1. New Ways of Looking at Binary Data Fitting in R Yoon G Kim, ygk1@humboldt.edu Colloquium Talk

2. 2 Can we “stabilize” this? Appetizer

3. 3 After taking LOG … > y1 <- rep(c(100,200),times=10) > y2 <- rep(c(10,20),times=10) > x <- c(1:20) > data <- cbind(x,y1,y2) > data[1:3,] x y1 y2 [1,] 1 100 10 [2,] 2 200 20 [3,] 3 100 10 > par(mfrow=c(1,2)) > plot(y1~x,type="l",ylim=c(0,250),col="blue",ylab="") > lines(y2~x,type="l",col="red") > plot(log(y1)~x,type="l",ylim=c(0,6),col="blue",ylab="") > lines(log(y2)~x,type="l",col="red") Log transformed

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5. 5 Outline Exploring options available when assumptions of classical linear models are untenable. In this talk: What can we do when observations are not continuous and the residuals are not normally distributed nor identically distributed ?

6. 6 Defined by three assumptions: (1) the response variable is continuous. (2) the residuals ( ε ) are normally distributed and... (3)... independently (3a) and identically distributed (3b). Today, we will consider a range of options available when assumptions (1) (2) and/or (3b) are not verified. Classical Linear Models

7. 7 Many situations exist: The response variable could be (1) a count (number of individuals in a population) (number of species in a community) (2) a proportion (proportion "cured" after treatment) (proportion of threatened species) (3) a categorical variable (breeding/non-breeding) (different phenotypes) (4) a strictly positive value (esp. time to success) (or time to failure) (... ) and so forth Non-continuous response variable

8. 8 These types of non-continuous variables also tend to deviate from the assumptions of Normality (assumption #2) and Homoscedasticity (assumption #3b) (1) A count variable often follows a Poisson distribution (where the variance increases linearly with the mean) (2) A proportion often follows a Binomial distribution (where the variance reaches a maximum for intermediate values and a minimum at either end: 0% or 100%) Added difficulties

9. 9 These types of non-continuous variables also tend to deviate from the assumptions of Normality (assumption #2) and Homoscedasticity (assumption #3b). (3) A categorical variable tends to follow a Binomial distribution (when the variable has only two levels) or a Multinomial distribution (when the variable has more than two levels) (4) Time to success/failure can follow an exponential distribution or an inverse Gaussian distribution (the latter having a variance increasing much more quickly than the mean). Added difficulties

10. 10 Many of these situations can be unified under a central framework. Since all these distributions (and a few more) belong to the exponential family of distributions. Fortunately Probability density function (if y is continuous) Probability mass function (if y is discrete) Canonical (location) parameter Dispersion parameter Canonical form mean variance

11. 11 The Normal distribution Probability density function Canonical form   Canonical (location) parameter Dispersion parameter

12. 12 The Poisson distribution Probability mass function Canonical form   = 1 = 1 Canonical (location) parameter Dispersion parameter

13. 13 The Binomial distribution Probability mass function Canonical form   = 1 = 1 Canonical (location) parameter Dispersion parameter

14. 14 Why is that remotely useful? 1) A single algorithm (maximum likelihood) will cope with all these situations. 2) Different types of Variance can be accommodated When Var is constant -> Normal (Gaussian) When Var increases linearly with the mean -> Poisson When Var has a humped back shape -> Binomial When Var increases as the square of the mean -> Gamma (means the coefficient of variation remains constant) When Var increases as the cube of the mean -> inverse Gaussian 3) Most types of data are thus effectively covered

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16. 16 Two ways to cope with non-independent observations When design is balanced ( "equal sample size" ) We can use factors to partition our observations in different "groups" and analyze them as an ANOVA or ANCOVA. … when factors are "crossed" or when they are “nested" When design is unbalanced ( "uneven sample size" ) Mixed effect models are then called for. Non-independent Observations

17. 17 How does it work? 1) You need to specify the family of distribution to use 2) You need to specify the link function linear predictor link function For each type of variable the "natural" link function to use is indicated by the canonical parameter Link NormalIdentity Poisson Log Binomial Logit GammaInverse Inv.Gaussian Inverse square

18. 18 Binary variable The response variable contains only 0’s and 1’s. The probability that a place is “occupied” is p, and we write The objective is to determine how Y influences p. The family to use is Binomial and the canonical link is logit. Example: The response is occupation of territories and the explanatory variable is the resource availability in each territory > occupy <- read.table("D:\\STAT999\\RBook\\occupation.txt",header=T) > dim(occupy) [1] 150 2 > occupy[1:3,] resources occupied 1 14.18154 0 2 18.68306 0 3 20.22156 0 > attach(occupy) Crawley, M.J. (2007) The R Book: 597-598

