Hi, I'm trying to use some form of survival analysis (e.g. accelerated failure time or proportional hazards) to study seed germination time. My data is interval censored (also called "grouped-time") and has a random effect (also called "frailty"). How can I do this in SAS? My seeds were on petri dishes that were censused on day 1, 3, 5, 8, and 14. These are interval censored, with a seed that germinated on d8 really germinated between day 6 and day 8. PROC LIFEREG can handle interval censoring, but as far as i know, it doesn't handle random effects. Seeds on the same petri dish are not independent, so I need to be able to include the dish as a random effect. Is there an approach that allows interval censoring as well as random effects? If so , how should I modify the SAS code below to include random effects? Will I need a more general procedure like PROC NLMIXED? What is the best/easiest way to do this? PROC LIFEREG DATA=survGeno2 plots=probplot; TITLE "Interval censored AFT"; CLASS genotype; MODEL (start,fin)=genotype; /*how do i include a petri dish random effect?*/ RUN; Thanks! Tarek ... View more

I am trying to estimate covariance components from PROC MIXED and having trouble with convergence problems in the REML approach when I use the NOBOUNDS option. Let me explain why I am using NOBOUNDS in case this is the problem, I am trying to estimate genetic covariances among 12 traits in a group of self-fertilizing plants.  There are ~400 plants and 160 genotypes.  The easy approach to estimating covariances is to calculate the covariance among genotype means.  This approach introduces some bias into covariance estimates, so I want to compare estimates of variance and covariance components.  I’ll refer to these approaches as “Among-genotype covariance” and “covariance components” approaches. To calculate between trait covariance components, I organized my data (for 160 genotypes) like this: Genotype         Trait                 Value 1                      Biomass           1.28 1                      Biomass           1.07 1                      Biomass           1.97 1                      TLength           1070.7 1                      TLength           875.14 1                      TLength           1547.93 To get the (genetic) covariance component between Biomass and TLength, I used this code: PROC MIXED DATA=multi56 covtest cl;       WHERE trait IN ("Biomass","TLength")       CLASS genotype trait;       MODEL value=trait;       RANDOM trait/SUBJECT=genotype TYPE=un;       /* The RANDOM statement tells SAS to estimate the 2 x 2 among genotype variance component matrix for the two traits listed in the where statement */       RUN; The resulting covariance parameter output was: Covariance Parameter Estimates Cov Parm         Subject Estimate       Standard Error Z Value   Pr Z   Alpha   Lower  Upper UN(1,1)           Genotype         0          .           .           .           .           .           . UN(2,1)           Genotype         50450       19630       2.57     0.0102 0.05     11975  88924     UN(2,2)           Genotype         316010      90306       3.50     0.0002 0.05     193511   606860 Residual                                55618       3437.23     16.18   <.0001 0.05     49451  63021 The covariance estimate of 50450 is waaay bigger than the estimate I got from the covariance among the genotype means for Biomass and TLength (the Among genotype approach) which equals 257.1.  SO I figured the 0 variance component above could be the problem.  I ran the same PROC MIXED code above only adding the NOBOUND option: PROC MIXED DATA=multi56 covtest cl NOBOUND ; Now my covariance component estimates nearly match as I expect them to: Iteration History Iteration           Evaluations      -2 Res Log Like  Criterion 0                    1                    10109.62259781                   WARNING: Stopped because of infinite likelihood. Covariance Parameter Values At Last Iteration Cov Parm         Subject   Estimate UN(1,1)           genotype  -34470 UN(2,1)           genotype   249.06 UN(2,2)           genotype  188855 Residual                         83814 NOTE: 125 observations are not included because of missing values. WARNING: Stopped because of infinite likelihood. NOTE: PROCEDURE MIXED used (Total process time): real time           0.11 seconds       cpu time            0.03 seconds The covariance component estimate of 249 I got by allowing negative variances is much more reasonable than 50450, considering the estimate from the among genotype covariance was 257.  The problem is that the REML approach stopped and gave me warnings.  Does this mean I cannot trust this covariance component output? I thoguth about running PROC MIXED with METHOD=TYPE3, but I can’t do that because I have SUBJECT=genotype TYPE=un as part of my random statement.  What is the best way to get the correct between trait covariance components? ... View more