Event History Modeling: a Guide for Social Scientists

Event History Modeling: A Guide for Social Scientists by Janet M. Box-Steffensmeier and Bradford S. JonesRelevant Files for Stata Output Data file: UNFINAL2.DTA (Stata 7.0 Dataset; already stset) “DO” file: UNpeacekeeping1.do (Contains code) Log file: UNpeacekeeping1.doc (Log file in WORD format of output from “do” file; does not contain graphs) UNpeacekeeping1.rtf (Log file in RTF format of output from “do” file; does not contain graphs)The following is a Stata log file containing the output generated from the file “Unpeacekeeping1.do”. This Stata DO file will generate Kaplan-Meier estimates, Weibull (and exponential) estimates, and Cox estimates using the U.N. Peacekeeping data. Results from parametric and Cox models are published in Chapters 3 and 4 of Event History Modeling: A Guide for Social Scientists. Graphs generated from the “do” file are not reproduced in this file (though if you execute the file, the graphs will be saved on your computer). log: c:\ehbook\UNpeacekeeping.log. do c:\ehbook\Unpeacekeeping1. * DATA SET USED IS UNFINAL2.DTA* . *Note that all variables that are created in this file are DROPPED; edit as necessary.* . . use c:\data\statadata\UNFINAL2. . *Note that directory may be different depending on the location of your data files* . . * 1 Produce Kaplan-Meier Estimates:* . sts list failure _d: failed analysis time _t: durationBeg. Net Survivor Std. Time Total Fail Lost Function Error [95% Conf. Int.] ------2 54 1 0 0.9815 0.0183 0.8757 0.9974 4 53 1 0 0.9630 0.0257 0.8599 0.9906 5 52 2 0 0.9259 0.0356 0.8146 0.9715 7 50 3 0 0.8704 0.0457 0.7472 0.9360 10 47 1 0 0.8519 0.0483 0.7255 0.9230 11 46 1 0 0.8333 0.0507 0.7042 0.9096 12 45 2 0 0.7963 0.0548 0.6624 0.8816 13 43 1 1 0.7778 0.0566 0.6420 0.8672 15 41 2 0 0.7398 0.0598 0.6005 0.8369 16 39 1 1 0.7209 0.0612 0.5802 0.8214 18 37 1 0 0.7014 0.0626 0.5594 0.8053 19 36 1 0 0.6819 0.0638 0.5389 0.7889 21 35 1 0 0.6624 0.0649 0.5186 0.7723 23 34 2 1 0.6235 0.0667 0.4789 0.7385 25 31 3 1 0.5631 0.0687 0.4185 0.6848 26 27 1 0 0.5423 0.0693 0.3980 0.6660 28 26 1 0 0.5214 0.0697 0.3777 0.6469 29 25 1 1 0.5005 0.0699 0.3577 0.6276 30 23 2 0 0.4570 0.0703 0.3164 0.5870 31 21 1 0 0.4353 0.0702 0.2962 0.5663 34 20 1 0 0.4135 0.0700 0.2764 0.5453 46 19 3 0 0.3482 0.0684 0.2189 0.4807 48 16 2 0 0.3047 0.0664 0.1823 0.4362 49 14 1 0 0.2829 0.0651 0.1645 0.4134 59 13 0 1 0.2829 0.0651 0.1645 0.4134 66 12 1 0 0.2593 0.0638 0.1453 0.3890 70 11 1 0 0.2358 0.0622 0.1266 0.3642 71 10 0 1 0.2358 0.0622 0.1266 0.3642 99 9 0 1 0.2358 0.0622 0.1266 0.3642 127 8 0 2 0.2358 0.0622 0.1266 0.3642 128 6 1 0 0.1965 0.0630 0.0912 0.3310 284 5 0 1 0.1965 0.0630 0.0912 0.3310 319 4 0 1 0.1965 0.0630 0.0912 0.3310 452 3 0 1 0.1965 0.0630 0.0912 0.3310 634 2 0 1 0.1965 0.0630 0.0912 0.3310 641 1 0 1 0.1965 0.0630 0.0912 0.3310 ------. . * 2 Produce Graph of K-M Estimates given from above command* . sts graph, t1("Kaplan-Meier Survival Function Estimates") b2("Duration of U.N. Mission") noborder saving(UN > KMgraph1, replace) failure _d: failed analysis time _t: duration. . * 3 Produces Kaplan-Meier Estimates for the three conflict types* . sts list, by(contype) failure _d: failed analysis time _t: durationBeg. Net Survivor Std. Time Total Fail Lost Function Error [95% Conf. Int.] ------contype=1 4 14 1 0 0.9286 0.0688 0.5908 0.9896 5 13 1 0 0.8571 0.0935 0.5394 0.9622 12 12 1 0 0.7857 0.1097 0.4725 0.9254 13 11 1 1 0.7143 0.1207 0.4063 0.8819 15 9 1 0 0.6349 