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! 

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