phantom variables

ya

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Jun 27, 2015, 8:45:45 AM6/27/15

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Dear list,

I am trying to add phantom variables to the weak invariance model with categorical indicators. With or without phantom variables, the models worked fine, however, the model with phantom variables had 6 more df. Could you give me some suggestions on what I missed please? Thank you very much. Please see my code below:

Without phatom variables:

crossmod="
#measurement model
bul1=~bul1_1+bul2_1+bul3_1+bul4_1+bul5_1+bul6_1
bul2=~bul1_2+bul2_2+bul3_2+bul4_2+bul5_2+bul6_2
bul3=~bul1_3+bul2_3+bul3_3+bul4_3+bul5_3+bul6_3
"

fitweakinvar=sem(crossmod,data=grades1to9,missing="pairwise",group="Boy",group.equal=c("loadings"),
ordered=c("bul1_1","bul2_1","bul3_1","bul4_1","bul5_1","bul6_1",
          "bul1_2","bul2_2","bul3_2","bul4_2","bul5_2","bul6_2",
          "bul1_3","bul2_3","bul3_3","bul4_3","bul5_3","bul6_3")
                )
summary(fitweakinvar,fit.measures=T,standardized=T)

lavaan (0.5-18) converged normally after 32 iterations

Used Total

Number of observations per group

0 15666 17627

1 15395 18007

Estimator DWLS Robust

Minimum Function Test Statistic 2576.839 2668.714

Degrees of freedom 279 279

P-value (Chi-square) 0.000 0.000

Scaling correction factor 1.004

Shift parameter for each group:

0 51.777

1 50.881

for simple second-order correction (Mplus variant)

With phantom variables:

> crossmodphan="#when latent var is not equal across groups
+ #measurement model
+ bul1=~bul1_1+bul2_1+bul3_1+bul4_1+bul5_1+bul6_1
+ bul2=~bul1_2+bul2_2+bul3_2+bul4_2+bul5_2+bul6_2
+ bul3=~bul1_3+bul2_3+bul3_3+bul4_3+bul5_3+bul6_3
+ #structure model
+ bul1ph=~bul1;bul2ph=~bul2;bul3ph=~bul3;
+ bul1~~0*bul1;bul2~~0*bul2;bul3~~0*bul3;
+ bul1ph~~1*bul1ph;bul2ph~~1*bul2ph;bul3ph~~1*bul3ph;
+ bul1~~0*bul2;bul1~~0*bul3;bul2~~0*bul3;
+ bul1~~0*bul1ph;bul1~~0*bul2ph;bul1~~0*bul3ph;bul2~~0*bul1ph;bul2~~0*bul2ph;bul2~~0*bul3ph;bul3~~0*bul1ph;bul3~~0*bul2ph;bul3~~0*bul3ph;
+ bul1ph~~bul2ph;bul1ph~~bul3ph;bul2ph~~bul3ph;
+ bul1~0*1;bul2~0*1;bul3~0*1;bul1ph~0*1;bul2ph~0*1;bul3ph~0*1;
+ "
> fitlatentcovinvarphan=sem(crossmodphan,data=grades1to9,missing="pairwise",group="Boy",group.equal=c("loadings"),
+ ordered=c("bul1_1","bul2_1","bul3_1","bul4_1","bul5_1","bul6_1",
+           "bul1_2","bul2_2","bul3_2","bul4_2","bul5_2","bul6_2",
+           "bul1_3","bul2_3","bul3_3","bul4_3","bul5_3","bul6_3")
+                 )

lavaan (0.5-18) converged normally after  14 iterations

                                                  Used       Total
  Number of observations per group
  0                                              15666       17627
  1                                              15395       18007

  Estimator                                       DWLS      Robust
  Minimum Function Test Statistic             7701.290    5856.872
  Degrees of freedom                               285         285
  P-value (Chi-square)                           0.000       0.000
  Scaling correction factor                                  1.345
  Shift parameter for each group:
    0                                                       66.157
    1                                                       65.013
    for simple second-order correction (Mplus variant)

ya

Edward Rigdon

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Jun 27, 2015, 9:57:53 AM6/27/15

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In the phantom variable (pv) code, the variances of your 3 factors are fixed, but in the no-pv code they are free. In the pv code, intercepts of the 3 factors are set to 0, while in the no-pv code, they are not constrained. You are using the Sem function, so you could say that auto-constraints are really responsible for the 3+3 = 6 additional free parameters.

