1. 1 BINARY CHOICE MODELS: LINEAR PROBABILITY MODEL Economists are often interested in the factors behind the decision-making of individuals or enterprises, examples being shown above. Why do some people go to college while others do not? Why do some women enter the labor force while others do not? Why do some people buy houses while others rent? Why do some people migrate while others stay put?

2. 2 The models that have been developed for this purpose are known as qualitative response or binary choice models, with the outcome, which we will denote Y, being assigned a value of 1 if the event occurs and 0 otherwise. BINARY CHOICE MODELS: LINEAR PROBABILITY MODEL Why do some people go to college while others do not? Why do some women enter the labor force while others do not? Why do some people buy houses while others rent? Why do some people migrate while others stay put?

3. Why do some people go to college while others do not? Why do some women enter the labor force while others do not? Why do some people buy houses while others rent? Why do some people migrate while others stay put? 3 Models with more than two possible outcomes have also been developed, but we will confine our attention to binary choice models. BINARY CHOICE MODELS: LINEAR PROBABILITY MODEL

4. 4 The simplest binary choice model is the linear probability model where, as the name implies, the probability of the event occurring, p, is assumed to be a linear function of a set of explanatory variables. BINARY CHOICE MODELS: LINEAR PROBABILITY MODEL

5. 5 XXiXi 1 0  1 +  2 X i y, p Graphically, the relationship is as shown, if there is just one explanatory variable. 11 BINARY CHOICE MODELS: LINEAR PROBABILITY MODEL

6. 6 Of course p is unobservable. One has data on only the outcome, Y. In the linear probability model this is used like a dummy variable for the dependent variable. BINARY CHOICE MODELS: LINEAR PROBABILITY MODEL

7. Why do some people graduate from high school while others drop out? 7 As an illustration, we will take the question shown above. We will define a variable GRAD which is equal to 1 if the individual graduated from high school, and 0 otherwise. BINARY CHOICE MODELS: LINEAR PROBABILITY MODEL

8. . g GRAD = 0. replace GRAD = 1 if S > 11 (509 real changes made) 8 The Stata output above shows the construction of the variable GRAD. It is first set to 0 for all respondents, and then changed to 1 for those who had more than 11 years of schooling. BINARY CHOICE MODELS: LINEAR PROBABILITY MODEL

9. 9 Here is the result of regressing GRAD on ASVABC. It suggests that every additional point on the ASVABC score increases the probability of graduating by 0.007, that is, 0.7%. BINARY CHOICE MODELS: LINEAR PROBABILITY MODEL. g GRAD = 0. replace GRAD = 1 if S > 11 (509 real changes made). reg GRAD ASVABC Source | SS df MS Number of obs = 540 -------------+------------------------------ F( 1, 538) = 49.59 Model | 2.46607893 1 2.46607893 Prob > F = 0.0000 Residual | 26.7542914 538.049729166 R-squared = 0.0844 -------------+------------------------------ Adj R-squared = 0.0827 Total | 29.2203704 539.05421219 Root MSE =.223 ------------------------------------------------------------------------------ GRAD | Coef. Std. Err. t P>|t| [95% Conf. Interval] -------------+---------------------------------------------------------------- ASVABC |.0070697.0010039 7.04 0.000.0050976.0090419 _cons |.5794711.0524502 11.05 0.000.4764387.6825035 ------------------------------------------------------------------------------

10. . g GRAD = 0. replace GRAD = 1 if S > 11 (509 real changes made). reg GRAD ASVABC Source | SS df MS Number of obs = 540 -------------+------------------------------ F( 1, 538) = 49.59 Model | 2.46607893 1 2.46607893 Prob > F = 0.0000 Residual | 26.7542914 538.049729166 R-squared = 0.0844 -------------+------------------------------ Adj R-squared = 0.0827 Total | 29.2203704 539.05421219 Root MSE =.223 ------------------------------------------------------------------------------ GRAD | Coef. Std. Err. t P>|t| [95% Conf. Interval] -------------+---------------------------------------------------------------- ASVABC |.0070697.0010039 7.04 0.000.0050976.0090419 _cons |.5794711.0524502 11.05 0.000.4764387.6825035 ------------------------------------------------------------------------------ 10 The intercept has no sensible meaning. Literally it suggests that a respondent with a 0 ASVABC score has a 58% probability of graduating. However a score of 0 is not possible. BINARY CHOICE MODELS: LINEAR PROBABILITY MODEL

