Statistical modelling
Gil McVean, Department of StatisticsTuesday 24th Jan 2012
Questions to ask…
• How can we use models in statistical inference?
• Where do commonly-used models come from?
• What is a likelihood?
• How can we use likelihood to estimate parameters, measure certainty and test hypotheses?
• How do we add covariates to simple models?
• How do you construct and use linear models?
• What’s so general about generalised linear modelling?
2
3
What is a random variable?
• A random variable is a number associated with the outcome of a stochastic process– Waiting time for next bus– Average number hours sunshine in May– Age of current prime-minister
• In statistics, we want to take observations of random variables and use this to make statements about the underlying stochastic process
– Did this vaccine have any effect?– Which genes contribute to disease susceptibility?– Will it rain tomorrow?
• Parametric models provide much power in the analysis of variation (parameter estimation, hypothesis testing, model choice, prediction)
– Statistical models of the random variables– Models of the underlying stochastic process
4
What is a distribution?
• A distribution characterises the probability (mass) associated with each possible outcome of a stochastic process
• Distributions of discrete data characterised by probability mass functions
• Distributions of continuous data are characterised by probability density functions (pdf)
• For RVs that map to the integers or the real numbers, the cumulative density function (cdf) is a useful alternative representation
1)( x
xXP
1)(
dxxf
)( xXP
x
)(xf
x
0 1 2 3
5
Expectations and variances
• Suppose we took a large sample from a particular distribution, we might want to summarise something about what observations look like ‘on average’ and how much variability there is
• The expectation of a distribution is the average value of a random variable over a large number of samples
• The variance of a distribution is the average squared difference between randomly sampled observations and the expected value
x
dxxxfxXxPXE )(or )()(
x
dxxfxExxXPxExXVar )()(or )()()( 22
6
iid
• In most cases, we assume that the random variables we observe are independent and identically distributed
• The iid assumption allows us to make all sorts of statements both about what we expect to see and how much variation to expect
• Suppose X, Y and Z are iid random variables and a and b are constants
)(3)()()()( XEZEYEXEZYXE
)(Var3)(Var)(Var)(Var)(Var XZYXZYX
bXaEbaXE )()(
)(Var)(Var 2 XabaX
)(VarVar 11 XX ni
in
7
Where do ‘commonly-used’ distributions come from?
• At the core of much statistical theory and methodology lie a series of key distributions (e.g. Normal, Poisson, Exponential, etc.)
• These distributions are closely related to each other and can be ‘derived’ as the limit of simple stochastic processes when the random variable can be counted or measured
• In many settings, more complex distributions are constructed from these ‘simple’ distributions
– Ratios: E.g. Beta, Cauchy– Compound: E.g. Geometric, Beta– Mixture models
8
The simplest model
• Bernoulli trials– Outcomes that can take only two values: (0 and 1) with probabilities q and 1 - q
respectively. E.g. coin flipping, indicator functions
• The likelihood function calculates the probability of the data
• What is the probability of observing the sequence (if q = 0.5)– 01001101100111101001?– 11111111111000000000?
• Are they both equally probable?
