1. Linear Modelling I Richard Mott Wellcome Trust Centre for Human Genetics

2. Synopsis Linear Regression Correlation Analysis of Variance Principle of Least Squares

3. Correlation

4. Correlation and linear regression Is there a relationship? How do we summarise it? Can we predict new obs? What about outliers?

5. Correlation Coefficient r -1 < r < 1 r=0 no relationship r=0.6 r=1 perfect positive linear r=-1 perfect negative linear

6. Examples of Correlation (taken from Wikipedia)

7. Calculation of r Data

8. Correlation in R > cor(bioch$Biochem.Tot.Cholesterol,bioch$Biochem.HDL,use="complete") [1] 0.2577617 > cor.test(bioch$Biochem.Tot.Cholesterol,bioch$Biochem.HDL,use="complete") Pearson's product-moment correlation data: bioch$Biochem.Tot.Cholesterol and bioch$Biochem.HDL t = 11.1473, df = 1746, p-value < 2.2e-16 alternative hypothesis: true correlation is not equal to 0 95 percent confidence interval: 0.2134566 0.3010088 sample estimates: cor 0.2577617 > pt(11.1473,df=1746,lower.tail=FALSE) # T distribution on 1746 degrees of freedom [1] 3.154319e-28

9. Linear Regression Fit a straight line to data a intercept b slope e i error – Normally distributed – E(e i ) = 0 – Var(e i ) =  2

10. Example: simulated data R code > # simulate 30 data points > x <- rnorm(30) > e <- rnorm(30) > x <- 1:30 > e <- rnorm(30,0,5) > y <- 1 + 3*x + e > # fit the linear model > f <- lm(y ~ x) > # plot the data and the predicted line > plot(x,y) > abline(reg=f) > print(f) Call: lm(formula = y ~ x) Coefficients: (Intercept) x -0.08634 3.04747

11. Least Squares Estimate a, b by least squares Minimise sum of squared residuals between y and the prediction a+bx Minimise

12. Why least squares? LS gives simple formulae for the estimates for a, b If the errors are Normally distributed then the LS estimates are “optimal” In large samples the estimates converge to the true values No other estimates have smaller expected errors LS = maximum likelihood Even if errors are not Normal, LS estimates are often useful

13. Analysis of Variance (ANOVA) LS estimates have an important property: they partition the sum of squares (SS) into fitted and error components total SS = fitting SS + residual SS only the LS estimates do this Component Degrees of freedom Mean Square (ratio of SS to df) F-ratio (ratio of FMS/RMS) Fitting SS1 Residual SSn-2 Total SSn-1

14. ANOVA in R ComponentSS Degrees of freedom Mean SquareF-ratio Fitting SS20872.71 965 Residual SS605.62821.6 Total SS21478.329 > anova(f) Analysis of Variance Table Response: y Df Sum Sq Mean Sq F value Pr(>F) x 1 20872.7 20872.7 965 < 2.2e-16 *** Residuals 28 605.6 21.6 > pf(965,1,28,lower.tail=FALSE) [1] 3.042279e-23

15. Hypothesis testing no relationship between y and x Assume errors e i are independent and normally distributed N(0,  2 ) If H 0 is true then the expected values of the sums of squares in the ANOVA are degrees freedom Expectation F ratio = (fitting MS)/(residual MS) ~ 1 under H 0 F >> 1 implies we reject H 0 F is distributed as F(1,n-2)

16. Degrees of Freedom Suppose are iid N(0,1) Then ie n independent variables What about ? These values are constrained to sum to 0: Therefore the sum is distributed as if it comprised one fewer observation, hence it has n-1 df (for example, its expectation is n-1) In particular, if p parameters are estimated from a data set, then the residuals have p constraints on them, so they behave like n-p independent variables

17. The F distribution If e 1 ….e n are independent and identically distributed (iid) random variables with distribution N(0,  2 ), then: e 1 2 /  2 … e n 2 /  2 are each iid chi-squared random variables with chi-squared distribution on 1 degree of freedom    The sum S n =  i e i 2 /  2 is distributed as chi-squared  n  If T m is a similar sum distributed as chi-squared  m , but independent of S n, then (S n /n)/(T m /m) is distributed as an F random variable F(n,m) Special cases: – F(1,m) is the same as the square of a T-distribution on m df – for large m, F(n,m) tends to  n 

18. ANOVA – HDL example > ff <- lm(bioch$Biochem.HDL ~ bioch$Biochem.Tot.Cholesterol) > ff Call: lm(formula = bioch$Biochem.HDL ~ bioch$Biochem.Tot.Cholesterol) Coefficients: (Intercept) bioch$Biochem.Tot.Cholesterol 0.2308 0.4456 > anova(ff) Analysis of Variance Table Response: bioch$Biochem.HDL Df Sum Sq Mean Sq F value Pr(>F) bioch$Biochem.Tot.Cholesterol 1 149.660 149.660 1044 Residuals 1849 265.057 0.143 > pf(1044,1,28,lower.tail=FALSE) [1] 1.040709e-23 HDL = 0.2308 + 0.4456*Cholesterol

