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 -1 < r < 1 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] > 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 = , df = 1746, p-value < 2.2e-16 alternative hypothesis: true correlation is not equal to 0 95 percent confidence interval: sample estimates: cor > pt( ,df=1746,lower.tail=FALSE) # T distribution on 1746 degrees of freedom [1] e-28

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

Example: simulated data R code > # simulate 30 data points > x <- rnorm(30) > e <- rnorm(30) > x <- 1:30 > e <- rnorm(30,0,5) > y < *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

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 SS1 Residual SSn-2 Total SSn-1

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

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)

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 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 

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$Biochem.Tot.Cholesterol > anova(ff) Analysis of Variance Table Response: bioch$Biochem.HDL Df Sum Sq Mean Sq F value Pr(>F) bioch$Biochem.Tot.Cholesterol Residuals > pf(1044,1,28,lower.tail=FALSE) [1] e-23 HDL = *Cholesterol

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

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=0.8 (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 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)

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 < 2.2e-16 *** Residuals Signif. codes: 0 '***' '**' 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 Residual SS Total SS

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) a <- anova(f)

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 = , df = 28, p-value = alternative hypothesis: true correlation is not equal to 0 95 percent confidence interval: sample estimates: cor > cor.test(xx,y,method="spearman") Spearman's rank correlation rho data: xx and y S = , p-value = alternative hypothesis: true rho is not equal to 0 sample estimates: rho

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)