More on shrinkage plots

Published

2026-04-10

Code
using CairoMakie
using DataFrames
using LinearAlgebra
using MixedModels
using MixedModelsMakie
using Random
using ProgressMeter

const progress = isinteractive()
false

Load the kb07 data set (don’t tell Reinhold that I used these data).

kb07 = MixedModels.dataset(:kb07)
Arrow.Table with 1789 rows, 7 columns, and schema:
 :subj      String
 :item      String
 :spkr      String
 :prec      String
 :load      String
 :rt_trunc  Int16
 :rt_raw    Int16
contrasts = Dict(
  :spkr => HelmertCoding(),
  :prec => HelmertCoding(),
  :load => HelmertCoding(),
)
m1 = let
  form = @formula(
    rt_trunc ~
      1 +
      spkr * prec * load +
      (1 + spkr + prec + load | subj) +
      (1 + spkr + prec + load | item)
  )
  fit(MixedModel, form, kb07; contrasts, progress)
end
VarCorr(m1)
Column Variance Std.Dev Corr.
subj (Intercept) 91064.7403 301.7693
spkr: old 1850.5250 43.0177 +0.78
prec: maintain 3843.8825 61.9991 -0.59 +0.03
load: yes 4246.0532 65.1617 +0.36 +0.83 +0.53
item (Intercept) 131173.0381 362.1782
spkr: old 1656.7129 40.7027 +0.44
prec: maintain 60946.6098 246.8737 -0.69 +0.35
load: yes 1795.0255 42.3677 +0.32 +0.16 -0.14
Residual 446886.0133 668.4953
issingular(m1)
true
print(m1)
Linear mixed model fit by maximum likelihood
 rt_trunc ~ 1 + spkr + prec + load + spkr & prec + spkr & load + prec & load + spkr & prec & load + (1 + spkr + prec + load | subj) + (1 + spkr + prec + load | item)
    logLik   -2 logLik      AIC         AICc        BIC     
 -14318.5613  28637.1227  28695.1227  28696.1119  28854.3156

Variance components:
             Column       Variance  Std.Dev.   Corr.
subj     (Intercept)      91064.7403 301.7693
         spkr: old         1850.5250  43.0177 +0.78
         prec: maintain    3843.8825  61.9991 -0.59 +0.03
         load: yes         4246.0532  65.1617 +0.36 +0.83 +0.53
item     (Intercept)     131173.0381 362.1782
         spkr: old         1656.7129  40.7027 +0.44
         prec: maintain   60946.6098 246.8737 -0.69 +0.35
         load: yes         1795.0255  42.3677 +0.32 +0.16 -0.14
Residual                 446886.0133 668.4953
 Number of obs: 1789; levels of grouping factors: 56, 32

  Fixed-effects parameters:
───────────────────────────────────────────────────────────────────────────────
                                             Coef.  Std. Error      z  Pr(>|z|)
───────────────────────────────────────────────────────────────────────────────
(Intercept)                             2181.67        77.299   28.22    <1e-99
spkr: old                                 67.7485      18.2934   3.70    0.0002
prec: maintain                          -333.921       47.1493  -7.08    <1e-11
load: yes                                 78.7703      19.5383   4.03    <1e-04
spkr: old & prec: maintain               -21.9657      15.806   -1.39    0.1646
spkr: old & load: yes                     18.3843      15.806    1.16    0.2448
prec: maintain & load: yes                 4.53389     15.806    0.29    0.7742
spkr: old & prec: maintain & load: yes    23.6074      15.806    1.49    0.1353
───────────────────────────────────────────────────────────────────────────────

1 Expressing the covariance of random effects

Earlier today we mentioned that the parameters being optimized are from a “matrix square root” of the covariance matrix for the random effects. There is one such lower triangular matrix for each grouping factor.

