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 emmanuel bofah posted on Tuesday, May 27, 2014 - 2:08 pm
Will like to ask if is now possible to include covariates in the alignment optimization method in version 7.2.
 Bengt O. Muthen posted on Tuesday, May 27, 2014 - 6:10 pm
No, covariates are not yet allowed.
 Lim Jie Xin posted on Monday, February 20, 2017 - 5:47 am
Hi, I would like to know if there is any recommendation/literature on the performance of the alignment method in the presence of missing data (assuming MCAR/MAR). Can we use multiple imputation alongside this method?
 Tihomir Asparouhov posted on Tuesday, February 21, 2017 - 11:10 am
Alignment is likelihood based and assumes MAR. Imputation is also available. Missing data doesn't have any implications for alignment as far as I know.
 WEN Congcong posted on Thursday, March 02, 2017 - 5:15 am
Professors,

Hello! In a multiple group analysis, the model can be expressed as Y=vg+g+ (1)and =ag+(2). In your alignment paper published in 2014, the configural model M0 transforms the factor in each group to mean 0 and variance 1 and got g0=(g-ag)/̦g(3).

With equation 1 and 2, g-ag should equal g. I dont know how did you get the equation 3 and what does this equation 3 represents. Thanks in advance!
 Bengt O. Muthen posted on Thursday, March 02, 2017 - 5:46 pm
Please send a pdf with these formulas to Support so we can read them. Also give your license number.
 WEN Congcong posted on Wednesday, April 25, 2018 - 12:45 am
Professors,

Hello, in the 2014 paper of Asparouhov and Muthen, its said After minimizing the total loss function, 2G 1 of the group-specific factor mean and variance parameters can be estimated. Identification is achieved by estimating all groups factor means and variances except the first group.

I want to ask why the factor mean and variance of the first group cannot be identified? Can they be estimated by using the variance1*variance2**varianceg=1 simply only one constraint?
 Tihomir Asparouhov posted on Wednesday, April 25, 2018 - 9:11 am
You can use the following analysis command option
metric=product;
to estimate the model under the constraint
variance1*variance2**varianceg=1

The models are equivalent though, in one case you are fixing variance1 to 1, in the other to 1/(variance2**varianceg)
 WEN Congcong posted on Thursday, April 26, 2018 - 1:06 am
Sorry, I didnt ask clearly my questions. My questions are

1.When minimizing the total loss function, is the first group regarded as a reference group so we cannot estimate its variance and mean? I dont really know why the 2 parameters cannot be estimated, but it previously said that alignment can estimate 2g-1 parameters.

2. what is the purpose of this constraint variance1*variance2**varianceg=1, for estimating the factor mean of the first group?

I am not very familiar with the computation process, so please forgive me for asking elementary questions.
 Tihomir Asparouhov posted on Thursday, April 26, 2018 - 9:23 am
1. The fixed alignment estimates 2g-2 parameters and the reference group has 0 mean and variance=1 for the same reason you have to do that if you just had one group. The free alignment can in addition estimate the mean in the first group so you get 2g-1 parameters that can be estimated.

2. The purpose of that constraint is to replace the constraint of variance1=1, it doesn't have anything to do with the mean in the first group (which is controlled by the option alignment=free/fixed;). You get that constraint with the option metric=product option.

All the computational details are in
http://statmodel.com/examples/webnote18.pdf

The alignment options are described in the User's guide page 672-673
 WEN Congcong posted on Thursday, April 26, 2018 - 8:29 pm
Thanks for your detailed answer! Have a nice day!
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