1. 1 Section 3 Probit and Logit Models

2. 2 Dichotomous Data Suppose data is discrete but there are only 2 outcomes Examples –Graduate high school or not –Patient dies or not –Working or not –Smoker or not In data, y i =1 if yes, y i =0 if no

3. 3 How to model the data generating process? There are only two outcomes Research question: What factors impact whether the event occurs? To answer, will model the probability the outcome occurs Pr(Y i =1) when y i =1 or Pr(Y i =0) = 1- Pr(Y i =1) when y i =0

4. 4 Think of the problem from a MLE perspective Likelihood for i’th observation L i = Pr(Y i =1) Yi [1 - Pr(Y i =1)] (1-Yi) When y i =1, only relevant part is Pr(Y i =1) When y i =0, only relevant part is [1 - Pr(Y i =1)]

5. 5 L = Σ i ln[L i ] = = Σ i {y i ln[Pr(y i =1)] + (1-y i )ln[Pr(y i =0)] } Notice that up to this point, the model is generic. The log likelihood function will determined by the assumptions concerning how we determine Pr(y i =1)

6. 6 Modeling the probability There is some process (biological, social, decision theoretic, etc) that determines the outcome y Some of the variables impacting are observed, some are not Requires that we model how these factors impact the probabilities Model from a ‘latent variable’ perspective

7. 7 Consider a women’s decision to work y i * = the person’s net benefit to work Two components of y i * –Characteristics that we can measure Education, age, income of spouse, prices of child care –Some we cannot measure How much you like spending time with your kids how much you like/hate your job

8. 8 We aggregate these two components into one equation y i * = β 0 + x 1i β 1 + x 2i β 2 +… x ki β k + ε i = x i β + ε i x i β (measurable characteristics but with uncertain weights) ε i random unmeasured characteristics Decision rule: person will work if y i * > 0 (if net benefits are positive) y i =1 if y i *>0 y i =0 if y i * ≤0

9. 9 y i =1 if y i *>0 y i * = x i β + ε i > 0 only if ε i > - x i β y i =0 if y i * ≤0 y i * = x i β + ε i ≤ 0 only if ε i ≤ - x i β

10. 10 Suppose x i β is ‘big.’ –High wages –Low husband’s income –Low cost of child care We would expect this person to work, UNLESS, there is some unmeasured ‘variable’ that counteracts this

11. 11 Suppose a mom really likes spending time with her kids, or she hates her job. The unmeasured benefit of working has a big negative coefficient ε i If we observe them working, ε i must not have been too big, since y i =1 if ε i > - x i β

12. 12 Consider the opposite. Suppose we observe someone NOT working. Then ε i must not have been big, since y i =0 if ε i ≤ - x i β

13. 13 Logit Recall y i =1 if ε i > - x i β Since ε i is a logistic distribution Pr(ε i > - x i β) = 1 – F(- x i β) The logistic is also a symmetric distribution, so 1 – F(- x i β) = F(x i β) = exp(x i β)/(1+exp(x i β))

14. 14 When ε i is a logistic distribution Pr(y i =1) = exp(x i β)/(1+exp(x i β)) Pr(y i =0) = 1/(1+exp(x i β))

15. 15 Example: Workplace smoking bans Smoking supplements to 1991 and 1993 National Health Interview Survey Asked all respondents whether they currently smoke Asked workers about workplace tobacco policies Sample: workers Key variables: current smoking and whether they faced by workplace ban

16. 16 Data: workplace1.dta Sample program: workplace1.doc Results: workplace1.log

17. 17 Description of variables in data. desc; storage display value variable name type format label variable label ------------------------------------------------------------------------ > - smoker byte %9.0g is current smoking worka byte %9.0g has workplace smoking bans age byte %9.0g age in years male byte %9.0g male black byte %9.0g black hispanic byte %9.0g hispanic incomel float %9.0g log income hsgrad byte %9.0g is hs graduate somecol byte %9.0g has some college college float %9.0g -----------------------------------------------------------------------

