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Chuang Chen posted on Thursday, March 29, 2018 - 1:36 pm
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Hello, If I add a quadratic term to my unconditional model, I'd like to only have the fixed effect of the quadratic term. Can I ask how I write code? Is like: Model: i s q | y1@0 y2@1 y3@2 y4@3; [q@0]; If I have individual time score, can I write as: i s q | y1 - y4 AT t1 - t4; [q@0]; Thanks a lot!!! |
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In both cases you should say q@0; ! the variance Not [q@0]; ! the mean |
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Chuang Chen posted on Thursday, March 29, 2018 - 5:28 pm
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Hi Muthen, Thank you for your correcting. Can I ask more? If I add covariate into models, should I still write as: i s q | y1 - y4 AT t1 - t4; q@0; i on x1 x2; s on x1 x2; q on x1 x2; Thanks a lot! Best, Chuang |
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You can. In this case, it means that the residual variance of q is zero. There will be variation in q itself given that it is specified to be a function of x1, x2; that's ok. |
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Chuang Chen posted on Thursday, March 29, 2018 - 5:52 pm
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Hi Muthen, Thanks for your reply. Based on your interpretation, does it mean all the variance of q explain by x1 and x2? Based on my model selection result, growth curve model with the fixed effect quadratic term (also contain random s) shows the best fit. I'd like to add covariates, and to explain quadratic term is not different among individuals. (fixed effect, can I explain like this?) How can I modify my code? Thanks a lot! Best, Chuang |
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Q1: Yes. The quadratic term is different among individuals because it varies as a function of x's that vary among individuals. See the Raudenbush-Bryk multilevel book. I think they call this case a "varying" effect instead of a "random" effect. |
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