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

Synopsis Linear Regression Correlation Analysis of Variance Principle of Least Squares

Correlation

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

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

Examples of Correlation (taken from Wikipedia)

Calculation of r Data

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

Linear Regression Fit a straight line to data a intercept b slope ei error Normally distributed E(ei) = 0 Var(ei) = s2

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

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

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

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 SS 1 Residual SS n-2 Total SS n-1

ANOVA in R Component SS Degrees of freedom Mean Square F-ratio Fitting SS 20872.7 1 965 Residual SS 605.6 28 21.6 Total SS 21478.3 29 > 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

Hypothesis testing no relationship between y and x Assume errors ei are independent and normally distributed N(0,s2) If H0 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 H0 F >> 1 implies we reject H0 F is distributed as F(1,n-2)

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

The F distribution If e1….en are independent and identically distributed (iid) random variables with distribution N(0,s2), then: e12/s2 … en2/s2 are each iid chi-squared random variables with chi-squared distribution on 1 degree of freedom c12 The sum Sn = Si ei2/s2 is distributed as chi-squared cn2 If Tm is a similar sum distributed as chi-squared cm2, but independent of Sn, then (Sn/n)/(Tm/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 cn2

ANOVA – HDL example HDL = 0.2308 + 0.4456*Cholesterol > 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

correlation and ANOVA r2 = FSS/TSS = fraction of variance explained by the model r2 = 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

Effect of sample size on significance 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

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()

Problems with non-linearity All plots have r=0 Problems with non-linearity All plots have r=0.8 (taken from Wikipedia)

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

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)

One-Way ANOVA LS estimators average over group i

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

One-Way ANOVA Under Ho: no differences between groups F ~ F(p-1,n-p) Component SS Degrees of freedom Mean Square (ratio of SS to df) F-ratio (ratio of FMS/RMS) Fitting SS p-1 Residual SS n-p Total SS n-1 Under Ho: no differences between groups F ~ F(p-1,n-p)

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 > Component SS Degrees of freedom Mean Square (ratio of SS to df) F-ratio (ratio of FMS/RMS) Fitting SS 6.3870 173 0.0369 3.4375 Residual SS 18.5478 1727 0.0107 Total SS 24.9348 1900

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 are 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() BEWARE: If a numeric column contains non-numeric values (eg “N” being used instead of “NA” for a missing value, then the column is interpreted as a factor

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)

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

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)

Non-parametric Methods So far we have assumed the errors in the data are Normally distributed P-values and estimates can be inaccurate if this is not the case Non-parametric methods are a (partial) way round this problem Make fewer assumptions about the distribution of the data independent identically distributed

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”)

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 S = 2473.775, p-value = 0.01267 alternative hypothesis: true rho is not equal to 0 rho 0.4496607

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 )

Permutation Tests as non-parametric tests Example: One-way ANOVA: permute group identity between subjects count fraction of permutations in which the ANOVA p-value is smaller than the true p-value a = anova(lm( bioch$Biochem.HDL ~ bioch$Family)) p = a[1,5] pv = rep(0,1000) for( i in 1:1000) { perm = sample(bioch$Family,replace=FALSE) a = anova(lm( bioch$Biochem.HDL ~ perm )) pv[i] = a[1,5] } pval = mean(pv <p)