1 Lecture Twelve

2 Outline Failure Time Analysis Linear Probability Model Poisson Distribution

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

5 Duration of Post-War Economic Expansions in Months

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7 Estimated Survivor Function for Ten Post-War Expansions

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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 Interval hazard rate=#ending/#at risk

Cumulative Hazard Function In general: For the exponential,

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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 Weibull Distribution Cumulative Hazard Function

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22 Dependent Variable: LNCUMHAZ Method: Least Squares Sample: 2 11 Included observations: 10 VariableCoefficientStd. Errort-StatisticProb. LNDUR C R-squared Mean dependent var Adjusted R-squared S.D. dependent var S.E. of regression Akaike info criterion Sum squared resid Schwarz criterion Log likelihood F-statistic Durbin-Watson stat Prob(F-statistic)

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 = ( )/0.104 = 4.20t = ( )/0.104 = 4.20

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 Lab Seven

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28 Cumulative Hazard Rate for Fan Failure y = 4E-05x R 2 = Duration in Hours Cumulative Hazard

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

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

32 Accept that the hazard rate is constant

33 Part II: Linear Probability Model

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 LOTTERYAGE CHILDREN EDUCATION INCOME Data

OLS Regression

37 OLS slope biased toward zero

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 First Compare Tobit Slope to OLS Slope Tobit slope: OLS slope:

40 Dependent Variable: LOTTERY Method: ML - Censored Normal (TOBIT) Sample: Included observations: 100 Left censoring (value) at zero Convergence achieved after 5 iterations Covariance matrix computed using second derivatives CoefficientStd. Errorz-StatisticProb. C INCOME Error Distribution SCALE:C(3) R-squared Mean dependent var Adjusted R-squared S.D. dependent var S.E. of regression Akaike info criterion Sum squared resid Schwarz criterion Log likelihood Hannan-Quinn criter Avg. log likelihood Left censored obs23 Right censored obs0 Uncensored obs77 Total obs100 Tobit:Eviews

41 OLS Excel

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 BERNLOTTERY

44 Averages for Players and Non-Players

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46 Dependent Variable: BERN Method: Least Squares Sample: Included observations: 100 VariableCoefficientStd. Errort-StatisticProb. C INCOME R-squared Mean dependent var Adjusted R-squared S.D. dependent var S.E. of regression Akaike info criterion Sum squared resid Schwarz criterion Log likelihood F-statistic Durbin-Watson stat Prob(F-statistic)

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 Non-Linear Probability Models Probit (Normit)

50 Dependent Variable: BERN Method: ML - Binary Probit Sample: Included observations: 100 Convergence achieved after 4 iterations Covariance matrix computed using second derivatives VariableCoefficientStd. Errorz-StatisticProb. C INCOME Mean dependent var S.D. dependent var S.E. of regression Akaike info criterion Sum squared resid Schwarz criterion Log likelihood Hannan-Quinn criter Restr. log likelihood Avg. log likelihood LR statistic (1 df) McFadden R-squared Probability(LR stat)8.81E-07 Obs with Dep=023 Total obs100 Obs with Dep=177

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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 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 = Poisson, where the mean of the Poisson,  equals n*p = 0.1 p(k=2) = {exp[-1] 1 2 }/2! = 0.184