STA216: Generalized Linear Models. Lecture 1. Review and Introduction

Transcription

1 STA216: Generalized Linear Models Lecture 1. Review and Introduction Let y 1,..., y n denote n independent observations on a response Treat y i as a realization of a random variable Y i In the general linear model, we assume that Y i N(µ i, σ 2 ), and we further assume that the expected value µ i is a linear function µ i = x iβ, where x i = (x i1,..., x ip ) is a p 1 vector of predictors (covariates) and β is a vector of unknown parameters (regression coefficients). 1

2 The generalized linear model, generalizes both the random & systematic components Likelihood Function: The Exponential Family We assume that observations come from a distribution in the exponential family with the following probability density function: f(y i ; θ i, φ) = exp { y i θ i b(θ i ) a(φ) + c(y i, φ) }. (1) Here, θ i, φ are parameters and a( ), b( ) and c( ) are known functions. The θ i and φ are location and scale parameters, respectively. 2

3 For example, for the Normal distribution, we have f(y i ; θ i, φ) = 1 2πσ exp{ (y i µ) 2 /2σ 2 } = exp[(y i µ µ 2 /2)/σ 2 {y 2 i /σ 2 + log(2πσ 2 )}/2] so that θ i = µ, φ = σ 2, a i (φ) = φ, b(θ i ) = θ 2 i /2 and c(y i, φ) = 1 2 {y2 i /σ 2 + log(2πσ 2 )}. Thus, in this case θ i is the mean and φ is the variance Let l(θ i, φ; y i ) = logf(y i ; θ i, φ) denote the log-likelihood function We can derive the mean and variance for the general case using: E ( l ) ( 2 l ) ( l ) 2 = 0 and E + E = 0. θ i θi 2 θ i 3

4 Note that l(θ i ; y i ) = {y i θ i b(θ i )}/a(φ) + c(y i, φ) It follows that l θ i = {y i b (θ i )}/a(φ) and 2 l θ 2 i = b (θ i )/a(φ). Hence, from the previous equalities, we have 0 = E ( l ) = {E(yi ) b (θ i )}/a(φ), θ i which implies that E(y i ) = b (θ i ) 4

5 Similarly, we have 0 = E ( 2 l ) ( l ) 2 + E θi 2 θ i = b (θ i )/a(φ) + E[{y i b (θ i )} 2 /a(φ) 2 ] = b (θ i )a(φ) + E(y 2 i ) 2E(y i )b (θ i ) + b (θ i ) 2 = b (θ i )a(φ) + E(y 2 i ) E(y i ) 2 var(y i ) = b (θ i )a(φ) For most commonly used exponential family distributions, a(φ) = φ/w i, where φ is a dispersion parameter and w i is a weight (typically equal to one) Hence, the mean and variance will typically follow the form: µ i = b (θ i ) and σ 2 = b (θ i )φ. 5

6 Characteristics of common distributions in the exponential family Normal Poisson Binomial Gamma Notation N(µ i, σ 2 ) Pois(µ i ) Bin(n i, π i ) G(µ i, ν) Range of y i (, ) [0, ) [0, n i ] (0, ) Dispersion, φ σ 2 1 1/n i ν 1 Cumulant: b(θ i ) θ 2 i /2 exp(θ i ) log(1 + e θ i) log( θ i ) Mean function, µ(θ i ) θ i exp(θ i ) 1/(1 + e θ i) 1/θ i Canonical link: θ(µ i ) identify log logit reciprocal Variance function, V (µ i ) 1 µ µ(1 µ) µ 2 6

7 Systematic Component, Link Functions Instead of modeling the mean, µ i, as a linear function of predictors, x i, we introduce on one-to-one continuously differentiable transformation g( ) and focus on η i = g(µ i ), where g( ) will be called the link function and η i the linear predictor. We assume that the transformed mean follows a linear model, η i = x iβ. Since the link function is invertible and one-to-one, we have µ i = g 1 (η i ) = g 1 (x iβ). 7

8 Note that we are transforming the expected value, µ i, instead of the raw data, y i. For classical linear models, the mean is the linear predictor. In this case, the identity link is reasonable since both µ i and η i can take any value on the real line. This is not the case in general. 8

9 Link Functions for Poisson Data For example, if Y i Poi(µ i ) then µ i must be > 0. In this case, a linear model is not reasonable since for some values of x i µ i 0. By using the model, η i = log(µ i ) = x iβ, we are guaranteed to have µ i > 0 for all β R p and all values of x i. In general, a link function for count data should map the interval (0, ) R (i.e., from the + real numbers to the entire real line). The log link is a natural choice 9