18. 18 Summary statistics sum; Variable | Obs Mean Std. Dev. Min Max -------------+-------------------------------------------------------- smoker | 16258.25163.433963 0 1 worka | 16258.6851396.4644745 0 1 age | 16258 38.54742 11.96189 18 87 male | 16258.3947595.488814 0 1 black | 16258.1119449.3153083 0 1 -------------+-------------------------------------------------------- hispanic | 16258.0607086.2388023 0 1 incomel | 16258 10.42097.7624525 6.214608 11.22524 hsgrad | 16258.3355271.4721889 0 1 somecol | 16258.2685447.4432161 0 1 college | 16258.3293763.4700012 0 1

19. 19 Running a probit probit smoker age incomel male black hispanic hsgrad somecol college worka; The first variable after ‘probit’ is the discrete outcome, the rest of the variables are the independent variables Includes a constant as a default

20. 20 Running a logit logit smoker age incomel male black hispanic hsgrad somecol college worka; Same as probit, just change the first word

21. 21 Running linear probability reg smoker age incomel male black hispanic hsgrad somecol college worka, robust; Simple regression. Standard errors are incorrect (heteroskedasticity) robust option produces standard errors with arbitrary form of heteroskedasticity

22. 22 Probit Results Probit estimates Number of obs = 16258 LR chi2(9) = 819.44 Prob > chi2 = 0.0000 Log likelihood = -8761.7208 Pseudo R2 = 0.0447 ------------------------------------------------------------------------------ smoker | Coef. Std. Err. z P>|z| [95% Conf. Interval] -------------+---------------------------------------------------------------- age | -.0012684.0009316 -1.36 0.173 -.0030943.0005574 incomel | -.092812.0151496 -6.13 0.000 -.1225047 -.0631193 male |.0533213.0229297 2.33 0.020.0083799.0982627 black | -.1060518.034918 -3.04 0.002 -.17449 -.0376137 hispanic | -.2281468.0475128 -4.80 0.000 -.3212701 -.1350235 hsgrad | -.1748765.0436392 -4.01 0.000 -.2604078 -.0893453 somecol | -.363869.0451757 -8.05 0.000 -.4524118 -.2753262 college | -.7689528.0466418 -16.49 0.000 -.860369 -.6775366 worka | -.2093287.0231425 -9.05 0.000 -.2546873 -.1639702 _cons |.870543.154056 5.65 0.000.5685989 1.172487 ------------------------------------------------------------------------------

23. 23 How to measure fit? Regression (OLS) –minimize sum of squared errors –Or, maximize R 2 –The model is designed to maximize predictive capacity Not the case with Probit/Logit –MLE models pick distribution parameters so as best describe the data generating process –May or may not ‘predict’ the outcome well

24. 24 Pseudo R 2 LL k log likelihood with all variables LL 1 log likelihood with only a constant 0 > LL k > LL 1 so | LL k | < |LL 1 | Pseudo R 2 = 1 - |LL 1 /LL k | Bounded between 0-1 Not anything like an R 2 from a regression

25. 25 Predicting Y Let b be the estimated value of β For any candidate vector of x i, we can predict probabilities, P i P i = Ф(x i b) Once you have P i, pick a threshold value, T, so that you predict Y p = 1 if P i > T Y p = 0 if P i ≤ T Then compare, fraction correctly predicted

26. 26 Question: what value to pick for T? Can pick.5 –Intuitive. More likely to engage in the activity than to not engage in it –However, when the  is small, this criteria does a poor job of predicting Y i =1 –However, when the  is close to 1, this criteria does a poor job of picking Y i =0

27. 27 *predict probability of smoking; predict pred_prob_smoke; * get detailed descriptive data about predicted prob; sum pred_prob, detail; * predict binary outcome with 50% cutoff; gen pred_smoke1=pred_prob_smoke>=.5; label variable pred_smoke1 "predicted smoking, 50% cutoff"; * compare actual values; tab smoker pred_smoke1, row col cell;

28. 28. sum pred_prob, detail; Pr(smoker) ------------------------------------------------------------- Percentiles Smallest 1%.0959301.0615221 5%.1155022.0622963 10%.1237434.0633929 Obs 16258 25%.1620851.0733495 Sum of Wgt. 16258 50%.2569962 Mean.2516653 Largest Std. Dev..0960007 75%.3187975.5619798 90%.3795704.5655878 Variance.0092161 95%.4039573.5684112 Skewness.1520254 99%.4672697.6203823 Kurtosis 2.149247

