Introduction to Econometrics, 5th edition Type author name/s here Dougherty Introduction to Econometrics, 5th edition Chapter heading Chapter 10: Binary Choice and Limited Dependent Variable Models, and Maximum Likelihood Estimation © Christopher Dougherty, 2016. All rights reserved.

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? Economists are often interested in the factors behind the decision-making of individuals or enterprises, examples being shown above. 1

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? 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. 2

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? Models with more than two possible outcomes have also been developed, but we will confine our attention to binary choice models. 3

BINARY CHOICE MODELS: LINEAR PROBABILITY MODEL 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. 4

BINARY CHOICE MODELS: LINEAR PROBABILITY MODEL y, p 1 b1 +b2Xi b1 Xi X Graphically, the relationship is as shown, if there is just one explanatory variable. 5

BINARY CHOICE MODELS: LINEAR PROBABILITY MODEL 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. 6

BINARY CHOICE MODELS: LINEAR PROBABILITY MODEL Why do some people graduate from high school while others drop out? 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. 7

BINARY CHOICE MODELS: LINEAR PROBABILITY MODEL . gen GRAD = 0 . replace GRAD = 1 if S > 11 (452 real changes made) 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. 8

BINARY CHOICE MODELS: LINEAR PROBABILITY MODEL . gen GRAD = 0 . replace GRAD = 1 if S > 11 (452 real changes made) . reg GRAD ASVABC ---------------------------------------------------------------------------- Source | SS df MS Number of obs = 500 -----------+------------------------------ F( 1, 498) = 58.01 Model | 4.52721798 1 4.52721798 Prob > F = 0.0000 Residual | 38.864782 498 .078041731 R-squared = 0.1043 -----------+------------------------------ Adj R-squared = 0.1025 Total | 43.392 499 .086957916 Root MSE = .27936 GRAD | Coef. Std. Err. t P>|t| [95% Conf. Interval] -----------+---------------------------------------------------------------- ASVABC | .1060002 .0139173 7.62 0.000 .0786564 .133344 _cons | .87522 .0130523 67.06 0.000 .8495757 .9008643 Here is the result of regressing GRAD on ASVABC. It suggests that an increase of one unit in the ASVABC score increases the probability of graduating by 0.106, that is, 10.6%. ASVABC is scaled so that it has mean 0 and its units are standard deviations. 9

BINARY CHOICE MODELS: LINEAR PROBABILITY MODEL . gen GRAD = 0 . replace GRAD = 1 if S > 11 (452 real changes made) . reg GRAD ASVABC ---------------------------------------------------------------------------- Source | SS df MS Number of obs = 500 -----------+------------------------------ F( 1, 498) = 58.01 Model | 4.52721798 1 4.52721798 Prob > F = 0.0000 Residual | 38.864782 498 .078041731 R-squared = 0.1043 -----------+------------------------------ Adj R-squared = 0.1025 Total | 43.392 499 .086957916 Root MSE = .27936 GRAD | Coef. Std. Err. t P>|t| [95% Conf. Interval] -----------+---------------------------------------------------------------- ASVABC | .1060002 .0139173 7.62 0.000 .0786564 .133344 _cons | .87522 .0130523 67.06 0.000 .8495757 .9008643 The intercept implies that an individual with the mean ASVABC score, 0, would have an 88 percent probability of graduating. 10

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

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

BINARY CHOICE MODELS: LINEAR PROBABILITY MODEL 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 pi and 0 with probability (1 – pi) The expected value in observation i is therefore b1 + b2Xi. 13

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

BINARY CHOICE MODELS: LINEAR PROBABILITY MODEL Y, p 1 b1 +b2Xi b1 Xi X The probability function is thus also the nonstochastic component of the relationship between Y and X. 15

BINARY CHOICE MODELS: LINEAR PROBABILITY MODEL In observation i, for Yi to be 1, ui must be (1 – b1 – b2Xi). For Yi to be 0, ui must be (– b1 – b2Xi). 16

BINARY CHOICE MODELS: LINEAR PROBABILITY MODEL Y, p A 1 1 – b1 – b2Xi b1 +b2Xi b1 b1 + b2Xi B Xi X 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. 17

BINARY CHOICE MODELS: LINEAR PROBABILITY MODEL Y, p A 1 1 – b1 – b2Xi b1 +b2Xi b1 b1 + b2Xi B Xi X Further, it can be shown that the population variance of the disturbance term in observation i is given by (b1 + b2Xi)(1 – b1 – b2Xi). This changes with Xi, and so the distribution is heteroskedastic. 18

