Class: LinearMixedModel
Extract covariance parameters of linear mixedeffects model
[
returns
the covariance parameters and related statistics in psi
,mse
,stats
]
= covarianceParameters(lme
,Name,Value
)stats
with
additional options specified by one or more Name,Value
pair
arguments.
For example, you can specify the confidence level for the confidence limits of covariance parameters.
lme
— Linear mixedeffects modelLinearMixedModel
objectLinear mixedeffects model, specified as a LinearMixedModel
object constructed using fitlme
or fitlmematrix
.
Specify optional
commaseparated pairs of Name,Value
arguments. Name
is
the argument name and Value
is the corresponding value.
Name
must appear inside quotes. You can specify several name and value
pair arguments in any order as
Name1,Value1,...,NameN,ValueN
.
'Alpha'
— Significance levelSignificance level, specified as the commaseparated pair consisting of
'Alpha'
and a scalar value in the range 0 to 1. For a value α,
the confidence level is 100*(1–α)%.
For example, for 99% confidence intervals, you can specify the confidence level as follows.
Example: 'Alpha',0.01
Data Types: single
 double
psi
— Estimate of covariance parametersEstimate of covariance parameters that parameterize the prior
covariance of the random effects, returned as a cell array of length R,
such that psi{r}
contains the covariance matrix
of random effects associated with grouping variable g_{r}, r =
1, 2, ..., R. The order of grouping variables is
the same order you enter when you fit the model.
mse
— Residual variance estimateResidual variance estimate, returned as a scalar value.
stats
— Covariance parameter estimates and related statisticsCovariance parameter estimates and related statistics, returned as a cell array of length (R + 1) containing dataset arrays with the following columns.
Group  Grouping variable name 
Name1  Name of the first predictor variable 
Name2  Name of the second predictor variable 
Type 

Estimate 
Standard deviation of the random effect associated
with predictor Correlation between the random effects associated with
predictors 
Lower  Lower limit of a 95% confidence interval for the covariance parameter 
Upper  Upper limit of a 95% confidence interval for the covariance parameter 
stats{r}
is a dataset array containing statistics
on covariance parameters for the rth grouping variable, r =
1, 2, ..., R. stats{R+1}
contains
statistics on the residual standard deviation. The dataset array for
the residual error has the fields Group
, Name
, Estimate
, Lower
,
and Upper
.
Load the sample data.
load('fertilizer.mat');
The dataset array includes data from a splitplot experiment, where soil is divided into three blocks based on the soil type: sandy, silty, and loamy. Each block is divided into five plots, where five different types of tomato plants (cherry, heirloom, grape, vine, and plum) are randomly assigned to these plots. The tomato plants in the plots are then divided into subplots, where each subplot is treated by one of four fertilizers. This is simulated data.
Store the data in a dataset array called ds
, for practical purposes, and define Tomato
, Soil
, and Fertilizer
as categorical variables.
ds = fertilizer; ds.Tomato = nominal(ds.Tomato); ds.Soil = nominal(ds.Soil); ds.Fertilizer = nominal(ds.Fertilizer);
Fit a linear mixedeffects model, where Fertilizer
is the fixedeffects variable, and the mean yield varies by the block (soil type), and the plots within blocks (tomato types within soil types) independently. This model corresponds to
$${y}_{ijk}={\beta}_{0}+\sum _{j=2}^{5}{\beta}_{2j}I[T{]}_{ij}+{b}_{0jk}(S*T{)}_{jk}+{\u03f5}_{ijk},$$
where $$i$$ = 1, 2, ..., 60 corresponds to the observations, $$j$$ = 2, ..., 5 corresponds to the tomato types, and $$k$$ = 1, 2, 3 corresponds to the blocks (soil). $${S}_{k}$$ represents the $$k$$ th soil type, and $$(S*T{)}_{jk}$$ represents the $$j$$ th tomato type nested in the $$k$$ th soil type. $$I[T{]}_{ij}$$ is the dummy variable representing the level $$j$$ of the tomato type.