0.1308 0.3312 0.8297 16 8 1 0 0.5556 0.1364 0.2636 0.7717 23 7 1 0 0.4762 0.1381 0.2026 0.7083 25 6 1 1 0.3968 0.1360 0.1478 0.6396 28 4 1 0 0.2976 0.1334 0.0820 0.5559 30 3 1 0 0.1984 0.1203 0.0343 0.4603 34 2 1 0 0.0992 0.0924 0.0061 0.3504 46 1 1 0 0.0000 . . . contype=2 7 10 1 0 0.9000 0.0949 0.4730 0.9853 16 9 0 1 0.9000 0.0949 0.4730 0.9853 26 8 1 0 0.7875 0.1340 0.3809 0.9426 31 7 1 0 0.6750 0.1551 0.2906 0.8825 70 6 1 0 0.5625 0.1651 0.2094 0.8092 127 5 0 1 0.5625 0.1651 0.2094 0.8092 128 4 1 0 0.4219 0.1737 0.1110 0.7126 319 3 0 1 0.4219 0.1737 0.1110 0.7126 634 2 0 1 0.4219 0.1737 0.1110 0.7126 641 1 0 1 0.4219 0.1737 0.1110 0.7126 contype=3 2 30 1 0 0.9667 0.0328 0.7861 0.9952 5 29 1 0 0.9333 0.0455 0.7589 0.9829 7 28 2 0 0.8667 0.0621 0.6828 0.9478 10 26 1 0 0.8333 0.0680 0.6450 0.9270 11 25 1 0 0.8000 0.0730 0.6080 0.9048 12 24 1 0 0.7667 0.0772 0.5720 0.8813 15 23 1 0 0.7333 0.0807 0.5369 0.8567 18 22 1 0 0.7000 0.0837 0.5026 0.8312 19 21 1 0 0.6667 0.0861 0.4692 0.8047 21 20 1 0 0.6333 0.0880 0.4365 0.7775 23 19 1 1 0.6000 0.0894 0.4045 0.7495 25 17 2 0 0.5294 0.0918 0.3378 0.6889 29 15 1 1 0.4941 0.0922 0.3059 0.6573 30 13 1 0 0.4561 0.0926 0.2716 0.6232 46 12 2 0 0.3801 0.0915 0.2070 0.5521 48 10 2 0 0.3041 0.0876 0.1477 0.4766 49 8 1 0 0.2661 0.0845 0.1202 0.4371 59 7 0 1 0.2661 0.0845 0.1202 0.4371 66 6 1 0 0.2217 0.0812 0.0884 0.3924 71 5 0 1 0.2217 0.0812 0.0884 0.3924 99 4 0 1 0.2217 0.0812 0.0884 0.3924 127 3 0 1 0.2217 0.0812 0.0884 0.3924 284 2 0 1 0.2217 0.0812 0.0884 0.3924 452 1 0 1 0.2217 0.0812 0.0884 0.3924 ------. . *4 Produce Graph of K-M Estimates by Conflict Type* . sts graph, by(contype) t1("Kaplan-Meier Survival Function Estimates") b2("Duration of U.N. Mission") nobor > der nolabel saving(UNKMgraph2, replace) failure _d: failed analysis time _t: duration. *Note that graphs were modified in Corel Draw; I added labels to graph lines using Corel Draw; omitting the > nolabel command* . *will allow Stata to apply labels* . . *Using UN data for exponential and weibull models* . . * 5 Estimate Exponential Model using proportional hazards estimates.* . streg civil interst, dist(exp) nohr failure _d: failed analysis time _t: duration Iteration 0: log likelihood = -103.03289 Iteration 1: log likelihood = -90.211473 Iteration 2: log likelihood = -86.44131 Iteration 3: log likelihood = -86.354656 Iteration 4: log likelihood = -86.354481 Iteration 5: log likelihood = -86.354481 Exponential regression -- log relative-hazard form No. of subjects = 54 Number of obs = 54 No. of failures = 39 Time at risk = 3994 LR chi2(2) = 33.36 Log likelihood = -86.354481 Prob > chi2 = 0.0000------_t | Coef. Std. Err. z P>|z| [95% Conf. Interval] ------+------civil | 1.169344 .3588703 3.26 0.001 .4659714 1.872717 interst | -1.6401 .4954337 -3.31 0.001 -2.611132 -.6690679 _cons | -4.350864 .2132007 -20.41 0.000 -4.76873 -3.932999 ------. . * 6 Estimate Exponential Model using the accelerated failure time estimates.* . streg civil interst, dist(exp) time failure _d: failed analysis time _t: durationIteration 0: log likelihood = -103.03289 Iteration 1: log likelihood = -90.211473 Iteration 2: log likelihood = -86.44131 Iteration 3: log likelihood = -86.354656 Iteration 4: log likelihood = -86.354481 Iteration 5: log likelihood = -86.354481 Exponential regression -- accelerated failure-time form No. of subjects = 54 Number of obs = 54 No. of failures = 39 Time at risk = 3994 LR chi2(2) = 33.36 Log likelihood = -86.354481 Prob > chi2 = 0.0000------_t | Coef. Std. Err. z P>|z| [95% Conf. Interval] ------+------civil | -1.169344 .3588703 -3.26 0.001 -1.872717 -.4659714 interst | 1.6401 .4954337 3.31 0.001 .6690679 2.611132 _cons | 4.350864 .2132007 20.41 0.000 3.932999 4.76873 ------. . . *7 Using generate to create variable storing hazard rate for each conflict type (we're referencing coeffic > ients from AFT model)* . . gen haz_civil=exp(-(_b[_cons]+_b[civil])) . gen haz_interstate=exp(-(_b[_cons]+_b[interst])). gen haz_icw=exp(-(_b[_cons])). . *8 Using display to compute h(t) for each conflict type* . display exp(-(_b[_cons]+_b[civil])) .04152249. display exp(-(_b[_cons]+_b[interst])) .00250125. display exp(-(_b[_cons])) .01289566. . *9 We could also use predict option* . predict hazardrate, hazard (4 missing values generated). *Note that the predict option will compute h(t) for each observation.* . . *10 Graphing the hazard rates* . gr haz_civil haz_interstate haz_icw duration, c(lll) s(iii) xlab ylab t1("Hazard Rates") b2("Duration of U. > N. Peacekeeping Missions") saving(UNexp, replace). . *Again, as with all graphs, I label them in Corel. To get the legend attached to the above graph (so you* . * can figure out which line corresponds to which conflict type, omit the s(iii) and t1( ) options).* . . *11 Dropping newly created variables from dataset* . drop haz_* hazardrate. . *12 Using display to compute the expected survival time for each conflict type* . display exp(_b[_cons]+_b[civil]) 24.083333. display exp(_b[_cons]+_b[interst]) 399.8. display exp(_b[_cons]) 77.545455. . *13 We could also use predict option* . predict ETexp, mean (option mean time assumed; predicted mean time) (4 missing values generated). *Note that the predict option will compute E(T) for each observation.* . . *14 Using display to compute the median survival time for each conflict type* . display exp(_b[_cons]+_b[civil])*log(2) 16.693295. display exp(_b[_cons]+_b[interst])*log(2) 277.12024 . display exp(_b[_cons])*log(2) 53.750413. . *15 We could also use predict option* . predict MTexp, median (option median time assumed; predicted median time) (4 missing values generated). . *16 Dropping newly created variables from dataset* . drop ETexp MTexp. . *17 Estimate Weibull Model with Proportional Hazards Parameterization* . streg civil interst, dist(weib) nohr failure _d: failed analysis time _t: durationFitting constant-only model:Iteration 0: log likelihood = -103.03289 Iteration 1: log likelihood = -93.501426 Iteration 2: log likelihood = -93.488663 Iteration 3: log likelihood = -93.488663Fitting full model:Iteration 0: log likelihood = -93.488663 Iteration 1: log likelihood = -86.548564 Iteration 2: log likelihood = -84.667898 Iteration 3: log likelihood = -84.655162 Iteration 4: log likelihood = -84.655157 Weibull regression -- log relative-hazard form No. of subjects = 54 Number of obs = 54 No. of failures = 39 Time at risk = 3994 LR chi2(2) = 17.67 Log likelihood = -84.655157 Prob > chi2 = 0.0001------_t | Coef. Std. Err. z P>|z| [95% Conf. Interval] ------+------civil | .8879245 .3832017 2.32 0.020 .1368629 1.638986 interst | -1.401441 .5117817 -2.74 0.006 -2.404515 -.3983673 _cons | -3.459909 .4952858 -6.99 0.000 -4.430652 -2.489167 ------+------/ln_p | -.2145617 .1237889 -1.73 0.083 -.4571834 .02806 ------+------p | .806895 .0998846 .6330642 1.028457 1/p | 1.239319 .1534138 .97233 1.579619 ------. . *18 Estimate Weibull Model with Accelerated Failure Time Parameterization* . streg civil interst, dist(weib) time