knk
iixXPP )1()|()|( qqqqX
9
The binomial distribution
• Often, we don’t care about the exact order in which successes occurred. We might therefore want to ask about the probability of k successes in n trials. This is given by the binomial distribution
• For example, the probability of exactly 3 heads in 4 coins tosses = – P(HHHT)+P(HHTH)+P(HTHH)+P(THHH)– Each order has the same Bernoulli probability = (1/2)4
– There are 4 choose 3 = 4 orders
• Generally, if the probability of success is q, the probability of k successes in n trials
• The expected number of successes is nq and the variance is nq(1-q)
knk
kn
nkP
)1(),|( qqq n = 10
q = 0.2
10
The geometric distribution
• Bernoulli trials have a memory-less property– The probability of success (X = 1) next time is independent of the number of successes
in the preceding trials
• The number of trials between subsequent successes takes a geometric distribution– The probability that the first success occurs at the kth trial
• You can expect to wait an average of 1/q trials for a success, but the variance is
1)1()|( kkP qqq
2
1)(Varq
qk
q = 0.5 q = 0.05
20 1000
11
The Poisson distribution
• The Poisson distribution is often used to model ‘rare events’
• It can be derived in two ways– The limit of the Binomial distribution as q→0 and n→∞ (nq = m)– The number of events observed in a given time for a Poisson process (more later)
• It is parameterised by the expected number of events = m– The probability of k events is
• The expected number of events is m, and the variance is also m
• For large m, the Poisson is well approximated by the normal distribution
!)|(
kekP
kmmm
red = Poisson(5)blue = bin(100,0.05)
12
Going continuous
• In many situations while the outcome space of random variables may really be discrete (or at least measurably discrete), it is convenient to allow the random variables to be continuously distributed
• For example, the distribution of height in mm is actually discrete, but is well approximated by a continuous distribution (e.g. normal)
• Commonly-used continuous distributions arise as the limit of discrete processes
13
The Poisson process
• Consider a process when in every unit of time some event might occur
• E.g. every generation there is some chance of a gene mutating (with probability of approx 1 in 100,000 )
• The probability of exactly one change in a sufficiently small interval h ≡ 1/n is P = vh ≡ v/n, where P is the probability of one change and n is the number of trials.
• The probability of two or more changes in a sufficiently small interval h is essentially 0
• In the limit of the number of trials becoming large the total number of events (e.g. mutations) follows the Poisson distribution
Time
hh
14
The exponential distribution
• In the Poisson process, the time between successive events follows an exponential distribution
– This is the continuous analogue of the geometric distribution– It is memory-less. i.e. f(x + t | X > t) = f(x)
x
f(x) xexf )|( 2/1)(Var
/1)(
x
xE
15
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
0 0.5 1 1.5 2 2.5 3
The gamma distribution
• The gamma distribution arises naturally as the distribution of a series of iid random exponential variables
• The gamma distribution has expectation a/b and variance a/b2
• More generally, a need not be an integer (for example, the Chi-square distribution with one degree of freedom is a Gamma(½, ½) distribution)
) ,(Gamma~
)(~
21
nSXXXS
ExpX
n
xexxf baa
abba
1
)(),|(
a = b = 0.5
a = b = 2
a = b = 10
16
The beta distribution
• The beta distribution models random variables that take the value [0,1]
• It arises naturally as the proportional ratio of two gamma distributed random variables
• The expectation is a/(a + b)
• In Bayesian statistics, the beta distribution is the natural prior for binomial proportions (beta-binomial)
– The Dirichlet distribution generalises the beta to more than 2 proportions
0
1
2
3
4
5
6
7
8
9
10
0 0.2 0.4 0.6 0.8 1
a = b = 0.5
a = b = 2
a = b = 10),(Beta~
),(Gamma~),(Gamma~
21
2
1
aa
qaqa
YXX
YX
11 )1()()()(),|(
ba
bababa xxxf
17
The normal distribution
• The normal distribution is related to most distributions through the central limit theorem
• The normal distribution naturally describes variation of characters influenced by a large number of processes (height, weight) or the distribution of large numbers of events (e.g. limit of binomial with large np or Poisson with large m)
2
2)(21exp
21),;(
m
m xxf
0
0.005
0.01
0.015
0.02
0.025
0.03
0.035
0.04
0.045
50 100 150
blue = Poiss(100)red = N(100,10)
18
The exponential family of distributions
• Many of the distributions covered (e.g. normal, binomial, Poisson, gamma) belong to the exponential family of probability distributions
• a k-parameter member of the family has a density or frequency function of the form
• E.g. the Bernoulli distribution (x = 0 or 1) is
• Such distributions have the useful property that simple functions of the data, T(x), contain all the information about model parameter
– E.g. in Bernoulli case T(x) = x
k
iii xSdxTcxf
1
)()()()(exp);( qqq
)1ln(
1lnexp)1()( 1 q
qqqq xxXP xx
Estimating parameters from data
19
Parameter estimation
• We can formulate most questions in statistics in terms of making statements about underlying parameters
• We want to devise a framework for estimating those parameters and making statements about our certainty
• There are several different approaches to making such statements– Moment estimators– Likelihood– Bayesian estimation
20
Moment estimation
• Moment methods are the simplest way of estimating parameters
• In such techniques parameter values are found that match sample moments (mean, variance, etc.) to those expected
• E.g. for random variables X1, X2, etc. sampled from a N(m,2) distribution
212
1
212
1
1
)(
)(
m
nn
n
iin
n
iin
sE
XE
XXs
XX
21
2ˆ
ˆ
s
X
nn
m
21
Problems with moment estimation
• It is not always possible to exactly match sample moments with their expectation
• It is not clear when using moment methods how much of the information in the data about the parameters is being used
– Often not much..
