I am working with a multivariate model with a Gamma distribution and I would like to make use of the lme4 syntaxis deployed in glmmTMB, however, I have noticed something strange with my model. Apparently, the model easily converges when I use stats::glm, but seems to give an error in the glmmTMB framework. Here is a reproducible example:
d <- data.frame(gamlss.data::plasma) # Sample dataset
m4.1 <- glm(calories ~ fat*fiber, family = Gamma(link = "log"), data = d) # Dos parámetros con interacción
m4.2 <- glmmTMB(calories ~ fat*fiber, family = Gamma(link = "log"), data = d) # Dos parámetros con interacción
>Warning message:
In fitTMB(TMBStruc) :
Model convergence problem; non-positive-definite Hessian matrix. See vignette('troubleshooting')
I guess, the solution might lie on the control parameters, but after looking on the troubleshooting vignette, I am not sure on where to start.
One solution can be to scale variables (as long as they are numeric).
d <- data.frame(gamlss.data::plasma) # Sample dataset
m4.1 <- glm(calories ~ fat*fiber, family = Gamma(link = "log"), data = d)
m4.2 <- glmmTMB(calories ~ scale(fat)*scale(fiber), family = Gamma(link = "log"), data = d)
Here, the second model converges fine, whereas it didn't before.
However, note the difference in parameter estimates between the two models:
> summary(m4.1)
Call:
glm(formula = calories ~ fat * fiber, family = Gamma(link = "log"),
data = d)
Deviance Residuals:
Min 1Q Median 3Q Max
-0.42031 -0.07605 -0.00425 0.07011 0.60073
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 6.120e+00 5.115e-02 119.654 <2e-16 ***
fat 1.412e-02 6.693e-04 21.104 <2e-16 ***
fiber 5.108e-02 3.704e-03 13.789 <2e-16 ***
fat:fiber -4.092e-04 4.476e-05 -9.142 <2e-16 ***
(Dispersion parameter for Gamma family taken to be 0.0177092)
Null deviance: 40.6486 on 314 degrees of freedom
Residual deviance: 5.4494 on 311 degrees of freedom
AIC: 4307.2
Number of Fisher Scoring iterations: 4
______________________________________________________________
> summary(m4.2)
Family: Gamma ( log )
Formula: calories ~ scale(fat) * scale(fiber)
Data: d
AIC BIC logLik deviance df.resid
4307.2 4326.0 -2148.6 4297.2 310
Dispersion estimate for Gamma family (sigma^2): 0.0173
Conditional model:
Estimate Std. Error z value Pr(>|z|)
(Intercept) 7.458146 0.007736 964.0 <2e-16 ***
scale(fat) 0.300768 0.008122 37.0 <2e-16 ***
scale(fiber) 0.104224 0.007820 13.3 <2e-16 ***
scale(fat):scale(fiber) -0.073786 0.008187 -9.0 <2e-16 ***
This is because the estimates are based on scaled parameters, so they must be interpreted with caution, or 'un-scaled.' See:
Understanding `scale` in R for an understanding of what the scale() function does, and see: interpretation of scaled regression coefficients... for a more in-depth understanding of what this means in a model
As a last note, the fact that models converge doesn't mean that they are a good fit.
Related
I have fitted a mixed effects model considering both functions widely used in R, namely: the lme function from the nlme package and the lmer function from the lme4 package.
To readjust the model from lme to lme4, following the same reparametrization, I used the following information from this topic, being that is only possible to do this in lme4 in a hackable way.: Heterocesdastic model of mixed effects via lmer function
I apologize for hosting the data in a link, however, I couldn't find an internal R database that has variables that might match my problem.