The random effects and observation error have the following prior distributions: $${b}_{0k}\sim N(0,{\sigma}_{S}^{2})$$, $${b}_{0jk}\sim N(0,{\sigma}_{S*T}^{2})$$, and $${\u03f5}_{ijk}\sim N(0,{\sigma}^{2})$$.
lme = fitlme(ds,'Yield ~ Fertilizer + (1Soil) + (1Soil:Tomato)');
Compute the covariance parameter estimates (estimates of $${\sigma}_{S}^{2}$$ and $${\sigma}_{S*T}^{2}$$) of the randomeffects terms.
psi = covarianceParameters(lme)
psi=2×1 cell array
{[3.4041e16]}
{[ 352.8481]}
Compute the residual variance ($${\sigma}^{2}$$).
[~,mse] = covarianceParameters(lme)
mse = 151.9007
Load the sample data.
load('weight.mat');
weight
contains data from a longitudinal study, where 20 subjects are randomly assigned to 4 exercise programs, and their weight loss is recorded over six 2week time periods. This is simulated data.
Store the data in a dataset array. Define Subject
and Program
as categorical variables.
ds = dataset(InitialWeight,Program,Subject,Week,y); ds.Subject = nominal(ds.Subject); ds.Program = nominal(ds.Program);
Fit a linear mixedeffects model where the initial weight, type of program, week, and the interaction between the week and type of program are the fixed effects. The intercept and week vary by subject.
For 'reference'
dummy variable coding, fitlme
uses Program A as reference and creates the necessary dummy variables $$I[.]$$. This model corresponds to
$$\begin{array}{rrl}{y}_{im}& =& {\beta}_{0}+{\beta}_{1}I{W}_{i}+{\beta}_{2}{\text{Week}}_{i}+{\beta}_{3}I[PB{]}_{I}+{\beta}_{4}I[PC{]}_{i}\\ & & +{\beta}_{5}I[PD{]}_{i}+{b}_{0m}+{b}_{1m}{\text{Week}}_{im}+{\u03f5}_{im}\end{array}$$
where $$i$$ corresponds to the observation number, $$i=1,2,...,120$$, and $$m$$ corresponds to the subject number, $$m=1,2,...,20$$. $${\beta}_{j}$$ are the fixedeffects coefficients, $$j=0,1,...,8$$, and $${b}_{0m}$$ and $${b}_{1m}$$ are random effects. $$IW$$ stands for initial weight and $$I[.]$$ is a dummy variable representing a type of program. For example, $$I[PB{]}_{i}$$ is the dummy variable representing Program B.
The random effects and observation error have the following prior distributions:
$$\left(\begin{array}{c}{b}_{0m}\\ {b}_{1m}\end{array}\right)\sim N(0,\left(\begin{array}{cc}{\sigma}_{0}^{2}& {\sigma}_{0,1}\\ {\sigma}_{0,1}& {\sigma}_{1}^{2}\end{array}\right))$$
and
$${\u03f5}_{im}\sim N(0,{\sigma}^{2}).$$
lme = fitlme(ds,'y ~ InitialWeight + Program + (WeekSubject)');
Compute the estimates of covariance parameters for the random effects.
[psi,mse,stats] = covarianceParameters(lme)
psi = 1x1 cell array
{2x2 double}
mse = 0.0105
stats=2×1 cell array
{3x7 classreg.regr.lmeutils.titleddataset}
{1x5 classreg.regr.lmeutils.titleddataset}
mse
is the estimated residual variance. It is the estimate for $${\sigma}^{2}$$.
To see the covariance parameters estimates for the randomeffects terms ($${\sigma}_{0}^{2}$$, $${\sigma}_{1}^{2}$$, and $${\sigma}_{0,1}$$), index into psi
.
psi{1}
ans = 2×2
0.0572 0.0490
0.0490 0.0624
The estimate of the variance of the random effects term for the intercept, $${\sigma}_{0}^{2}$$, is 0.0572. The estimate of the variance of the random effects term for week, $${\sigma}_{1}^{2}$$, is 0.0624. The estimate for the covariance of the random effects terms for the intercept and week, $${\sigma}_{0,1}$$, is 0.0490.
stats
is a 2by1 cell array. The first cell of stats
contains the confidence intervals for the standard deviation of the random effects and the correlation between the random effects for intercept and week. To display them, index into stats
.
stats{1}
ans = Covariance Type: FullCholesky Group Name1 Name2 Type Subject {'(Intercept)'} {'(Intercept)'} {'std' } Subject {'Week' } {'(Intercept)'} {'corr'} Subject {'Week' } {'Week' } {'std' } Estimate Lower Upper 0.23927 0.14365 0.39854 0.81971 0.38663 0.95658 0.2497 0.18303 0.34067
The display shows the name of the grouping parameter (Group
), the randomeffects variables (Name1
, Name2
), the type of the covariance parameters (Type
), the estimate (Estimate
) for each parameter, and the 95% confidence intervals for the parameters (Lower
, Upper
). The estimates in this table are related to the estimates in psi
as follows.