failure _d: failed analysis time _t: durationFitting constant-only model:Iteration 0: log likelihood = -103.03289 Iteration 1: log likelihood = -93.501426 Iteration 2: log likelihood = -93.488663 Iteration 3: log likelihood = -93.488663Fitting full model:Iteration 0: log likelihood = -93.488663 Iteration 1: log likelihood = -86.548564 Iteration 2: log likelihood = -84.667898 Iteration 3: log likelihood = -84.655162 Iteration 4: log likelihood = -84.655157 Weibull regression -- accelerated failure-time form No. of subjects = 54 Number of obs = 54 No. of failures = 39 Time at risk = 3994 LR chi2(2) = 17.67 Log likelihood = -84.655157 Prob > chi2 = 0.0001------_t | Coef. Std. Err. z P>|z| [95% Conf. Interval] ------+------civil | -1.100421 .4457861 -2.47 0.014 -1.974146 -.2266966 interst | 1.736832 .6165459 2.82 0.005 .5284242 2.94524 _cons | 4.28793 .2652436 16.17 0.000 3.768062 4.807798 ------+------/ln_p | -.2145617 .1237889 -1.73 0.083 -.4571834 .02806 ------+------p | .806895 .0998846 .6330642 1.028457 1/p | 1.239319 .1534138 .97233 1.579619 ------. . . *19 Computing estimated hazard rates for each conflict type (we're referencing AFT estimates)* . . *20 First, note that lambda=exp(-beta'x). We can compute lambda for* . *each confict type:* . gen lambda_civil=exp(-(_b[_cons]+_b[civil])). gen lambda_interstate=exp(-(_b[_cons]+_b[interst])). gen lambda_icw=exp(-(_b[_cons])). . *21 Second, note that h(t)=lambda*p*(lambda*t)^(p-1). We can generate the* . *hazard rate for each covariate profile:* . gen haz_civil=lambda_civil*e(aux_p)*(lambda_civil*duration)^(e(aux_p)-1) (4 missing values generated). gen haz_interstate=lambda_interstate*e(aux_p)*(lambda_interstate*duration)^(e(aux_p )-1) (4 missing values generated) . gen haz_icw=lambda_icw*e(aux_p)*(lambda_icw*duration)^(e(aux_p)-1) (4 missing values generated). . *22 Now we can graph the hazard rates* . gr haz_civil haz_interstate haz_icw duration, c(lll) s(iii) ylab xlab t1("Hazard Rates") b2("Duration of U. > N. Peacekeeping Missions") saving(UNweibhaz, replace). . *23 We could also use the predict option to compute h(t) for every observation* . predict ht, hazard (4 missing values generated). . drop lambda_* haz_* ht. . *24 Displaying the expected survival times for each conflict type* . display exp(lngamma(1+(1/e(aux_p))))*exp(_b[_cons]+_b[civil]) 27.283894. display exp(lngamma(1+(1/e(aux_p))))*exp(_b[_cons]+_b[interst]) 465.70383. display exp(lngamma(1+(1/e(aux_p))))*exp(_b[_cons]) 81.999892. . *25 We could also use the predict option to compute E(T) for every observation* . predict ETweibull, mean (option mean time assumed; predicted mean time) (4 missing values generated). . *26 Displaying the median survival times for each conflict type* . display exp(_b[_cons]+_b[civil])*log(2)^(1/e(aux_p)) 15.383298. display exp(_b[_cons]+_b[interst])*log(2)^(1/e(aux_p)) 262.57471. display exp(_b[_cons])*log(2)^(1/e(aux_p)) 46.233457. . *27 We could also use the predict option to compute M(T) for every observation* . predict MTweibull, median (option median time assumed; predicted median time) (4 missing values generated). . drop ETweibull MTweibull. . *28 Computing the extimated survivor functions for each conflict type* . *First, note that lambda=exp(-beta'x). We can compute lambda for* . *each confict type:* . gen lambda_civil=exp(-(_b[_cons]+_b[civil])) . gen lambda_interstate=exp(-(_b[_cons]+_b[interst])). gen lambda_icw=exp(-(_b[_cons])). . *29 Second, note that S(t)=exp^-(lambda*t)^p. We can generate