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Linear Modelling II

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Linear Modelling II

Richard Mott

Wellcome Trust Centre for Human Genetics

- So far, we have learnt about
- Correlation
- Linear Regression
- One-Way Analysis of Variance
- Non-parametric alternatives

- In this lecture we will cover
- Relationship between Least Squares and Max Likelihood
- Fitting complex linear models
- Testing the fits of models
- Comparing Models
- Checking Goodness of Fit
- Prediction

- Likelihood = “probability of the data given the model”
- Basis of parametric statistical inference
- Different hypotheses can often be expressed in terms of different values of the parameters q
- Hypothesis testing equivalent to comparing the likelihoods for different q

- General Framework for constructing hypothesis tests:
S = Likelihood( data | H1) / Likelihood( data | H0)

Reject H0 if S > s(H0, H1)

Threshold s is chosen such that there is a probability a of making a false rejection under H0.

- is the size of the test (or false positive rate or Type I error) e.g. a=0.05
- is the power of the test, the probability of correctly rejecting H0 when H1 is true e.g. b=0.8
1- b is the type II error, the false negative rate

Generally, for fixed sample size n, if we fix a then we can’t fix b

If we fix a and b then we must let n vary.

The Neyman Pearson Lemma states that the likelihood ratio test is the most powerful of all tests of a given size

Type III error: "correctly rejecting the null hypothesis for the wrong reason".

- Often a hypothesis does not completely specify all the parameters in the model
- eg comparing mean of a sample to 0
- don’t specify the variance in the model
- H0 : data are distributed as N(0,s2)
- H1 : data are distributed as N(m,s2)
- assume Gaussian (Normal) errors
- Have to estimate the free parameters s, mto perform test

- Maximum Likelihood:
- first choose estimates for any free parameters to maximise the likelihood
- Use the maximised likelihood in the LR test
- ML estimates of parameters have good properties in large samples

- H0 is a submodel of H1
- parameters q
- are fixed qo under H0
- unknown q1 under H1, estimated by max likelihood

- parameters q
- max likelihood estimates under H1 are
- ie at

a quadratic form with rank k (= difference in number of free parameters)

Asymptotic distribution of 2 * log-likelihood ratio is

- Objects are classified into K classes
- Probability of class i is pi

- Ho: Probability of each class is the same = 1/K
- H1: Probability of classes are different
- Data: Ni observed in class i

Likelihood

log-Likelihood

log-Likelihood under Ho

maximised log-Likelihood under H1

2*log-Likelihood ratio

Compare to chi-square test, both distributed as

Density function of Normal distribution

Likelihood for linear regression model

Collect terms, converting to a sum

Take logarithms

Maximising the log Likelihood with respect to a, b is

identical to minimising the residual sum of squares (RSS)

Least Squares = Maximum Likelihood for Normal Data

MLE’s for a, b are the same as LS estimates.

The MLE for s2 = RSS/n (slightly biased estimate)

Residual Mean Square = RSS/(n-1)

- A linear model with multiple explanatory variables
- e.g. in the Biochemistry data set, HDL (y) may depend on Tot.Cholesterol (x) and GENDER (z)
yi = a + b1xi + b2zi + ei

- In general:
yi = b0 + b1x1i + b2 x2i + ….. + bpxpi + eib0 is the intercept (grand mean)

E(yi)= b0 + b1x1i + b2 x2i + ….. + bpxpilinear predictor

ei ~ N(0,s2) errors are iid Normal

- e.g. in the Biochemistry data set, HDL (y) may depend on Tot.Cholesterol (x) and GENDER (z)

In original formulation, data are grouped in a table

Rewrite the data as a single vector of numbers y1 …. yn

and define p indicator variables, one for each group k=1..p:

In new formulation, data are a vector but there are p+1 variables

Linear Models are best understood using matrix notation.

The Design Matrix, X captures all the information about the explanatory variables

n rows(subjects)

p+1 columns(variables)

Then, using matrix multiplication, the linear model is written as

The least squares estimators in the general linear model minimise

There is a “solution”: if the square matrix is invertible then

is the transpose of

If the matrix is not invertible then the model is over-parameterised,

ie some of the explanatory variables can be constructed from combinations

of the other variables. Then it is necessary to reduce the design matrix by

removing columns until is invertible.