11. 11 Unfortunately, the linear probability model has some serious shortcomings. First, there are problems with the disturbance term. BINARY CHOICE MODELS: LINEAR PROBABILITY MODEL

12. 12 As usual, the value of the dependent variable Y i in observation i has a nonstochastic component and a random component. The nonstochastic component depends on X i and the parameters. The random component is the disturbance term. BINARY CHOICE MODELS: LINEAR PROBABILITY MODEL

13. 13 The nonstochastic component in observation i is its expected value in that observation. This is simple to compute, because it can take only two values. It is 1 with probability p i and 0 with probability (1 – p i ) The expected value in observation i is therefore  1 +  2 X i. BINARY CHOICE MODELS: LINEAR PROBABILITY MODEL

14. 14 This means that we can rewrite the model as shown. BINARY CHOICE MODELS: LINEAR PROBABILITY MODEL

15. XXiXi 1 0  1 +  2 X i Y, p 15 The probability function is thus also the nonstochastic component of the relationship between Y and X. 11 BINARY CHOICE MODELS: LINEAR PROBABILITY MODEL

16. 16 In observation i, for Y i to be 1, u i must be (1 –  1 –  2 X i ). For Y i to be 0, u i must be (–  1 –  2 X i ). BINARY CHOICE MODELS: LINEAR PROBABILITY MODEL

17. XXiXi 1 0  1 +  2 X i Y, p 11 17 The two possible values, which give rise to the observations A and B, are illustrated in the diagram. Since u does not have a normal distribution, the standard errors and test statistics are invalid. Its distribution is not even continuous. A B  1 +  2 X i 1 –  1 –  2 X i BINARY CHOICE MODELS: LINEAR PROBABILITY MODEL

18. XXiXi 1 0  1 +  2 X i Y, p 11 A B 18 Further, it can be shown that the population variance of the disturbance term in observation i is given by (  1 +  2 X i )(1 –  1 –  2 X i ). This changes with X i, and so the distribution is heteroscedastic. BINARY CHOICE MODELS: LINEAR PROBABILITY MODEL  1 +  2 X i 1 –  1 –  2 X i

19. XXiXi 1 0  1 +  2 X i Y, p 11 A B 19 Yet another shortcoming of the linear probability model is that it may predict probabilities of more than 1, as shown here. It may also predict probabilities less than 0. BINARY CHOICE MODELS: LINEAR PROBABILITY MODEL  1 +  2 X i 1 –  1 –  2 X i

20. 20 The Stata command for saving the fitted values from a regression is predict, followed by the name that you wish to give to the fitted values. We are calling them PROB. BINARY CHOICE MODELS: LINEAR PROBABILITY MODEL. g GRAD = 0. replace GRAD = 1 if S > 11 (509 real changes made). reg GRAD ASVABC Source | SS df MS Number of obs = 540 -------------+------------------------------ F( 1, 538) = 49.59 Model | 2.46607893 1 2.46607893 Prob > F = 0.0000 Residual | 26.7542914 538.049729166 R-squared = 0.0844 -------------+------------------------------ Adj R-squared = 0.0827 Total | 29.2203704 539.05421219 Root MSE =.223 ------------------------------------------------------------------------------ GRAD | Coef. Std. Err. t P>|t| [95% Conf. Interval] -------------+---------------------------------------------------------------- ASVABC |.0070697.0010039 7.04 0.000.0050976.0090419 _cons |.5794711.0524502 11.05 0.000.4764387.6825035 ------------------------------------------------------------------------------. predict PROB