Data: https://drive.google.com/file/d/1jKFhs4MGaVxh-OPErvLDfMNmQBouywoM/view?usp=sharing
The fitted models were:
library(nlme)
library(lme4)
ModLME = lme(Var1~I(Var2)+I(Var2^2),
random = ~1|Var3,
weights = varIdent(form=~1|Var4),
Dataone, method="REML")
ModLMER = lmer(Var1~I(Var2)+I(Var2^2)+(1|Var3)+(0+dummy(Var4,"1")|Var5),
Dataone, REML = TRUE,
control=lmerControl(check.nobs.vs.nlev="ignore",
check.nobs.vs.nRE="ignore"))
Which are equivalent, see:
all.equal(REMLcrit(ModLMER), c(-2*logLik(ModLME)))
[1] TRUE
all.equal(fixef(ModLME), fixef(ModLMER), tolerance=1e-7)
[1] TRUE
> summary(ModLME)
Linear mixed-effects model fit by REML
Data: Dataone
AIC BIC logLik
-209.1431 -193.6948 110.5715
Random effects:
Formula: ~1 | Var3
(Intercept) Residual
StdDev: 0.05789852 0.03636468
Variance function:
Structure: Different standard deviations per stratum
Formula: ~1 | Var4
Parameter estimates:
0 1
1.000000 5.641709
Fixed effects: Var1 ~ I(Var2) + I(Var2^2)
Value Std.Error DF t-value p-value
(Intercept) 0.9538547 0.01699642 97 56.12093 0
I(Var2) -0.5009804 0.09336479 97 -5.36584 0
I(Var2^2) -0.4280151 0.10038257 97 -4.26384 0
summary(ModLMER)
Linear mixed model fit by REML. t-tests use Satterthwaites method [lmerModLmerTest]
Formula: Var1 ~ I(Var2) + I(Var2^2) + (1 | Var3) + (0 + dummy(Var4, "1") |
Var5)
Data: Dataone
Control: lmerControl(check.nobs.vs.nlev = "ignore", check.nobs.vs.nRE = "ignore")
REML criterion at convergence: -221.1
Scaled residuals:
Min 1Q Median 3Q Max
-4.1151 -0.5891 0.0374 0.5229 2.1880
Random effects:
Groups Name Variance Std.Dev.
Var3 (Intercept) 6.466e-12 2.543e-06
Var5 dummy(Var4, "1") 4.077e-02 2.019e-01
Residual 4.675e-03 6.837e-02
Number of obs: 100, groups: Var3, 100; Var5, 100
Fixed effects:
Estimate Std. Error df t value Pr(>|t|)
(Intercept) 0.95385 0.01700 95.02863 56.121 < 2e-16 ***
I(Var2) -0.50098 0.09336 92.94048 -5.366 5.88e-07 ***
I(Var2^2) -0.42802 0.10038 91.64017 -4.264 4.88e-05 ***
However, when observing the residuals of these models, note that they are not similar. See that in the model adjusted by lmer, mysteriously appears a residue with the shape of a few points close to a straight line. So, how could you solve such a problem so that they are identical? I believe the problem is in the lme4 model.
aa=plot(ModLME, main="LME")
bb=plot(ModLMER, main="LMER")
gridExtra::grid.arrange(aa,bb,ncol=2)
I can tell you what's going on and what should in principle fix it, but at the moment the fix doesn't work ...
The residuals being plotted take all of the random effects into account, which in the case of the lmer fit includes the individual-level random effects (the (0+dummy(Var4,"1")|Var5) term), which leads to weird residuals for the Var4==1 group. To illustrate this:
plot(ModLMER, col = Dataone$Var4+1)
i.e., you can see that the weird residuals are exactly the ones in red == those for which Var4==1.
In theory we should be able to get the same residuals via:
res <- Dataone$Var1 - predict(ModLMER, re.form = ~(1|Var3))
i.e., ignore the group-specific observation-level random effect term. However, it looks like there is a bug at the moment ("contrasts can be applied only to factors with 2 or more levels").