• Why should MSE be the best way of measuring the value of an estimator?
22
Is there an optimal way to estimate parameters?
• For any model the maximum information about model parameters is obtained by considering the likelihood function
• The likelihood function is proportional to the probability of observing the data given a specified parameter value
• One natural choice for point estimation of parameters is the maximum likelihood estimate, the parameter values that maximise the probability of observing the data
• The maximum likelihood estimate (mle) has some useful properties (though is not always optimal in every sense )
23
An intuitive view on likelihood
1 ,0 2 m
1 ,2 2 m
4 ,0 2 m
24
An example
• Suppose we have data generated from a Poisson distribution. We want to estimate the parameter of the distribution
• The probability of observing a particular random variable is
• If we have observed a series of iid Poisson RVs we obtain the joint likelihood by multiplying the individual probabilities together
!);(
XeXP
Xmmm
Xnn
i
X
n
XXX
n
eL
eL
Xe
Xe
XeXXXP
i
n
mm
mm
mmmm
m
m
mmm
);(
);(
!!!);,,,(
2121
21
X
X
25
Comments
• Note in the likelihood function the factorials have disappeared. This is because they provide a constant that does not influence the relative likelihood of different values of the parameter
• It is usual to work with the log likelihood rather than the likelihood. Note that maximising the log likelihood is equivalent to maximising the likelihood
• We can find the mle of the parameter analytically
X
Xnndd
XnneL Xnn
mmm
mmmmm m
ˆ
log);();(
XX
Note that here the mle is the same as the moment estimator
Find where the derivative of the log likelihood is zero
Take the natural log of the likelihood function
26
Sufficient statistics
• In this example we could write the likelihood as a function of a simple summary of the data – the mean
• This is an example of a sufficient statistic. These are statistics that contain all information about the parameter(s) under the specified model
• For example, support we have a series of iid normal RVs
222
2
2
2/)(212
21
2log);,(
);,(
),;,,,(
2
2)(22
1
22
mmm
m
m
m
m
XXn
eL
eXXXP
n
n
i
Xn
i
iX
i
XX
Mean square Mean 27
Properties of the maximum likelihood estimate
• The maximum likelihood estimate can be found either analytically or by numerical maximisation
• The mle is consistent in that it converges to the truth as the sample size gets infinitely large
• The mle is asymptotically efficient in that it achieves the minimum possible variance (the Cramér-Rao Lower Bound) as n→∞
• However, the mle is often biased for finite sample sizes– For example, the mle for the variance parameter in a normal distribution is the sample
variance
28
Comparing parameter estimates
• Obtaining a point estimate of a parameter is just one problem in statistical inference
• We might also like to ask how good different parameter values are
• One way of comparing parameters is through relative likelihood
• For example, suppose we observe counts of 12, 22, 14 and 8 from a Poisson process
• The maximum likelihood estimate is 14. The relative likelihood is given by
Xnne
LL
mm
mm mm
ˆ);ˆ();( )ˆ(
XX
29
Using relative likelihood
• The relative likelihood and log likelihood surfaces are shown below
30
Interval estimation
• In most cases the chance that the point estimate you obtain for a parameter is actually the correct one is zero
• We can generalise the idea of point estimation to interval estimation
• Here, rather than estimating a single value of a parameter we estimate a region of parameter space
– We make the inference that the parameter of interest lies within the defined region
• The coverage of an interval estimator is the fraction of times the parameter actually lies within the interval
• The idea of interval estimation is intimately linked to the notion of confidence intervals
31
Example
• Suppose I’m interested in estimating the mean of a normal distribution with known variance of 1 from a sample of 10 observations
• I construct an interval estimator
• The chart below shows how the coverage properties of this estimator vary with a
aXaX ,q̂
If I choose a to be 0.62 I would have coverage of 95%
32
Confidence intervals
• It is a short step from here to the notion of confidence intervals
• We find an interval estimator of the parameter that, for any value of the parameter that might be possible, has the desired coverage properties
• We then apply this interval estimator to our observed data to get a confidence interval
• We can guarantee that among repeat performances of the same experiment the true value of the parameter would be in this interval 95% of the time
• We cannot say ”There is a 95% chance of the true parameter being in this interval”
33
Example – confidence intervals for normal distribution
• Creating confidence intervals for the mean of normal distributions is relatively easy because the coverage properties of interval estimators do not depend on the mean (for a fixed variance)
• For example, the interval estimator below has 95% coverage properties for any mean
• There is an intimate link between confidence intervals and hypothesis testing
na
aXaX
m
96.1
,ˆ
34
Confidence intervals and likelihood
• Thanks to the central limit theorem there is another useful result that allows us to define confidence intervals from the log-likelihood surface
• Specifically, the set of parameter values for which the log-likelihood is not more than 1.92 less than the maximum likelihood will define a 95% confidence interval
– In the limit of large sample size the LRT is approximately chi-squared distributed under the null
• This is a very useful result, but shouldn’t be assumed to hold – i.e. Check with simulation
35
Making simple models complex: Adding covariates
36
What is a covariate?
• A covariate is a quantity that may influence the outcome of interest– Genotype at a SNP– Age of mice when measurement was taken– Batch of chips from which gene expression was measured
• Previously, we have looked at using likelihood to estimate parameters of underlying distributions
• We want to generalise this idea to ask how covariates might influence the underlying parameters
• Much statistical modelling is concerned with considering linear effects of covariates on underlying parameters
37
What is a linear model?
• In a linear model, the expectation of the response variable is defined as a linear combination of explanatory variables
• Explanatory variables can include any function of the original data
• But the link between E(Y) and X (or some function of X) is ALWAYS linear and the error is ALWAYS Gaussian
bbbb ...3210 iiiii ZXZXYResponse variable
Intercept Linear relationships with explanatory variables
Interaction term
Gaussian error
bbbb ...322
10iZ
ii
Zii XeXY
38
What is a GLM?