The standard deviation of the randomeffects term for intercept is 0.23927 = sqrt(0.0527). Likewise, the standard deviation of the random effects term for week is 0.2497 = sqrt(0.0624). Finally, the correlation between the randomeffects terms of intercept and week is 0.81971 = 0.0490/(0.23927*0.2497).
Note that this display also shows which covariance pattern you use when fitting the model. In this case, the covariance pattern is FullCholesky
. To change the covariance pattern for the randomeffects terms, you must use the 'CovariancePattern'
namevalue pair argument when fitting the model.
The second cell of stats
includes similar statistics for the residual standard deviation. Display the contents of the second cell.
stats{2}
ans = Group Name Estimate Lower Upper Error {'Res Std'} 0.10261 0.087882 0.11981
The estimate for residual standard deviation is the square root of mse
, 0.10261 = sqrt(0.0105).
Load the sample data.
load carbig
Fit a linear mixedeffects model for miles per gallon (MPG), with fixed effects for acceleration and weight, a potentially correlated random effect for intercept and acceleration grouped by model year, and an independent random effect for weight, grouped by the origin of the car. This model corresponds to
$${\text{MPG}}_{imk}={\beta}_{0}+{\beta}_{1}{\text{Acc}}_{i}+{\beta}_{2}{\text{Weight}}_{i}+{b}_{10m}+{b}_{11m}{\text{Acc}}_{i}+{b}_{21k}{\text{Weight}}_{i}+{\u03f5}_{imk}$$
where $$m=1,2,...,13$$ represents the levels for the variable Model_Year
, and $$k=1,2,...,8$$ represents the levels for the variable Origin
. $$MP{G}_{imk}$$ is the miles per gallon for the ith observation,m th model year, andk th origin that correspond to the ith observation. The randomeffects terms and the observation error have the following prior distributions:
$${b}_{1m}=\left(\begin{array}{c}{b}_{10m}\\ {b}_{11m}\end{array}\right)\sim N(0,\left(\begin{array}{cc}{\sigma}_{10}^{2}& {\sigma}_{10,11}\\ {\sigma}_{10,11}& {\sigma}_{11}^{2}\end{array}\right)),$$
$${b}_{2k}\sim N(0,{\sigma}_{2}^{2}),$$
$${\u03f5}_{imk}\sim N(0,{\sigma}^{2}).$$
Here, the randomeffects term $${b}_{1m}$$ represents the first random effect at level $$m$$ of the first grouping variable. The randomeffects term $${b}_{10m}$$ corresponds to the first random effects term (1), for the intercept (0), at the $$m$$ th level ( $$m$$ ) of the first grouping variable. Likewise $${b}_{11m}$$ is the level $$m$$ for the first predictor (1) in the first randomeffects term (1).
Similarly, $${b}_{2k}$$ stands for the second random effectsterm at level $$k$$ of the second grouping variable.
$${\sigma}_{10}^{2}$$ is the variance of the randomeffects term for the intercept, $${\sigma}_{11}^{2}$$ is the variance of the random effects term for the predictor acceleration, and $${\sigma}_{10,11}$$ is the covariance of the randomeffects terms for the intercept and the predictor acceleration. $${\sigma}_{2}^{2}$$ is the variance of the second randomeffects term, and $${\sigma}^{2}$$ is the residual variance.
First, prepare the design matrices for fitting the linear mixedeffects model.