19. 19 Binary variable > table(occupied) occupied 0 1 58 92 > modell <- glm(occupied~resources, family=binomial) > > plot(resources, occupied, type="n") > rug(jitter(resources[occupied==0])) > rug(jitter(resources[occupied==1]),side=3) > xv <- 0:1000 > yv <- predict(modell, list(resources=xv),type="response") by default the link for a Binomial is logistic

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21. 21 cutr <- cut(resources,5) tapply(occupied,cutr,sum) (13.2,209] (209,405] (405,600] (600,796] (796,992] 0 10 25 26 31 table(cutr) cutr (13.2,209] (209,405] (405,600] (600,796] (796,992] 31 29 30 29 31 probs <- tapply(occupied,cutr,sum)/table(cutr) probs (13.2,209] (209,405] (405,600] (600,796] (796,992] 0.0000000 0.3448276 0.8333333 0.8965517 1.0000000 attr(,"class") [1] "table" probs <- as.vector(probs) resmeans <- tapply(resources,cutr,mean) resmeans <- as.vector(resmeans) points(resmeans,probs,pch=16,cex=2) se <- sqrt(probs*(1-probs)/table(cutr)) up <- probs + as.vector(se) down <- probs - as.vector(se) for(i in 1:5) { lines(c(resmeans[i],resmeans[i]),c(up[i],down[i]))}

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23. 23 > grid_x <- seq(10,990,by=0.5) > modell_p <- predict(modell,new=data.frame(resources=grid_x),type="response") > modelp <- glm(occupied~resources, family=binomial(link=probit)) > modelp_p <- predict(modelp,new=data.frame(resources=grid_x),type="response") > modelcl <- glm(occupied~resources, family=binomial(link=cloglog)) > modelcl_p <- predict(modelcl,new=data.frame(resources=grid_x),type="response") > modelca <- glm(occupied~resources, family=binomial(link=cauchit)) > modelca_p <- predict(modelca,new=data.frame(resources=grid_x),type="response") Various Link Functions

24. 24 To draw … > newdata <- data.frame(grid_x,modell_p,modelp_p,modelcl_p,modelca_p) > library(lattice) > print(xyplot(modell_p+modelp_p+modelcl_p+ modelca_p ~ grid_x, + data=newdata, type ="l", xlab="resources", + ylab="p",lwd=1.5, lty=c(1,2,3,4), col=c(1:4), + panel = function(x, y,...) { + panel.xyplot(x, y,...) + panel.text(occupy$resources,occupy$probs,"x", cex=1.5, type="p",...) + })) > legend("topleft", legend=c("logit","probit","cloglog","cauchit"),lty=c(1:4), col=c(1:4), lwd=1.5) > > par(new=F) > points(resmeans,probs,pch=16,cex=2) > for (i in 1:5){ + lines(c(resmeans[i],resmeans[i]),c(up[i],down[i]))}

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26. 26 Binary variable > summary(modell) Call: glm(formula = occupied ~ resources, family = binomial) Coefficients: Estimate Std. Error z value Pr(>|z|) (Intercept) -3.744592 0.669923 -5.590 2.28e-08 *** resources 0.009762 0.001568 6.227 4.77e-10 *** --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 (Dispersion parameter for binomial family taken to be 1) Null deviance: 200.170 on 149 degrees of freedom Residual deviance: 97.152 on 148 degrees of freedom AIC: 101.15 Number of Fisher Scoring iterations: 6 Only valid if the Response variable is indeed a binomial also called G-statistic

27. 27 Binary variable > (dp <- sum(residuals(modell, type="pearson")^2)/modell$df.res) [1] 0.8472199) Pearson's residuals This dispersion parameter (  ) must be calculated. Residual degrees of freedom Suggests that the Variance is 0.85 times the Mean. In statistical terms there is no overdispersion. In biological terms, it suggests that the counts are independent from each other and are not Aggregated (i.e. Clumped). Typically Overdispersed count data follow a Negative Binomial distribution, which is not part of the Exponential families of distribution. It won't be covered here, but it can be approximated as a quasi-binomial (family="quasibinomial"). If you need it in your future work, you can also try glm.nb (in MASS package)

28. 28 Binary variable > summary(modell, dispersion=dp) Call: glm(formula = occupied ~ resources, family = binomial) Coefficients: Estimate Std. Error z value Pr(>|z|) (Intercept) -3.744592 0.616628 -6.073 1.26e-09 *** resources 0.009762 0.001443 6.765 1.33e-11 *** --- (Dispersion parameter for binomial family taken to be 0.8472199) Null deviance: 200.170 on 149 degrees of freedom Residual deviance: 97.152 on 148 degrees of freedom AIC: 101.15 Number of Fisher Scoring iterations: 6 The summary table can be adjusted with the dispersion parameter These Values can now be taken at face value How good is the model? 1 – (Res. Dev. / Null Dev.) = 51.47 %