• There are many settings where the error is non-Gaussian and/or the link between E(Y) and X is not necessarily linear
– Discrete data (e.g. counts in multinomial or Poisson experiments)– Categorical data (e.g. Disease status)– Highly-skewed data (e.g. Income, ratios)
• Generalised linear models keep the notion of linearity, but enable the use of non-Gaussian error models
• g is called the link function– In linear models, the link function is the identity
• The response variable can be drawn from any distribution of interest (the distribution function)
– In linear models this is Gaussian
...)( 32101
iiiiii ZXZXgYE bbbbm
39
Poisson regression
• In Poisson regression the expected value of the response variable is given by the exponent of the linear term
• The link function is the log
• Note that several distribution functions are possible (normal, Poisson, binomial counts), though in practice Poisson regression is typically used to model count data (particularly when counts are low)
...exp)( 3210 iiiiii ZXZXYE bbbbm
40
Example: Caesarean sections in public and private hospitals
41
Boxplots of rates of C sections
42
Fitting a model without covariates
43
> analysis<-glm(d$Caes ~ d$Births, family = "poisson")> summary(analysis)
Call:glm(formula = d$Caes ~ d$Births, family = "poisson")
Deviance Residuals: Min 1Q Median 3Q Max -2.81481 -0.73305 -0.08718 0.74444 2.19103
Coefficients: Estimate Std. Error z value Pr(>|z|) (Intercept) 2.132e+00 1.018e-01 20.949 < 2e-16 ***d$Births 4.406e-04 5.395e-05 8.165 3.21e-16 ***---Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
(Dispersion parameter for poisson family taken to be 1)
Null deviance: 99.990 on 19 degrees of freedomResidual deviance: 36.415 on 18 degrees of freedomAIC: 127.18
Number of Fisher Scoring iterations: 4
Implies an average of 13 per 1000 births
Fitting a model with covariates
• Unexpectedly, this indicates that public hospitals actually have a higher rate of Caesarean sections than private ones
44
glm(formula = d$Caes ~ d$Births + d$Hospital, family = "poisson")
Deviance Residuals: Min 1Q Median 3Q Max -2.3270 -0.6121 -0.0899 0.5398 1.6626
Coefficients: Estimate Std. Error z value Pr(>|z|) (Intercept) 1.351e+00 2.501e-01 5.402 6.58e-08 ***d$Births 3.261e-04 6.032e-05 5.406 6.45e-08 ***d$Hospital 1.045e+00 2.729e-01 3.830 0.000128 ***---Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Implies an average of 15.2 per 1000 births in public hospitals and 5.4 per 1000 births in private ones
Checking model fit
• Look a distribution of residuals and how well observed values are predicted
45
What’s going on?
• Initial summary suggested opposite result to GLM analysis. Why?
• Relationship between no. Births and no. C sections does not appear to be linear
• Accounting for this removes most of the apparent differences between hospital types
• There is also one quite influential outlier
46
Normal Caesarean
Private 412 16
Public 21121 285
Relative risk Private (compared to public) = 2.8
Simpson’s paradox
• Be careful about adding together observations – this can be misleading
• E.g. Berkeley sex-bias case
