Supervised Learning. HT2015: SC4 Statistical Data Mining and Machine Learning. Regression Example: Boston Housing. Regression Example: Boston Housing

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1 HT2015: SC4 Statistical Data Mining and Machine Learning Dino Sejdinovic Department of Statistics Oford Unsupervised learning: To etract structure and postulate hypotheses about data generating process from unlabelled observations 1,..., n. Visualize, summarize and compress data. Supervised learning: In addition to the observations of X, we have access to their response variables / labels Y Y: we observe {( i, y i } n i=1. Types of supervised learning: Classification: discrete responses, e.g. Y = {+1, 1} or {1,..., K}. Regression: a numerical value is observed and Y = R. The goal is to accurately predict the response Y on new observations of X, i.e., to learn a function f : R p Y, such that f (X will be close to the true response Y. Regression Eample: Boston Housing Regression Eample: Boston Housing The original data are 506 observations on 13 variables X; medv is the response variable Y. crim per capita crime rate by town zn proportion of residential land zoned for lots over 25,000 sq.ft indus proportion of non-retail business acres per town chas Charles River dummy variable (= 1 if tract bounds river; 0 otherwise no nitric oides concentration (parts per 10 million rm average number of rooms per dwelling age proportion of owner-occupied units built prior to 1940 dis weighted distances to five Boston employment centers rad inde of accessibility to radial highways ta full-value property-ta rate per USD 10,000 ptratio pupil-teacher ratio by town b 1000(B ^2 where B is the proportion of blacks by town lstat percentage of lower status of the population medv median value of owner-occupied homes in USD 1000 s > str(x data.frame : 506 obs. of 13 variables: $ crim : num $ zn : num $ indus : num $ chas : int $ no : num $ rm : num $ age : num $ dis : num $ rad : int $ ta : num $ ptratio: num $ black : num $ lstat : num > str(y num[1:506] Goal: predict median house price Y given 13 predictor variables X of a new district.

Supervised learning: In addition to the observations of X, we have access to their response variables / labels Y Y: we observe {( i, y i } n i=1.

2 Classification Eample: Lymphoma Loss function We have gene epression measurements X of n = 62 patients for p = 4026 genes. For each patient, Y {0, 1} denotes one of two subtypes of cancer. Goal: predict cancer subtype given gene epressions of a new patient. > str(x data.frame : 62 obs. of 4026 variables: $ Gene 1 : num $ Gene 2 : num $ Gene 3 : num $ Gene 4 : num $ Gene 5 : num $ Gene 6 : num $ Gene 7 : num $ Gene 8 : num $ Gene 9 : num $ Gene 10 : num $ Gene 11 : num $ Gene 12 : num > str(y num [1:62] Suppose we made a prediction Ŷ = f (X Y based on observation of X. How good is the prediction? We can use a loss function L : Y Y R + to formalize the quality of the prediction. Typical loss functions: Misclassification loss (or 0-1 loss for classification { 0 f (X = Y L(Y, f (X = 1 f (X Y. Squared loss for regression L(Y, f (X = (f (X Y 2. Many other choices are possible, e.g., weighted misclassification loss. In classification, if estimated probabilities ˆp(k for each class k Y are returned, log-likelihood loss (or log loss L(Y, ˆp = log ˆp(Y is often used. Risk The Bayes Classifier Risk paired observations {( i, y i } n i=1 viewed as i.i.d. realizations of a random variable (X, Y on X Y with joint distribution P XY For a given loss function L, the risk R of a learned function f is given by the epected loss R(f = E PXY [L(Y, f (X], where the epectation is with respect to the true (unknown joint distribution of (X, Y. The risk is unknown, but we can compute the empirical risk: R n (f = 1 n n L(y i, f ( i. i=1 What is the optimal classifier if the joint distribution (X, Y were known? The density g of X can be written as a miture of K components (corresponding to each of the classes: g( = π k g k (, where, for k = 1,..., K, P(Y = k = π k are the class probabilities, g k ( is the conditional density of X, given Y = k. The Bayes classifier f Bayes : {1,..., K} is the one with minimum risk: [ R(f =E [L(Y, f (X] = E X EY X [L(Y, f (X X] ] = E [L(Y, f (X X = ] g(d X The minimum risk attained by the Bayes classifier is called Bayes risk. Minimizing E[L(Y, f (X X = ] separately for each suffices.

657-1.328-1.330... $ Gene 3 : num -0.776-0.588 0.409-0.991-1.517... $ Gene 4 : num -0.474-1.588 0.219 0.978-1.604... $ Gene 5 : num -1.896-1.960-1.695-0.348-0.595... $ Gene 6 : num -2.075-2.117 0.

