Binary outcome Multiple membership model

Welcome to the forum for R2MLwiN users. Feel free to post your question about R2MLwiN here. The Centre for Multilevel Modelling take no responsibility for the accuracy of these posts, we are unable to monitor them closely. Do go ahead and post your question and thank you in advance if you find the time to post any answers!

Go to R2MLwiN: Running MLwiN from within R >> http://www.bris.ac.uk/cmm/software/r2mlwin/
dtaylor7
Posts: 8
Joined: Tue Dec 12, 2023 5:08 pm

Re: Binary outcome Multiple membership model

Post by dtaylor7 »

One last thing about the specifics of this model - my model is a binary outcome multiple membership model, with level 1 units (tests) being examined (equally weighted) by two level two units (readers), which are strictly nested within level three units (centres). Effectively like the Three-level multiple membership data structure in LEMMA C13.1.2, but without student 8 changing local authority. Does Level 2 being strictly nested within Level 3 despite being a multiple membership model require a change to the model specification, and if not does it require any different interpretation than a regular 3-level hierarchical multilevel model?
billb
Posts: 161
Joined: Fri May 21, 2010 1:21 pm

Re: Binary outcome Multiple membership model

Post by billb »

Hi,
I think the fact level 2 is nested within level 3 is fine here as for MCMC nested is simply a special case of the more general cross classified model - just need to make sure that your IDs are unique i.e. if you have a reader 1 that this is the same person throughout and if say you have a reader in the second centre they are not also numbered 1.
The only other thing that comes to mind that is relevant for all multiple membership models is that it is not so straightforward to work out things like VPCs as level 2 here will have weights involved so the variance of the random effects at this level will not translate directly into the VPC formula
For example if you had an observation with equally weighted readers 1 and 2 (weights 0.5 each) and centre 6 say then the variance for the observation
V = var(centre) + 0.5^2 var(reader1) + 0.5^2 var (reader2) + var (residual)
V = var (centre) + 0.5 var (reader) + var (residual)
Hope my rough notation makes sense here so the importance of var (reader) will be less than just comparing the variances at each level and obviously will not be constant if the weights are not equal or some observations have more than 2 readers etc.
I have probably thrown up more questions here than answers but hope it is helpful.
Bill.
Post Reply