47
Applicants % admitted
Men 8442 44%
Women 4321 35%
Major Men WomenApplicants % admitted Applicants % admitted
A 825 62% 108 82%B 560 63% 25 68%C 325 37% 593 34%D 417 33% 375 35%E 191 28% 393 24%F 272 6% 341 7%
Appears that women have lower success
But actually women are typically more successful at the Departmental level, just apply to more competitive subjects
48
Finding MLEs in GLM
• In linear modelling we can use the beautiful compactness of linear algebra to find MLEs and estimates of the variance for parameters
• Consider an n by k+1 data matrix, X, where n is the number of observations and k is the number of explanatory variables, and a response vector Y
– the first column is ‘1’ for the intercept term
• The MLEs for the coefficients (b) can be estimated using
• In GLMs, there is usually no such compact analytical expression for the MLEs– Use numerical methods to maximise the likelihood
YXXXβ TT 1ˆ ))(,(~ˆ 21XXββ TN
49
Testing hypotheses in GLMs
• For the parameters we are interested in we typically want to ask how much evidence there is that these are different from zero
• For this we need to construct confidence intervals for the parameter estimates
• We could estimate the confidence interval by finding all parameter values with log-likelihood no greater than 1.92 units worse than the MLE (2-unit support interval)
• Alternatively, we might appeal to the central limit theorem (CLT) and use bootstrap techniques to estimate the variance of parameter estimates
• However, we can also appeal to theoretical considerations of likelihood (again based on the CLT) that show that parameter estimates are asymptotically normal with variance described by the Fisher information matrix
• Informally, the information matrix describes the sharpness of the likelihood curve around the MLE and the extent to which parameter estimates are correlated
50
Logistic regression
• When only two types of outcome are possible (e.g. disease/not-disease) we can model counts by the binomial
• If we want to perform inference about the factors that influence the probability of ‘success’ it is usual to use the logistic model
• The link function here is the logit
...exp1
...exp)(210
210
ii
iii ZX
ZXYEbbb
bbb
m
mm1
log)(g
51
Example: testing for genotype association
• In a cohort study, we observe the number of individuals in a population that get a particular disease
• We want to ask whether a particular genotype is associated with increased risk
• The simplest test is one in which we consider a single coefficient for the genotypic value
Genotype AA Aa aa
Genotypic value
0 1 2
Frequency in population
2 21 12
Probability of disease
p0 p1 p2
][ 1011
iGi ep bb
01
2
b0 = -4b1 = 2
52
A note on the model
• Note that each copy of the risk allele contribute in an additive way to the exponent
• This does not mean that each allele ‘adds’ a fixed amount to the probability of disease
• Rather, each allele contributes a fixed amount to the log-odds
• This has the effect of maintaining Hardy-Weinberg equilibrium within both the cases and controls
iGP
P10)diseaseNot (
)disease(logoddslog bb
53
Cont.