21. . tab PROB if PROB > 1 Fitted | values | Freq. Percent Cum. ------------+----------------------------------- 1.000381 | 6 4.76 4.76 1.002308 | 9 7.14 11.90 1.004236 | 7 5.56 17.46 1.006163 | 3 2.38 19.84 ********************************************* 1.040855 | 11 8.73 93.65 1.042783 | 3 2.38 96.03 1.04471 | 2 1.59 97.62 1.046638 | 3 2.38 100.00 ------------+----------------------------------- Total | 126 100.00 21 tab is the Stata command for tabulating the values of a variable, and for cross-tabulating two or more variables. We see that there are 126 observations where the fitted value is greater than 1. (The middle rows of the table have been omitted.) BINARY CHOICE MODELS: LINEAR PROBABILITY MODEL

22. 22 In this example there were no fitted values of less than 0.. tab PROB if PROB > 1 Fitted | values | Freq. Percent Cum. ------------+----------------------------------- 1.000381 | 6 4.76 4.76 1.002308 | 9 7.14 11.90 1.004236 | 7 5.56 17.46 1.006163 | 3 2.38 19.84 ********************************************* 1.040855 | 11 8.73 93.65 1.042783 | 3 2.38 96.03 1.04471 | 2 1.59 97.62 1.046638 | 3 2.38 100.00 ------------+----------------------------------- Total | 126 100.0 0. tab PROB if PROB < 0 no observations BINARY CHOICE MODELS: LINEAR PROBABILITY MODEL

23. 23 The main advantage of the linear probability model over logit and probit analysis, the alternatives considered in the next two sequences, is that it is much easier to fit. For this reason it used to be recommended for initial, exploratory work. BINARY CHOICE MODELS: LINEAR PROBABILITY MODEL. tab PROB if PROB > 1 Fitted | values | Freq. Percent Cum. ------------+----------------------------------- 1.000381 | 6 4.76 4.76 1.002308 | 9 7.14 11.90 1.004236 | 7 5.56 17.46 1.006163 | 3 2.38 19.84 ********************************************* 1.040855 | 11 8.73 93.65 1.042783 | 3 2.38 96.03 1.04471 | 2 1.59 97.62 1.046638 | 3 2.38 100.00 ------------+----------------------------------- Total | 126 100.0 0. tab PROB if PROB < 0 no observations

24. 24 However, this consideration is no longer relevant, now that computers are so fast and powerful, and logit and probit are typically standard features of regression applications. BINARY CHOICE MODELS: LINEAR PROBABILITY MODEL. tab PROB if PROB > 1 Fitted | values | Freq. Percent Cum. ------------+----------------------------------- 1.000381 | 6 4.76 4.76 1.002308 | 9 7.14 11.90 1.004236 | 7 5.56 17.46 1.006163 | 3 2.38 19.84 ********************************************* 1.040855 | 11 8.73 93.65 1.042783 | 3 2.38 96.03 1.04471 | 2 1.59 97.62 1.046638 | 3 2.38 100.00 ------------+----------------------------------- Total | 126 100.0 0. tab PROB if PROB < 0 no observations

25. Copyright Christopher Dougherty 2012. These slideshows may be downloaded by anyone, anywhere for personal use. Subject to respect for copyright and, where appropriate, attribution, they may be used as a resource for teaching an econometrics course. There is no need to refer to the author. The content of this slideshow comes from Section 10.1 of C. Dougherty, Introduction to Econometrics, fourth edition 2011, Oxford University Press. Additional (free) resources for both students and instructors may be downloaded from the OUP Online Resource Centre http://www.oup.com/uk/orc/bin/9780199567089/http://www.oup.com/uk/orc/bin/9780199567089/. Individuals studying econometrics on their own who feel that they might benefit from participation in a formal course should consider the London School of Economics summer school course EC212 Introduction to Econometrics http://www2.lse.ac.uk/study/summerSchools/summerSchool/Home.aspx http://www2.lse.ac.uk/study/summerSchools/summerSchool/Home.aspx or the University of London International Programmes distance learning course 20 Elements of Econometrics www.londoninternational.ac.uk/lsewww.londoninternational.ac.uk/lse. 2012.11.26