l1 = first(m1.λ)   # Cholesky factor of relative covariance for subj
4×4 LowerTriangular{Float64, Matrix{Float64}}:
  0.451416    ⋅          ⋅          ⋅ 
  0.0502725  0.0401697   ⋅          ⋅ 
 -0.0550321  0.0726579  0.01714     ⋅ 
  0.0351643  0.0850281  0.0321733  0.0

Notice the zero on the diagonal. A triangular matrix with zeros on the diagonal is singular.

l2 = last(m1.λ)    # this one is also singular
4×4 LowerTriangular{Float64, Matrix{Float64}}:
  0.541781    ⋅           ⋅          ⋅ 
  0.0268651  0.0546398    ⋅          ⋅ 
 -0.253039   0.268006    0.0229063   ⋅ 
  0.0200358  0.00129292  0.0601135  0.0

To regenerate the covariance matrix we need to know that the covariance is not the square of l1, it is l1 * l1' (so that the result is symmetric) and multiplied by σ̂²

Σ₁ = varest(m1) .* (l1 * l1')
4×4 Matrix{Float64}:
  91064.7   10141.5     -11101.7     7093.74
  10141.5    1850.52        67.9472  2316.37
 -11101.7      67.9472    3843.88    2142.48
   7093.74   2316.37      2142.48    4246.05
diag(Σ₁)  # compare to the variance column in the VarCorr output
4-element Vector{Float64}:
 91064.74027228545
  1850.5249616692
  3843.882496563724
  4246.0532111504635
sqrt.(diag(Σ₁))
4-element Vector{Float64}:
 301.76934945796836
  43.017728457802136
  61.9990523844012
  65.16174653238251

2 Shrinkage plots

Code
shrinkageplot(m1)
Figure 1: Shrinkage plot of model m1

The upper left panel shows the perfect negative correlation for those two components of the random effects.

shrinkageplot(m1, :item)
X1 = Int.(m1.X')
8×1789 Matrix{Int64}:
  1   1   1   1   1  1   1   1   1   1  …   1   1   1   1   1   1   1  1   1
 -1   1   1  -1  -1  1   1  -1  -1   1      1  -1  -1   1   1  -1  -1  1   1
 -1   1  -1   1  -1  1  -1   1  -1   1     -1   1  -1   1  -1   1  -1  1  -1
  1  -1  -1  -1  -1  1   1   1   1  -1      1   1   1  -1  -1  -1  -1  1   1
  1   1  -1  -1   1  1  -1  -1   1   1     -1  -1   1   1  -1  -1   1  1  -1
 -1  -1  -1   1   1  1   1  -1  -1  -1  …   1  -1  -1  -1  -1   1   1  1   1
 -1  -1   1  -1   1  1  -1   1  -1  -1     -1   1  -1  -1   1  -1   1  1  -1
  1  -1   1   1  -1  1  -1  -1   1  -1     -1  -1   1  -1   1   1  -1  1  -1
X1 * X1'
8×8 Matrix{Int64}:
 1789    -1    -1     3    -3     1     1     3
   -1  1789    -3     1    -1     3     3     1
   -1    -3  1789     1    -1     3     3     1
    3     1     1  1789     3    -1    -1    -3
   -3    -1    -1     3  1789     1     1     3
    1     3     3    -1     1  1789    -3    -1
    1     3     3    -1     1    -3  1789    -1
    3     1     1    -3     3    -1    -1  1789

3 How to interpret a shrinkage plot

  • Extreme shrinkage (shrunk to a line or to a point) is easy to interpret – the term is not providing benefit and can be removed.
  • When the range of the blue dots (shrunk values) is comparable to those of the red dots (unshrunk) it indicates that the term after shrinkage is about as strong as without shrinkage.
  • By itself, this doesn’t mean that the term is important. In some ways you need to get a feeling for the absolute magnitude of the random effects in addition to the relative magnitude.
  • Small magnitude and small relative magnitude indicate you can drop that term