• Suppose in a given study we observe the following counts
• We can fit a GLM using the logit link function and binomial probabilities
• We have genotype data stored in the vector gt and disease status in the vector status
• Using R, this is specified by the command– glm(formula = status ~ gt, family = binomial)
Genotype 0 1 2
Counts with disease
26 39 21
Counts without disease
1298 567 49
54
Interpreting results
Call:glm(formula = status ~ gt, family = binomial)
Deviance Residuals: Min 1Q Median 3Q Max -0.8554 -0.4806 -0.2583 -0.2583 2.6141
Coefficients: Estimate Std. Error z value Pr(>|z|) (Intercept) -4.6667 0.2652 -17.598 <2e-16 ***gt 1.2833 0.1407 9.123 <2e-16 ***---Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
(Dispersion parameter for binomial family taken to be 1)
Null deviance: 967.36 on 1999 degrees of freedomResidual deviance: 886.28 on 1998 degrees of freedomAIC: 890.28
Number of Fisher Scoring iterations: 6
MLE for coefficients
Standard error for estimates
Estimate/std. error
P-value from normal distribution
Measure of contribution of individual observations to overall goodness of fit (for MLE model)
Measure of goodness of fit of null (compared to saturated model)
Measure of goodness of fit of fitted model
Penalised likelihood used in model choiceNumber of iterations used to find MLE
55
Adding in extra covariates
• Adding in additional explanatory variables in GLM is essentially the same as in linear model analysis
• Likewise, we can look at interactions
• In the disease study we might want to consider age as a potentially important covariate
glm(formula = status ~ gt + age, family = binomial)
Coefficients: Estimate Std. Error z value Pr(>|z|) (Intercept) -7.00510 0.79209 -8.844 <2e-16 ***gt 1.46729 0.17257 8.503 <2e-16 ***age 0.04157 0.01903 2.185 0.0289 *
56
Adding model complexity
• In the disease status analysis we might want to generalise the fitted model to one in which each genotype is allowed its own risk
][ 2111011
iGiG IIi ep bbb
glm(formula = status ~ g1 + g2 + age, family = binomial)
Coefficients: Estimate Std. Error z value Pr(>|z|) (Intercept) -5.46870 0.73155 -7.476 7.69e-14 ***g1TRUE 1.25694 0.26278 4.783 1.72e-06 ***g2TRUE 3.00816 0.33871 8.881 < 2e-16 ***age 0.04224 0.01915 2.206 0.0274 * ---Null deviance: 684.44 on 1999 degrees of freedomResidual deviance: 609.09 on 1996 degrees of freedomAIC: 617.09
57
Model choice
• It is worth remembering that we cannot simply identify the ‘significant’ parameters and put them in our chosen model
• Significance for a parameter tests the marginal null hypothesis that the coefficient associated with that parameter is zero
• If two explanatory variables are highly correlated then marginally neither may be significant, yet a linear model contains only one would be highly significant
• There are several ways to choose appropriate models from data. These typically involve adding in parameters one at a time and adding some penalty to avoid over-fitting.
58
Top Related