Why is not confint.default which is based on asymptotic normality doesn't work for lmer model ?
fit <- lmer(y~(1|operator)+(1|part),data=dat)
Linear mixed model fit by REML ['lmerMod']
Formula: y ~ (1 | operator) + (1 | part)
Data: dat
REML criterion at convergence: 409.3913
Random effects:
Groups Name Std.Dev.
part (Intercept) 3.2018
operator (Intercept) 0.1031
Residual 0.9398
Number of obs: 120, groups: part, 20; operator, 3
Fixed Effects:
(Intercept)
22.39
confint.default(fit)
Error in as.integer(x) :
cannot coerce type 'S4' to vector of type 'integer'
What is the error saying ? How can I get confidence interval based on asymptotic normality for lmer model ?
Don't use confint.default(), just use confint(). The methods to calculate confidence intervals are different for the different model types. You can see the different methods with methods(confint). The "correct" version of the function is called based on the class of the first object you pass to the function. Directly calling one of the methods usually isn't a good idea.
There are options for how to calculate the bounds for the lmer objects. Look at the help page for ?confint.merMod to see the options unique to that model type.
#MrFlick is correct, but it may be worth adding that confint.merMod() gives likelihood profile CIs by default; confint(.,method="Wald") will give the confidence intervals based on asymptotic normality:
‘"Wald"’: approximating the confidence intervals (of fixed-effect
parameters only; all variance-covariance parameters CIs will
be returned as ‘NA’) based on the estimated local curvature
of the likelihood surface;
(this is obvious from the help page, but is probably worth restating here).
Related
I am trying to run a multi-level model to account for the fact that votes for a country's presidential elections may be nested within groups (depending of voters' mother tongues, places of residence etc.). In order to do so, I use the glmer function of the lme4 package.
m1<-glmer(vote_DPP ~ 1 + (1 | county_city),
family = binomial(link="logit"), data = d3)
Here, my vote variable is binary, representing whether people vote for a given party (1) or not (0). Since I believe results may change depending on people's state of residence, I want to allow intercepts to vary across states. However, I see no variation of intercept when I run my code.
Generalized linear mixed model fit by maximum likelihood (Laplace Approximation) ['glmerMod']
Family: binomial ( logit )
Formula: vote_DPP ~ 1 + (1 | county_city)
Data: d3
AIC BIC logLik deviance df.resid
1746.7918 1757.2001 -871.3959 1742.7918 1343
Random effects:
Groups Name Std.Dev.
county_city (Intercept) 0.2559
Number of obs: 1345, groups: county_city, 17
Fixed Effects:
(Intercept)
0.5937
What puzzles me here is the complete absence of variance column. I have seen other forums on the web regarding problems with variance = 0, but I cannot seem to find anything about the complete disappearance of this column (which makes me think it's probably something very simple I missed). First time posting in here, and quite a beginner in R and Stats, so any help would be appreciated :)
If you're concerned about seeing if the variance is zero, that's equivalent to seeing if the standard deviation is zero (similarly for "is (std dev/variance) small, although in this case they will be on different scales"). Furthermore, if the std dev/variance are zero or nearly zero you should get a "singular fit" message as well.
#Roland's comment is correct that summary() will print both the standard deviation and the variance by default. You can ask for both (or either) in the output of print() as well by specifying the ranef.comp (random effect component) argument:
library(lme4)
gm1 <- glmer(incidence/size ~ period + (1|herd),
data = cbpp,
weight = size,
family = binomial)
print(gm1, ranef.comp = c("Std.Dev.", "Variance"))
## ...
## Random effects:
## Groups Name Std.Dev. Variance
## herd (Intercept) 0.6421 0.4123
## ...
(You can similarly modify which components are shown in the summary printout: for example if you only want to see the variance, you can specify print(summary(gm1), ranef.comp = c("Variance")).)
For a bit more context: the standard deviation and variance are essentially redundant information (the standard error of the estimates of the random effects are not shown because they can be unreliable estimates of uncertainty in this case). Which form is more useful depends on the application: standard deviations are easier to compare to the corresponding fixed effects, variances can sometimes be used to make conclusions about partitioning of variance across effects (although doing this is more complicated than in the classic linear, balanced ANOVA case).
I am doing model averaging with MuMIn and trying to interpret the results.
Everything works fine, but I am wondering about the names of my coefficients in the results:
Model-averaged coefficients:
(full average)
Estimate Std. Error Adjusted SE
cond((Int)) 0.9552775 0.0967964 0.0969705
cond(Distanzpunkt) -0.0001217 0.0001451 0.0001453
cond(area_km2) 0.0022712 0.0030379 0.0030422
cond(prop) 0.0487036 0.1058994 0.1060808
Does someone know, what "cond()" tells me and why it appears in the model output?
