1 BINARY CHOICE MODELS: LINEAR PROBABILITY MODEL Economists are often interested in the factors behind the decision-making of individuals or enterprises,

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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?

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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.










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


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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.


iii XYpp 21)1(

5

XXi

1

0

1 +2Xi

y, p

Graphically, the relationship is as shown, if there is just one explanatory variable.

iii XYpp 21)1(


1

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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.


iii XYpp 21)1(

• Why do some people graduate from high school while others drop out?

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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.


. 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------------------------------------------------------------------------------


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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.

. g GRAD = 0


. reg GRAD ASVABC



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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%.


. g GRAD = 0


. reg GRAD ASVABC



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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.

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Unfortunately, the linear probability model has some serious shortcomings. First, there are problems with the disturbance term.


iii XYpp 21)1(

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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.


iii XYpp 21)1(

iii uYEY )(

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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 1 + 2Xi.


iii XYpp 21)1(

iii uYEY )(

iiiii XpppYE 21)1(01)(

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This means that we can rewrite the model as shown.


iii XYpp 21)1(

iii uYEY )(


iii uXY 21

XXi

1

0

1 +2Xi

Y, p iii XYpp 21)1(


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The probability function is thus also the nonstochastic component of the relationship between Y and X.

1

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iii uXY 21

iii XuY 2111

iii XuY 210

In observation i, for Yi to be 1, ui must be (1 – 1 – 2Xi). For Yi to be 0, ui must be (– 1 – 2Xi).

iii uYEY )(


iii XYpp 21)1(

XXi

1

0

1 +2Xi

Y, p iii XYpp 21)1(


1

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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 + 2Xi

1 – 1 – 2Xi

XXi

1

0

1 +2Xi

Y, p


1

A

1 – 1 – 2Xi

B

1 + 2Xi

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Further, it can be shown that the population variance of the disturbance term in observation i is given by (1 + 2Xi)(1 – 1 – 2Xi). This changes with Xi, and so the distribution is heteroscedastic.

)1)(( 21212

iiu XXi

XXi

1

0

1 +2Xi

Y, p


1

A

B

1 + 2Xi

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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.

1 – 1 – 2Xi

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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.


. g GRAD = 0


. reg GRAD ASVABC



. predict PROB

. 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

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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.)



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In this example there were no fitted values of less than 0.


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

. tab PROB if PROB < 0no observations


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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.


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



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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.


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


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1 BINARY CHOICE MODELS: LINEAR PROBABILITY MODEL Economists are often interested in the factors behind the decision-making of individuals or enterprises,