Distribution of variable

[BvnHU]

Hello,

I want to use latent class growth analysis, but my dependent variable is distributed like this: .

Another variable we want to use is distributed as followed:

Based on the information from the vignettes and articles, i'm not quite sure if lcmm can handle this kind of data. Can you tell me whether this is possible and if not, what would be a solution?

Thanks in advance,

Bas

[VivianePhilipps]

Hello Bas,

I'm not sure, by looking at your first histogram, if your outcome is ordinal. If you have only 8 modalities, it's a bit small to consider it as a continuous variable. So, the lcmm function with link="thresholds" should be preferred. For the second, it is asymmetric, but it is close to the MMSE's distribution in the paquid sample (look at hist(paquid$MMSE)), so you can try with splines link. You won't be able to capture all variations as in your pink density, but according to the context it may be sufficient.

Best,

Viviane

[BvnHU]

Thanks @VivianePhilipps! I managed to perform the analysis. What i don't understand from the output, is that there is no intercept for class 1 under the Fixed effects in the longitudinal model overview. When i use predictY and plot the model, class 1 is represented, but based on which intercept? So my question is: How can i determine the intercept for class 1?

Maximum Likelihood Estimates:

Fixed effects in the class-membership model: (the class of reference is the last class)

                 coef      Se    Wald p-value

intercept class1 0.11220 0.22408 0.501 0.61658 intercept class2 2.08575 0.16005 13.032 0.00000 intercept class3 -1.07169 0.27167 -3.945 0.00008

Fixed effects in the longitudinal model:

                                 coef      Se    Wald p-value

intercept class1 (not estimated) 0
intercept class2 -5.47246 0.36667 -14.925 0.00000 intercept class3 -0.20171 0.60383 -0.334 0.73834 intercept class4 -6.06015 0.45287 -13.382 0.00000 Time class1 -1.43401 0.13921 -10.301 0.00000 Time class2 0.01049 0.03193 0.329 0.74250 Time class3 0.14260 0.20109 0.709 0.47824 Time class4 1.42951 0.12765 11.199 0.00000

Variance-covariance matrix of the random-effects: intercept Time intercept 0.12302
Time 0.01760 0.00252

Residual standard error (not estimated) = 1

Parameters of the link function:

                     coef      Se    Wald p-value

Linear 1 (intercept) 54.09246 2.65251 20.393 0.00000 Linear 2 (std err) 7.90029 0.18885 41.834 0.00000

[CecileProust-Lima]

Hi, because you use lcmm function (rather than hlme) function, a transformation is included (last part linear1 linear2). To make the model identifiable given this transformation, we add 2 constraints on the underlying level (after transformation):

• the intercept of the first class is 0
• • the variance of the error is 1. this transformation is useful when you do not have a Gaussian outcome. here you assume a linear transformation which corresponds exactly to the Gaussian case (linear 1 replaces intercept not estimated, linear 2 replaces the standard error not estimated). So if this is what you want to do, then use hlme, it will be easier. Cécile