Page 1Page 2Page 3Page 4Page 5Page 6Page 7Page 8Page 9Page 10Page 11Page 12Page 13Page 14Page 15Page 16Page 17Page 18Page 19Page 20Page 21Page 22Page 23Page 24Page 25Page 26Page 27Page 28Page 29Page 30Page 31Page 32Page 33Page 34Page 35Page 36Page 37Page 381parametric estimation, univariate:by maximum likelihoodestimation in standard softwareno special procedures for univariate estimationhow might one do univariate estimation using regression program?2use regression program, no regressorsSAS syntax:proc lifereg;model ttr28*rcsr28(1) = onee / covb;onee is variable equal to 1default is Weibull distributioncan compute survival from regression parameterscan also compute confidence intervals, etc.3Stata:burn data from KM. streg, dist(w) failure _d: d_st analysis time _t: t_stFitting constant-only model:Iteration 0: log likelihood = -151.73168Iteration 1: log likelihood = -150.72726Iteration 2: log likelihood = -150.72516Iteration 3: log likelihood = -150.72516Fitting full model:Iteration 0: log likelihood = -150.725164Weibull regression -- log relative-hazard form No. of subjects = 154 Number of obs = 154No. of failures = 48Time at risk = 3357 LR chi2(0) = -0.00Log likelihood = -150.72516 Prob > chi2 = .------------------------------------------------------------------------------ _t | Haz. Ratio Std. Err. z P>|z| [95% Conf. Interval]-------------+-----------------------------------------------------------------------------+---------------------------------------------------------------- /ln_p | -.1698046 .1246513 -1.36 0.173 -.4141166 .0745075-------------+---------------------------------------------------------------- p | .8438297 .1051845 .6609239 1.077353 1/p | 1.185073 .1477209 .9282005 1.513034------------------------------------------------------------------------------5. stcurve, survivalproduces survival curve:6compare with nonparametric:. sts graphcan also use statato obtainestimates ofmean, mediansurvival undermodelsamplingvariabilityassessment notdone7asthma data again:3 estimates: KM, 2 parametricexponentialWeibull8describe differences between curves, confidence intervals9artificial example:exponential survival, hazard = 1potential censoring times distributed uniformly on [1,3]10comment on differences in confidence intervals, etc.11Weibull parameters: parameter point estimate standard error2 parameterWeibull model:Intercept ( ) 0.01788 0.05062Shape (6) 1.02810 0.04242exponential (1parameter) model:Intercept ( ) 0.02151 0.0489712pure left censoring (i.e., left censored observations and uncensored observationsonly)assume observations are random sample of all subjectsassume left censoring randomcan estimate survival curve using following algorithm:get follow-up times for various subjects13consider students learning to read in 1 gradestassume all children learn to read by end of gradetime scale-agedata: T/ time at which reading status first knownAge of entry to 1 grade for kids already readingstage of learning to read for other 1 gradersstL* / indicator of left censoring14algorithm for estimation:pick large time J Jcompute “reversed” failure-time variable T / J - TJ Lapply KM estimator to “reversed” data {T ,* }estimate on new scalecompute15example:data generated from Weibull distribution (shape parameter = 1.5, scaleparameter = 1)entry time uniform on [0,1]16note that 1 observation time and failure time are unassociated; thus censoring isstrandom, and estimate is unbiased17when entry time (left censoring time) is associated with outcome, estimatorbiased:18more general censoring pattern: left and right censored (double censored)use self-consistency algorithm (we saw before that KM estimate can be viewedas resulting from this)jgrid of time points t at which subjects observednumber of subjects observed at time jright censored: rjleft censored: cjfailing dexample: data from section 1.17-time at which California high school boys firstsmoked marijuana19start off with initial guess: ignore left censored observationswill compute KM estimator making this assumption# of subjects at risk at each time:20KM estimator:21from survival estimates, compute relative probabilities of failing at given age():22for someone left censored at given age, compute probability of failure at givenage given previous estimates ( )23these are proportional to probabilities in previous graph; divide probability iniprevious graph by total probability of failing by age t (jestimate number of events at time t by ; i.e., add to the observed number of events at j the estimated number of events atj among subjects left censored at i2425note that number failing at earlier times increases more than number failing atlater times (since subject left censored at t failed at earlier time)26estimate the number of subjects at risk: # at risk atearlier timeincreasesmore thannumber atlater time(evenproportion-ately)27updated survival estimate using product-limit estimator with time-specific risks28final/first iteration survival probability is somewhat lower than initial one, asexpectedfirst iteration survival probability not consistent with initial estimatessubsequent iterations don’t change things muchidea: as before, stop when updates due to current survival estimates areconsistent with survival estimates29same idea used for interval censoringsteps:iidivide time into a grid including all left- and right- endpoints (L , R ],iiweight if is in (L , R ], 0 otherwisecompute initial survival estimate30iterative steps:jcompute probability of event occurring at time J ; jestimate the number of events at J by jfor each individual, probability of event occuring at J is proportion ofjevents occuring in permissible interval (denominator) that occur at Jii ijif have subject with failure-time known exactly (i.e., L =R ), " will be 1 foronly 1 time (i.e., the failure-time); that individual’s contribution to the sumwill not change with changes in the presumed survival function31jcompute the estimated # at risk at time J by (i.e., estimated number ofsubjects dying at time j or later; here formula is apparently for case whereno right censored subjects, unless let m = 4)look at example in book (5.2); direction