19. correlation and ANOVA r 2 = FSS/TSS = fraction of variance explained by the model r 2 = F/(F+n-2) – correlation and ANOVA are equivalent – Test of r=0 is equivalent to test of b=0 – T statistic in R cor.test is the square root of the ANOVA F statistic – r does not tell anything about magnitudes of estimates of a, b – r is dimensionless

20. Effect of sample size Total Cholesterol vs HDL data Example R session to sample subsets of data and compute correlations seqq <- seq(10,300,5) corr <- matrix(0,nrow=length(seqq),ncol=2) colnames(corr) <- c( "sample size", "P-value") n <- 1 for(i in seqq) { res <- rep(0,100) for(j in 1:100) { s <- sample(idx,i) data <- bioch[s,] co <- cor.test(data$Biochem.Tot.Cholesterol, data$Biochem.HDL,na="pair") res[j] <- co$p.value } m <- exp(mean(log(res))) cat(i, m, "\n") corr[n,] <- c(i, m) n <- n+1 }

21. Calculating the right sample size n The R library “pwr” contains functions to compute the sample size for many problems, including correlation pwr.r.test() and ANOVA pwr.anova.test()

22. Problems with non-linearity All plots have r=0.8 (taken from Wikipedia)

23. Non-Parametric Correlation Spearman Rank Correlation Coefficient Replace observations by their ranks eg x= ( 5, 1, 4, 7 ) -> rank(x) = (3,1,2,4) Compute sum of squared differences between ranks in R: – cor( x, y, method=“spearman”) – cor.test(x,y,method=“spearman”)

24. Spearman Correlation > cor.test(xx,y, method=“pearson”) Pearson's product-moment correlation data: xx and y t = 0.0221, df = 28, p-value = 0.9825 alternative hypothesis: true correlation is not equal to 0 95 percent confidence interval: -0.3566213 0.3639062 sample estimates: cor 0.004185729 > cor.test(xx,y,method="spearman") Spearman's rank correlation rho data: xx and y S = 2473.775, p-value = 0.01267 alternative hypothesis: true rho is not equal to 0 sample estimates: rho 0.4496607

25. Multiple Correlation The R cor function can be used to compute pairwise correlations between many variables at once, producing a correlation matrix. This is useful for example, when comparing expression of genes across subjects. Gene coexpression networks are often based on the correlation matrix. in R mat <- cor(df, na=“pair”) – computes the correlation between every pair of columns in df, removing missing values in a pairwise manner – Output is a square matrix correlation coefficients

26. One-Way ANOVA Model y as a function of a categorical variable taking p values – eg subjects are classified into p families – want to estimate effect due to each family and test if these are different – want to estimate the fraction of variance explained by differences between families – ( an estimate of heritability)

27. One-Way ANOVA LS estimators average over group i

28. One-Way ANOVA Variance is partitioned in to fitting and residual SS total SS n-1 fitting SS between groups p-1 residual SS with groups n-p degrees of freedom

29. One-Way ANOVA ComponentSS Degrees of freedom Mean Square (ratio of SS to df) F-ratio (ratio of FMS/RMS) Fitting SSp-1 Residual SSn-p Total SSn-1 Under H o : no differences between groups F ~ F(p-1,n-p)

30. One-Way ANOVA in R fam <- lm( bioch$Biochem.HDL ~ bioch$Family ) > anova(fam) Analysis of Variance Table Response: bioch$Biochem.HDL Df Sum Sq Mean Sq F value Pr(>F) bioch$Family 173 6.3870 0.0369 3.4375 < 2.2e-16 *** Residuals 1727 18.5478 0.0107 --- Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 > ComponentSS Degrees of freedom Mean Square (ratio of SS to df) F-ratio (ratio of FMS/RMS) Fitting SS 6.3870 1730.03693.4375 Residual SS 18.547817270.0107 Total SS24.93481900

31. Non-Parametric One-Way ANOVA Kruskall-Wallis Test Useful if data are highly non-Normal – Replace data by ranks – Compute average rank within each group – Compare averages – kruskal.test( formula, data )

32. Factors in R Grouping variables in R are called factors When a data frame is read with read.table() – a column is treated as numeric if all non-missing entries are numbers – a column is boolean if all non-missing entries are T or F (or TRUE or FALSE) – a column is treated as a factor otherwise – the levels of the factor is the set of distinct values – A column can be forced to be treated as a factor using the function as.factor(), or as a numeric vector using as.numeric()

33. Linear Modelling in R The R function lm() fits linear models It has two principal arguments (and some optional ones) f <- lm( formula, data ) – formula is an R formula – data is the name of the data frame containing the data (can be omitted, if the variables in the formula include the data frame)

34. formulae in R Biochem.HDL ~ Biochem$Tot.Cholesterol – linear regression of HDL on Cholesterol – 1 df Biochem.HDL ~ Family – one-way analysis of variance of HDL on Family – 173 df The degrees of freedom are the number of independent parameters to be estimated

35. ANOVA in R f <- lm(Biochem.HDL ~ Tot.Cholesterol, data=biochem) [OR f <- lm(biochem$Biochem.HDL ~ biochem$Tot.Cholesterol)] a <- anova(f) f <- lm(Biochem.HDL ~ Family, data=biochem) a <- anova(f)