I. IntroductionII. Failure Time Models and the Weibull DistributionIII. TransformationsII. Cumulative Hazard FunctionDuration“Interval” Hazard RateCumulative Hazard RateIII. Poisson DistributionOct. 27, 2005 LEC #11 ECON 240A-1 L. Phillips Weibull Distribution; Transformations; Poisson DistributionI. IntroductionIn Lecture Ten we introduced the exponential distribution as a parametric approach to estimating the distribution of time until failure. This distribution has one parameter, lambda, and the reciprocal of lambda is the mean time until failure. So the exponential is parsimonious in parameters to estimate but this simplicity came at a price of two assumptions. First, the hazard rate is constant for the exponential, which is restrictive. Second, the exponential has the no memory feature, which means that the survival time to date does not affect the expected time remaining before failure. The Weibull is a distribution that permits a little more flexibility, but at a price of two parameters. The survivor function is also nonlinear in these parameters, which raises a question about whether we can linearize the function through transformation. We can not. That leaves the question of how to estimate the equation.Lastly, we turn to another distribution, the Poisson, which can be used as an approximation to the binomial for rare events. It is useful for modeling problems such as the number of defects on a foot of magnetic recording tape and other applications to quality control.II. Failure Time Models and the Weibull DistributionThe Weibull Distribution has the cumulative distribution function:F(t) = 1 – exp[-(t/)] , (1)And so the survivor function is:S(t) = 1 – F(t) = exp[-(t/)] (2)Taking the derivative of the cumulative distribution function yields the density function,dF(t)/dt = f(t) = (/) (t/)-1 exp[-(t/)], (3)Oct. 27, 2005 LEC #11 ECON 240A-2 L. Phillips Weibull Distribution; Transformations; Poisson Distributionand so the hazard rate is:h(t) = f(t)/S(t) = (/) (t/)-1 (4)Thus the hazard rate is a power function of the duration, t. If beta equals one, then the hazard rate is a constant, 1/. If beta is greater than one, then the hazard rate increases with survival time, t. If beta is less than one, then the hazard rate is a decreasing function of t. So, depending on the value of this parameter, three different patterns of behavior for the hazard rate can be explained. This does not cover all possibilities such as a situation where the hazard rate may first increase and then decrease with survival time, but it is more flexible than the exponential.III. TransformationsWe saw in the previous lecture that taking the logarithms of both sides of the equation for the Weibull’s survivor function did not result in an equation linear in the parameters. We could try this transformation on the hazard rate:ln[h(t)] = ln[] - ln[] + ( - 1)ln[t] (5)This is not linear in alpha and beta, but it is close if we do not try and separately identify those two parameters, and let the intercept equal ln[] - ln[], and the slope equal beta minus one, i.e. ln[h(t)] = a + b ln[t] (6)We are interested in the slope. If it is not significantly different from zero, then we can accept the null hypothesis that beta equals one. The logarithm of the hazard rate is plottedagainst the logarithm of survival time (duration) in Figure1.Oct. 27, 2005 LEC #11 ECON 240A-3 L. Phillips Weibull Distribution; Transformations; Poisson Distribution------------------------------------------------------------------------------------------------------------The coefficient of determination is only 5.5%. The F-distribution statistic calculated fromthis R2 is not significant and Student’s t-statistic for the null hypothesis that the slope is zero is only 0.64, so we can not reject the hypothesis that beta equals one and the hazard rate is constant. Of course there are only a few observations, but the exponential seems appropriate.II. Cumulative Hazard FunctionThere is another test for whether the exponential distribution is appropriate for theduration of post-war expansions. It is the cumulative hazard function, H(t), which is the sum of the hazard function, h(t):H(t) = t0h(u) du (7)Figure 1: Log Hazard Rate Vs. Log Durationy = 0.366x - 5.1181R2 = 0.0549-6-5-4-3-2-100 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5Log DurationLog Hazard RateLog Hazard RateLinear RegressionOct. 27, 2005 LEC #11 ECON 240A-4 L. Phillips Weibull Distribution; Transformations; Poisson DistributionAnd applying this to the exponential, where from Eq. (20) of chapter ten, the hazard rate is the constant lambda:H(t) = t0h(u) du = = t0 du = t, (8)So the cumulative hazard function for the exponential is a linear function of the time untilfailure. We can calculate the cumulative hazard rate using the values we calculated for theinterval hazard rate in Table 5 of the previous chapter, reproduced in part in Table 1 below. The cumulative hazard rate is just the running sum of the hazard rate, so at-------------------------------------------------------------------------------------------------------Table 1: Estimated Hazard Rate and Cumulative Hazard Rate, Post-War ExpansionsDuration # Ending # at Risk “ Interval” Hazard Rate Cumulative Hazard Rate0 0 1012 1 10 0.1000 0.100024 1 9 0.1111 0.211136 1 8 0.1250 0.336137 1 7 0.1429 0.479039 1 6 0.1667 0.645645 1 5 0.2000 0.845658 1 4 0.2500 1.095692 1 3 0.3333 1.4290106 1 2 0.5000 1.9290125+ 1the expansion of duration 12 months, the cumulative hazard rate equals the hazard rate, atthe expansion of duration 24 months, the cumulative hazard rate is equal to the hazard rate at that duration plus the previous hazard rate, i.e. 0.1111 + 0.10000, and so on.This cumulative hazard rate is plotted against duration in Figure 2.-----------------------------------------------------------------------------------------------------------Oct. 27, 2005 LEC #11 ECON 240A-5 L. Phillips Weibull Distribution; Transformations; Poisson Distribution--------------------------------------------------------------------------------------------Note that the linear regression is a good fit, with an R2 of 0.96, supporting evidence that the expansion durations are distributed exponentially. Note also that the slope, which is an estimate of the hazard rate or lambda, is 0.0192. Since the reciprocal of lambda is the mean time to failure, we get an estimate of 52 months. So
View Full Document
Unlocking...