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| Sanjoy posted on Monday, March 21, 2005 - 5:57 pm
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Prof. Linda and Bengt Muthen
Dear Madam and/or Sir
this is my model situation, "f's " are the three latent variables and their corresponding indicator variables are Y's ... X's share some common elements between them
f1 = function of (f2, f3 and X1)
f2 = function of (f3 and X2)
f3 = function of (f2 and X3)
f1 has only one indicator Y1 which is binary
f2 has only one indicator Y2 which is ordered categorical (five scale)
f3 has only one indicator Y3 which is ordered categorical (five scale)
I have three quick questions ... kindly suggest me what to do or rectify if I'm wrong
Q1. Is it OK if I fit three equations using only "ON" statement ... I have tried to fit "f1 by Y1, f2 by Y2, f3 by Y3" and then "f1 on f2 f3 X1" and so on ... then in the output it's showing warning no convergence ... but if you write "Y1 ON Y2 Y3 X1", "Y2 ON Y3 X2" and so on ... it runs nicely, and the result is OK too
Q2. why is it happening such ...though it's all single indicator variable, we still have "measurement section" of the SEM, don't we, why can't we mention them by "BY" statement ... and that's primarily how Statistics of SEM differs from Econometrics, where we usually don't model dependent variable with measurement error, errors in Econometrics come from the explanatory side (please correct me if I'm wrong)
Q3. our system is Non-recursive, I need to check whether it's been identified or not ...I suppose MPlus would have informed me if it wasn't (please correct me if I'm wrong)... nonetheless, I need the algebra behind this, I have to report it in my dissertation ...please suggest me a comprehensive reading
Thanks and regards ... sincerely, Sanjoy |
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| bmuthen posted on Monday, March 21, 2005 - 6:14 pm
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| With single indicators you cannot estimate separately a measurement error variance and a factor variance. This is described in SEM books. This implies that when you use BY you try to estimate a factor variance and it is not identified. When you use ON you do not estimate a factor variance and the model is identified. You should use ON here, but it would be equivalent to using BY if you also fix the factor variance (e.g, to 1). |
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| Sanjoy posted on Tuesday, March 22, 2005 - 8:36 pm
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Thank you Sir ... below is the model Statement for my second model where I'm using three latent variable, one of them "Y1" has one indicator and two of them have three indicator variables
CATEGORICAL ARE Y1 R7-R9 B6-B8;
ANALYSIS: PARAMETERIZATION=THETA;
ESTIMATOR=WLSMV;
MODEL: B by B6-B8; !equation 1
R by R7-R9; !equation 2
Y1 on B R X7-X12; !equation 3
B on R X2 X8 X9 X11 X15 X9 X10; !equation 4
R on B X5 X9 X10 X12; !equation 5
up to this result is OK ... But
my QUESTION is I want to put some parameter constriant, following page 423 of Mplus User's Guide I have tried but it didn't work ...could u kindly rectify me please
I WANT TO CHECK, say for equation 3, parameter coefficients associated with B and R both are equal to zero ... this is what I have done
MODEL: B by B6-B8;
R by R7-R9;
Y1 on B(p1) R(p2) X7-X12;
B on R(p3) X2 X8 X9 X11 X15 X9 X10;
R on B(p4) X5 X9 X10 X12;
MODEL CONSTRAINT: p1 EQ 0;
p2 EQ 0;
Nothing is happening, I guess I grossly misunderstood the "labeling thing" ... besides it'a also saying X7 to X12 aren't being used
thanks and regards ...Sanjoy |
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You can only have one label per line. Try the following.
MODEL: B by B6-B8;
R by R7-R9;
Y1 on B(p1)
R(p2)
X7-X12;
B on R(p3)
X2 X8 X9 X11 X15 X9 X10;
R on B(p4)
X5 X9 X10 X12;
MODEL CONSTRAINT: p1 EQ 0;
p2 EQ 0; |
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| Sanjoy posted on Wednesday, March 23, 2005 - 11:18 am
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Thank u madam ... But... the problem remains, it's saying
"
*** ERROR in Model Constraint command
Unknown parameter label in MODEL CONSTRAINT: 0
in assignment: 0 =
"
this is what I have written
MODEL: B by B6-B8;
R by R7-R9;
Y1 on B(C1)
R(C2)
X7-X12;
B on R X2 X8 X9 X11 X15 X9 X10;
R on B X5 X9 X10 X12;
! C1 and C2 are the model constraints on parameters
MODEL CONSTRAINT: C1 EQ 0;
C2 EQ 0;
I can send u the data set if u think that would be required
thanks and regards ... sanjoy |
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| I think you should be saying c1 = 0; and c2 = 0; although you could just say in the MODEL command y1 ON b20; and r@0; |
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| I should have said y on b@0; |
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| Sanjoy posted on Wednesday, March 23, 2005 - 7:11 pm
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Thanks ..."@0" worked nicely
also, under model constraint I put different model constraints like b1=-b2; that ran perfectly too
however, for the SAME model where "@0" worked nicely I have tried to fit the same constraint on the SAME parameters writing c1=0
it does not work then
the below one works NICELY
MODEL: B by B6-B8;
R by R7-R9;
Y1 on B@0
R
X7-X12;
B on R X2 X8 X9 X11 X15 X9 X10;
R on B X5 X9 X10 X12;
the below version did NOT work
MODEL: B by B6-B8;
R by R7-R9;
Y1 on B(C1)
R X7-X12;
B on R X2 X8 X9 X11 X15 X9 X10;
R on B X5 X9 X10 X12;
MODEL CONSTRAINT: C1 = 0;
it's saying
"
THE MODEL ESTIMATION TERMINATED NORMALLY
THE STANDARD ERRORS OF THE MODEL PARAMETER ESTIMATES COULD NOT BE COMPUTED. THE MODEL MAY NOT BE IDENTIFIED. CHECK YOUR MODEL.
"
thanks and regards ...Sanjoy |
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| If you are interested in me looking at this further, you need to send full outputs of that which worked and did not work along with your license number to support@statmodel.com. |
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| Simon O. F. posted on Wednesday, November 04, 2009 - 11:04 am
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Hi,
I'm testing some path models (single item variables) and for some models, I got no degree of freedom left. I was wondering if it is normal in path models or should I report something else then the fit indices like R-square ?
Thanks in advance, |
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| If you have no degrees of freedom, the model is saturated and model fit is not relevant. |
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