3 The Bayes Classifier The Bayes Classifier: Eample Consider the 0-1 loss. The risk simplifies to: [ E L(Y, f (X ] X = = L(k, f (P(Y = k X = =1 P(Y = f ( X = The risk is minimized by choosing the class with the greatest posterior probability: f Bayes ( = arg ma P(Y = k X = = arg ma π k g k ( K j=1 π = arg ma jg j ( π k g k (. The functions π k g k ( are called discriminant functions. The discriminant function with maimum value determines the predicted class of. A simple two Gaussians eample: Suppose X N (µ Y, 1, where µ 1 = 1 and µ 2 = 1 and assume equal priors π 1 = π 2 = 1/2. g 1 ( = 1 ( + 12 ep ( and g 2 ( = 1 ( 12 ep (. 2π 2 2π Optimal classification is f Bayes ( = arg ma marginal density { 1 if < 0, π k g k ( = 2 if 0. The Bayes Classifier: Eample Plug-in Classification How do you classify a new observation if now the standard deviation is still 1 for class 1 but 1/3 for class 2? e 32 1e 25 1e 18 1e 11 1e 04 Looking at density in a log-scale, optimal classification is to select class 2 if and only if [0.34, 2.16]. The Bayes Classifier chooses the class with the greatest posterior probability f Bayes ( = arg ma π k g k (. We know neither the g k nor the class probabilities π k! The plug-in classifier chooses the class f ( = arg ma ˆπ k ĝ k (, where we plugged in estimates ˆπ k of π k and k = 1,..., K and estimates ĝ k ( of, is an eample of plug-in classification.

= arg ma π k g k ( K j=1 π = arg ma jg j ( π k g k (. The functions π k g k ( are called discriminant functions. The discriminant function with maimum value determines the predicted class of.

4 LDA is the most well-known and simplest eample of plug-in classification. Assume multivariate normal conditional density g k ( for each class k: X Y = k N (µ k, Σ, g k ( =(2π p/2 Σ 1/2 ep ( 1 2 ( µ k Σ 1 ( µ k, each class can have a different mean µ k, all classes share the same covariance Σ. For an observation, the k-th log-discriminant function is log π k g k ( = c + log π k 1 2 ( µ k Σ 1 ( µ k The quantity ( µ k Σ 1 ( µ k is the squared Mahalanobis distance between and µ k. If Σ = I p and π k = 1 K, LDA simply chooses the class k with the nearest (in the Euclidean sense class mean. Epanding the term ( µ k Σ 1 ( µ k, log π k g k ( = c + log π k 1 2 ( µ k Σ 1 µ k 2µ k Σ 1 + Σ 1 = c + log π k 1 2 µ k Σ 1 µ k + µ k Σ 1 Setting a k = log(π k 1 2 µ k Σ 1 µ k and b k = Σ 1 µ k, we obtain log π k g k ( = c + a k + b k i.e. a linear discriminant function in. Consider choosing class k over k : a k + b k > a k + b k a + b > 0 where a = a k a k and b = b k b k. The Bayes classifier thus partitions X into regions with the same class predictions via separating hyperplanes. The Bayes classifier under these assumptions is more commonly known as the LDA classifier. Parameter Estimation How to estimate the parameters of the LDA model? We can achieve this by maimum likelihood (EM algorithm is not needed here since the class variables y i are observed!. Let n k = #{j : y j = k} be the number of observations in class k. l(π, (µ k K, Σ = log p ( ( i, y i n i=1 π, (µ k K, Σ = n i=1 log π yi g yi ( i =c + log π k 1 2 ( log Σ + ( j µ k Σ 1 ( j µ k ML estimates: ˆπ k = n k n ˆµ k = 1 n k j ˆΣ = 1 n ( j ˆµ k ( j ˆµ k Note: the ML estimate of Σ is biased. For an unbiased estimate we need to divide by n K. library(mass data(iris ##save class labels ct <- unclass(iris$species ##pairwise plot pairs(iris[,1:4],col=ct Sepal.Length Sepal.Width

5 Just focus on two predictor variables. iris.data <- iris[,3:4] plot(iris.data,col=ct,pch=20,ce=1.5,ce.lab=1.4 Computing and plotting the LDA boundaries. ##fit LDA iris.lda <- lda(=iris.data,grouping=ct ##create a grid for our plotting surface <- seq(0,8,0.02 y <- seq(0,3,0.02 m <- length( n <- length(y z <- as.matri(epand.grid(,y,0 ##classes are 1,2 and 3, so set contours at 1.5 and 2.5 iris.ldp <- predict(iris.lda,z$class contour(,y,matri(iris.ldp,m,n, levels=c(1.5,2.5, add=true, d=false, lty=2

data,grouping=ct ##create a grid for our plotting surface <- seq(0,8,0.02 y <- seq(0,3,0.

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