An extremely hacky solution is to construct the random-effect predictions without the observation-level term yourself:
## fixed-effect predictions
p0 <- predict(ModLMER, re.form = NA)
## construct RE prediction, Var3 term only:
Z <- getME(ModLMER, "Z")
b <- drop(getME(ModLMER, "b"))
## zero out observation-level components
b[101:200] <- 0
## add RE predictions to fixed predictions
p1 <- drop(p0 + Z %*% b)
## plot fitted vs residual
plot(p1, Dataone$Var1 - p1)
For what it's worth, this also works:
library(glmmTMB)
ModGLMMTMB <- glmmTMB(Var1~I(Var2)+I(Var2^2)+(1|Var3),
dispformula = ~factor(Var4),
REML = TRUE,
data = Dataone)
I checked my linear regression model (WMAN = Species, WDNE = sea surface temp) and found auto-correlation so instead, I am trying generalized least squares with the following script;
library(nlme)
modelwa <- gls(WMAN ~WDNE, data=dat,
correlation = corAR1(form=~MONTH),
na.action=na.omit)
summary(modelwa)
I compared both models;
> library(MuMIn)
> model.sel(modelw,modelwa)
Model selection table
(Intrc) WDNE class na.action correlation df logLik AICc delta
modelwa 31.50 0.1874 gls na.omit crAR1(MONTH) 4 -610.461 1229.2 0.00
modelw 11.31 0.7974 lm na.excl 3 -658.741 1323.7 94.44
weight
modelwa 1
modelw 0
Abbreviations:
na.action: na.excl = ‘na.exclude’
correlation: crAR1(MONTH) = ‘corAR1(~MONTH)’
Models ranked by AICc(x)
I believe the results suggest I should use gls as the AIC is lower.
My problem is, I have been reporting F-value/R²/p-value, but the output from the gls does not have these?
I would be very grateful if someone could assist me in interpreting these results?
> summary(modelwa)
Generalized least squares fit by REML
Model: WMAN ~ WDNE
Data: mp2017.dat
AIC BIC logLik
1228.923 1240.661 -610.4614
Correlation Structure: ARMA(1,0)
Formula: ~MONTH
Parameter estimate(s):
Phi1
0.4809973
Coefficients:
Value Std.Error t-value p-value
(Intercept) 31.496911 8.052339 3.911524 0.0001
WDNE 0.187419 0.091495 2.048401 0.0424
Correlation:
(Intr)
WDNE -0.339
Standardized residuals:
Min Q1 Med Q3 Max
-2.023362 -1.606329 -1.210127 1.427247 3.567186
Residual standard error: 18.85341
Degrees of freedom: 141 total; 139 residual
>
I have now overcome the problem of auto-correlation so I can use lm()
Add lag1 of residual as an X variable to the original model. This can be done using the slide function in DataCombine package.
library(DataCombine)
econ_data <- data.frame(economics, resid_mod1=lmMod$residuals)
econ_data_1 <- slide(econ_data, Var="resid_mod1",
NewVar = "lag1", slideBy = -1)
econ_data_2 <- na.omit(econ_data_1)
lmMod2 <- lm(pce ~ pop + lag1, data=econ_data_2)
This script can be found here
I am trying to calculate the residual standard error of a linear regression model using the survey package. I am working with a complex design, and the sampling weight of the complex design is given by "weight" in the code below.
fitM1 <- lm(med~x1+x2,data=pop_sample,weight=weight)
fitM2 <- svyglm(med~x1+x2,data=pop_sample,design=design)
First, if I call "summary(fitM1)", I get the following:
Call: lm(formula=med~x1+x2,data=pop_sample,weights=weight)
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 0.001787 0.042194 0.042 0.966
x1 0.382709 0.061574 6.215 1.92e-09 ***
x2 0.958675 0.048483 19.773 < 2e-16 ***
Residual standard error: 9.231 on 272 degrees of freedom
Multiple R-squared: 0.8958, Adjusted R-squared: 0.8931
F-statistic: 334.1 on 7 and 272 DF, p-value: < 2.2e-16
Next, if I call "summary(fitM2)", I get the following:
summary(fitM2)
Call: svyglm(formula=med~x1+x2,data=pop_sample,design=design)
Survey design: svydesign(id=~id_cluster,strat=~id_stratum,weight=weight,data=pop_sample)
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 0.001787 0.043388 0.041 0.967878
x1 0.382709 0.074755 5.120 0.000334 ***
x2 0.958675 0.041803 22.933 1.23e-10 ***
When using "lm", I can extract the residual standard error by calling:
fitMvariance <- summary(fitM1)$sigma^2
However, I can't find an analogous function for "svyglm" anywhere in the survey package. The point estimates are the same when comparing the two approaches, but the standard errors of the coefficients (and, presumably, the residual standard error of the model) are different.