the* . *survivor functions for each covariate profile:* . gen surv_civil=exp(-(lambda_civil*duration)ê(aux_p)) (4 missing values generated). gen surv_interstate=exp(-(lambda_interstate*duration)ê(aux_p)) (4 missing values generated). gen surv_icw=exp(-(lambda_icw*duration)ê(aux_p)) (4 missing values generated). . *30 Now we can graph the survivor functions* . gr surv_civil surv_interstate surv_icw duration, c(sss) s(iii) ylab xlab t1("Survivor Function") b2("Durati > on of U.N. Peacekeeping Missions") saving(UNweibsurv, replace) . . *31 Note we can also use predict to generate the survivor probability for each observation:* . predict survival_time, surv (4 missing values generated). . drop lambda_* surv_* survival_time. . *32 Estimating Gompertz Model* . . streg civil interst, dist(gompertz) nohr failure _d: failed analysis time _t: durationFitting constant-only model:Iteration 0: log likelihood = -103.03818 Iteration 1: log likelihood = -88.564878 Iteration 2: log likelihood = -80.451626 Iteration 3: log likelihood = -79.833846 Iteration 4: log likelihood = -79.827491 Iteration 5: log likelihood = -79.827491 Fitting full model:Iteration 0: log likelihood = -79.827491 Iteration 1: log likelihood = -76.219033 Iteration 2: log likelihood = -75.749534 Iteration 3: log likelihood = -75.749466 Iteration 4: log likelihood = -75.749466 Gompertz regression -- log relative-hazard form No. of subjects = 54 Number of obs = 54 No. of failures = 39 Time at risk = 3994 LR chi2(2) = 8.16 Log likelihood = -75.749466 Prob > chi2 = 0.0169------_t | Coef. Std. Err. z P>|z| [95% Conf. Interval] ------+------civil | .630388 .3680023 1.71 0.087 -.0908832 1.351659 interst | -.8770961 .5050953 -1.74 0.082 -1.867065 .1128726 _cons | -3.640878 .252668 -14.41 0.000 -4.136099 -3.145658 ------+------gamma | -.0134251 .0044672 -3.01 0.003 -.0221806 -.0046696 ------. . *33 We could use predict to generate the hazard rates for each conflict type* . . predict haz_civil, haz, if civil==1 (44 missing values generated). predict haz_icw, haz, if icw==1 (28 missing values generated). predict haz_interst, haz, if interst==1 (48 missing values generated). . *34 And then graph them* . . gr haz_civil haz_icw haz_interst duration, s(iii) c(sss) ylab xlab t1("Hazard Rates")b2("Duration of U.N. P > eacekeeping Missions") . . drop haz_*. . *Or we can generate them ourselves using the formula for the Gompertz h(t)* . . *35 First we create lambda, noting that lambda=xb* . gen lambda_civil=_b[_cons]+_b[civil]. gen lambda_interstate=_b[_cons]+_b[interst]. gen lambda_icw=_b[_cons]. . *36 Second, we create h(t):* . gen haz_civil=exp(e(gamma)*duration)*exp(lambda_civil) (4 missing values generated). gen haz_interst=exp(e(gamma)*duration)*exp(lambda_interst) (4 missing values generated). gen haz_icw=exp(e(gamma)*duration)*exp(lambda_icw) (4 missing values generated). . gr haz_civil haz_icw haz_interst duration, s(iii) c(sss) ylab xlab t1("Hazard Rates")b2("Duration of U.N. P > eacekeeping Missions") saving(UNgomphaz, replace) . . drop haz_* . . *37 We can compute the survivor function for the Gompertz model* . . gen surv_civil=exp((-(exp(lambda_civil))/e(gamma))*(exp(e(gamma)*duration)-1)) (4 missing values generated). gen surv_interst=exp((- (exp(lambda_interst))/e(gamma))*(exp(e(gamma)*duration)-1)) (4 missing values generated). gen surv_icw=exp((-(exp(lambda_icw))/e(gamma))*(exp(e(gamma)*duration)-1)) (4 missing values generated). . *38 Now we graph them:* . gr surv_civil surv_interst surv_icw