Random Coefficients Model Examples

Transcription

1 Random Coefficients Model Examples STAT:5201 Week 15 - Lecture 2 1 / 26

2 Each subject (or experimental unit) has multiple measurements (this could be over time, or it could be multiple measurements on a continuous variable x). Random effects are included in the model as to allow for a random intercept and a random slope, essentially allowing for a separate line for each subject. with b i = ( b0i b 1i Y ij = β 0 + β 1 x ij + b 0i + b 1i x ij + ɛ ) N ( [ 0 0, d11 d 12 d 21 d 22 b and ɛ independent from each other. Individual: E[Y ij b 0i, b 1i ] = (β 0 + b 0i ) + (β 1 + b 1i )x ij ]) and ɛ ij N(0, σ 2 ) Marginal: E[Y ij ] = β 0 + β 1 x ij 2 / 26

3 Adapted from John Fox example in Linear Mixed Models Appendix found in An R and S-PLUS Companion to Applied Regression. Eating disorders can be difficult to treat as many patients do not feel the need for treatment, when friends & family recognize the severity. Even after patients with eating disorders are hospitalized, they can continue behaviors that can be detrimental to their health. Here, data were collected recording the amount of exercise of 138 teenage girls hospitalized for eating disorders, and on a group of 93 control subjects. Variables subject: a factor with subject id codes. age: age in years. exercise: hours per week of exercise. group: factor indicating patient or control.. 3 / 26

4 We will consider the example in R and in SAS. > library(car) > data(blackmore) > head(blackmore) subject age exercise group patient patient patient patient patient patient > dim(blackmore) [1] > length(unique(blackmore$subject[blackmore$group=="patient"])) [1] 138 > length(unique(blackmore$subject[blackmore$group=="control"])) [1] 93 4 / 26

5 Fox transformed the response variable for numerous reasons (described in text) as log 2 (y + 5/60). > Blackmore$log.exercise <- log(blackmore$exercise + 5/60, 2) > attach(blackmore) Investigating the data with plots (in R). Use a random sample of 20 girls from each group for trend plotting. The groupeddata object from the nlme package is used to form the trellis plots. > library(nlme) > chosen.pat.ids=sample(unique(subject[group=="patient"]), 20) > chosen.pat.20=groupeddata(log.exercise ~ age subject, data=blackmore[is.element(subject,chosen.pat.ids),]) > chosen.con.ids=sample(unique(subject[group=="control"]), 20) > chosen.con.20=groupeddata(log.exercise ~ age subject, data=blackmore[is.element(subject,chosen.con.ids),]) 5 / 26

6 > print(plot(chosen.con.20, main="control Subjects", xlab="age",ylab="log2 Exercise", ylim=1.2*range(chosen.con.20$log.exercise, chosen.pat.20$log.exercise), layout=c(5,4),aspect=1), position=c(0, 0, 0.5, 1), more=t) > print(plot(chosen.pat.20, main="patients", xlab="age",ylab="log2 Exercise", ylim=1.2*range(chosen.con.20$log.exercise, chosen.pat.20$log.exercise), layout=c(5,4),aspect=1), position=c(0.5, 0, 1, 1)) 6 / 26

7 The groupeddata object is automatically plotted in order by average exercise. The subjects with the highest exercise values are in the top row, the subjects with the lowest exercise values are in the bottom row. Control Subjects Patients log2 Exercise log2 Exercise Age Age 7 / 26

8 Investigating the data with plots (in SAS). After I created subsetted data sets of patients from each group called control1 and patient1 (8 subjects per group), I used the PROC SGPANEL procedure to plot the individual trajectories. Here I ve asked for a linear regression line for each subject, but you can simply connect the observed points using the vline option instead of the reg option. proc sgpanel data=control1; title Control Subjects ; panelby subject/columns=4 rows=2; reg x=age y=log_exercise; rowaxis min=-4 max=4; colaxis values=(8, 10, 12, 14, 16); run; proc sgpanel data=patient1; <similar coding for the patient group as control group> 8 / 26

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11 You can also plot the overlay of these individual lines using PROC SGPLOT... proc sgplot data=control1; title Subset of Control Subjects ; reg x=age y=log_exercise/group=subject; run; 11 / 26

12 Investigating the subject-specific parameter estimates (in R). Fox formally fits a linear regression to each subject (231 separately fit models) in order to investigate the variability and correlation in the slopes and intercept estimates from a graphical perspective. The predictor age is transformed to represent age after the start of the study or age-8. He points out that the random coefficients model (fitted to all the data) fits a unified model that considers slopes and intercepts as random effects, and in that case, the estimated random effects or û are estimated using BLUPs (best linear unbiased predictors). 12 / 26

13 Investigating the subject-specific parameter estimates (in R). For a model with independent random subject effects (i.e. just a random intercept, as in the gene expression line example from earlier), the BLUPs are actually shrinkage estimators and fall between the individual observed values and the overall mean values. Formally, BLUPs are estimated as û = ĜZ ˆΣ 1 (y X ˆβ) where Σ = var(y) = ZGZ + R. 13 / 26