29. 29 Notice two things –Sample mean of the predicted probabilities is close to the sample mean outcome –99% of the probabilities are less than.5 –Should predict few smokers if use a 50% cutoff

30. 30 | predicted smoking, is current | 50% cutoff smoking | 0 1 | Total -----------+----------------------+---------- 0 | 12,153 14 | 12,167 | 99.88 0.12 | 100.00 | 74.93 35.90 | 74.84 | 74.75 0.09 | 74.84 -----------+----------------------+---------- 1 | 4,066 25 | 4,091 | 99.39 0.61 | 100.00 | 25.07 64.10 | 25.16 | 25.01 0.15 | 25.16 -----------+----------------------+---------- Total | 16,219 39 | 16,258 | 99.76 0.24 | 100.00 | 100.00 100.00 | 100.00 | 99.76 0.24 | 100.00

31. 31 Check on-diagonal elements. The last number in each 2x2 element is the fraction in the cell The model correctly predicts 74.75 + 0.15 = 74.90% of the obs It only predicts a small fraction of smokers

32. 32 Do not be amazed by the 75% percent correct prediction If you said everyone has a  chance of smoking (a case of no covariates), you would be correct Max[( ,(1-  )] percent of the time

33. 33 In this case, 25.16% smoke. If everyone had the same chance of smoking, we would assign everyone Pr(y=1) =.2516 We would be correct for the 1 -.2516 = 0.7484 people who do not smoke

34. 34 Key points about prediction MLE models are not designed to maximize prediction Should not be surprised they do not predict well In this case, not particularly good measures of predictive capacity

35. 35 Translating coefficients in probit: Continuous Covariates Pr(y i =1) = Φ[β 0 + x 1i β 1 + x 2i β 2 +… x ki β k ] Suppose that x 1i is a continuous variable d Pr(y i =1) /d x 1i = ? What is the change in the probability of an event give a change in x 1i?

36. 36 Marginal Effect d Pr(y i =1) /d x 1i = β 1 φ[β 0 + x 1i β 1 + x 2i β 2 +… x ki β k ] Notice two things. Marginal effect is a function of the other parameters and the values of x.

37. 37 Translating Coefficients: Discrete Covariates Pr(y i =1) = Φ[β 0 + x 1i β 1 + x 2i β 2 +… x ki β k ] Suppose that x 2i is a dummy variable (1 if yes, 0 if no) Marginal effect makes no sense, cannot change x 2i by a little amount. It is either 1 or 0. Redefine the variable of interest. Compare outcomes with and without x 2i

38. 38 y 1 = Pr(y i =1 | x 2i =1) = Φ[β 0 + x 1i β 1 + β 2 + x 3i β 3 +… ] y 0 = Pr(y i =1 | x 2i =0) = Φ[β 0 + x 1i β 1 + x 3i β 3 … ] Marginal effect = y 1 – y 0. Difference in probabilities with and without x 2i?

39. 39 In STATA Marginal effects for continuous variables, STATA picks sample means for X’s Change in probabilities for dichotomous outcomes, STATA picks sample means for X’s

40. 40 STATA command for Marginal Effects mfx compute; Must be after the outcome when estimates are still active in program.

41. 41 Marginal effects after probit y = Pr(smoker) (predict) =.24093439 ------------------------------------------------------------------------------ variable | dy/dx Std. Err. z P>|z| [ 95% C.I. ] X ---------+-------------------------------------------------------------------- age | -.0003951.00029 -1.36 0.173 -.000964.000174 38.5474 incomel | -.0289139.00472 -6.13 0.000 -.03816 -.019668 10.421 male*|.0166757.0072 2.32 0.021.002568.030783.39476 black*| -.0320621.01023 -3.13 0.002 -.052111 -.012013.111945 hispanic*| -.0658551.01259 -5.23 0.000 -.090536 -.041174.060709 hsgrad*| -.053335.01302 -4.10 0.000 -.07885 -.02782.335527 somecol*| -.1062358.01228 -8.65 0.000 -.130308 -.082164.268545 college*| -.2149199.01146 -18.76 0.000 -.237378 -.192462.329376 worka*| -.0668959.00756 -8.84 0.000 -.08172 -.052072.68514 ------------------------------------------------------------------------------ (*) dy/dx is for discrete change of dummy variable from 0 to 1

