viewed). 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 is of great practical importance. In the situation with our example, the equation allows you to predict what kind of turnover ("yig") will be at the store with one or another value of the selling area (one or another meaning of "x"). Yes, the resulting forecast will be only 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 of the school curriculum in grades 7-8. 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 for the optimal hyperbola, exponent, and some other functions.

In fact, it remains to distribute the promised goodies - so that you learn how to solve such examples not only accurately, but also quickly. We carefully study the standard:

A 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, in a Cartesian rectangular coordinate system, plot experimental points and a graph of the approximating function . Find the sum of squared deviations between empirical and theoretical values. Find out if the function is better (in terms of the least squares method) approximate experimental points.

Note that "x" values ​​are natural values, and this has a characteristic meaningful meaning, which I will talk about a little later; but they, of course, can be fractional. In addition, depending on the content of a particular task, both "X" and "G" values ​​can be completely or partially negative. Well, we have been given a “faceless” task, and we start it solution:

We find the coefficients of the optimal function as a solution to the system:

For the purposes of a more compact notation, 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 a 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 gifted, and in such cases it saves Cramer's method:
, so the system has a unique solution.

Let's do a check. I understand that I don’t want to, but why skip mistakes where you can absolutely not miss them? Substitute the found solution into the left side of each equation of the system:

The right parts of the corresponding equations are obtained, which means that the system is solved correctly.

Thus, the desired approximating function: – from all linear functions experimental data is best approximated by it.

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 angular coefficient. Function informs 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 sold.

To plot the approximating function, we find two of its values:

and execute the drawing:


The constructed 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.

Calculate the sum of squared deviations between empirical and theoretical values. Geometrically, this is the sum of the squares of the lengths of the "crimson" segments (two of which are so small you can't even see them).

Let's summarize the calculations in a table:


They can again be carried out manually, just in case I will give an example for the 1st point:

but it is much more efficient to do the already known way:

Let's repeat: what is the meaning of the result? From all linear functions function the exponent is the smallest, that is, it is the best approximation in its family. And here, by the way, the final question of the problem is not accidental: what if the proposed exponential function will it be better to approximate the experimental points?

Let's find the corresponding sum of squared deviations - to distinguish them, I will designate them with the letter "epsilon". The technique is exactly the same:


And again for every fire calculation for the 1st point:

In Excel, we use the standard function EXP (Syntax can be found in Excel Help).

Conclusion: , so the exponential function approximates the experimental points worse than the straight line .

But it should be noted here that "worse" is doesn't mean yet, what is wrong. Now I built a graph of this exponential function - and it also passes close to the points - so much so that without an analytical study it is difficult to say which function is more accurate.

This completes the solution, and I return to the question of the natural values ​​of the argument. In various studies, as a rule, economic or sociological, months, years or other equal time intervals are numbered with natural "X". Consider, for example, such a problem.

The method of least squares (LSM) allows you to estimate various quantities using the results of many measurements containing random errors.

Characteristic MNC

The main idea of ​​this method is that the sum of squared errors is considered as a criterion for the accuracy of the solution of the problem, which is sought to be minimized. When using this method, both numerical and analytical approaches can be applied.

In particular, as a numerical implementation, the least squares method implies making as many measurements of an unknown random variable as possible. Moreover, the more calculations, the more accurate the solution will be. On this set of calculations (initial data), another set of proposed solutions is obtained, from which the best one is then selected. If the set of solutions is parametrized, then the least squares method will be reduced to finding the optimal value of the parameters.

As an analytical approach to the implementation of the LSM on the set of initial data (measurements) and the proposed set of solutions, some (functional) is defined, which can be expressed by a formula obtained as a certain hypothesis that needs to be confirmed. In this case, the least squares method is reduced to finding the minimum of this functional on the set of squared errors of the initial data.

Note that not the errors themselves, but the squares of the errors. Why? The fact is that often the deviations of measurements from the exact value are both positive and negative. When determining the average, simple summation can lead to an incorrect conclusion about the quality of the estimate, since the mutual cancellation of positive and negative values ​​will reduce the sampling power of the set of measurements. And, consequently, the accuracy of the assessment.

To prevent this from happening, the squared deviations are summed up. Even more than that, in order to equalize the dimension of the measured value and the final estimate, the sum of squared errors is used to extract

Some applications of MNCs

MNC is widely used in various fields. For example, in probability theory and mathematical statistics, the method is used to determine such a characteristic of a random variable as the standard deviation, which determines the width of the range of values ​​of a random variable.

After alignment, we get a function of the following form: g (x) = x + 1 3 + 1 .

We can approximate this data with a linear relationship y = a x + b by calculating the appropriate parameters. To do this, we will need to apply the so-called least squares method. You will also need to make a drawing to check which line will best align the experimental data.

