Christopher Dougherty EC220 - Introduction to econometrics (chapter 3) Slideshow: exercise 3.5 Original citation: Dougherty, C. (2012) EC220 - Introduction to econometrics (chapter 3). [Teaching Resource] © 2012 The Author This version available at: Available in LSE Learning Resources Online: May 2012 This work is licensed under a Creative Commons Attribution-ShareAlike 3.0 License. This license allows the user to remix, tweak, and build upon the work even for commercial purposes, as long as the user credits the author and licenses their new creations under the identical terms

3.5 Explain why the intercept in the regression of EEARN on ES is equal to zero. 1 EXERCISE 3.5 This exercise relates to the Frisch–Lovell–Waugh procedure for graphing the relationship between the dependent variable and one of the explanatory variables in a multiple regression model.

The model was an earnings function with earnings hypothesized to be determined by years of schooling and years of work experience. 2 EXERCISE 3.5. reg EARNINGS S EXP Source | SS df MS Number of obs = F( 2, 537) = Model | Prob > F = Residual | R-squared = Adj R-squared = Total | Root MSE = EARNINGS | Coef. Std. Err. t P>|t| [95% Conf. Interval] S | EXP | _cons |

3 To represent the relationship between EARNINGS and S graphically, we first regress EARNINGS on EXP and save the residuals, which we call EEARN. EXERCISE 3.5. reg EARNINGS EXP Source | SS df MS Number of obs = F( 1, 538) = 2.98 Model | Prob > F = Residual | R-squared = Adj R-squared = Total | Root MSE = EARNINGS | Coef. Std. Err. t P>|t| [95% Conf. Interval] EXP | _cons | predict EEARN, resid

4 Then we do the same with S. We regress it on EXP and save the residuals as ES. EXERCISE 3.5. reg S EXP Source | SS df MS Number of obs = F( 1, 538) = Model | Prob > F = Residual | R-squared = Adj R-squared = Total | Root MSE = S | Coef. Std. Err. t P>|t| [95% Conf. Interval] EXP | _cons | predict ES, resid

5 A plot of EEARN and ES then gives a proper graphical representation of the relationship between EARNINGS and S, controlling for the effects of EXP. EXERCISE 3.5

Here is the regression of EEARN on ES. The intercept is 8.10e–09, which means 8.10x10 –9, or – , effectively zero. What is the reason for this? 6 EXERCISE 3.5. reg EEARN ES Source | SS df MS Number of obs = F( 1, 538) = Model | Prob > F = Residual | R-squared = Adj R-squared = Total | Root MSE = EEARN | Coef. Std. Err. t P>|t| [95% Conf. Interval] ES | _cons | 8.10e

. reg EEARN ES Source | SS df MS Number of obs = F( 1, 568) = Model | Prob > F = Residual | R-squared = Adj R-squared = Total | Root MSE = EEARN | Coef. Std. Err. t P>|t| [95% Conf. Interval] ES | _cons | -5.99e EXERCISE In the simple regression model, the intercept is calculated as the mean of the dependent variable minus b 2 times the mean of the explanatory variable.

. reg EEARN ES Source | SS df MS Number of obs = F( 1, 568) = Model | Prob > F = Residual | R-squared = Adj R-squared = Total | Root MSE = EEARN | Coef. Std. Err. t P>|t| [95% Conf. Interval] ES | _cons | -5.99e EXERCISE In this regression the Y variable is the EARNINGS residuals and the X variable is the S residuals.

. reg EEARN ES Source | SS df MS Number of obs = F( 1, 568) = Model | Prob > F = Residual | R-squared = Adj R-squared = Total | Root MSE = EEARN | Coef. Std. Err. t P>|t| [95% Conf. Interval] ES | _cons | -5.99e The means of both of these are zero because residuals from OLS regressions always have zero means. Hence the intercept must be zero. EXERCISE 4.5

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