This process is done automatically in R, but can result in some parameters

being thrown out. A factor with q levels will only require q-1 parameters.

One area of difficulty is that two apparently different models can produce

identical fits to the data, even though the parameter estimates may look

completely different

Mathematically, this can happen if the design matrix for one model can be

derived from the other by forming appropriate linear combinations of the columns

This is called re-parameterisation:

One example is the reduction in columns sometimes necessary to make invertible

Another example: fitting a factor with just two levels (eg sex) is equivalent to

coding sex numerically by the number of X chromosomes (ie 1 or 2).

The parameter estimates will appear different but the ANOVA and fitted values

will be identical.

Another example: fitting a factor with q levels usually requires q-1 parameters because

one level can be constructed from the others plus the overall mean.

However, one can omit the mean, in which case the number of parameters for the factor increases to q

If X is an R data frame with columns named “y”, “A”, “B” “C” etc,

then we can form model formulae

y ~ A

y ~ A + B

y ~ A + B + C

y ~ B + C

etc

If a column is numeric then adding a term for the column will mean

an additional parameter must be estimated

If the column is a factor with q levels then the column is implicitly expanded

as q-1 indicator variables, and q-1 additional parameters must be estimated

Warning: if an factor column is accidentally interpreted as numeric then

the model fit will almost certainly be wrong !

By default, the overall mean is included in an R model formula.

It can be included explicitly:

y ~ 1(just fit a model with the overall mean, the null model)

y ~ 1 + A(fit the overall mean plus A)

or excluded:

y ~ A -1

In the latter case:

if A is numeric, this corresponds to a linear regression

with intercept=0, ie the line of fit is forced to pass through the origin (0,0)

if A is a factor, this model is a reparameterisation of y ~ A. the fit is the same

but the parameter estimates look different.

Consider a model with two factors

A (Sex: male vs female) and

B (Treatment: treated vs untreated):

y ~ A + B

This means that the effects of sex and treatment act additively and independently

of each other, for example

being female might add +1 unit to the average response compared to males,

regardless of treatment,

being in the treatment group might add +2 units, regardless of sex.

Overall, two parameters are needed for the model

Now consider a model with two factors

A (Sex: male vs female) and

B (Treatment: treated vs untreated):

y ~ A * B

This means that the effects of sex and treatment interact with each other, for example

being female with treatment adds +2 units to the average response

being female with no treatment adds 1 unit

being male with treatment adds +4 units

being male with no treatment adds 0 units

Now three parameters are required to model the data

the additive model

y ~ A + B

f.additive <- lm(y ~ A + B )

is a submodel of (nested in) the interaction model

y ~ A * B

f.interaction <- lm( y ~ A * B )

This means that the additive model is a special case of the interaction model,

corresponding to setting certain parameters to be 0

R will let us compare nested models, to see if there is evidence for an interaction:

a <- anova(f.additive, f.interaction)

- A model M1 is nested within M2 (M1 is a submodel of M2) if
- M1 corresponds to forcing some of the parameters in M2 to 0
- [ More generally, the n x p1 design matrix X1 for M1 satisfies

- X1 = A X2
- where X2 is the n x p2 design matrix for M2 and A is a p2 x p1 matrix ]
- The partial F test is used to compare the models:
- Fit the models M1, M2with fitting SS FSS1 <= FSS2
- If M1 is true then the additional p2-p1 parameters in the model contribute
- nothing of substance, and the extra fitting SS = FSS2 – FSS1 is “stolen” from
- RSS1 and should be distributed as a residual sum of squares with p2-p1 df
- Therefore Fpartial = [(FSS2 – FSS1)/(p2-p1)]/[ RSS2/(n-p2) ] should be distributed
- like an F(p2-p1, n-p2 ) random variable.
- The partial F test is an equivalent of the likelihood-ratio test; it is exact if the
- errors are Normally distributed.

