The block quotation below, from leaders in the field of mixed effect modeling, claims that coordinate shifts in models with zero correlation between random effects ('ZCP' models) changes model predictions. But, can someone elaborate on or further justify their claims?

The statements in question are from Bates et al's 2015 paper on lme4, Fitting Linear Mixed-Effects Models Using lme4, page 7, second paragraph (download link). $\newcommand{\slope}{\text{slope}} \newcommand{\int}{\text{int}} \newcommand{\intercept}{\text{intercept}}$

Here is a paraphrasing of what they wrote:

Although zero correlation parameter models are used to reduce complexity of random-slopes models, they have one drawback. Models in which slopes and intercepts are allowed to have non-zero correlation are invariant to additive shifts of a continuous predictor.

This invariance breaks down when the correlation is constrained to zero; any shift in the predictor will necessarily lead to a change in the estimated correlation, and in the likelihood and predictions of the model.1 For example, we can eliminate the correlation in fm1 simply by shifting Days [the predictor accompanying $\slope$] by an amount equal to the ratio of the estimated among-subject standard deviations multiplied by the estimated correlation, i.e.2,


The use of such models should ideally be restricted to cases where the predictor is measured on a ratio scale (i.e., the zero point on the scale is meaningful, not just a location defined by convenience or convention).


Numbered in accord with the superscripts above...

  1. I can see that any shift in the coordinate system by which the predictor is measured will lead to a change in the estimated correlation, thereby leading to non-zero correlation. This supports the statement that zero correlation parameter models are not invariant under shifts in predictor coordinate systems, and therefore that any model with non-zero random effects correlations can be transformed into a model with zero correlations by a suitable shift in coordinates. I think it also supports the third paragraph in the paraphrasing above: ZCP models (and zero intercept models—see below; but please check me on this) are only valid for models using certain, special, coordinate systems. But why should a coordinate shift change predictions for such models?

    For example, a shift in coordinates will also change the fixed-effect intercept term for group averages (see below), but only by an amount appropriate to the change in origin for the predictor's coordinate system. Such a change does not impact model predictions, so long as the new coordinate system is used for the shifted predictor.

    To elaborate, if the fixed-effect slope associated with the shifted predictor is positive, and the origin for the predictor's coordinate system is shifted in the negative direction, then the fixed-effect intercept will decrease, and any associated random effect intercepts will also change correspondingly, reflecting the new definition of 'origin' (and therefore intercept) in the shifted coordinate system. By the way, I think this reasoning also implies that a zero intercept model is also not invariant under such shifts.

  1. I think I have a reasonable way of working this out, but have derived an answer slightly different than Bates et al. Am I going wrong somewhere?

    Below is my answer. Following that is the description of how I arrived at my result. In summary, I find that if I shift the $x$ origin negatively by $\delta > 0$, so that in the new coordinate system the predictor takes on values $x' = x + \delta$, then the correlation $\rho'$ in the new coordinate system is zero if:


    This differs from Bates et al's result.

Description of my method (Optional Reading): Let's say we have the correlation of two random effects, $\slope$ and $\intercept$ ($\int$ for short), both corresponding to the same grouping factor with $k$ levels (numbered by $i$, ranging from $1$ to $k$). Let's also say that the continuous predictor with which the random $\slope$ is paired is called $x$, defined such that the product $x\times\slope_i$ generates the conditional contribution to the fitted value $\hat y_{obs}$ for level $i$ of the associated grouping factor. Although in reality the MLE algorithm determines the value of $\rho$ to maximize the likelihood, I would expect that the expression below should be a dimensionally correct way of determining the effects of a uniform translation in $x$, the multiplier of the random effect for $\slope$.

$$\rho_{\slope:\int} = \frac{E_{i}\big[(\slope_i -\overline {\slope_i})(\int_i -\overline {\int_i})\big]}{\sqrt{E_{i}\big[(\slope_i -\overline {\slope_i})^2\big]E_{i}\big[(\int_i-\overline {\int_i})^2\big]}}$$

To arrive at my result, I first rewrote the old value for the intercept in terms of a new value for the intercept, $\int' = -\delta \times \slope + \int$ (here, $\delta > 0$, the 'leftward' shift in origin for predictor $x$). Then, I substituted the resulting expression into the numerator of the above formula for $\rho$, calculating the value of $\delta$ that resulted in a zero covariance in the new coordinate system. Note that as stated in Question 1 above, the fixed-effect intercept term will also change in an analogous manner: $\beta_0' =- \delta\times\beta_x + \beta_0$. (Here, $\beta_x$ is the fixed-effect predictor associated with the shifted predictor $x.$)

  • 1
    A few rough ideas. $\hat{y}$ changes if (1) the fixed slope changes or (2) the random slopes change. For (1): the fixed slope can be viewed as a weighted mean of the cluster-specific slopes, where the weight depends in part on the estimated variance components. Omitting the covariance alters the var. estimates, changing the weights, changing the fixed slope. For (2): the random slopes are the cluster-specific slopes "shrunk" toward the fixed slope in proportion to the same weights. Omitting the covariance alters the var. estimates, changing the degree of shrinkage, changing the random slopes. – Jake Westfall Apr 10 '16 at 23:46
  • I'm a little disappointed this hasn't gotten more attention, @clarpaul. You might just put your own answer in. If no one else answers, I'll just give the bounty to you. – gung Apr 14 '16 at 18:40
  • Thanks @gung, my answer would be closely aligned with my "Edits" above. The bounty would be nice, but I may not have time before it expires. I encourage anyone to take my "Edits" and turn them into an answer, if they agree with the basic reasoning, and are willing to take the time to polish them a bit. – clarpaul Apr 14 '16 at 22:28
up vote 4 down vote accepted
  1. The answer to this question turns out to be rather definitional. If one shifted the coordinates of the independent variables of a ZCP model and allowed correlations to develop in an unconstrained manner, predictions would not change, because linear mixed effects models with unconstrained correlations are translation invariant (one can show this with a bit of math). But, by definition, a ZCP model has correlations constrained to $0$. On shifting coordinates, correlations would not be allowed to develop as required in an unconstrained LME model. Therefore, ZCP models are not translation invariant, and a coordinate shift would change model predictions. And (if you expect LME models to be translation invariant to sensible coordinate shifts) only models in which such coordinate shifts don't make sense are theoretically sensible as ZCP models (i.e., the 'special' ones mentioned in the third paragraph of the paraphrase of Bates et al above). [Note: I will embellish this answer in the future to include formulas I have derived for the correlation that develops when coordinate-shifting an initially ZCP model, and for the proof that LME models with unconstrained correlations are translation invariant.]
  2. Bates et al's result is simply a typo. The answer, $\delta$, must have the same dimensions as the predictor, $x$ (Days), which is shifted. Since, w.l.o.g., $\sigma_{intercept}$ and $\rho$ can be considered to have dimensions of unity, $\sigma_{slope}$, which has dimensions $1/x$ (the same dimensions as $slope$), must be in the denominator in order for $\delta$ to have the correct dimensions.

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