Survey Analysis
use the library survey in the r to perform survey analysis, it offers a wide range of functions to calculate the statistics like Percentage, Lower CI, Upper CI, population and RSE.
RSE
we can use thesvyby function in the survey package to get all the statistics including the Root squared error
library("survey")
Survey design: svydesign(id=~id_cluster,strat=~id_stratum,weight=weight,data=pop_sample)
svyby(~med, ~x1+x2, design, svytotal, deff=TRUE, verbose=TRUE,vartype=c("se","cv","cvpct","var"))
The cvpct will give the root squared error
Refer for further information svyby
Because svyglm is built on glm not lm, the variance estimate is called $dispersion rather than $sigma
> data(api)
> dstrat<-svydesign(id = ~1, strata = ~stype, weights = ~pw, data = apistrat,
+ fpc = ~fpc)
> model<-svyglm(api00~ell+meals+mobility, design=dstrat)
> summary(model)$dispersion
variance SE
[1,] 5172 492.28
This is the estimate of $\sigma^2$, which is the population residual variance. In this example we actually have the whole population, so we can compare
> popmodel<-lm(api00~ell+meals+mobility, data=apipop)
> summary(popmodel)$sigma
[1] 70.58365
> sqrt(5172)
[1] 71.91662
I'm an R noob, I hope you can help me:
I'm trying to analyse a dataset in R, but I'm not sure how to interpret the output of summary(glmer(...)) and the documentation isn't a big help:
> data_chosen_stim<-glmer(open_chosen_stim~closed_chosen_stim+day+(1|ID),family=binomial,data=chosenMovement)
> summary(data_chosen_stim)
Generalized linear mixed model fit by maximum likelihood (Laplace Approximation) ['glmerMod']
Family: binomial ( logit )
Formula: open_chosen_stim ~ closed_chosen_stim + day + (1 | ID)
Data: chosenMovement
AIC BIC logLik deviance df.resid
96.7 105.5 -44.4 88.7 62
Scaled residuals:
Min 1Q Median 3Q Max
-1.4062 -1.0749 0.7111 0.8787 1.0223
Random effects:
Groups Name Variance Std.Dev.
ID (Intercept) 0 0
Number of obs: 66, groups: ID, 35
Fixed effects:
Estimate Std. Error z value Pr(>|z|)
(Intercept) 0.4511 0.8715 0.518 0.605
closed_chosen_stim2 0.4783 0.5047 0.948 0.343
day -0.2476 0.5060 -0.489 0.625
Correlation of Fixed Effects:
(Intr) cls__2
clsd_chsn_2 -0.347
day -0.916 0.077
I understand the GLM behind it, but I can't see the weights of the independent variables and their error bounds.
update: weights.merMod already has a type argument ...
I think what you're looking for weights(object,type="working").
I believe these are the diagonal elements of W in your notation?
Here's a trivial example that matches up the results of glm and glmer (since the random effect is bogus and gets an estimated variance of zero, the fixed effects, weights, etc etc converges to the same value).
Note that the weights() accessor returns the prior weights by default (these are all equal to 1 for the example below).
Example (from ?glm):
d.AD <- data.frame(treatment=gl(3,3),
outcome=gl(3,1,9),
counts=c(18,17,15,20,10,20,25,13,12))
glm.D93 <- glm(counts ~ outcome + treatment, family = poisson(),
data=d.AD)
library(lme4)
d.AD$f <- 1 ## dummy grouping variable
glmer.D93 <- glmer(counts ~ outcome + treatment + (1|f),
family = poisson(),
data=d.AD,
control=glmerControl(check.nlev.gtr.1="ignore"))
Fixed effects and weights are the same:
all.equal(fixef(glmer.D93),coef(glm.D93)) ## TRUE
all.equal(unname(weights(glm.D93,type="working")),
weights(glmer.D93,type="working"),
tol=1e-7) ## TRUE
I am trying to understand the difference between two different fitting methods for a data set with a bounded response variable. The response variable is a fraction and therefore has a range of [0,1]. I have uncovered through my Google searching that there are a lot of different methods out there as this is a common operation. I am currently interested in the difference between the stock R GLM fit and the Beta regression offered in the betareg package. I am using the GasolineYield data set from the "betareg" package as my sample data set. Before I post the code and the results my two questions are the following:
Am I performing the Logistic Regression fit in R using the builtin R GLM correctly?
