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 Patrick Malone posted on Monday, April 28, 2003 - 8:22 am
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
 Katherine Masyn posted on Thursday, May 01, 2003 - 9:23 am
Pat,

I don't have a specific example--that 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 co-occurring 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
 Patrick Malone posted on Monday, May 05, 2003 - 6:44 am
Thanks. I'll chew on that.
 Rolf Gjestad posted on Tuesday, March 08, 2011 - 2:09 am
Hi,
I want to do a latent level-difference 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 Time-discrete 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 non-significant, which could indicate a Level factor only. Is it possible to analyze level and difference based on several f-factors?

I tried: MODEL:
f by D1_0-D7_0@1 ; f on Age_0 ; f@0 ;
g by D1_1-D7_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 Level-only 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
 Bengt O. Muthen posted on Tuesday, March 08, 2011 - 8:44 am
I am not familiar with that approach. What is the "Level-only model" and what is it that gets "0 variation"?

Models with f variance free may be hard to estimate.
 Rolf Gjestad posted on Tuesday, March 08, 2011 - 11:52 am
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 (pre-specified loadings 1 and 1) and the slope factor capturing the difference between parent and child (pre-specified 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 time-discrete 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
 Rolf Gjestad posted on Tuesday, March 08, 2011 - 11:56 am
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
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