What exactly is OLS (least squares method)

The main thing we need to do is to find such linear dependence coefficients at which the value of the function of two variables F (a, b) = ∑ i = 1 n (y i - (a x i + b)) 2 will be the smallest. In other words, for certain values ​​of a and b, the sum of the squared deviations of the presented data from the resulting straight line will have a minimum value. This is the meaning of the least squares method. All we have to do to solve the example is to find the extremum of the function of two variables.

How to derive formulas for calculating coefficients

In order to derive formulas for calculating the coefficients, it is necessary to compose and solve a system of equations with two variables. To do this, we calculate the partial derivatives of the expression F (a , b) = ∑ i = 1 n (y i - (a x i + b)) 2 with respect to a and b and equate them to 0 .

δ F (a , b) δ a = 0 δ F (a , b) δ b = 0 ⇔ - 2 ∑ i = 1 n (y i - (a x i + b)) x i = 0 - 2 ∑ i = 1 n ( y i - (a x i + b)) = 0 ⇔ a ∑ i = 1 n x i 2 + b ∑ i = 1 n x i = ∑ i = 1 n x i y i a ∑ i = 1 n x i + ∑ i = 1 n b = ∑ i = 1 n y i ⇔ a ∑ i = 1 n x i 2 + b ∑ i = 1 n x i = ∑ i = 1 n x i y i a ∑ i = 1 n x i + n b = ∑ i = 1 n y i

To solve a system of equations, you can use any methods, such as substitution or Cramer's method. As a result, we should get formulas that calculate the coefficients using the least squares method.

n ∑ i = 1 n x i y i - ∑ i = 1 n x i ∑ i = 1 n y i n ∑ i = 1 n - ∑ i = 1 n x i 2 b = ∑ i = 1 n y i - a ∑ i = 1 n x i n

We have calculated the values ​​of the variables for which the function
F (a , b) = ∑ i = 1 n (y i - (a x i + b)) 2 will take the minimum value. In the third paragraph, we will prove why it is so.

This is the application of the least squares method in practice. His formula, which is used to find the parameter a , includes ∑ i = 1 n x i , ∑ i = 1 n y i , ∑ i = 1 n x i y i , ∑ i = 1 n x i 2 , and the parameter
n - it denotes the amount of experimental data. We advise you to calculate each amount separately. The coefficient value b is calculated immediately after a .

Let's go back to the original example.

Example 1

Here we have n equal to five. To make it more convenient to calculate the required amounts included in the coefficient formulas, we fill out the table.

Solution

The fourth row contains the data obtained by multiplying the values ​​from the second row by the values ​​of the third for each individual i . The fifth line contains the data from the second squared. The last column shows the sums of the values ​​of the individual rows.

Let's use the least squares method to calculate the coefficients a and b we need. To do this, substitute the desired values ​​from the last column and calculate the sums:

n ∑ i = 1 n x i y i - ∑ i = 1 n x i ∑ i = 1 n y i n ∑ i = 1 n - ∑ i = 1 n x i 2 b = ∑ i = 1 n y i - a ∑ i = 1 n x i n ⇒ a = 5 33 , 8 - 12 12, 9 5 46 - 12 2 b = 12, 9 - a 12 5 ⇒ a ≈ 0, 165 b ≈ 2, 184

We got that the desired approximating straight line will look like y = 0 , 165 x + 2 , 184 . Now we need to determine which line will best approximate the data - g (x) = x + 1 3 + 1 or 0 , 165 x + 2 , 184 . Let's make an estimate using the least squares method.

To calculate the error, we need to find the sums of squared deviations of the data from the lines σ 1 = ∑ i = 1 n (y i - (a x i + b i)) 2 and σ 2 = ∑ i = 1 n (y i - g (x i)) 2 , the minimum value will correspond to a more suitable line.

σ 1 = ∑ i = 1 n (y i - (a x i + b i)) 2 = = ∑ i = 1 5 (y i - (0 , 165 x i + 2 , 184)) 2 ≈ 0 , 019 σ 2 = ∑ i = 1 n (y i - g (x i)) 2 = = ∑ i = 1 5 (y i - (x i + 1 3 + 1)) 2 ≈ 0 , 096

Answer: since σ 1< σ 2 , то прямой, наилучшим образом аппроксимирующей исходные данные, будет
y = 0 , 165 x + 2 , 184 .

The least squares method is clearly shown in the graphic illustration. The red line marks the straight line g (x) = x + 1 3 + 1, the blue line marks y = 0, 165 x + 2, 184. Raw data are marked with pink dots.

Let us explain why exactly approximations of this type are needed.

They can be used in problems that require data smoothing, as well as in those where the data needs to be interpolated or extrapolated. For example, in the problem discussed above, one could find the value of the observed quantity y at x = 3 or at x = 6 . We have devoted a separate article to such examples.