Within the models, the coefficients are named "Distanzpunkt", "area_km2" and "prop".
Were you fitting a zero inflation model with glmmTMB? If so, then cond() is referring to the terms in the conditional model, rather than the zero-inflation model.
I want to report the results of an one factorial lme from the nlme package. I want to know the overall effect of A on y. To do so I would compare the model with a Null model:
m1 <- lme(y~A,random=~1|B/C,data=data,weights=varIdent(form = ~1|A),method="ML")
m0 <- lme(y~1,random=~1|B/C,data=data,weights=varIdent(form = ~1|A),method="ML")
I am using maximum likelihood because I am comparing models with different main effects.
stats::anova(m0,m1) gives me a significant p value, meaning that there is a significant effect of A on y. However, in contrast to lmer models made with lme4, no Chi2 values are given. First: Is this approach valid? And second: What is the best way to report the result?
Thanks for your answers
An anova with lme should give you the same information as with lmer. Both use what's called a deviance test or likelihood ratio test. The L.ratio part in the table returned by anova is simply the difference in the loglikelihood of the two models multiplied by -2. A deviance test tests this value against a Chi2 distribution with the difference in model parameters (in your case 1) degrees of freedom. So the value reported under L.ratio for lme models is the same as the Chi2 value reported for lmer models (assuming the models are the same of course, and lmer rounds the value to a decimal).
The approach is valid and you could report the value under L.ratio along with the degrees of freedom and p-value, but I would add more information in your report such as the fixed and random coefficients of both models and other parameters that you've added (such as the difference in variance for levels of A specified under weights). If you're only interested in the fixed effect of A than a Wald test should also be appropriate though REML estimates are recommended in cases with a small number of groups (Snijders & Bosker, 2012). The test statistic is the t-value and associated p-value in the model summary output summary(m1). Chapter 6 in Snijders & Bosker (2012) gives a great explanation on tests for fixed and random parameters. Along with reporting examples.
I'm trying to understand the function lmer. I've found plenty of information about how to use the command, but not much about what it's actually doing (save for some cryptic comments here: http://www.bioconductor.org/help/course-materials/2008/PHSIntro/lme4Intro-handout-6.pdf). I'm playing with the following simple example:
library(data.table)
library(lme4)
options(digits=15)
n<-1000
m<-100
data<-data.table(id=sample(1:m,n,replace=T),key="id")
b<-rnorm(m)
data$y<-rand[data$id]+rnorm(n)*0.1
fitted<-lmer(b~(1|id),data=data,verbose=T)
fitted
I understand that lmer is fitting a model of the form Y_{ij} = beta + B_i + epsilon_{ij}, where epsilon_{ij} and B_i are independent normals with variances sigma^2 and tau^2 respectively. If theta = tau/sigma is fixed, I computed the estimate for beta with the correct mean and minimum variance to be
c = sum_{i,j} alpha_i y_{ij}
where
alpha_i = lambda/(1 + theta^2 n_i)
lambda = 1/[\sum_i n_i/(1+theta^2 n_i)]
n_i = number of observations from group i
I also computed the following unbiased estimate for sigma^2:
s^2 = \sum_{i,j} alpha_i (y_{ij} - c)^2 / (1 + theta^2 - lambda)
These estimates seem to agree with what lmer produces. However, I can't figure out how log likelihood is defined in this context. I calculated the probability density to be
pd(Y_{ij}=y_{ij}) = \prod_{i,j}[f_sigma(y_{ij}-ybar_i)]
* prod_i[f_{sqrt(sigma^2/n_i+tau^2)}(ybar_i-beta) sigma sqrt(2 pi/n_i)]
where
ybar_i = \sum_j y_{ij}/n_i (the mean of observations in group i)
f_sigma(x) = 1/(sqrt{2 pi}sigma) exp(-x^2/(2 sigma)) (normal density with sd sigma)
But log of the above is not what lmer produces. How is log likelihood computed in this case (and for bonus marks, why)?
Edit: Changed notation for consistency, striked out incorrect formula for standard deviation estimate.
The links in the comments contained the answer. Below I've put what the formulae simplify to in this simple example, since the results are somewhat intuitive.
lmer fits a model of the form , where and are independent normals with variances and respectively. The joint probability distribution of and is therefore
where
.
The likelihood is obtained by integrating this with respect to (which isn't observed) to give
where is the number of observations from group , and is the mean of observations from group . This is somewhat intuitive since the first term captures spread within each group, which should have variance , and the second captures the spread between groups. Note that is the variance of .