--Ed Rigdon

Sent from my iPad

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ya

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Jun 27, 2015, 10:37:53 AM6/27/15

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Dear Ed,

Thank you very much for the response. I have checked the output of both models, the intercept of the three factors were fixed at 0 in both models. The variances for the factors were fixed at 0 and for the phantom variables at 1 as intended so that the latent covariance can be standardized(if I understand right). So maybe there are some other reasons about these 6 df?

I noticed that the varianceS of the first indicator of each factor were fixed at 0 in all groups in the phantom variable model, whereas they were freely estimated in the non-phantom variable model. Which seems like 3*2=6?

Best regards,

ya

Terrence Jorgensen

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Jun 29, 2015, 8:49:44 AM6/29/15

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I noticed that the varianceS of the first indicator of each factor were fixed at 0 in all groups in the phantom variable model, whereas they were freely estimated in the non-phantom variable model. Which seems like 3*2=6?

In the PhantomVar code, the residual variances of the first indicators are fixed to zero? I don't think that should happen, so that sounds like the discrepancy in df. If that's what you meant, try manually freeing them in the syntax, because they should not be constrained.

Terry

ya

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Jun 29, 2015, 9:19:09 AM6/29/15

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Dear Terry,

Thank you very much for the response.

I manually freed the residual variances of the first indicators in the phantom variable model, but they are still 0 in both groups. Besides, the residual variances were somehow constrained to be equal across groups although I only asked for weak invariance. It seems something is wrong in the model. Please see the code and the residual variance part of the output below:

crossmodphan="#when latent var is not equal across groups

#measurement model
bul1=~bul1_1+bul2_1+bul3_1+bul4_1+bul5_1+bul6_1
bul2=~bul1_2+bul2_2+bul3_2+bul4_2+bul5_2+bul6_2
bul3=~bul1_3+bul2_3+bul3_3+bul4_3+bul5_3+bul6_3

#structure model
bul1ph=~bul1;bul2ph=~bul2;bul3ph=~bul3;
bul1~~0*bul1;bul2~~0*bul2;bul3~~0*bul3;
bul1ph~~1*bul1ph;bul2ph~~1*bul2ph;bul3ph~~1*bul3ph;
bul1~~0*bul2;bul1~~0*bul3;bul2~~0*bul3;

bul1~~0*bul1ph;bul1~~0*bul2ph;bul1~~0*bul3ph;bul2~~0*bul1ph;bul2~~0*bul2ph;bul2~~0*bul3ph;bul3~~0*bul1ph;bul3~~0*bul2ph;bul3~~0*bul3ph;

bul1ph~~bul2ph;bul1ph~~bul3ph;bul2ph~~bul3ph;

bul1~0*1;bul2~0*1;bul3~0*1;bul1ph~0*1;bul2ph~0*1;bul3ph~0*1;

bul1_1~~bul1_1;bul1_2~~bul1_2;bul1_3~~bul1_3 # freely estimate the residual variance

"
fitlatentcovinvarphan=sem(crossmodphan,data=grades1to9,missing="pairwise",group="Boy",group.equal=c("loadings"),

ordered=c("bul1_1","bul2_1","bul3_1","bul4_1","bul5_1","bul6_1",
"bul1_2","bul2_2","bul3_2","bul4_2","bul5_2","bul6_2",
"bul1_3","bul2_3","bul3_3","bul4_3","bul5_3","bul6_3")
)

summary(fitlatentcovinvarphan,fit.measures=T,standardized=T)

Group 1 [0]:

