27 Day 27 (July 12)

27.1 Announcements

  • Peer review assignment is posted
  • Please email me and request a 20 min time period during the final week of class (July 22 - 26) to give your final presentation. When you email me please suggest three times/dates that work for you.

27.2 Distributional assumptions

27.3 Constant variance

  • If \(\mathbf{y}\sim\text{N}(\mathbf{X\boldsymbol{\beta}},\sigma^{2}\mathbf{I})\), then each \(\varepsilon_i\) follows a normal distribution with the same value of \(\sigma^2\).

    • A linear model with constant variance is the simplest possible model we could assume. It is like the intercept only model in that regard.
    • Example: Hubble data
    library(gamair)
data(hubble)

m1 <- lm(y ~ x - 1, data = hubble)
confint(m1)
    ##      2.5 %   97.5 %
## x 68.37937 84.78297
    # Plot data and E(y)
plot(hubble$x,hubble$y,xlab="Distance (Mpc)",
ylab=expression("Velocity (km"*s^{-1}*")"),pch=20)
abline(m1)

    # Plot residuals vs. x
e.hat <- residuals(m1)
hist(e.hat,col="grey",breaks=15,main="",xlab=expression(hat(epsilon)))  

    plot(hubble$x,e.hat,xlab="Distance(Mpc)",ylab=expression(hat(epsilon)),pch=20)

    # Plot prediction intervals
y.pi <- predict(m1,newdata=data.frame(x = 0:25),interval = "predict")
plot(hubble$x,hubble$y,xlab="Distance (Mpc)",
ylab=expression("Velocity (km"*s^{-1}*")"),xlim=c(0,25),ylim=c(0,2500),pch=20)
points(0:25,y.pi[,2],typ="l",lwd=2,col="purple")
points(0:25,y.pi[,3],typ="l",lwd=2,col="purple")

    • A test for equal variance (see pg. 77 in Faraway (2014))
    • What causes non-constant variance (also called heteroskedasticity).

  • What can be done?

    • A transformation may work (see pg. 77 Faraway (2014))
    • Build a more appropriate model for the variance. For example, \[\mathbf{y}\sim\text{N}(\mathbf{X\boldsymbol{\beta}},\mathbf{V})\] where the \(i^\text{th}\) and \(j^{th}\) entry of the matrix \(\mathbf{V}\) equals \[v_{ij}(\theta)=\begin{cases} \sigma^{2}e^{2\theta x_i} &\text{if}\ i=j\\ 0 &\text{if}\ i\neq j \end{cases}\]
    library(nlme)
m2 <- gls(y~x-1,weights = varExp(form=~x),data=hubble)
summary(m2)
    ## Generalized least squares fit by REML
##   Model: y ~ x - 1 
##   Data: hubble 
##       AIC      BIC   logLik
##   322.432 325.8385 -158.216
## 
## Variance function:
##  Structure: Exponential of variance covariate
##  Formula: ~x 
##  Parameter estimates:
##     expon 
## 0.1060336 
## 
## Coefficients:
##     Value Std.Error  t-value p-value
## x 76.2548  3.854585 19.78288       0
## 
## Standardized residuals:
##        Min         Q1        Med         Q3        Max 
## -2.4272668 -0.4726915 -0.1548023  0.7929118  1.8437320 
## 
## Residual standard error: 55.17695 
## Degrees of freedom: 24 total; 23 residual
    confint(m2)
    ##      2.5 %   97.5 %
## x 68.69995 83.80965
    # Plot prediction intervals
X <- model.matrix(y~x-1,data=hubble)
sigma2.hat <- summary(m1)$sigma^2
sigma2.hat*solve(t(X)%*%X)
    ##          x
## x 15.71959
    vcov(m1)
    ##          x
## x 15.71959
    X <- model.matrix(y~x-1,data=hubble)
beta.hat <- coef(m2)
sigma2.hat <- summary(m2)$sigma^2
theta.hat <- summary(m2)$modelStruct$varStruct
V <- diag(c(sigma2.hat*exp(2*X%*%theta.hat)))
X.new <- model.matrix(~x-1,data=data.frame(x = 1:25))
y.pi <- cbind(X.new%*%beta.hat - qt(0.975,24-1)*sqrt(X.new^2%*%solve(t(X)%*%solve(V)%*%X) + sigma2.hat*exp(2*X.new%*%theta.hat)),
X.new%*%beta.hat + qt(0.975,24-1)*sqrt(X.new^2%*%solve(t(X)%*%solve(V)%*%X) + sigma2.hat*exp(2*X.new%*%theta.hat)))