4 Conclusions from these plots

  • Only the intercept for the subj appears to be contributing explanatory power
  • For the item both the intercept and the spkr appear to be contributing
m2 = let
  form = @formula(
    rt_trunc ~
      1 + prec * spkr * load + (1 | subj) + (1 + prec | item)
  )
  fit(MixedModel, form, kb07; contrasts, progress)
end
VarCorr(m2)
Column Variance Std.Dev Corr.
item (Intercept) 133027.438 364.729
prec: maintain 63841.835 252.669 -0.70
subj (Intercept) 88870.014 298.111
Residual 460948.573 678.932
Code
shrinkageplot(m2)
Figure 2: Shrinkage plot of model m2
m3 = let
  form = @formula(
    rt_trunc ~
      1 + prec + spkr + load + (1 | subj) + (1 + prec | item)
  )
  fit(MixedModel, form, kb07; contrasts, progress)
end
VarCorr(m3)
Column Variance Std.Dev Corr.
item (Intercept) 133014.905 364.712
prec: maintain 63768.184 252.524 -0.70
subj (Intercept) 88819.302 298.026
Residual 462443.261 680.032
rng = Random.seed!(1234321);
m3btstrp = parametricbootstrap(rng, 2000, m3);
DataFrame(shortestcovint(m3btstrp))
9×5 DataFrame
Row type group names lower upper
String String? String? Float64 Float64
1 β missing (Intercept) 2022.91 2334.0
2 β missing prec: maintain -430.239 -239.802
3 β missing spkr: old 34.0592 96.72
4 β missing load: yes 46.5349 109.526
5 σ item (Intercept) 270.065 451.979
6 σ item prec: maintain 181.739 325.127
7 ρ item (Intercept), prec: maintain -0.907255 -0.490094
8 σ subj (Intercept) 233.834 364.504
9 σ residual missing 657.341 702.655
ridgeplot(m3btstrp)
Figure 3: Ridge plot of the fixed-effects coefficients from the bootstrap sample
ridgeplot(m3btstrp; show_intercept=false)
Figure 4: Ridge plot of the fixed-effects coefficients from the bootstrap sample (with the intercept)
m4 = let
  form = @formula(
    rt_trunc ~
      1 + prec + spkr + load + (1 + prec | item) + (1 | subj)
  )
  fit(MixedModel, form, kb07; contrasts, progress)
end
m4bstrp = parametricbootstrap(rng, 2000, m4);
ridgeplot(m4bstrp; show_intercept=false)
DataFrame(shortestcovint(m4bstrp))
9×5 DataFrame
Row type group names lower upper
String String? String? Float64 Float64
1 β missing (Intercept) 2034.15 2335.45
2 β missing prec: maintain -427.769 -248.337
3 β missing spkr: old 35.6938 97.8081
4 β missing load: yes 45.3095 107.368
5 σ item (Intercept) 260.498 451.649
6 σ item prec: maintain 179.495 315.414
7 ρ item (Intercept), prec: maintain -0.901831 -0.471356
8 σ subj (Intercept) 236.684 360.363
9 σ residual missing 657.178 702.739
VarCorr(m4)
Column Variance Std.Dev Corr.
item (Intercept) 133014.905 364.712
prec: maintain 63768.184 252.524 -0.70
subj (Intercept) 88819.302 298.026
Residual 462443.261 680.032
Code
let mods = [m1, m2, m4]
  DataFrame(;
    geomdof=(sum  leverage).(mods),
    npar=dof.(mods),
    deviance=deviance.(mods),
    AIC=aic.(mods),
    BIC=bic.(mods),
    AICc=aicc.(mods),
  )
end
3×6 DataFrame
Row geomdof npar deviance AIC BIC AICc
Float64 Int64 Float64 Float64 Float64 Float64
1 131.599 29 28637.1 28695.1 28854.3 28696.1
2 107.543 13 28658.5 28684.5 28755.8 28684.7
3 103.478 9 28663.9 28681.9 28731.3 28682.0
scatter(fitted(m4), residuals(m4))
Figure 5: Residuals versus fitted values for model m4

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