X = [ones(406,1) Acceleration Weight]; Z = {[ones(406,1) Acceleration],[Weight]}; Model_Year = nominal(Model_Year); Origin = nominal(Origin); G = {Model_Year,Origin};
Fit the model using the design matrices.
lme = fitlmematrix(X,MPG,Z,G,'FixedEffectPredictors',.... {'Intercept','Acceleration','Weight'},'RandomEffectPredictors',... {{'Intercept','Acceleration'},{'Weight'}},'RandomEffectGroups',{'Model_Year','Origin'});
Compute the estimates of covariance parameters for the random effects.
[psi,mse,stats] = covarianceParameters(lme)
psi=2×1 cell array
{2x2 double }
{[6.8209e08]}
mse = 9.0706
stats=3×1 cell array
{3x7 classreg.regr.lmeutils.titleddataset}
{1x7 classreg.regr.lmeutils.titleddataset}
{1x5 classreg.regr.lmeutils.titleddataset}
The residual variance mse
is 9.0755. psi
is a 2by1 cell array, and stats
is a 3by1 cell array. To see the contents, you must index into these cell arrays.
First, index into the first cell of psi
.
psi{1}
ans = 2×2
8.5184 0.8270
0.8270 0.1074
The first cell of psi
contains the covariance parameters for the correlated random effects for intercept $${\sigma}_{10}^{2}$$ as 8.5160, and for acceleration $${\sigma}_{11}^{2}$$ as 0.1087. The estimate for the covariance of the randomeffects terms for the intercept and acceleration $${\sigma}_{10,11}$$ is 0.8387.
Now, index into the second cell of psi
.
psi{2}
ans = 6.8209e08
The second cell of psi
contains the estimate for the variance of the randomeffects term for weight $${\sigma}_{2}^{2}$$.
Index into the first cell of stats
.
stats{1}
ans = Covariance Type: FullCholesky Group Name1 Name2 Type Model_Year {'Intercept' } {'Intercept' } {'std' } Model_Year {'Acceleration'} {'Intercept' } {'corr'} Model_Year {'Acceleration'} {'Acceleration'} {'std' } Estimate Lower Upper 2.9186 1.1636 7.3206 0.86452 0.98113 0.28666 0.32777 0.18707 0.5743
This table shows the standard deviation estimates for the randomeffects terms for intercept and acceleration. Note that the standard deviations estimates are the square roots of the diagonal elements in the first cell of psi
. Specifically, 2.9182 = sqrt(8.5160) and 0.32968 = sqrt(0.1087). The correlation is a function of the covariance of intercept and acceleration, and the standard deviations of intercept and acceleration. The covariance of intercept and acceleration is the offdiagonal value in the first cell of psi, 0.8387. So, the correlation is .8387/(0.32968*2.92182) = 0.87.
The grouping variable for intercept and acceleration is Model_Year
.
Index into the second cell of stats
.
stats{2}
ans = Covariance Type: FullCholesky Group Name1 Name2 Type Estimate Origin {'Weight'} {'Weight'} {'std'} 0.00026117 Lower Upper 9.2451e05 0.00073778
The second cell of stats
has the standard deviation estimate and the 95% confidence limits for the standard deviation of the randomeffects term for Weight
. The grouping variable is Origin
.
Index into the third cell of stats
.
stats{3}
ans = Group Name Estimate Lower Upper Error {'Res Std'} 3.0117 2.8022 3.237
The third cell of stats
contains the estimate for residual standard deviation and the 95% confidence limits. The estimate for residual standard deviation is the square root of mse
, sqrt(9.0755) = 3.0126.
Construct 99% confidence intervals for the covariance parameters.
[~,~,stats] = covarianceParameters(lme,'Alpha',0.01);
stats{1}
ans = Covariance Type: FullCholesky Group Name1 Name2 Type Model_Year {'Intercept' } {'Intercept' } {'std' } Model_Year {'Acceleration'} {'Intercept' } {'corr'} Model_Year {'Acceleration'} {'Acceleration'} {'std' } Estimate Lower Upper 2.9186 0.87161 9.7732 0.86452 0.98999 0.024344 0.32777 0.15684 0.68497
stats{2}
ans = Covariance Type: FullCholesky Group Name1 Name2 Type Estimate Origin {'Weight'} {'Weight'} {'std'} 0.00026117 Lower Upper 6.6711e05 0.0010225
stats{3}
ans = Group Name Estimate Lower Upper Error {'Res Std'} 3.0117 2.7394 3.3111
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