plot(hubble$x,hubble$y,xlab="Distance (Mpc)",
 ylab=expression("Velocity (km"*s^{-1}*")"),xlim=c(0,25),ylim=c(0,2500),pch=20)
points(1:25,y.pi[,1],typ="l",lwd=2,col="purple")
points(1:25,y.pi[,2],typ="l",lwd=2,col="purple")

    • Example: Schizophrenia data
    url <- "http://www.stat.columbia.edu/~gelman/book/data/schiz.asc"
df1 <- data.frame(time = scan(url,skip=5),schizophrenia = c(rep("no",30*11),rep("yes",30*6)))
head(df1)
    ##   time schizophrenia
## 1  312            no
## 2  272            no
## 3  350            no
## 4  286            no
## 5  268            no
## 6  328            no
    boxplot(time~schizophrenia,data=df1,xlab = "Schizophrenic",ylab="Reaction time (in milliseconds)")

  • Example

    • Linear model with constant variance
    m1 <- lm(time ~ schizophrenia,data=df1)
summary(m1)
    ## 
## Call:
## lm(formula = time ~ schizophrenia, data = df1)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -280.87  -70.17  -16.52   37.66 1207.13 
## 
## Coefficients:
##                  Estimate Std. Error t value Pr(>|t|)    
## (Intercept)       310.170      9.057   34.25   <2e-16 ***
## schizophreniayes  196.697     15.245   12.90   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 164.5 on 508 degrees of freedom
## Multiple R-squared:  0.2468, Adjusted R-squared:  0.2453 
## F-statistic: 166.5 on 1 and 508 DF,  p-value: < 2.2e-16
    • Plot residuals vs. predictor to check assumptions
    e.hat <- residuals(m1)
x <- model.matrix(m1)[,2]
plot(x,e.hat,ylab=expression(hat(epsilon)))

    • Plot residuals for each category
    par(mfrow=c(2,1))
hist(e.hat[which(x==0)],col="grey",xlab=expression(hat(epsilon)),main=expression(hat(epsilon)*" for x = 0"),xlim=c(range(e.hat)),breaks=seq(min(e.hat),max(e.hat),length.out=40))
hist(e.hat[which(x==1)],col="grey",xlab=expression(hat(epsilon)),main=expression(hat(epsilon)*" for x = 1"),xlim=c(range(e.hat)),breaks=seq(min(e.hat),max(e.hat),length.out=40))

