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Do y'all have any examples of simultaneous modeling of survival for two outcomes? The possibility is suggested in Muthen & Masyn, but I'm having trouble wrapping my mind around it. Thanks, Pat 


Pat, I don't have a specific examplethat sort of model is still in development. However, there are a couple of ways to think about this. One is as a traditional competing risk model, where you have more than one possible event during each time period but only one of the events may be experienced by a given individual during any one time period. This fits into the current modeling framework as an LCA with unordered multinomial indicators. In this way, you can obtain estimates for the hazard probabilities for each competing risk. Another way to think of modeling two separate outcomes that may be cooccurring is with essentially two parallel LCA models, i.e., two sets of of indicators and two latent class variables. However, it's not completely clear to me yet how to model what may be a recurrsive relationship between the two event processes. Best, Katherine Masyn 


Thanks. I'll chew on that. 


Hi, I want to do a latent leveldifference analysis with residual variance in the level 1 variables set equal in order to compute ICC (Newsom, 2002). With a continuous variable this is completely OK. However, I wish to analyze Timediscrete survival for addicted women with matched control. With one control, this would be two survival processes, and a Level and Difference factor to capture similarity and difference. The f factor (survival factor), as described in the Mplus manual, is regressed on Age. Due to matching I would expect the Difference variable to be very small and nonsignificant, which could indicate a Level factor only. Is it possible to analyze level and difference based on several ffactors? I tried: MODEL: f by D1_0D7_0@1 ; f on Age_0 ; f@0 ; g by D1_1D7_1@1 ; g on Age_1 ; g@0 ; L by f@1 g@1 ; !D by f@1 g@0 ; !D with L@0 ; f (1) ; g (1) ; The Level and Difference model did not work and the Levelonly model gave 0 variation (Variances: Level=0.000, P=0.838). If such a model is possible, should the threshold be estimated freely or set equal in all survival models? Best, Rolf 


I am not familiar with that approach. What is the "Levelonly model" and what is it that gets "0 variation"? Models with f variance free may be hard to estimate. 


Thanks for fast response, Sorry that I did not explain this as good as I should. Newsom used the latent growth curve model in order to analyze dyadic data (e.g. man and wife, parent  child) with the intercept factor capturing the common level in the dyad (prespecified loadings 1 and 1) and the slope factor capturing the difference between parent and child (prespecified 0 and 1). Then you would get the cluster mean level and variation (intercept factor) and the mean difference within dyads and variation (slope factor). I used this model for the continuous variable AGE and all worked very well. However, I wondered if this framework also could be used to analyze survival models for alcoholics and their matched controls in this way in order to account for the clustering effects. I have allready used the two level approach to test within and between cluster level of age and the relation between age and survival (both with cox regression and timediscrete survival). But I do not believe that this approach is accounting for the matching between alcoholics and controls, only for the dependency for the controls. Regards, Rolf 


I am sorry. I forgot to answer that zero variation was on the intercept factor (common level for the dyad). And this was found in a model without any difference factor (Slope). Rolf 

WenHsu Lin posted on Monday, January 05, 2015  10:52 pm



Hi, I want to run a recurrent event discretetime survival model. Is that doable in Mplus? Do I simply use the regular discretetime survival syntax? Thank you 


You can run it as twolevel model, type=complex (with ID as the cluster variable) or a multivariate model where each recurrence is modeled as a separate sequence of binary variables for discretetime survival. If the number of recurrences is subject specific you will need to add missing data for the subjects with less than the max number of recurrences in the multivariate model. 

WenHsu Lin posted on Tuesday, January 06, 2015  10:56 pm



Hi, Asparouhov What you mean is that if I have 5 time points max and 5 events max, I will need to set up five separate discretetime survival. Is this correct? And people who drop out or who do not experience the max number of recurrences are coded as missing. This part is a little confusing. Do I set up missing or 0 (event not happened)? Thank you. 


It should be setup as missing. That way it will not be a part of the likelihood (it shouldn't). 

WenHsu Lin posted on Thursday, January 22, 2015  5:56 am



Hi, I still do not get it. DO you have any example? Thank you 


User's guide example 6.19: Discretetime survival analysis. You also might want to do some modeling for these variables treating them as continuous. That might help you understand how to model data with different number of observations for each individual. Also look at the Mplus commands WIDETOLONG and LONGTOWIDE and look at example 9.16. Although that example has the same number of observations in each cluster it doesn't have to. This is just for illustration purposes to understand how to model data with different number of observations for each individual. We don't have exactly the example you are asking for. 


Hi, the syntax for my recurrent event was as follow, is this correct (6 time point and 4 events)? variable: names are age income sex track sd1sd6; usevariables are sex track income age; sd11sd16 !event1 sd21 sd26 !event2( first recurrent) sd31sd36 !event3(second recurrent) sd41sd46; !event4 (third recurrent) categorical are sd11sd16 sd21 sd26 sd31sd36 sd41sd46; missing is blank; analysis: iteration = 50000; estimator=mlr; model: f1 by sd11sd16@1; f1@0; f2 by sd21sd26@1; f2@0; f3 by sd31sd36@1; f3@0; f4 by sd41sd46@1; f4@0; f1 f2 f3 f4 on income sex age track; 


You want to make sure the 4 processes are correlated, so add f by f1f4@1; 

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