Variances:

bul1 0.000 0.000 0.000

bul2 0.000 0.000 0.000

bul3 0.000 0.000 0.000

bul1ph 1.000 1.000 1.000

bul2ph 1.000 1.000 1.000

bul3ph 1.000 1.000 1.000

bul1_1 0.000 0.000 0.000

bul1_2 0.000 0.000 0.000

bul1_3 0.000 0.000 0.000

bul2_1 0.598 0.598 0.598

bul3_1 0.438 0.438 0.438

bul4_1 0.409 0.409 0.409

bul5_1 0.292 0.292 0.292

bul6_1 0.305 0.305 0.305

bul2_2 0.476 0.476 0.476

bul3_2 0.372 0.372 0.372

bul4_2 0.308 0.308 0.308

bul5_2 0.277 0.277 0.277

bul6_2 0.228 0.228 0.228

bul2_3 0.398 0.398 0.398

bul3_3 0.305 0.305 0.305

bul4_3 0.222 0.222 0.222

bul5_3 0.128 0.128 0.128

bul6_3 0.172 0.172 0.172

Group 2 [1]:

Variances:

bul1 0.000 0.000 0.000

bul2 0.000 0.000 0.000

bul3 0.000 0.000 0.000

bul1ph 1.000 1.000 1.000

bul2ph 1.000 1.000 1.000

bul3ph 1.000 1.000 1.000

bul1_1 0.000 0.000 0.000

bul1_2 0.000 0.000 0.000

bul1_3 0.000 0.000 0.000

bul2_1 0.598 0.598 0.598

bul3_1 0.438 0.438 0.438

bul4_1 0.409 0.409 0.409

bul5_1 0.292 0.292 0.292

bul6_1 0.305 0.305 0.305

bul2_2 0.476 0.476 0.476

bul3_2 0.372 0.372 0.372

bul4_2 0.308 0.308 0.308

bul5_2 0.277 0.277 0.277

bul6_2 0.228 0.228 0.228

bul2_3 0.398 0.398 0.398

bul3_3 0.305 0.305 0.305

bul4_3 0.222 0.222 0.222

bul5_3 0.128 0.128 0.128

bul6_3 0.172 0.172 0.172

ya

Yves Rosseel

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Jul 16, 2015, 5:44:40 AM7/16/15

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Two general comments:

1) study the parameter table of both models. There you should be able to
spot any discrepancies.

2) when using phantom variables, I would prefer to use the lavaan() function

Yves.

ya

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Jul 17, 2015, 1:47:27 PM7/17/15

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Dear Yves,

Thank you very much for the suggestions. Now I have three follow up questions:

1. I have located the problem using parTable(). The loading of the three latent variables on the phantom variables were fixed at 1 by default. I think this is for the identification purpose as the latent variable was the only "indicator" for the phantom variable, right?

2. I tried to free the loadings by:

bul1ph=~NA*bul1;bul2ph=~NA*bul2;bul3ph=~NA*bul3;

although the loadings were freely estimated, they were still constrained to be equal across groups. Is it because I asked for the weak invariance (group.equal=c(loadings))?

3. How could the loadings of the latent variables on the phantom variables be freely estimated and not constrained to be equal across groups? I have 3 phantom variables, I think this is why there were 6 more DF in the phantom model than in the weak invariance model. 3 loadings * 2 groups = 6.

Please find my code below and give me some suggestions (with this code, there were only 3 more DF):

crossmodphan="#when latent var is not equal across groups
#measurement model
bul1=~bul1_1+bul2_1+bul3_1+bul4_1+bul5_1+bul6_1
bul2=~bul1_2+bul2_2+bul3_2+bul4_2+bul5_2+bul6_2
bul3=~bul1_3+bul2_3+bul3_3+bul4_3+bul5_3+bul6_3
#structure model

bul1ph=~NA*bul1;bul2ph=~NA*bul2;bul3ph=~NA*bul3;

bul1~~0*bul1;bul2~~0*bul2;bul3~~0*bul3;
bul1ph~~1*bul1ph;bul2ph~~1*bul2ph;bul3ph~~1*bul3ph;
bul1~~0*bul2;bul1~~0*bul3;bul2~~0*bul3;
bul1~~0*bul1ph;bul1~~0*bul2ph;bul1~~0*bul3ph;bul2~~0*bul1ph;bul2~~0*bul2ph;bul2~~0*bul3ph;bul3~~0*bul1ph;bul3~~0*bul2ph;bul3~~0*bul3ph;
bul1ph~~bul2ph;bul1ph~~bul3ph;bul2ph~~bul3ph;

