1. 1 Lecture Twelve

2. 2 Outline Failure Time Analysis Linear Probability Model Poisson Distribution

3. 3 Failure Time Analysis Example: Duration of Expansions Issue: does the probability of an expansion ending depend on how long it has lasted? Exponential distribution: assumes the answer since the hazard rate is constant Weibull distribution allows a test to be performed

4. 4 Part II: Failure Time Analysis Exponential –survival function –hazard rate Weibull Exploratory Data Analysis, Lab Seven

5. 5 Duration of Post-War Economic Expansions in Months

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7. 7 Estimated Survivor Function for Ten Post-War Expansions

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12. Exponential Distribution Hazard rate: ratio of density function to the survivor function: h(t) = f(t)/S(t) measure of probability of failure at time t given that you have survived that long for the exponential it is a constant: h(t) =

13. 13 Interval hazard rate=#ending/#at risk

14. Cumulative Hazard Function In general: For the exponential,

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18. Weibull Distribution F(t) = 1 - exp[ S(t) = ln S(t) = - (t/   h(t) = f(t)/S(t) f(t) = dF(t)/dt = - exp[-(t/    t/   h(t) = (  t/   if  h(t) = constant if  h(t) is increasing function if  h(t) is a decreasing function

19. 19 Weibull Distribution Cumulative Hazard Function

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22. 22 Dependent Variable: LNCUMHAZ Method: Least Squares Sample: 2 11 Included observations: 10 VariableCoefficientStd. Errort-StatisticProb. LNDUR1.4366620.10355813.873030.0000 C -5.9207400.403303-14.680610.0000 R-squared0.960092 Mean dependent var-0.409591 Adjusted R-squared0.955103 S.D. dependent var1.038386 S.E. of regression 0.220022 Akaike info criterion-0.013326 Sum squared resid0.387276 Schwarz criterion0.047191 Log likelihood 2.066628 F-statistic192.4609 Durbin-Watson stat1.210695 Prob(F-statistic)0.000001

23. 23 Is Beta More Than One? H 0 : beta=1 A : beta>1, and hazard rate is increasing with time, i.e. expansions are more likely to end the longer they lastH A : beta>1, and hazard rate is increasing with time, i.e. expansions are more likely to end the longer they last t = ( 1.437 - 1)/0.104 = 4.20t = ( 1.437 - 1)/0.104 = 4.20

24. 24 Conclude Economic expansions are at increasing risk the longer they last the business cycle is not dead so much for the new economics maybe Karl Marx was right, capitalism is an inherently unstable system, subject to cycles

25. 25 Lab Seven

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28. 28 Cumulative Hazard Rate for Fan Failure y = 4E-05x + 0.0089 R 2 = 0.9816 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 010002000300040005000600070008000900010000 Duration in Hours Cumulative Hazard

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30. 30 Lambda= 3.89 x 10 -5, mean=25,707 hrs Regress Cumulative Hazard on Duration

31. 31 Fan Failure Is the hazard rate really constant? Regress ln cumulative hazard on ln duration

32. 32 Accept that the hazard rate is constant

33. 33 Part II: Linear Probability Model

34. 34 Lab Six Lottery what effect do the zeros have on the regression of % income spent on the lottery versus explanatory variables such as household income?

35. 35 LOTTERYAGE CHILDREN EDUCATION INCOME 5.000000 50.00000 2.000000 15.00000 41.00000 7.000000 26.00000 0.000000 10.00000 22.00000 0.000000 40.00000 3.000000 13.00000 24.00000 10.00000 46.00000 2.000000 9.000000 20.00000 5.000000 40.00000 3.000000 14.00000 32.00000 5.000000 39.00000 2.000000 15.00000 42.00000 3.000000 36.00000 3.000000 8.000000 18.00000 0.000000 44.00000 1.000000 16.00000 47.00000 0.000000 47.00000 4.000000 20.00000 85.00000 6.000000 52.00000 1.000000 10.00000 23.00000 0.000000 51.00000 2.000000 18.00000 61.00000 0.000000 41.00000 2.000000 17.00000 70.00000 12.00000 42.00000 2.000000 9.000000 22.00000 7.000000 53.00000 1.000000 12.00000 27.00000 11.00000 72.00000 1.000000 9.000000 25.00000 Data

36. OLS Regression

37. 37 OLS slope biased toward zero

38. 38 Effect of the Zeros on Regression Bias the OLS regression slope towards zero can’t throw away the zeros, this is the mistake the NASA engineers made with Challenger. They threw away the launches where no o-rings had failed. Start with a simpler model: does the household play the lottery or not?

