Introduction to regression

Transcription

1 Itroductio to regressio Regressio Bria Caffo, Jeff Leek ad Roger Peg Johs Hopkis Bloomberg School of Public Health

2 A famous motivatig example (Perhaps surprisigly, this example is still relevat) Predictig height: the Victoria approach beats moder geomics 2/15

3 Questios for this class Cosider tryig to aswer the followig kids of questios: To use the parets' heights to predict childres' heights. To try to fid a parsimoious, easily described mea relatioship betwee paret ad childre's heights. To ivestigate the variatio i childres' heights that appears urelated to parets' heights (residual variatio). To quatify what impact geotype iformatio has beyod paretal height i explaiig child height. To figure out how/whether ad what assumptios are eeded to geeralize fidigs beyod the data i questio. Why do childre of very tall parets ted to be tall, but a little shorter tha their parets ad why childre of very short parets ted to be short, but a little taller tha their parets? (This is a famous questio called 'Regressio to the mea'.) 3/15

4 Galto's Data Let's look at the data first, used by Fracis Galto i Galto was a statisticia who iveted the term ad cocepts of regressio ad correlatio, fouded the joural Biometrika, ad was the cousi of Charles Darwi. You may eed to ru istall.packages("usigr")if the UsigRlibrary is ot istalled. Let's look at the margial (parets disregardig childre ad childre disregardig parets) distributios first. Paret distributio is all heterosexual couples. Correctio for geder via multiplyig female heights by Overplottig is a issue from discretizatio. 4/15

5 Code library(usigr);data(galto) par(mfrow=c(1,2)) hist(galto$child,col="blue",breaks=100) hist(galto$paret,col="blue",breaks=100) 5/15

6 Fidig the middle via least squares Cosider oly the childre's heights. How could oe describe the "middle"? Oe defiitio, let Y i be the height of child i for i = 1,, = 928, the defie the middle as the value of μ that miimizes ( Y i μ) 2 This is physical ceter of mass of the histrogram. You might have guessed that the aswer μ = Xˉ. 6/15

7 Experimet Use R studio's maipulate to see what value of μ miimizes the sum of the squared deviatios. library(maipulate) myhist<-fuctio(mu){ hist(galto$child,col="blue",breaks=100) lies(c(mu,mu),c(0,150),col="red",lwd=5) mse<-mea((galto$child-mu)^2) text(63,150,paste("mu=",mu)) text(63,140,paste("mse=",roud(mse,2))) } maipulate(myhist(mu),mu=slider(62,74,step=0.5)) 7/15

8 The least squares estimate is the empirical mea hist(galto$child,col="blue",breaks=100) meachild<-mea(galto$child) lies(rep(meachild,100),seq(0,150,legth=100),col="red",lwd=5) 8/15

9 The math follows as: ( Y i μ) 2 = ( Y i Ȳ + Ȳ μ) 2 = ( Y i Ȳ) ( Y )( μ) + ( μ i Ȳ Ȳ Ȳ ) 2 = ( Y i Ȳ) 2 + 2( Ȳ μ) ( Y i Ȳ) + ( Ȳ μ) 2 = ( Y i Ȳ) 2 + 2( Ȳ μ)( Y i Ȳ) + ( Ȳ μ) 2 = ( Y i Ȳ) 2 + ( Ȳ μ) 2 ( Y i Ȳ) 2 9/15

10 Comparig childres' heights ad their parets' heights plot(galto$paret,galto$child,pch=19,col="blue") 10/15

11 Size of poit represets umber of poits at that (X, Y) combiatio (See the Rmd file for the code). 11/15

12 Regressio through the origi Suppose that X i are the parets' heights. Cosider pickig the slope β that miimizes ( Y i X i β) 2 This is exactly usig the origi as a pivot poit pickig the lie that miimizes the sum of the squared vertical distaces of the poits to the lie Use R studio's maipulate fuctio to experimet Subtract the meas so that the origi is the mea of the paret ad childre's heights 12/15

13 myplot<-fuctio(beta){ y<-galto$child-mea(galto$child) x<-galto$paret-mea(galto$paret) freqdata<-as.data.frame(table(x,y)) ames(freqdata)<-c("child","paret","freq") plot( as.umeric(as.vector(freqdata$paret)), as.umeric(as.vector(freqdata$child)), pch=21,col="black",bg="lightblue", cex=.15*freqdata$freq, xlab="paret", ylab="child" ) ablie(0,beta,lwd=3) poits(0,0,cex=2,pch=19) mse<-mea((y-beta*x)^2) title(paste("beta=",beta,"mse=",roud(mse,3))) } maipulate(myplot(beta),beta=slider(0.6,1.2,step=0.02)) 13/15

14 The solutio I the ext few lectures we'll talk about why this is the solutio lm(i(child-mea(child))~i(paret-mea(paret))-1,data=galto) Call: lm(formula=i(child-mea(child))~i(paret-mea(paret))- 1,data=galto) Coefficiets: I(paret-mea(paret)) /15

15 Visualizig the best fit lie Size of poits are frequecies at that X, Y combiatio 15/15