What it means for rho to be 0 in lme() when using compound symmetry
The 'default' structure is generally the unstructured positive definite matrix. (In lme4, this constraint is loosened to positive semi-definite). Starting from compound symmetric is already placing a constraint and so forcing all off-diagonal elements to zero may be the best solution *under that constraint*. (And a multiple of the identity matrix is a stricter constraint than a diagonal matrix, even though both have all off diagonal elements set to zero.) A more modern take would be using something like rePCA on the full unstructured matrix (https://arxiv.org/abs/1506.04967 or https://doi.org/10.33016/nextjournal.100002) to see what the effective dimensionality is. Of course, you can fit a model with a diagonal RE covariance matrix even if the data have some correlation, at the cost of changing how shrinkage works (https://doingbayesiandataanalysis.blogspot.com/2019/07/shrinkage-in-hierarchical-models-random.html). That may be an acceptable (variance-bias) tradeoff -- less efficient shrinkage but also less overparameterization. All of these comments without looking at your data.
On 4/12/20 4:23 pm, Simon Harmel wrote:
Thanks, Phillip. Given the estimated rho of 0 obtained from the default
correlation structure in lme(), can we say that for this dataset there
is no dependence left after fitting the 2-level model shown in my
original post?
In other words, once getting a rho of 0 from the default correlation
structure for this model, then one doesn't need to think of alternative
correlation structures, because even the default correlation structure
has shown that there is no dependence to model.
Is this a reasonable conclusion?
On Fri, Dec 4, 2020, 9:11 AM Phillip Alday <me at phillipalday.com
<mailto:me at phillipalday.com>> wrote:
From
https://stat.ethz.ch/R-manual/R-devel/library/nlme/html/pdCompSymm.html
:
"This function is a constructor for the pdCompSymm class, representing a
positive-definite matrix with compound symmetry structure (constant
diagonal and constant off-diagonal elements)."
Any multiple of the identity matrix is technically compound symmetric,
because all the off-diagonal elements are the same (0).
Phillip
On 28/11/20 2:30 am, Simon Harmel wrote:
> Hello All,
>
> Below, I'm using corCompSymm() (compound symmetry) for my simple
model.
>
> The rho is estimated to be 0. I was wondering what it means for
rho in the
> var-covariance matrix to be "0"? Is my var-covariance matrix below
valid?
> -- Thank you all, Simon
> #----------------------------------------------------------------
> library(nlme)
> data <-
>
> m <- lme(Achieve ~ time, random = ~1|subid, data = data, correlation =
> corCompSymm())
>
>? ?aa <- corMatrix(m$modelStruct$corStruct)[[1]]
>? ?aa * sigma(m)^2
>
>? ? ? ? ? [,1]? ? ?[,2]? ? ?[,3]? ? ?[,4]
> [1,] 112.5003? ?0.0000? ?0.0000? ?0.0000
> [2,]? ?0.0000 112.5003? ?0.0000? ?0.0000
> [3,]? ?0.0000? ?0.0000 112.5003? ?0.0000
> [4,]? ?0.0000? ?0.0000? ?0.0000 112.5003
>
>? ? ? ?[[alternative HTML version deleted]]
>
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