42. 42 Interpret results 10% increase in income will reduce smoking by 2.9 percentage points 10 year increase in age will decrease smoking rates.4 percentage points Those with a college degree are 21.5 percentage points less likely to smoke Those that face a workplace smoking ban have 6.7 percentage point lower probability of smoking

43. 43 Do not confuse percentage point and percent differences –A 6.7 percentage point drop is 29% of the sample mean of 24 percent. –Blacks have smoking rates that are 3.2 percentage points lower than others, which is 13 percent of the sample mean

44. 44 Comparing Marginal Effects VariableLPProbitLogit age-0.00040-0.00048 incomel-0.0289-0.0287-0.0276 male 0.0167 0.0168 0.0172 Black-0.0321-0.0357-0.0342 hispanic-0.0658-0.0706-0.0602 hsgrad-0.0533-0.0661-0.0514 college-0.2149-0.2406-0.2121 worka-0.0669-0.0661-0.0658

45. 45 When will results differ? Normal and logit CDF look –Similar in the mid point of the distribution –Different in the tails You obtain more observations in the tails of the distribution when –Samples sizes are large –  approaches 1 or 0 These situations will produce more differences in estimates

46. 46 Some nice properties of the Logit Outcome, y=1 or 0 Treatment, x=1 or 0 Other covariates, x Context, –x = whether a baby is born with a low weight birth –x = whether the mom smoked or not during pregnancy

47. 47 Risk ratio RR = Prob(y=1|x=1)/Prob(y=1|x=0) Differences in the probability of an event when x is and is not observed How much does smoking elevate the chance your child will be a low weight birth

48. 48 Let Y yx be the probability y=1 or 0 given x=1 or 0 Think of the risk ratio the following way Y 11 is the probability Y=1 when X=1 Y 10 is the probability Y=1 when X=0 Y 11 = RR*Y 10

49. 49 Odds Ratio OR=A/B = [Y 11 /Y 01 ]/[Y 10 /Y 00 ] A = [Pr(Y=1|X=1)/Pr(Y=0|X=1)] = odds of Y occurring if you are a smoker B = [Pr(Y=1|X=0)/Pr(Y=0|X=0)] = odds of y happening if you are not a smoker What are the relative odds of Y happening if you do or do not experience X

50. 50 Suppose Pr(Y i =1) = F(β o + β 1 X i + β 2 Z) and F is the logistic function Can show that OR = exp(β 1 ) = e β1 This number is typically reported by most statistical packages

51. 51 Details Y 11 = exp(β o + β 1 + β 2 Z) /(1+ exp(β o + β 1 + β 2 Z) ) Y 10 = exp(β o + β 2 Z)/(1+ exp(β o +β 2 Z)) Y 01 = 1 /(1+ exp(β o + β 1 + β 2 Z) ) Y 00 = 1/(1+ exp(β o +β 2 Z) [Y 11 /Y 01 ] = exp(β o + β 1 + β 2 Z) [Y 10 /Y 00 ] = exp(β o + β 2 Z) OR=A/B = [Y 11 /Y 01 ]/[Y 10 /Y 00 ] = exp(β o + β 1 + β 2 Z)/ exp(β o + β 2 Z) = exp(β 1 )

52. 52 Suppose Y is rare,  close to 0 –Pr(Y=0|X=1) and Pr(Y=0|X=0) are both close to 1, so they cancel Therefore, when  is close to 0 –Odds Ratio = Risk Ratio Why is this nice?