duration, c(sss) s(iii) ylab xlab t1("Survivor Function") b2("Duration > of U.N. Peacekeeping Missions") saving(UNgompsurv, replace) . . drop surv_* lambda_*. . *++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ ++++++++++++* . . *Cox Model of U.N. Peacekeeping Missions* . . *39 Estimating Cox PH model using nohr option* . . stcox civil interst, nohr failure _d: failed analysis time _t: durationIteration 0: log likelihood = -131.62249 Iteration 1: log likelihood = -127.43384 Iteration 2: log likelihood = -127.15768 Iteration 3: log likelihood = -127.15763 Refining estimates: Iteration 0: log likelihood = -127.15763Cox regression -- Breslow method for tiesNo. of subjects = 54 Number of obs = 54 No. of failures = 39 Time at risk = 3994 LR chi2(2) = 8.93 Log likelihood = -127.15763 Prob > chi2 = 0.0115------_t | _d | Coef. Std. Err. z P>|z| [95% Conf. Interval] ------+------civil | .7348046 .3781278 1.94 0.052 -.0063122 1.475921 interst | -.8556111 .5042314 -1.70 0.090 -1.843886 .1326643 ------. . *40 This gives coefficients estimates and not hazard ratios. To see the hazard ratios, we can redisplay the > results by typing:* . . stcox, hrCox regression -- Breslow method for tiesNo. of subjects = 54 Number of obs = 54 No. of failures = 39 Time at risk = 3994 LR chi2(2) = 8.93 Log likelihood = -127.15763 Prob > chi2 = 0.0115------_t | _d | Haz. Ratio Std. Err. z P>|z| [95% Conf. Interval] ------+------civil | 2.085075 .7884246 1.94 0.052 .9937077 4.375065 interst | .4250234 .2143101 -1.70 0.090 .1582014 1.141867 ------. . *The results from the first model are the log of the estimates from this model* . . * Retreiving Baseline Functions* . . *41 Estimate Cox Model* . . stcox civil interst, nohr basech(inthaz) basesurv(surv) basehc(haz) failure _d: failed analysis time _t: durationIteration 0: log likelihood = -131.62249 Iteration 1: log likelihood = -127.43384 Iteration 2: log likelihood = -127.15768 Iteration 3: log likelihood = -127.15763 Refining estimates: Iteration 0: log likelihood = -127.15763Cox regression -- Breslow method for tiesNo. of subjects = 54 Number of obs = 54 No. of failures = 39 Time at risk = 3994 LR chi2(2) = 8.93 Log likelihood = -127.15763 Prob > chi2 = 0.0115------_t | _d | Coef. Std. Err. z P>|z| [95% Conf. Interval] ------+------civil | .7348046 .3781278 1.94 0.052 -.0063122 1.475921 interst | -.8556111 .5042314 -1.70 0.090 -1.843886 .1326643 ------. . *42 Now obtain hazard ratios:* . . predict hr, hr. . *which is exp(xb)* . . *43 The last three commands produce the estimates of the baseline functions,* . *which we can graph:* . . gr surv durat, ylab xlab c(J) t1("Baseline Survivor Function") b2("Duration of U.N. Peacekeeping Missions") > s(ii) saving(UNcoxst, replace). . *44 To compute the survivor function for each observation we generate new variable:* . . gen surv_i=surv^hr (4 missing values generated). . *45 From this, we can compute the conflict-specific survivor function:* . . gen surv_civil=surv_i if civil==1 (44 missing values generated). gen surv_icw=surv_i if icw==1 (28 missing values generated). gen surv_ic=surv_i if interst==1 (48 missing values generated). . *46 Now we can graph them* . . gr surv_civil surv_icw surv_ic durat, c(JJJ) s(iii) ylab xlab t1("Survivor Functions") b2("Duration of U.N. > Peacekeeping Missions") saving(UNcoxst2, replace). . drop surv_civil surv_icw surv_ic surv_i hr haz inthaz surv. log close

Event History Modeling: a Guide for Social Scientists

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