14 Before moving to a unified mixed model, we consider truly fitting a separate line to each subject (so, not a random coefficients model). Again, the nlme package is utilized when employing the lmlist function: > pat.list=lmlist(log.exercise ~ I(age - 8) subject, subset = group=="patient", data=blackmore) > con.list=lmlist(log.exercise ~ I(age - 8) subject, subset = group=="control", data=blackmore) > pat.coef=coef(pat.list) > con.coef=coef(con.list) > par(mfrow=c(1,2)) > boxplot(pat.coef[,1], con.coef[,1], main="intercepts", names=c("patients","controls")) > boxplot(pat.coef[,2], con.coef[,2], main="slope", names=c("patients","controls")) 14 / 26

15 Intercepts Slope Patients Controls Patients Controls The intercept represents the level of exercise at the start of the study. As expected, there is a great deal of variation in both the intercepts and the slopes. The median intercepts are fairly similar for patients and controls, but there is somewhat more variation among patients. The slopes are higher on average for patients than for controls and the slopes tend to be positive (suggesting their exercise increases over time). 15 / 26

16 It makes sense to also plot the relationship between the estimated intercept and slope parameters. The dataellipse function is in the car library. > plot(c(-5,4),c(-1.2,1.2),xlab="intercept",ylab="slope",type="n", main="(individual) Estimates of slope and intercept") > points(con.coef[,1],con.coef[,2],col=1) > points(pat.coef[,1],pat.coef[,2],col=2) > abline(v=0) > abline(h=0) > legend(-4.5,-.7,c("controls","patients"),col=c(1,2),pch=c(1,1)) > dataellipse(con.coef[,1],con.coef[,2],levels=c(.5,.95),add=true, plot.points=false,col=1) > dataellipse(pat.coef[,1],pat.coef[,2],levels=c(.5,.95),add=true, plot.points=false,col=2) 16 / 26

17 (Individual) Estimates of slope and intercept slope Controls Patients intercept Recall that we are on the log-scale base 2 for our response, so y = 0 coincides with 1 hour of exercise a week. It looks like the two groups have a reasonably similar correlation structure for the slope and intercept. It also looks like the patients have a shifted distribution such that they tend to have higher slopes. 17 / 26

18 Fitting the random coefficients model in SAS This model allows for a random slope and random intercept for each subject (which are allowed to be correlated). The population-level mean structure allows for separate lines for each treatment group (control and patient). The predictor age is transformed to represent age after the start of the study or age-8. data Blackmore; set Blackmore; age_trans = age-8; run; proc mixed data=blackmore; class subject group; model log_exercise = group age_trans group*age_trans/solution ddfm=satterth; random intercept age_trans/subject=subject type=un gcorr; run; 18 / 26

19 The Mixed Procedure Dimensions Covariance Parameters 4 Columns in X 6 Columns in Z Per Subject 2 Subjects 231 Max Obs Per Subject 5 Estimated G Correlation Matrix Row Effect subject Col1 Col2 1 Intercept age_trans We see that the correlation between b 0i and b 1i is estimated to be negative (ρ = ). 19 / 26

20 The Mixed Procedure Covariance Parameter Estimates Standard Z Cov Parm Subject Estimate Error Value Pr Z UN(1,1) subject <.0001 UN(2,1) subject UN(2,2) subject Residual <.0001 We see that the correlation between b 0i and b 1i is estimated to be negative (ρ = ) and marginally significant with a p= / 26

21 Solution for Fixed Effects Standard Effect group Estimate Error DF t Value Pr > t Intercept <.0001 group control group patient age_trans <.0001 age_trans*group control <.0001 age_trans*group patient Type 3 Tests of Fixed Effects Num Den Effect DF DF F Value Pr > F group age_trans <.0001 age_trans*group <.0001 The groups do not have significantly different intercepts (average exercise values at start of study, at age 8), but they do have significantly different slopes with the patient group having a higher slope than the control group. 21 / 26

22 I can capture the estimated BLUPs or û = ĜZ ˆΣ 1 (y X ˆβ) where Σ = var(y) = ZGZ + R using the ODS output and the solution option in the random statement: ods output SolutionR=blups; proc mixed data=blackmore covtest; class subject group; model log_exercise = group age_trans group*age_trans/ddfm=satterth; random intercept age_trans/subject=subject type=un gcorr solution; run; /* Solution for the random effects are BLUPs*/ ods output close; 22 / 26

23 proc print data=blups (obs=10); run; StdErr Obs Effect subject Estimate Pred DF tvalue Probt 1 Intercept age_trans Intercept age_trans Intercept age_trans Intercept age_trans Intercept age_trans / 26

24 Below I ve plotted the estimated BLUPs for the random slopes against the estimated slopes from the separately fit regression lines (in absolute values). Absolute values of slopes BLUP of slope separately fit slope (individual regression) 24 / 26

25 Fitting the random coefficients model (in R) Using the lme function in the nlme package, we see the same estimates for the covariance parameters as in SAS: > lme.1=lme(log.exercise~i(age-8)*group, random=~i(age-8) subject, data=blackmore) > summary(lme.1) Linear mixed-effects model fit by REML Data: Blackmore Random effects: Formula: ~I(age - 8) subject Structure: General positive-definite, Log-Cholesky parametrization StdDev Corr (Intercept) (Intr) I(age - 8) Residual / 26

26 Fitting the random coefficients model (in R) Square the estimates to match the SAS estimates: Var(Intercept)= = Var(slope)= = Corr(Intercept, slope)= = Var(Residual)= = / 26