    • Looks like there are some outliers due to skewness in the distribution of \(\mathbf{y}\) and the variance is not constant. Let’s tackle the problem of non-constant variance first.
    • Some possible solutions
    • Build a more appropriate model for the variance. For example, \[\mathbf{y}\sim\text{N}(\mathbf{X\boldsymbol{\beta}},\mathbf{V})\] where the \(i^\text{th}\) and \(j^{th}\) entry of the matrix \(\mathbf{V}\) equals \[v_{ij}(\theta)=\begin{cases} \sigma^{2}e^{2\theta x_i} &\text{if}\ i=j\\ 0 &\text{if}\ i\neq j \end{cases}\]
    • For additional options see ?varClasses after loading the nlme package in R.
    • Fit model using ?gls function
    library(nlme)
x <- ifelse(df1$schizophrenia=="no",0,1)
m2 <- gls(time ~ schizophrenia, weights = varExp(form =~x), method = "REML",data=df1)
summary(m2)
    ## Generalized least squares fit by REML
##   Model: time ~ schizophrenia 
##   Data: df1 
##        AIC      BIC    logLik
##   6200.793 6217.715 -3096.396
## 
## Variance function:
##  Structure: Exponential of variance covariate
##  Formula: ~x 
##  Parameter estimates:
##    expon 
## 1.399026 
## 
## Coefficients:
##                     Value Std.Error  t-value p-value
## (Intercept)      310.1697  3.571554 86.84447       0
## schizophreniayes 196.6970 19.914372  9.87714       0
## 
##  Correlation: 
##                  (Intr)
## schizophreniayes -0.179
## 
## Standardized residuals:
##        Min         Q1        Med         Q3        Max 
## -1.6363885 -0.6461803 -0.1875710  0.4166345  7.2106463 
## 
## Residual standard error: 64.8805 
## Degrees of freedom: 510 total; 508 residual
    # 95% CI for theta/2
intervals(m2)$varStruct
    ##          lower     est.    upper
## expon 1.270298 1.399026 1.527755
## attr(,"label")
## [1] "Variance function:"
    # See ?varExp
theta.hat <- m2$model[1]$varStruct*2
sigma2.hat <- summary(m2)$sigma^2

sigma2.hat*exp(theta.hat*0) 
    ## Variance function structure of class varExp representing
##    expon 
## 4209.479
    sigma2.hat*exp(theta.hat*1) 
    ## Variance function structure of class varExp representing
##    expon 
## 69088.72
    # Check compared to sample variance
var(df1$time[which(df1$schizophrenia=="no")])
    ## [1] 4209.479
    var(df1$time[which(df1$schizophrenia=="yes")])
    ## [1] 69088.72
    # p-value from glm should be the same as t-test with unequal variances
t.test(time ~ schizophrenia,alternative = c("two.sided"),mu = 0, var.equal = FALSE,data = df1)
    ## 
##  Welch Two Sample t-test
## 
## data:  time by schizophrenia
## t = -9.8771, df = 190.98, p-value < 2.2e-16
## alternative hypothesis: true difference in means between group no and group yes is not equal to 0
## 95 percent confidence interval:
##  -235.9773 -157.4166
## sample estimates:
##  mean in group no mean in group yes 
##          310.1697          506.8667
    # Conclusiongs from var.test should match up with m2
var.test(time ~ schizophrenia,alternative = c("two.sided"),data = df1)
    ## 
##  F test to compare two variances
## 
## data:  time by schizophrenia
## F = 0.060929, num df = 329, denom df = 179, p-value < 2.2e-16
## alternative hypothesis: true ratio of variances is not equal to 1
## 95 percent confidence interval:
##  0.04683408 0.07847916
## sample estimates:
## ratio of variances 
##          0.0609286
    • Plot residuals vs. predictor to check assumption
    e.hat <- residuals(m2)
plot(x,e.hat,ylab=expression(hat(epsilon)))

    • Plot residuals for each category
    par(mfrow=c(2,1))
hist(e.hat[which(x==0)],col="grey",xlab=expression(hat(epsilon)),main=expression(hat(epsilon)*" for x = 0"),xlim=c(range(e.hat)),breaks=seq(min(e.hat),max(e.hat),length.out=40))
hist(e.hat[which(x==1)],col="grey",xlab=expression(hat(epsilon)),main=expression(hat(epsilon)*" for x = 1"),xlim=c(range(e.hat)),breaks=seq(min(e.hat),max(e.hat),length.out=40))

    • Normal distribution is symmetric, the support of reaction times is always positive. We probably won’t be able to alleviate the issue of skewness using the normal distribution.

Literature cited

Faraway, J. J. 2014. Linear Models with r. CRC Press.