BINARY CHOICE MODELS: LINEAR PROBABILITY MODEL Y, p A 1 1 – b1 – b2Xi b1 +b2Xi b1 b1 + b2Xi B Xi X 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. 19

BINARY CHOICE MODELS: LINEAR PROBABILITY MODEL . gen GRAD = 0 . replace GRAD = 1 if S > 11 (452 real changes made) . reg GRAD ASVABC ---------------------------------------------------------------------------- Source | SS df MS Number of obs = 500 -----------+------------------------------ F( 1, 498) = 58.01 Model | 4.52721798 1 4.52721798 Prob > F = 0.0000 Residual | 38.864782 498 .078041731 R-squared = 0.1043 -----------+------------------------------ Adj R-squared = 0.1025 Total | 43.392 499 .086957916 Root MSE = .27936 GRAD | Coef. Std. Err. t P>|t| [95% Conf. Interval] -----------+---------------------------------------------------------------- ASVABC | .1060002 .0139173 7.62 0.000 .0786564 .133344 _cons | .87522 .0130523 67.06 0.000 .8495757 .9008643 . predict PROB 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. 20

BINARY CHOICE MODELS: LINEAR PROBABILITY MODEL . tab PROB if PROB > 1 Fitted | values | Freq. Percent Cum. ------------+----------------------------------- 1.001353 | 1 1.30 1.30 1.001776 | 1 1.30 2.60 1.002197 | 1 1.30 3.90 1.004591 | 1 1.30 5.19 ********************************************* 1.13173 | 1 1.30 96.10 1.139848 | 1 1.30 97.40 1.144364 | 1 1.30 98.70 1.155131 | 1 1.30 100.00 Total | 77 100.00 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 77 observations where the fitted value is greater than 1. (The middle rows of the table have been omitted.) 21

BINARY CHOICE MODELS: LINEAR PROBABILITY MODEL . tab PROB if PROB > 1 Fitted | values | Freq. Percent Cum. ------------+----------------------------------- 1.001353 | 1 1.30 1.30 1.001776 | 1 1.30 2.60 1.002197 | 1 1.30 3.90 1.004591 | 1 1.30 5.19 ********************************************* 1.13173 | 1 1.30 96.10 1.139848 | 1 1.30 97.40 1.144364 | 1 1.30 98.70 1.155131 | 1 1.30 100.00 Total | 77 100.00 . tab PROB if PROB < 0 no observations In this example there were no fitted values of less than 0. 22

BINARY CHOICE MODELS: LINEAR PROBABILITY MODEL . tab PROB if PROB > 1 Fitted | values | Freq. Percent Cum. ------------+----------------------------------- 1.001353 | 1 1.30 1.30 1.001776 | 1 1.30 2.60 1.002197 | 1 1.30 3.90 1.004591 | 1 1.30 5.19 ********************************************* 1.13173 | 1 1.30 96.10 1.139848 | 1 1.30 97.40 1.144364 | 1 1.30 98.70 1.155131 | 1 1.30 100.00 Total | 77 100.00 . tab PROB if PROB < 0 no observations 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. 23

BINARY CHOICE MODELS: LINEAR PROBABILITY MODEL . tab PROB if PROB > 1 Fitted | values | Freq. Percent Cum. ------------+----------------------------------- 1.001353 | 1 1.30 1.30 1.001776 | 1 1.30 2.60 1.002197 | 1 1.30 3.90 1.004591 | 1 1.30 5.19 ********************************************* 1.13173 | 1 1.30 96.10 1.139848 | 1 1.30 97.40 1.144364 | 1 1.30 98.70 1.155131 | 1 1.30 100.00 Total | 77 100.00 . tab PROB if PROB < 0 no observations However, this consideration is no longer relevant. logit and probit are now standard features of regression applications. 24

Copyright Christopher Dougherty 2016. 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, fifth edition 2016, Oxford University Press. Additional (free) resources for both students and instructors may be downloaded from the OUP Online Resource Centre http://www.oxfordtextbooks.co.uk/orc/dougherty5e/. 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 or the University of London International Programmes distance learning course 20 Elements of Econometrics www.londoninternational.ac.uk/lse. 2016.05.18