bul1~0*1;bul2~0*1;bul3~0*1;bul1ph~0*1;bul2ph~0*1;bul3ph~0*1;#no need for means in weak invariance model
"

fitlatentcovinvarphan=sem(crossmodphan,data=grades1to9,missing="pairwise",group="Boy",group.equal=c("loadings"),
ordered=c("bul1_1","bul2_1","bul3_1","bul4_1","bul5_1","bul6_1",
          "bul1_2","bul2_2","bul3_2","bul4_2","bul5_2","bul6_2",
          "bul1_3","bul2_3","bul3_3","bul4_3","bul5_3","bul6_3")
                )

Best regards,

ya

Terrence Jorgensen

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Jul 18, 2015, 6:51:04 AM7/18/15

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1. I have located the problem using parTable(). The loading of the three latent variables on the phantom variables were fixed at 1 by default. I think this is for the identification purpose as the latent variable was the only "indicator" for the phantom variable, right?

They won't be fixed by default if you use the lavaan() function instead of the sem() function, as Yves suggested you should do.

2. I tried to free the loadings by:

bul1ph=~NA*bul1;bul2ph=~NA*bul2;bul3ph=~NA*bul3;

although the loadings were freely estimated, they were still constrained to be equal across groups. Is it because I asked for the weak invariance (group.equal=c(loadings))?

There are two groups, so specify NA for both groups:

bul1ph =~ c(NA, NA)*bul1

Again, use the lavaan() function, so you can specify all fixed/free parameters (and equality constraints) using the model syntax instead of the "group.equal" argument. That way, you'll know exactly what model you are fitting.

Terry

ya

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Jul 18, 2015, 7:12:47 AM7/18/15

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Dear Terry,

Thank you very much for the response.

I tried this

bul1ph=~c(NA,NA)*bul1;bul2ph=~c(NA,NA)*bul2;bul3ph=~c(NA,NA)*bul3;

Got an error:

Error in getModifier(rhs[[2L]]) :

lavaan ERROR: can not parse modifier:cNANA

ya

From: Terrence Jorgensen
Date: 2015-07-18 13:51
To: lavaan
Subject: Re: Re: phantom variables

1. I have located the problem using parTable(). The loading of the three latent variables on the phantom variables were fixed at 1 by default. I think this is for the identification purpose as the latent variable was the only "indicator" for the phantom variable, right?

Yves Rosseel

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Jul 20, 2015, 8:04:33 AM7/20/15

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On 07/18/2015 01:12 PM, ya wrote:
> Dear Terry,
>
> Thank you very much for the response.
>
> I tried this
>
> bul1ph=~c(NA,NA)*bul1;bul2ph=~c(NA,NA)*bul2;bul3ph=~c(NA,NA)*bul3;
>
> Got an error:
>
> Error in getModifier(rhs[[2L]]) :
> lavaan ERROR: can not parse modifier:cNANA

Yes. That is a bug in the parser of 0.5-18 (actually due to change in
R). This is already fixed in dev 0.5-19.

But if you are using the lavaan() function, you can just write

bul1ph =~ bul1; bul2ph =~ bul2; bul3ph =~ bul3

and the means/intercepts will be free in all groups.

Yves.

ya

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Jul 20, 2015, 9:55:57 AM7/20/15

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Dear Yves and Teery,

Thank you very much for all the suggestions.

The reason I did not use lavaan() is because I am not quite sure about myself on all the parameters I should specify. I still need more practice to get deeper understanding about the whole SEM modeling. For now, I am worred I may get the DF more wrong by using lavaan(). However, I actually found a way to solve the problem by using NA* multiplication in combination with the group.partial arguement. So now I have got the correct DF for both of my models. Thank you very much for all the help.

Best regards,

ya

From: Yves Rosseel
Date: 2015-07-20 15:04
To: lavaan
Subject: Re: phantom variables

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