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| Anonymous posted on Monday, September 27, 1999 - 4:44 pm
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| Why don't the degrees of freedom for MLMV behave as expected? |
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| The degrees of freedom are not calculated in the regular way for MLMV. They are instead estimated according the formula 109 on page 281 of the Mplus User's Guide. |
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| Anonymous posted on Monday, January 10, 2000 - 4:39 pm
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For scholars doing applied research the formula 109 is hard to understand.
Could you explain in plain English what effects the degrees of freedom? For instance, I noticed that using the same model specification but different data (with the same number of cases) can result in different degrees of freedom. What is the reason for that?
Thanks a lot in advance |
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| I haven't found a good intuitive way to explain the degrees of freedom with the mean- and variance-adjusted approach. I don't think you will find one in Satorra's work either. The d.f. is data dependent because it draws on the estimates, their derivatives, and the asymptotic covariance matrix of the sample statistics with the aim of choosing the degrees of freedom that gives a trustworthy chi-square-based p value. But that's not very intuitive. We'll post a good intuitive explanation if and when we find one. Others? |
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I have a situation where I have a group of latent factors (Group A) that are descriptive of the work environment. There is a second group of factors (Group B) that are descriptive of safety levels in the workplace. I have a group of 3 criterion (Group C) variables. The pattern of relationships in the model is A->B->C.
In my model, I have some non-significant paths between Group B and Group C. When I drop one of the paths between Group B and Group C, I actually lose degrees of freedom - it appears as though Mplus is automatically correlating residuals among my Group C variables, and adding some correlations among other variables. Any thoughts on what is going on? |
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I would have to see the output to say exactly. This could be due to the estimator. WLSMV and MLMV do not calculate degrees of freedom in the regular way. Or as you suggest, it could be due to certain default residual covariances. In the output, it shows how many parameters are in the model. You can check that in both cases and then look in the results section to see which parameters differ between the two runs.
Mplus defaults are described on pages 158-160 of the Mplus User's Guide. If you cannot figure it out, please send the two outputs to support@statmodel.com.
If there are parameters in the model by default, you can fix them to zero by, for example,
y1 WITH y2@0; |
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| Anonymous posted on Friday, November 05, 2004 - 9:09 am
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Hi Linda,
I have a path model with 10 continous variables.
due to non-nomrality of the variables, I am using mlmv estimator. from my output the number of free parameters is 38. but i have only 37 parameters estimated. Am I missing something? how do you caluclate free parameters for MLMV estimator.
thanks. |
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| LMuthen posted on Friday, November 05, 2004 - 10:22 am
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| The degrees of freedom for the MLMV estimator are not computed in the regular way. See formula 110 in Technical Appendix 4. |
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| Anonymous posted on Thursday, June 30, 2005 - 10:17 am
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I have a sample size of 24 and have estimated a model with 9 dependent variables
and 6 independent variables. Using ML and fitting a particular model, the DF is 70. It seems MPlus disregards the sample size when calculating DF. I realize such a complicated model should not be fit to such a small sample size, but why does Mplus allow it? |
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| In covariance structure models, degrees of freedom are basd on the number of restrictions imposed on the covariance matrix not on sample size. Unless, the covariance cannot be inverted, a model will be estimated. However, with such a small sample size, power would be low and standard errors large. |
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Hi,
I am running a path analytic model with two IVs and one DV to test for mediation.
What does it mean when I get a chi-square value of 0, df=0, and fit stats that appear perfect (CFI=1 and RMSEA=0)? Is there somewhere I can look to understand what I assume is a very simple issue?
Thank you!! |
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| It means that the model is just-identified so you can not test the fit of the model. You have the same number of H0 and H1 parameters. |
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Dear Linda,
Can a path analysis be correctly termed "saturated" or "fully recursive" even though the exogenous variables are not specified as correlated?
I have:
Y1 ON X1 - X10;
Y2 ON X1 - X10;
Y3 ON X1 - X10 Y1 Y2;
Y4 ON X1 - X10 Y1 Y2;
Y5 ON X1 - X10 Y1 Y2;
Y1 WITH Y2;
Y3 WITH Y4 Y5;
Y4 WITH Y5;
Is this saturated? Can I expect the fit of the model to be perfect (just-identified)?
Thank you! |
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Yes to question one.
It looks like you have 65 parameters in both your H0 and H1 models so the model would be just-identified. |
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Thank you for your response. I have a further question related to this model.
I am trying to use the MODEL TEST option to test for measurement invariance by gender. In order to do so I wish to label all parameters. When I specify labels in my model command, my model is no longer just-identified. Why is this so?
Here is my model:
Y1 ON X1 - X10 (p1 - p10);
Y2 ON X1 - X10 (p11 - p20);
Y3 ON X1 - X10 (p21 - p30)
Y1 (p31)
Y2 (p32);
Y4 ON X1 - X10 (p33 - p42)
Y1 (p43)
Y2 (p44);
Y5 ON X1 - X10 (p45 - p54)
Y1 (p55)
Y2 (p56);
Y1 WITH Y2 (p57);
Y3 WITH Y4 (p58)
Y5 (p59);
Y4 WITH Y5 (p60);
Thank you for your help, I am new to this. |
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| I would need more information to answer your question. Please send your input, data, output, and license number to support@statmodel.com. Also, please explain to me what you mean by measurement invariance. This usually refers to latent variables. |
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Dear list,
Chapter 3 of the manual does not help me understand why i have 5 degrees of freedom with this model:
V3 ON V2 V2_1 V3_1;
V2 ON V1 V1_1 V2_1;
V1 ON V1_1;
On a general level: What variances/covariances are estimated by default?
Many thanks,
Bjorn |
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| You can see which parameters are estimated as the default by looking at the results. In your example, you have five degrees of freedom because you have 15 sample statistics and 10 parameters. The fifteen sample statistics come about because you have 6 variances and covariances among the 3 dependent variables and 9 covariances among the 3 dependent and 3 independent variables. The parameters estimated are 7 regression coefficients and three residual variances. |
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Hi,
I have the following path model:
y ON x1 x2 x3;
x1 ON x2 x3;
Mplus tells me that I have 9 free parameters and 0 degrees of freedom. The covariance matrix of variables y and x1-x3 includes 10 elements, so I would have expected to have 1 df. Can you please explain why for Mplus degrees of freedom equal 0? |
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1 df would imply 1 restriction in the model. But this model has no restrictions - all paths are drawn.
The 9 parameters Mplus reports are for the distribution of y given x2, x3. The covariates x2 and x3 do not contribute parameters because that their distribution is not part of the model (just like in regression). So you should not consider 10 elements, but instead 3 variances-covariances among y and x1 plus 4 covariances between y, x1 and x2, x3. Which gives 7 covariance matrix elements. So that's eliminating the 3 covariance matrix elements corresponding to x2, x3. The Mplus parameters are the 5 slopes, the 2 residual variances, and the 2 intercepts. So 9 parameters. There are actually a total of 7 + 2 elements because you add the means of y1 and x1 since those are part of their distribution. Therefore you get 0 df. |
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| Wei Chun posted on Tuesday, May 11, 2010 - 11:15 pm
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I am running a path model and I get a chi-square value of 0, df=0, and CFI=1 and RMSEA=0.
It means that the model is just-identified. How to make it to be identified?
Thanks for your advice. |
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| A just-identified model is identified. Fit cannot be assessed. To obtain an overidentified model for which fit can be assessed, some paths need to be fixed at zero. Theory should decide on which of these paths are fixed at zero. |
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