53. 53 Population attributable risk Average outcome in the population  = (1- ) Y 10 + Y 11 = (1- )Y 10 + (RR)Y 10 Average outcomes are a weighted average of outcomes for X=0 and X=1 What would the average outcome be in the absence of X (e.g., reduce smoking rates to 0) Y a = Y 10

54. 54 Population Attributable Risk PAR Fraction of outcome attributed to X The difference between the current rate and the rate that would exist without X, divided by the current rate PAR = (  – Y a )/  = (RR – 1) /[(1- ) + RR ]

55. 55 Example: Maternal Smoking and Low Weight Births 6% births are low weight –< 2500 grams ( –Average birth is 3300 grams (5.5 lbs) Maternal smoking during pregnancy has been identified as a key cofactor –13% of mothers smoke –This number was falling about 1 percentage point per year during 1980s/90s –Doubles chance of low weight birth

56. 56 Natality detail data Census of all births (4 million/year) Annual files starting in the 60s Information about –Baby (birth weight, length, date, sex, plurality, birth injuries) –Demographics (age, race, marital, educ of mom) –Birth (who delivered, method of delivery) –Health of mom (smoke/drank during preg, weight gain)

57. 57 Smoking not available from CA or NY ~3 million usable observations I pulled.5% random sample from 1995 About 12,500 obs Variables: birthweight (grams), smoked, married, 4-level race, 5 level education, mothers age at birth

58. 58 ------------------------------------------------------------------------------ > - storage display value variable name type format label variable label ------------------------------------------------------------------------------ > - birthw int %9.0g birth weight in grams smoked byte %9.0g =1 if mom smoked during pregnancy age byte %9.0g moms age at birth married byte %9.0g =1 if married race4 byte %9.0g 1=white,2=black,3=asian,4=other educ5 byte %9.0g 1=0-8, 2=9-11, 3=12, 4=13-15, 5=16+ visits byte %9.0g prenatal visits ------------------------------------------------------------------------------

59. 59 dummy | variable, | =1 | =1 if mom smoked ifBW<2500 | during pregnancy grams | 0 1 | Total -----------+----------------------+---------- 0 | 11,626 1,745 | 13,371 | 86.95 13.05 | 100.00 | 94.64 89.72 | 93.96 | 81.70 12.26 | 93.96 -----------+----------------------+---------- 1 | 659 200 | 859 | 76.72 23.28 | 100.00 | 5.36 10.28 | 6.04 | 4.63 1.41 | 6.04 -----------+----------------------+---------- Total | 12,285 1,945 | 14,230 | 86.33 13.67 | 100.00 | 100.00 100.00 | 100.00 | 86.33 13.67 | 100.00

60. 60 Notice a few things –13.7% of women smoke –6% have low weight birth Pr(LBW | Smoke) =10.28% Pr(LBW |~ Smoke) = 5.36% RR = Pr(LBW | Smoke)/ Pr(LBW |~ Smoke) = 0.1028/0.0536 = 1.92

61. 61 Logit results Log likelihood = -3136.9912 Pseudo R2 = 0.0330 ------------------------------------------------------------------------------ lowbw | Coef. Std. Err. z P>|z| [95% Conf. Interval] -------------+---------------------------------------------------------------- smoked |.6740651.0897869 7.51 0.000.4980861.8500441 age |.0080537.006791 1.19 0.236 -.0052564.0213638 married | -.3954044.0882471 -4.48 0.000 -.5683654 -.2224433 _Ieduc5_2 | -.1949335.1626502 -1.20 0.231 -.5137221.1238551 _Ieduc5_3 | -.1925099.1543239 -1.25 0.212 -.4949791.1099594 _Ieduc5_4 | -.4057382.1676759 -2.42 0.016 -.7343769 -.0770994 _Ieduc5_5 | -.3569715.1780322 -2.01 0.045 -.7059081 -.0080349 _Irace4_2 |.7072894.0875125 8.08 0.000.5357681.8788107 _Irace4_3 |.386623.307062 1.26 0.208 -.2152075.9884535 _Irace4_4 |.3095536.2047899 1.51 0.131 -.0918271.7109344 _cons | -2.755971.2104916 -13.09 0.000 -3.168527 -2.343415 ------------------------------------------------------------------------------