> df <- read.table(“int.txt”, h=TRUE)

> f.add.add <- lm( y.add ~ sex + treat ) # simulated to only have additive effects

> anova(f.add.add)

Analysis of Variance Table

Response: y.add

Df Sum Sq Mean Sq F value Pr(>F)

sex 1 27.043 27.043 9.5686 0.002586 **

treat 1 312.262 312.262 110.4870 < 2.2e-16 ***

Residuals 97 274.144 2.826

> f.add.int <- lm( y.add ~ sex * treat ) # fitting an interaction

> anova(f.add.int)

Analysis of Variance Table

Response: y.add

Df Sum Sq Mean Sq F value Pr(>F)

sex 1 27.043 27.043 9.5124 0.002666 **

treat 1 312.262 312.262 109.8378 < 2.2e-16 ***

sex:treat 1 1.223 1.223 0.4300 0.513538 # interaction not significant

Residuals 96 272.921 2.843

> anova(f.add.add, f.add.int)# comparison of models

Analysis of Variance Table

Model 1: y.add ~ sex + treat

Model 2: y.add ~ sex * treat

Res.Df RSS Df Sum of Sq F Pr(>F)

1 97 274.144

2 96 272.921 1 1.223 0.43 0.5135

>

>

> df <- read.table(“int.txt”, h=TRUE)

> f.int.int <- lm( y.int ~ sex * treat ) # simulated to have an interaction

> anova(f.int.int)

Analysis of Variance Table

Response: y.int

Df Sum Sq Mean Sq F value Pr(>F)

sex 1 169.34 169.34 46.783 7.376e-10 *** # main effect for sex

treat 1 54.74 54.74 15.121 0.0001857 *** # main effect for treatment

sex:treat 1 252.93 252.93 69.876 4.909e-13 *** # interaction for sex and treatment

Residuals 96 347.49 3.62

f.add.int <- lm( y.add ~ sex * treat ) # simulated to only have additive effects

> anova(f.add.int)

Analysis of Variance Table

Response: y.add

Df Sum Sq Mean Sq F value Pr(>F)

sex 1 27.043 27.043 9.5124 0.002666 **

treat 1 312.262 312.262 109.8378 < 2.2e-16 ***

sex:treat 1 1.223 1.223 0.4300 0.513538

Residuals 96 272.921 2.843

> anova(f.int.add,f.int.int) # comparison of additive and interaction

Analysis of Variance Table

Model 1: y.int ~ sex + treat

Model 2: y.int ~ sex * treat

Res.Df RSS Df Sum of Sq F Pr(>F)

1 97 600.42

2 96 347.49 1 252.93 69.876 4.909e-13 ***

- The two factor model is a simple example of an experimental design,
- the objective is to distribute the individuals across different factor combinations so that we can estimate effects efficiently at minimum cost
- In this example, equal numbers were assigned to each factor combination, a balanced orthogonal design
- Balance = equal numbers of replicates in each factor combination
- Orthogonal = estimates of a factor’s main effects are unaffected by adding other factors or interactions (in the example, the columns representing sex and treatment are uncorrelated)

- Experimental design is a vast subject –
- Optimal design depends on what is wanted
- Have to distinguish between factors we can control and random factors.

Consider the R models

y ~ A + X(1)

y ~ A * X(2)

where A is a factor (with p levels) and X is numeric.

What do they represent?

will fit a series of p parallel linear regression lines

will fit a series of p linear regression lines, with no

constraint on being parallel

Model (1) is nested inside (2) so they can be compared

with a partial F-test

anova(fit1,fit2) tests the hypothesis of parallel lines

- Suppose we fit the models (X1, X2 are numeric)
Y ~ X1(1)

Y ~ X2(2)

Y ~ X1 + X2(3)

- Often we find
- FSS(Y ~ X1 + X2) < FSS(Y ~ X1) + FSS(Y ~ X2)
- The discrepancy is because in general X1, X2 are correlated with each other: only when cor(X1, X2) = 0 (we say they are orthogonal does
- FSS(Y ~ X1 + X2) = FSS(Y ~ X1) + FSS(Y ~ X2)
- And in extreme case when |cor(X1, X2)| = 1
- FSS(Y ~ X1 + X2) = FSS(Y ~ X1) = FSS(Y ~ X2)
- This makes the interpretation of model comparisons difficult: in general there will be a part of the FSS which could be explained by either X1, X2 and it is impossible to determine which is correct.
- Multicolinearity becomes a serious problem in models with many terms
- Another way of thinking about this problem is that it is good to design an experiment such that the X’s are orthogonal, although this is not possible if the X’s are observed quantities.

