I am studying the linear mixed model $y=X\beta + Z\gamma + \epsilon$ where $\beta$ is a vector of fixed effects, and it seems that $\gamma$ is usually specified as $\gamma \sim N(0,I\sigma^2)$.

My question is why does $\gamma$ have a mean of zero?

If I specify $\gamma \sim N(\alpha,I\sigma^2)$, where $\alpha$ is a vector of non-zero constants, then $y=X\beta + Z\alpha + Z\gamma^* + \epsilon$ where $\gamma^* \sim N(0,I\sigma^2)$. Is there anything incoherent with this model, can $Z$ wield both fixed and random effects?

  • 2
    $\begingroup$ In general the columns of the $Z$ matrix correspond to some or all of the columns of the $X$ matrix, consequently the non-zero means of the random effects can be found in as fixed effects in $\beta$ $\endgroup$ – user83346 Sep 15 '16 at 5:39

The mean of $\gamma$ is already played by the intercept. If you include both an intercept and the parameter $\alpha$ ($\gamma \sim N(\alpha, \sigma^2 I)$), then the model will be overparametrised and there will be confounding. The model is said to be non-identifiable.


$$y = \alpha + \gamma,\quad \gamma \sim \mathcal N(\color{green}{0}, \sigma^2_\gamma)$$ is the same as $$y = \gamma,\quad \gamma \sim \mathcal N(\color{green}{\alpha}, \sigma^2_\gamma)$$

EDIT: As noted by @fcop in his comment below, the above discussion concerns a random intercept. In the same way, the mean of a random effect is played by the corresponding fixed effect coefficient.

  • $\begingroup$ You can have random effects on slopes also I think ? $\endgroup$ – user83346 Sep 15 '16 at 8:40
  • $\begingroup$ Similary, the mean of the random effect of the slope is played by the fixed effect of the slope. $\endgroup$ – ocram Sep 15 '16 at 8:46
  • $\begingroup$ This is what I said in my comment below the question $\endgroup$ – user83346 Sep 15 '16 at 9:07
  • $\begingroup$ Indeed. I ve just tried to make it more explicit ;-) $\endgroup$ – ocram Sep 15 '16 at 9:16
  • $\begingroup$ So based on the above discussion, if I don't include the fixed effect term $Z\alpha$ explicitly in the model, then I can have $\gamma \sim N(\alpha,I\sigma^2)$ without the identification issue? $\endgroup$ – user77873 Sep 22 '16 at 9:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.