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The variable $Y$ is measured at time points $t_1$, $\ldots$, $t_9$ for each of five objects. Also available for each object is the value of $Y$ at time $t_0 = 0$ (baseline). Thus, the sample size is $n = 50$. I would like to fit a regression line of $Y$ versus time. Further, it is important for me to include the baseline measure in the model.

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Is it okay to include a random slope but no random intercept? Thus, my model would be: $$y_{ij} = (\beta_0 + \beta_1 t_{0i}) + (\beta_2 + \gamma_i) t_{ij} + \epsilon_{ij}$$ with $\beta_0$ an overall intercept, $\beta_1$ the baseline effect, $\beta_2$ an overall slope, and $\gamma_i$ the random deviation from the overall slope for object $i$.

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My concern is that $\beta_1 t_{0i}$ already results in an object-specific intercept and therefore I do not see the point to further include a random intercept in the model.

user7064
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2 Answers2

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First of all, I think your model specification is somewhat confusing. I would use $x_i=y_{i0}$ instead of $t_{0i}$ as $t_{ij}$ was used to represent time $j$ for individual $i$. You probably want to use baseline measurement rather than baseline time as covariate. So your model should be

$$y_{ij} = \beta_0 + \beta_1 x_i +(\beta_2 + \gamma_i)t_{ij} + \epsilon_{ij}$$ for $i= 1\ldots n$ and $j=1\ldots 9$. The purpose of random intercept is to describe the difference at baseline, which is carried forward to later time points. i.e A person with higher measurement at baseline is more likely to have higher measurement later.

Here comes the challenge: The above model cannot be used to describe what happened at baseline as the baseline measurement is already included in the model as a covariate. So using the above model to predict baseline outcome is meaningless. In this case, it is really hard to interpret the random effect and you shouldn't include it in the model.

Having said that, it does make sense to include random intercept if you shift your time variable so $t_1 = 0$. In this case, you avoid interpreting regression parameters at a time point that your model doesn't apply.

Peter

Peter
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  • Thx for the insight @Peter. What do you mean by "using the above model to predict baseline outcome"? – user7064 Oct 10 '14 at 17:28
  • Using your model setting, you cannot set $t_{ij}=0$ and predict $y_{i0}$ using the model. $y_{i0}$ is used as covariate in the model so your $j>0$. Although technically, you can set $t_{ij}$ to any number, putting in a negative number or zero is obviously meaningless. – Peter Oct 10 '14 at 17:41
  • OK, I see the point. In my particular case, however, I do not care about predicting the baseline outcome. Rather, I would like to be able to answer questions like "Given that the baseline outcome equals so much, how much is the outcome at time t?". If I understand you well, you agree with the fact that I shouldn't include a random intercept, isnt'it? Thank you again. – user7064 Oct 11 '14 at 05:52
  • After confirmation, I will accept ;-) – user7064 Oct 12 '14 at 05:38
  • No, you shouldn't include a random effect. – Peter Oct 14 '14 at 22:40
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I assume $j=1,...,9$.

Yes, you can treat $\beta_1t_{0i}$ as object-specific intercept. The difference from random intercept is that $\beta_1t_{0i}$ is fixed. It is kind of like the difference between fixed and random effects in econometrics. All objects share the same $\beta_1$, which is then weighted by the baseline value of each object. If the baseline measurement is important, it is fine to do so.

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