10 Link Functions for Binomial Data For the binomial distribution, 0 < µ i < 1. Therefore, the link function should map from (0, 1) R Standard choices: 1. logit: η i = log{µ i /(1 µ i )}. 2. probit: η i = Φ 1 (µ i ), where Φ( ) is the N(0, 1) cdf. 3. complementary log-log: η i = log{ log(1 µ i )}. Each of these choices is important in applications & will be considered in detail later in the course 10

11 Canonical Links and Sufficient Statistics Each of the distributions we have considered has a special, canonical, link function for which there exists a sufficient statistic equal in dimension to β. Canonical links occur when θ i = η i, with θ i the canonical parameter As a homework exercise, please show for next class that the following distributions are in the exponential family and have the listed canonical links: Normal η i = µ i Poisson η i = logµ i binomial η i = log{µ i /(1 µ i )} gamma η i = µ 1 i For the canonical links, the sufficient statistic is X y, with components i x ij y i, for j = 1,..., p. 11

12 Although canonical links often nice properties, selection of the link function should be based on prior expectation and model fit Example: Logistic Regression Suppose y i Bin(1, p i ), for i = 1,..., n, are independent 0/1 indicator variables of an adverse response (e.g., preterm birth) and let x i denote a p 1 vector of predictors for individual i (e.g., dose of dde exposure, race, age, etc). The likelihood is as follows: f(y β) = n = n = exp [ n p y i i (1 p i ) 1 y i = n ( p i ) y i (1 p i ) 1 p i exp { y i log ( p ) ( i 1 )} log 1 p i 1 p i {y i θ i log(1 + e θ i )} ]. 12

13 Choosing the canonical link, θ i = log ( p i 1 p i the likelihood has the following form: ) = x i β, f(y β) = exp[ n {y i x iβ log(1 + e x iβ )}]. This is logistic regression, which is widely used in epidemiology and other applications for modeling of binary response data. In general, if f(y i ; θ i, φ) is in the exponential family and θ i = θ(η i ), η i = x iβ, then the model is called a generalized linear model (GLM) 13

14 Maximum Likelihood Estimation of GLMs Unlike for the general linear model, there is no closed form expression for the MLE of β in general for GLMs. However, all GLMs can be fit using the same algorithm, a form of iteratively re-weighted least squares: 1. Given an initial value for β, calculate the estimated linear predictor η i = x i β and use that to obtain the fitted values µ i = g 1 ( η i ). Calculate the adjusted dependent variable, z i = η i + (y i µ i ) ( dη ) i dµ, 0 i where the derivative is evaluated at µ i. 2. Calculate the iterative weights W 1 i = ( dη ) i dµ V 0 i. i where V i is the variance function evaluated at µ i. 3. Regress z i on x i with weight W i to give new estimates of β 14

15 Justification for the IWLS procedure Note that the log-likelihood can be expressed as l = n {y i θ i b(θ i )}/a(φ) + c(y i, φ). To maximize this log-likelihood we need l/ β j, l β j = n = n = n l i dθ i dµ i θ i dµ i dη i (y i µ i ) a(φ) (y i µ i ) W i a(φ) η i β j 1 V i dµ i dη i x ij, dη i dµ i x ij since µ i = b (θ i ) and b (θ i ) = V i implies dµ i /dθ i = V i. With constant dispersion (a(φ) = φ), the MLE equations for β j : n W i (y i µ i ) dη i dµ i x ij = 0. 15

16 Fisher s scoring method uses the gradient vector, l/ β = u, and minus the expected value of the Hessian matrix E ( 2 l ) = A. β r β s Given the current estimate b of β, choose the adjustment δb so Aδb = u. Excluding φ, the components of u are u r = n so we have A rs = E( u r / β s ) = E n [ (yi µ i ) β s W i (y i µ i ) dη i dµ i x ir, { dη } i dη i Wi x ir + Wi x ir (y i µ i ) ]. dµ i dµ i β s The expectation of the first term is 0 and the second term is n W i dη i dµ i x ir µ i β s = n W i dη i dµ i x ir dµ i dη i η i β s = n W i x ir x is. 16

17 The new estimate b = b + δb of β thus satisfies Ab = Ab + Aδb = Ab + u, where (Ab) r = s A rs b s = n W i x ir η i. Thus, the new estimate b satisfies (Ab ) r = n W i x ir {η i + (y i µ i )dη i /dµ i }. These equations have the form of linear weighted least squares equation with weight W i and dependent variable z i. 17

18 Next Class Topic: Frequentist inference for GLMs Have homework exercise completed and written up Complete the following exercise in S-PLUS: 1. Simulate x i N(0, 1) and y i N( 1 + 2x i, 0.5), for i = 1,..., Fit the linear regression model E(y i x i ) = β 1 + β 2 x i using both the lm and glm functions - use help(lm) and help(glm) in S-PLUS to get details on implementation. 3. Answer questions: (a) What are the estimates? (b) Is there any difference in the output? 18