Message/Author 

Tim Seifert posted on Wednesday, February 21, 2007  7:09 am



Hello, The data I would like to model is children's recognition of printed words. At pretest (t=0), posttest (2 months), and followup (8 months) the number of words children could identify was recorded. The pretest data was normally distributed, a little less so at the other time points. Two questions. 1) Should the data be modelled as Poisson or continuous distributions? If modelled as a continuous (normal) distribution, the intercept = the meannumber of words at pretest (5.5). When modelled as Poisson, I am not sure what the intercept represents since its value is about 1.3. 2) Examination of the data suggests a quadratic term would be useful. I noticed in previous posts that four time points are preferred to three, and in adding a quadratic term the number of free parameters is zero. But I tried constraining the intercept and linear means, variances and covariances, but still have no free parameters. Can a quadratic term be modelled with only three time points? 


1) With high counts, Poisson is approximately normal so I would go with normal since it is simpler to work with. The Poisson means are on a log rate scale where rate is the "lambda" parameter of a Poisson distribution (see stat texts), 2)You can have a quadratic even with 3 time points, but you can't have all of its potential varcov parameters free. You can for example have a random intercept and random linear slope but a fixed quadratic. That is justidentified. There is no use in restraining the model just to get df's. 


Dear all, I have estimated a "Fixed Quadratic, Random Linear Model" in Mplus using the following multilevel syntax: ! timelevel model; %WITHIN% linear  y ON t_linear; quadratic  y ON t_quadratic; ! personlevel model; %BETWEEN% y* linear*; y WITH linear; quadratic@0; However, I didn't figure out how to specify this model using the Mplus growth curve syntax (i.e. i s q  y1@0 y2@1 y@2) . I would greatly appreciate your help. Thanks, Johannes Meier 


i s q  y1@0 y2@1 y@2 is the way to specify a quadratic growth model. The problem is that it is not identified with three time points. You need at least four. 


Hi Linda, Thanks for your answer. I have over 20 time points and I just wondered how to specify a "Fixed Quadratic, Random Linear Model" in the regular MPlus growth curve syntax. In my example above I did it the multilevel way (which I am used to as a former Stata user). If I would use just i s q  y1@0 y2@1 y3@2 ... y20@19 I would estimate a "Random Quadratic, Random Linear Model". Do you have any hint for me? Best regards, Johannes 


i s q  y1@0 y2@1 y3@2 ... y20@19 q@0; I would divide the time scores by ten given that the quadratic times scores are the squared value of the linear time scores. 

EFried posted on Monday, February 06, 2012  8:04 am



Dear Dr Muthén, when adding a quadratic term i s q  y1@0 y2@1 y3@2 Does the term "q" also have to be added to the i s (Q) ON x1 x2 x3; and the c%1 y1 y2 y3 i s (Q?); statements? Thank you 


Yes, in theory, but a quadratic growth model is not identified with three time points. 

EFried posted on Wednesday, February 08, 2012  6:20 am



I have 5 time points and just copied the example from above  apologies for the confusion. Thank you for the answer, I'll include the quadratic term in my ".. ON .." statements! 

sed48 posted on Monday, June 12, 2017  4:56 am



Dear Muthen, I have 6 timepoints in my dataset. I have tried a variety of models, including linear, quadratic, and piecewise models (with 1 breaking point). I have tried two types of piecewise models, either where both slopes are linear, or where one is quadratic and the other linear. My question concerns the piecewise model with both a linear segment and quadratic segment. i s1 q1  y0@0 y1@1 y2@2 y3@3 y4@3 y5@3 i s2  y0@0 y1@0 y2@0 y3@0 y4@4 y5@5 I am wondering whether this is really too complicated a model for the number of timepoints I have and it would result is overfitting. I would appreciate your advice on this. Many thanks! 


I would think that piecewise is more motivated when there is a substantive reason for a change in the growth. Perhaps you can instead try a cubic. 

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