Proof of the LSM method

For the function to take the minimum value for calculated a and b, it is necessary that at a given point the matrix of the quadratic form of the differential of the function of the form F (a, b) = ∑ i = 1 n (y i - (a x i + b)) 2 be positive definite. Let's show you how it should look.

Example 2

We have a second-order differential of the following form:

d 2 F (a ; b) = δ 2 F (a ; b) δ a 2 d 2 a + 2 δ 2 F (a ; b) δ a δ b d a d b + δ 2 F (a ; b) δ b 2 d 2b

Solution

δ 2 F (a ; b) δ a 2 = δ δ F (a ; b) δ a δ a = = δ - 2 ∑ i = 1 n (y i - (a x i + b)) x i δ a = 2 ∑ i = 1 n (x i) 2 δ 2 F (a ; b) δ a δ b = δ δ F (a ; b) δ a δ b = = δ - 2 ∑ i = 1 n (y i - (a x i + b) ) x i δ b = 2 ∑ i = 1 n x i δ 2 F (a ; b) δ b 2 = δ δ F (a ; b) δ b δ b = δ - 2 ∑ i = 1 n (y i - (a x i + b)) δ b = 2 ∑ i = 1 n (1) = 2 n

In other words, it can be written as follows: d 2 F (a ; b) = 2 ∑ i = 1 n (x i) 2 d 2 a + 2 2 ∑ x i i = 1 n d a d b + (2 n) d 2 b .

We have obtained a matrix of quadratic form M = 2 ∑ i = 1 n (x i) 2 2 ∑ i = 1 n x i 2 ∑ i = 1 n x i 2 n .

In this case, the values ​​of individual elements will not change depending on a and b . Is this matrix positive definite? To answer this question, let's check if its angular minors are positive.

Calculate the first order angular minor: 2 ∑ i = 1 n (x i) 2 > 0 . Since the points x i do not coincide, the inequality is strict. We will keep this in mind in further calculations.

We calculate the second-order angular minor:

d e t (M) = 2 ∑ i = 1 n (x i) 2 2 ∑ i = 1 n x i 2 ∑ i = 1 n x i 2 n = 4 n ∑ i = 1 n (x i) 2 - ∑ i = 1 n x i 2

After that, we proceed to the proof of the inequality n ∑ i = 1 n (x i) 2 - ∑ i = 1 n x i 2 > 0 using mathematical induction.

1. Let's check whether this inequality is valid for arbitrary n . Let's take 2 and calculate:

2 ∑ i = 1 2 (x i) 2 - ∑ i = 1 2 x i 2 = 2 x 1 2 + x 2 2 - x 1 + x 2 2 = = x 1 2 - 2 x 1 x 2 + x 2 2 = x 1 + x 2 2 > 0

We got the correct equality (if the values ​​x 1 and x 2 do not match).

1. Let's make the assumption that this inequality will be true for n , i.e. n ∑ i = 1 n (x i) 2 - ∑ i = 1 n x i 2 > 0 – true.
2. Now let's prove the validity for n + 1 , i.e. that (n + 1) ∑ i = 1 n + 1 (x i) 2 - ∑ i = 1 n + 1 x i 2 > 0 if n ∑ i = 1 n (x i) 2 - ∑ i = 1 n x i 2 > 0 .

We calculate:

(n + 1) ∑ i = 1 n + 1 (x i) 2 - ∑ i = 1 n + 1 x i 2 = = (n + 1) ∑ i = 1 n (x i) 2 + x n + 1 2 - ∑ i = 1 n x i + x n + 1 2 = = n ∑ i = 1 n (x i) 2 + n x n + 1 2 + ∑ i = 1 n (x i) 2 + x n + 1 2 - - ∑ i = 1 n x i 2 + 2 x n + 1 ∑ i = 1 n x i + x n + 1 2 = = ∑ i = 1 n (x i) 2 - ∑ i = 1 n x i 2 + n x n + 1 2 - x n + 1 ∑ i = 1 n x i + ∑ i = 1 n (x i) 2 = = ∑ i = 1 n (x i) 2 - ∑ i = 1 n x i 2 + x n + 1 2 - 2 x n + 1 x 1 + x 1 2 + + x n + 1 2 - 2 x n + 1 x 2 + x 2 2 + . . . + x n + 1 2 - 2 x n + 1 x 1 + x n 2 = = n ∑ i = 1 n (x i) 2 - ∑ i = 1 n x i 2 + + (x n + 1 - x 1) 2 + (x n + 1 - x 2) 2 + . . . + (x n - 1 - x n) 2 > 0

The expression enclosed in curly braces will be greater than 0 (based on what we assumed in step 2), and the rest of the terms will be greater than 0 because they are all squares of numbers. We have proven the inequality.

Answer: the found a and b will correspond to the smallest value of the function F (a, b) = ∑ i = 1 n (y i - (a x i + b)) 2, which means that they are the desired parameters of the least squares method (LSM).

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