Why are the standard errors reported in the Beta regression so much smaller than the standard errors for the R logistic regression?
R Setup Code
library(betareg)
data("GasolineYield", package = "betareg")
Beta Regression code from the "betareg" package
gy = betareg(yield ~ batch + temp, data = GasolineYield)
summary(gy)
Beta Regression summary output
Call:
betareg(formula = yield ~ batch + temp, data = GasolineYield)
Standardized weighted residuals 2:
Min 1Q Median 3Q Max
-2.8750 -0.8149 0.1601 0.8384 2.0483
Coefficients (mean model with logit link):
Estimate Std. Error z value Pr(>|z|)
(Intercept) -6.1595710 0.1823247 -33.784 < 2e-16 ***
batch1 1.7277289 0.1012294 17.067 < 2e-16 ***
batch2 1.3225969 0.1179020 11.218 < 2e-16 ***
batch3 1.5723099 0.1161045 13.542 < 2e-16 ***
batch4 1.0597141 0.1023598 10.353 < 2e-16 ***
batch5 1.1337518 0.1035232 10.952 < 2e-16 ***
batch6 1.0401618 0.1060365 9.809 < 2e-16 ***
batch7 0.5436922 0.1091275 4.982 6.29e-07 ***
batch8 0.4959007 0.1089257 4.553 5.30e-06 ***
batch9 0.3857930 0.1185933 3.253 0.00114 **
temp 0.0109669 0.0004126 26.577 < 2e-16 ***
Phi coefficients (precision model with identity link):
Estimate Std. Error z value Pr(>|z|)
(phi) 440.3 110.0 4.002 6.29e-05 ***
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Type of estimator: ML (maximum likelihood)
Log-likelihood: 84.8 on 12 Df
Pseudo R-squared: 0.9617
Number of iterations: 51 (BFGS) + 3 (Fisher scoring)
R GLM Logistic Regression code from stock R
glmfit = glm(yield ~ batch + temp, data = GasolineYield, family = "binomial")
summary(glmfit)
R GLM Logistic Regression summary output
Call:
glm(formula = yield ~ batch + temp, family = "binomial", data = GasolineYield)
Deviance Residuals:
Min 1Q Median 3Q Max
-0.100459 -0.025272 0.004217 0.032879 0.082113
Coefficients:
Estimate Std. Error z value Pr(>|z|)
(Intercept) -6.130227 3.831798 -1.600 0.110
batch1 1.720311 2.127205 0.809 0.419
batch2 1.305746 2.481266 0.526 0.599
batch3 1.562343 2.440712 0.640 0.522
batch4 1.048928 2.152385 0.487 0.626
batch5 1.125075 2.176242 0.517 0.605
batch6 1.029601 2.229773 0.462 0.644
batch7 0.540401 2.294474 0.236 0.814
batch8 0.497355 2.288564 0.217 0.828
batch9 0.378315 2.494881 0.152 0.879
temp 0.010906 0.008676 1.257 0.209
(Dispersion parameter for binomial family taken to be 1)
Null deviance: 2.34184 on 31 degrees of freedom
Residual deviance: 0.07046 on 21 degrees of freedom
AIC: 36.631
Number of Fisher Scoring iterations: 5
The standard errors are different because the variance assumptions in the two models are different.
Logistic regression assumes the response has a binomial distribution, while beta regression assumes it has a beta distribution.
The variance functions of the two are different. For the binomial, if you specify the mean (and $n$ is a given) the variance is determined. For the beta there's another free parameter, so it isn't determined by the mean and would presumably be estimated from the data.
This suggests that if you fit a quasibinomial GLM (adding a variance parameter) you might get closer to the same standard errors, but they still won't be the same, since they would weight the observations differently.
What you should actually do:
if your proportions are originally counts divided by some total count, then a binomial GLM would be an appropriate model to consider. (You would need the total counts, though.)
if your proportions are continuous fractions (the proportion of milk that's cream for example), then beta regression is an appropriate model to consider.