62. 62 Odds Ratios Smoked –exp(0.674) = 1.96 –Smokers are twice as likely to have a low weight birth _Irace4_2 (Blacks) –exp(0.707) = 2.02 –Blacks are twice as likely to have a low weight birth

63. 63 Asking for odds ratios Logistic y x1 x2; In this case xi: logistic lowbw smoked age married i.educ5 i.race4;

64. 64 Log likelihood = -3136.9912 Pseudo R2 = 0.0330 ------------------------------------------------------------------------------ lowbw | Odds Ratio Std. Err. z P>|z| [95% Conf. Interval] -------------+---------------------------------------------------------------- smoked | 1.962198.1761796 7.51 0.000 1.645569 2.33975 age | 1.008086.0068459 1.19 0.236.9947574 1.021594 married |.6734077.0594262 -4.48 0.000.5664506.8005604 _Ieduc5_2 |.8228894.1338431 -1.20 0.231.5982646 1.131852 _Ieduc5_3 |.8248862.1272996 -1.25 0.212.6095837 1.116233 _Ieduc5_4 |.6664847.1117534 -2.42 0.016.4798043.9257979 _Ieduc5_5 |.6997924.1245856 -2.01 0.045.4936601.9919973 _Irace4_2 | 2.028485.1775178 8.08 0.000 1.70876 2.408034 _Irace4_3 | 1.472001.4519957 1.26 0.208.8063741 2.687076 _Irace4_4 | 1.362817.2790911 1.51 0.131.9122628 2.035893 ------------------------------------------------------------------------------

65. 65 PAR PAR = (RR – 1) /[(1- ) + RR ] = 0.137 RR = 1.96 PAR = 0.116 11.6% of low weight births attributed to maternal smoking

66. 66 Hypothesis Testing in MLE models MLE are asymptotically normally distributed, one of the properties of MLE Therefore, standard t-tests of hypothesis will work as long as samples are ‘large’ What ‘large’ means is open to question What to do when samples are ‘small’ – table for a moment

67. 67 Testing a linear combination of parameters Suppose you have a probit model Φ[β 0 + x 1i β 1 + x 2i β 2 + x 3i β 3 +… ] Test a linear combination or parameters Simplest example, test a subset are zero β 1 = β 2 = β 3 = β 4 =0 To fix the discussion N observations K parameters J restrictions (count the equals signs, j=4)

68. 68 Wald Test Based on the fact that the parameters are distributed asymptotically normal Probability theory review –Suppose you have m draws from a standard normal distribution (z i ) –M = z 1 2 + z 2 2 + …. Z m 2 –M is distributed as a Chi-square with m degrees of freedom

69. 69 Wald test constructs a ‘quadratic form’ suggested by the test you want to perform This combination, because it contains squares of the true parameters, should, if the hypothesis is true, be distributed as a Chi square with j degrees of freedom. If the test statistic is ‘large’, relative to the degrees of freedom of the test, we reject, because there is a low probability we would have drawn that value at random from the distribution

70. 70 Reading values from a Table All stats books will report the ‘percentiles’ of a chi-square –Vertical axis (degrees of freedom) –Horizontal axis (percentiles) –Entry is the value where ‘percentile’ of the distribution falls below

71. 71 Example: Suppose 4 restrictions 95% of a chi-square distribution falls below 9.488. So there is only a 5% a number drawn at random will exceed 9.488 If your test statistic is below, cannot reject null If your test statistics is above, reject null

72. 72 Chi-square

73. 73 Wald test in STATA Default test in MLE models Easy to do. Look at program test hsgrad somecol college Does not estimate the ‘restricted’ model ‘Lower power’ than other tests, i.e., high chance of false negative

74. 74 -2 Log likelihood test * how to run the same tests with a -2 log like test; * estimate the unresticted model and save the estimates ; * in urmodel; probit smoker age incomel male black hispanic hsgrad somecol college worka; estimates store urmodel; * estimate the restricted model. save results in rmodel; probit smoker age incomel male black hispanic worka; estimates store rmodel; lrtest urmodel rmodel;

75. 75 I prefer -2 log likelihood test –Estimates the restricted and unrestricted model –Therefore, has more power than a Wald test In most cases, they give the same ‘decision’ (reject/not reject)