However, by default (REML=T) lmer maximises not the likelihood but the "REML criterion", obtained by additionally integrating this with respect to to give
where is given below.
Maximising likelihood (REML=F)
If is fixed, we can explicitly find the and which maximise likelihood. They turn out to be
Note has two terms for variation within and between groups, and is somewhere between the mean of and the mean of depending on the value of .
Substituting these into likelihood, we can express the log likelihood in terms of only:
lmer iterates to find the value of which minimises this. In the output, and are shown in the fields "deviance" and "logLik" (if REML=F) respectively.
Maximising restricted likelihood (REML=T)
Since the REML criterion doesn't depend on , we use the same estimate for as above. We estimate to maximise the REML criterion:
The restricted log likelihood is given by
In the output of lmer, and are shown in the fields "REMLdev" and "logLik" (if REML=T) respectively.
I am trying to take what I have read about multilevel modelling and merge it with what I know about glm in R. I am now using the height growth data from here.
I have done some coding shown below:
library(lme4)
library(ggplot2)
setwd("~/Documents/r_code/multilevel_modelling/")
rm(list=ls())
oxford.df <- read.fwf("oxboys/OXBOYS.DAT",widths=c(2,7,6,1))
names(oxford.df) <- c("stu_code","age_central","height","occasion_id")
oxford.df <- oxford.df[!is.na(oxford.df[,"age_central"]),]
oxford.df[,"stu_code"] <- factor(as.character(oxford.df[,"stu_code"]))
oxford.df[,"dummy"] <- 1
chart <- ggplot(data=oxford.df,aes(x=occasion_id,y=height))
chart <- chart + geom_point(aes(colour=stu_code))
# see if lm and glm give the same estimate
glm.01 <- lm(height~age_central+occasion_id,data=oxford.df)
glm.02 <- glm(height~age_central+occasion_id,data=oxford.df,family="gaussian")
summary(glm.02)
vcov(glm.02)
var(glm.02$residual)
(logLik(glm.01)*-2)-(logLik(glm.02)*-2)
1-pchisq(-2.273737e-13,1)
# lm and glm give the same estimation
# so glm.02 will be used from now on
# see if lmer without level2 variable give same result as glm.02
mlm.03 <- lmer(height~age_central+occasion_id+(1|dummy),data=oxford.df,REML=FALSE)
(logLik(glm.02)*-2)-(logLik(mlm.03)*-2)
# 1-pchisq(-3.408097e-07,1)
# glm.02 and mlm.03 give the same estimation, only if REML=FALSE
mlm.03 gives me the following output:
> mlm.03
Linear mixed model fit by maximum likelihood
Formula: height ~ age_central + occasion_id + (1 | dummy)
Data: oxford.df
AIC BIC logLik deviance REMLdev
1650 1667 -819.9 1640 1633
Random effects:
Groups Name Variance Std.Dev.
dummy (Intercept) 0.000 0.0000
Residual 64.712 8.0444
Number of obs: 234, groups: dummy, 1
Fixed effects:
Estimate Std. Error t value
(Intercept) 142.994 21.132 6.767
age_central 1.340 17.183 0.078
occasion_id 1.299 4.303 0.302
Correlation of Fixed Effects:
(Intr) ag_cnt
age_central 0.999
occasion_id -1.000 -0.999
You can see that there is a variance for the residual in the random effect section, which I have read from Applied Multilevel Analysis - A Practical Guide by Jos W.R. Twisk, that this represents the amount of "unexplained variance" from the model.
I wondered if I could arrive at the same residual variance from glm.02, so I tried the following:
> var(resid(glm.01))
[1] 64.98952
> sd(resid(glm.01))
[1] 8.061608
The results are slightly different from the mlm.03 output. Does this refer to the same "residual variance" stated in mlm.03?
Your glm.02 and glm.01 estimate a simple linear regression model using least squares. On the other hand, mlm.03 is a linear mixed model estimated through maximum likelihood.
I don't know your dataset, but it looks like you use the dummy variable to create a cluster structure at level-2 with zero variance.
So your question has basically two answers, but only the second answer is important in your case. The models glm.02 and mlm.03 do not contain the same residual variance estimate, because...
The models are usually different (mixed effects vs. classical regression). In your case, however, the dummy variable seems to supress the additional variance component in the mixed model. So for me the models seem to be equal.
The method used to estimate the residual variance is different. glm uses LS, lmer uses ML in your code. ML estimates for the residual variance are slightly biased (resulting in smaller variance estimates). This can be solved by using REML instead of ML to estimate variance components.
Using classic ML (instead of REML), however, is still necessary and correct for the likelihood-ratio test. Using REML the comparison of the two likelihoods would not be correct.
Cheers!