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11+Regression - A Zhensykbaev Regression Linear Regression...

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A. Zhensykbaev Regression Linear Regression Regression analysis studies the nature of the relationship between depen- dent and independent variables, estimates a regression equation that predict the value of the response variable from the explanatory variable. The linear regression is a statistical technique that uses a single indepen- dent variable X to predict the dependent variable Y , using a linear relation- ship. Simple linear regression . Suppose y = ( y 1 , . . . , y n ) is a realization of a sample Y = { Y 1 , . . . , Y n } ( y i are measurements of some property of interest in points x 1 , . . . , x n ). What can we say about the possible linear dependence of a property y on x ? Simple linear regression uses the simple linear regression model y = β 0 + β 1 x + ε it consists of the deterministic component β 0 + β 1 x and the random error ε . More detail y 1 = β 0 + β 1 x 1 + ε 1 y 2 = β 0 + β 1 x 2 + ε 2 . . . . . . . . . . . . . . . . . . . y n = β 0 + β 1 x n + ε n where intercept β 0 and slope β 1 are unknown parameters, ε = ( ε 1 , . . . , ε n ) is a random variable ( ε i is random error for observation y i ). - 6 " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " Pic. 1 1
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A. Zhensykbaev Regression The method of least squares . Basing on a realization y = ( y 1 , . . . , y n ) we have ε 1 = y 1 - β 0 - β 1 x 1 ε 2 = y 2 - β 0 - β 1 x 2 . . . . . . . . . . . . . . . . . . . ε n = y n - β 0 - β 1 x n The sum of squares of errors SSE is SSE = n X i =1 ε 2 i = n X i =1 ( y i - β 0 - β 1 x i ) 2 . The idea of the method of least squares is to estimate β 0 and β 1 such that the SSE will be minimal. According the extremum theory β 0 and β 1 we find from the system of equations ( SSE ) ∂β 0 = 0 ( SSE ) ∂β 1 = 0 or 0 + β 1 n X i =1 x i = n X i =1 y i β 0 n X i =1 x i + β 1 n X i =1 x 2 i = n X i =1 x i y i First equation meas that the sum of the errors ( SE ) equals to zero, SE = n X i =1 ε i = 0 . The solution of the system is ˆ β 0 = ¯ y - ˆ β 1 ¯ x ˆ β 1 = SS xy SS xx 2
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A. Zhensykbaev Regression where ¯ x = 1 n n X i =1 x i , ¯ y = 1 n n X i =1 x i , SS xy = n X i =1 ( x i - ¯ x )( y i - ¯ y ) = n X i =1 x i y i - 1 n n X i =1 x i n X i =1 y i , SS xx = n X i =1 ( x i - ¯ x ) 2 = n X i =1 x 2 i - 1 n n X i =1 x i 2 . Example 1. Suppose we want to model the monthly sales revenue as a function of the monthly advertising expenditure. We base on the following table Advertising Sales Month expenditure Revenue x i ($100) y i ($1000) x 2 i x i y i 1 1 1 1 1 2 2 1 4 2 3 3 2 9 6 4 4 2 16 8 5 5 4 25 20 Total x i = 15 y i = 10 x 2 i x i y i = 37 SS xy = 37 - 1 5 · 15 · 10 = 7 SS xx = 55 - 1 5 · 15 2 = 10 ˆ β 1 = SS xy SS xx = 7 10 = 0 . 7 ˆ β 0 = ¯ y - ˆ β 1 ¯ x = 1 5 · 10 - 0 . 7 · 1 5 · 15 = - 0 . 1 3
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A. Zhensykbaev Regression and fitted line is y = 0 . 7 x - 0 . 1 - 6 " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " " Pic. 1 We make four assumptions about the random error ε 1. The mean of the probability distribution of ε equals to 0. It implies that the mean value is equal to the deterministic component E y = β 0 + β 1 x .
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