In this example, negative residuals

are associated with small values of y

indicating poor model fit

- There is no equivalent to the partial F test when comparing non-nested models.
- The main issue is how to compensate for the fact that a model with more parameters will tend to fit the data better (in the sense of explaining more variance) than a model with fewer parameters
- But a model with many parameters tends to be a poorer predictor of new observations, because of over-fitting
- However, several criteria have been proposed for model comparison
- AIC (Akaike Information Criterion)
where L = the likelihood for the data, p= # parameters

For Linear Models,

where n is the number of observations and L is the maximized value of the likelihood function for the estimated model.

Among models with the same number of observations n, choose the model which minimises the simplified AIC

- BIC (Bayesian Information Criterion)

f <- lm( formula, data)

aic <- AIC(f)

bic <- AIC(f, k = log(nrow(data)))

- Example:
- We wish to find a set of SNPs that jointly explain variation in a quantitative trait
- There will be (usually) far more SNPs s than data points n
- There are a vast number 2s of possible models
- No model can contain more than n-1 parameters (model saturation)
- Forward Selection:
- Start with a small model and augment the model step by step so long as the improvement in fit satisfies a criterion (eg AIC or BIC, or partial F test)
- At each step, add the variable which maximises the improvement in fit

- Backwards Elimination:
- Start with a very large model and subtract terms from the model step by step
- At each step, delete the variable that minimises the decrease in fit

- Forward-Backward
- At each step, either add or delete a term depending on which optimises a criterion
- in R, stepAIC

- Model Averaging
- Rather than try to find a single model, integrate over many plausible models
- Bootstrapping
- Bayesian Model Averaging

- Rather than try to find a single model, integrate over many plausible models

- So far our models have had fixed effects
- each parameter can take any value independently of the others
- each parameter estimate uses up a degree of freedom
- models with large numbers of parameters have problems
- saturation - overfitting
- poor predictive properties
- numerically unstable and difficult to fit
- in some cases we can treat parameters as being sampled from a distribution – random effects
- only estimate the parameters needed to specify that distribution
- can result in a more robust model

- Testing genetic association across a large number of of families
- yi = trait value of i’th individual
- bi = genotype of individual at SNP of interest
- fi = family identifier (factor)
- y = m + b + f + e
- If we treat family effects f as fixed then we must estimate a large number of parameters
- Better to think of these effects as having a distribution N(0,t2) for some variance t
- genotype parameters b treated as fixed effects

- R will silently omit rows of data containing missing elements, and adjust the df accordingly
- R will only compare models using anova() if the models have been fitted to identical observations.
- Sporadic missing values in the explanatory variables can cause problems, because models may have a different numbers of complete cases
- Solution is to use the R function complete.cases() to identify those rows with complete data in the most general model, and specify these rows for modelling:
cc <- complete.cases(data.frame)

f <-lm( formula, data, subset=cc)

- Frequently data to be analysed are in several data frames, with a column of unique subject ids used to match the rows.
- Unfortunately, often the subjects differ slightly between data frames, eg frame1 might contain phenotype data and frame2 contain genotypes. Usually these two sets don’t agree perfectly because there will be subjects with genotypes but no phenotypes and vice versa
- Solution 1: Make a combined data frame containing the intersection of the subjects
intersect <- frame1$id[match(frame2$id, frame1$id, nomatch = 0)]

combined <- cbind( frame1[match(intersect,frame1$id),cols1], frame2[match(intersect,frame2$id),cols2])

- Solution 2: Make a combined data frame containing the union of the subjects
union <- unique(sort(c(as.character(frame1$id),as.character(frame2$id)))

combined <- cbind( frame1[match(union,frame1$id),cols1],frame2[match(union,frame2$id),cols2])

- cols1 is a list of the column names to include in frame1,
- cols2 is a list of the column names to include in frame2