39. 39 First Compare Tobit Slope to OLS Slope Tobit slope: -0.229 OLS slope: -0.142

40. 40 Dependent Variable: LOTTERY Method: ML - Censored Normal (TOBIT) Sample: 1 100 Included observations: 100 Left censoring (value) at zero Convergence achieved after 5 iterations Covariance matrix computed using second derivatives CoefficientStd. Errorz-StatisticProb. C12.216731.08061811.305310.0000 INCOME-0.2294600.033817-6.7853900.0000 Error Distribution SCALE:C(3)3.6352460.30720411.833320.0000 R-squared0.365088 Mean dependent var5.390000 Adjusted R-squared0.351997 S.D. dependent var3.786993 S.E. of regression 3.048478Akaike info criterion4.621338 Sum squared resid901.4421Schwarz criterion4.699493 Log likelihood-228.0669 Hannan-Quinn criter.4.652969 Avg. log likelihood-2.280669 Left censored obs23 Right censored obs0 Uncensored obs77 Total obs100 Tobit:Eviews

41. 41 OLS Excel

42. 42 Bernoulli Variable: Bern Bern = 0*(lottery=0) + 1*(lottery>0) Linear Probability Model: dummy dependent variable Bern(i) = c + a*income + b*age +d*children + f*education + e(i)

43. 43 BERNLOTTERY 15 170 110 15 1300 1600 112 17 111

44. 44 Averages for Players and Non-Players

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46. 46 Dependent Variable: BERN Method: Least Squares Sample: 1 100 Included observations: 100 VariableCoefficientStd. Errort-StatisticProb. C1.2261520.08522914.386530.0000 INCOME-0.0138560.002340-5.9219890.0000 R-squared 0.263545Mean dependent var0.770000 Adjusted R-squared0.256030 S.D. dependent var0.422953 S.E. of regression 0.364812 Akaike info criterion0.840929 Sum squared resid13.04261 Schwarz criterion0.893032 Log likelihood -40.04644 F-statistic 35.06996 Durbin-Watson stat2.043370 Prob(F-statistic)0.000000

47. 47 Linear Probability Model Note: in the linear probability model, income is determining the probability of playing the lottery not the % of income spent on the lottery, so the interpretation of the slope is different.

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49. 49 Non-Linear Probability Models Probit (Normit)

50. 50 Dependent Variable: BERN Method: ML - Binary Probit Sample: 1 100 Included observations: 100 Convergence achieved after 4 iterations Covariance matrix computed using second derivatives VariableCoefficientStd. Errorz-StatisticProb. C2.3732420.4096715.7930390.0000 INCOME-0.0467990.011017-4.2480300.0000 Mean dependent var0.770000 S.D. dependent var0.422953 S.E. of regression 0.362314 Akaike info criterion0.876836 Sum squared resid12.86458 Schwarz criterion0.928939 Log likelihood-41.84178 Hannan-Quinn criter.0.897923 Restr. log likelihood -53.92763 Avg. log likelihood -0.418418 LR statistic (1 df) 24.17171McFadden R-squared0.224112 Probability(LR stat)8.81E-07 Obs with Dep=023 Total obs100 Obs with Dep=177

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52. Part IV. Poisson Approximation to Binomial Conditions: f(x) = {exp[-  ]  x }/x! Assumptions: –the number of events occurring in non- overlapping intervals are independent –the probability of a single event occurring in a small interval is approximately proportional to the interval –the probability of more than one event in an interval is negligible

53. 53 Example Ten % of tools produced in a manufacturing process are defective. What is the probability of finding exactly two defectives in a random sample of 10? Binomial: p(k=2) = 10!/(8!2!)(0.1) 2 (0.9) 8 = 0.194 Poisson, where the mean of the Poisson,  equals n*p = 0.1 p(k=2) = {exp[-1] 1 2 }/2! = 0.184