29. 29 > summary(modell) Coefficients: Estimate Std. Error z value Pr(>|z|) (Intercept) -3.744592 0.669923 -5.590 2.28e-08 *** resources 0.009762 0.001568 6.227 4.77e-10 *** (Dispersion parameter for binomial family taken to be 1) Null deviance: 200.170 on 149 degrees of freedom Residual deviance: 97.152 on 148 degrees of freedom AIC: 101.15 > summary(modelp) Coefficients: Estimate Std. Error z value Pr(>|z|) (Intercept) -2.1437759 0.3448511 -6.217 5.08e-10 *** resources 0.0055046 0.0007811 7.047 1.82e-12 *** (Dispersion parameter for binomial family taken to be 1) Null deviance: 200.170 on 149 degrees of freedom Residual deviance: 97.024 on 148 degrees of freedom AIC: 101.02

30. 30 > summary(modelcl) Coefficients: Estimate Std. Error z value Pr(>|z|) (Intercept) -2.5902574 0.4293153 -6.033 1.60e-09 *** resources 0.0053519 0.0008337 6.419 1.37e-10 *** (Dispersion parameter for binomial family taken to be 1) Null deviance: 200.17 on 149 degrees of freedom Residual deviance: 102.30 on 148 degrees of freedom AIC: 106.30 > summary(modelca) Coefficients: Estimate Std. Error z value Pr(>|z|) (Intercept) -5.540198 1.644250 -3.369 0.000753 *** resources 0.014612 0.004205 3.475 0.000510 *** (Dispersion parameter for binomial family taken to be 1) Null deviance: 200.17 on 149 degrees of freedom Residual deviance: 99.69 on 148 degrees of freedom AIC: 103.69

31. 31 Bootstrapping > modell <- glm(occupied~resources,family=binomial) > bcoef <- matrix(0,1000,2) > for (i in 1:1000){ + indices <-sample(1:150,replace=T) + x <- resources[indices] + y <- occupied[indices] + modell <- glm(y~x, family=binomial) + bcoef[i,] <- modell$coef } > par(mfrow=c(1,2)) > plot(density(bcoef[,2]),xlab="Coefficient of x",main="") > abline(v=quantile(bcoef[,2],c(0.025,0.975)),lty=2, col=4) > plot(density(bcoef[,1]),xlab="Intercept",main="") > abline(v=quantile(bcoef[,1],c(0.025,0.975)),lty=2, col=4)

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33. 33 Jackknifing > jcoef <- matrix(0,150,2) > for (i in 1:150) { + modelj<-glm(occupied[-i]~resources[-i], family=binomial) + jcoef[i,] <- modelj$coef + } > par(mfrow=c(1,2)) > plot(density(jcoef[,2]),xlab="Coefficient of x",main="") > abline(v=quantile(jcoef[,2],c(0.025,0.975)),lty=2, col=4) > plot(density(jcoef[,1]),xlab="Intercept",main="") > abline(v=quantile(jcoef[,1],c(0.025,0.975)),lty=2, col=4)

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35. 35 C.I.’s > library(boot) > reg.boot<-function(regdat, index){ + x <- resources[index] + y <- occupied[index] + modell <- glm(y~x, family=binomial) + coef(modell) } > reg.model<-boot(occupy,reg.boot,R=10000) > boot.ci(reg.model,index=2) BOOTSTRAP CONFIDENCE INTERVAL CALCULATIONS Based on 10000 bootstrap replicates Intervals : Level Normal Basic 95% ( 0.0059, 0.0128 ) ( 0.0051, 0.0120 ) Level Percentile BCa 95% ( 0.0075, 0.0144 ) ( 0.0070, 0.0132 ) Calculations and Intervals on Original Scale

36. 36 > jack.after.boot(reg.model,index=2)

37. 37 108 th observation? > occupy[105:110,] resources occupied 105 703.1783 1 106 710.1274 1 107 716.7298 1 108 717.1994 0 109 733.3538 1 110 736.3060 1 > plot(resources, occupied) > text(resources[108],occupied[108],"Here",cex = 1.5,col="blue",pos=3) OR > fat.arrow <- function(size.x=0.5,size.y=0.5,ar.col="red"){ + size.x <- size.x*(par("usr")[2]-par("usr")[1])*0.1 + size.y <- size.y*(par("usr")[4]-par("usr")[3])*0.1 + pos <- locator(1) + xc <- c(0,1,0.5,0.5,-0.5,-0.5,-1,0) + yc <- c(0,1,1,6,6,1,1,0) + polygon(pos$x+size.x*xc,pos$y+size.y*yc,col=ar.col) } > fat.arrow()

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39. Yoon G Kim, ygk1@humboldt.edu Thank You!