CSC 411 Lecture 7: Linear Classification
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1 CSC 411 Lecture 7: Linear Classification Roger Grosse, Amir-massoud Farahmand, and Juan Carrasquilla University of Toronto UofT CSC 411: 07-Linear Classification 1 / 23
2 Overview Classification: predicting a discrete-valued target Binary classification: predicting a binary-valued target Examples predict whether a patient has a disease, given the presence or absence of various symptoms classify s as spam or non-spam predict whether a financial transaction is fraudulent UofT CSC 411: 07-Linear Classification 2 / 23
3 Overview Binary linear classification classification: predict a discrete-valued target binary: predict a binary target t {0, 1} Training examples with t = 1 are called positive examples, and training examples with t = 0 are called negative examples. Sorry. linear: model is a linear function of x, followed by a threshold: z = w T x + b { 1 if z r y = 0 if z < r UofT CSC 411: 07-Linear Classification 3 / 23
4 Some simplifications Eliminating the threshold We can assume WLOG that the threshold r = 0: Eliminating the bias w T x + b r w T x + b r }{{} b 0. Add a dummy feature x 0 which always takes the value 1. The weight w 0 is equivalent to a bias. Simplified model z = w T x { 1 if z 0 y = 0 if z < 0 UofT CSC 411: 07-Linear Classification 4 / 23
5 Examples NOT x 0 x 1 t b > 0 b + w < 0 b = 1, w = 2 UofT CSC 411: 07-Linear Classification 5 / 23
6 Examples AND x 0 x 1 x 2 t b < 0 b + w 2 < 0 b + w 1 < 0 b + w 1 + w 2 > 0 b = 1.5, w 1 = 1, w 2 = 1 UofT CSC 411: 07-Linear Classification 6 / 23
7 The Geometric Picture Input Space, or Data Space Here we re visualizing the NOT example Training examples are points Hypotheses are half-spaces whose boundaries pass through the origin The boundary is the decision boundary In 2-D, it s a line, but think of it as a hyperplane If the training examples can be separated by a linear decision rule, they are linearly separable. UofT CSC 411: 07-Linear Classification 7 / 23
8 The Geometric Picture Weight Space w 0 > 0 w 0 + w 1 < 0 Hypotheses are points Training examples are half-spaces whose boundaries pass through the origin The region satisfying all the constraints is the feasible region; if this region is nonempty, the problem is feasible UofT CSC 411: 07-Linear Classification 8 / 23
9 The Geometric Picture The AND example requires three dimensions, including the dummy one. To visualize data space and weight space for a 3-D example, we can look at a 2-D slice: The visualizations are similar, except that the decision boundaries and the constraints need not pass through the origin. UofT CSC 411: 07-Linear Classification 9 / 23
10 The Geometric Picture Visualizations of the AND example Data Space Weight Space Slice for x 0 = 1 Slice for w 0 = 1 What happened to the fourth constraint? UofT CSC 411: 07-Linear Classification 10 / 23
11 The Geometric Picture Some datasets are not linearly separable, e.g. XOR Proof coming next lecture... UofT CSC 411: 07-Linear Classification 11 / 23
12 Overview Recall: binary linear classifiers. Targets t {0, 1} z = w T x + b { 1 if z 0 y = 0 if z < 0 What if we can t classify all the training examples correctly? Seemingly obvious loss function: 0-1 loss { 0 if y = t L 0 1 (y, t) = 1 if y t = 1 y t. UofT CSC 411: 07-Linear Classification 12 / 23
13 Attempt 1: 0-1 loss As always, the cost J is the average loss over training examples; for 0-1 loss, this is the error rate: N J = 1 N i=1 1 y (i) t (i) UofT CSC 411: 07-Linear Classification 13 / 23
14 Attempt 1: 0-1 loss Problem: how to optimize? Chain rule: L 0 1 w j = L 0 1 z z w j But L 0 1 / z is zero everywhere it s defined! L 0 1 / w j = 0 means that changing the weights by a very small amount probably has no effect on the loss. The gradient descent update is a no-op. UofT CSC 411: 07-Linear Classification 14 / 23
15 Attempt 2: Linear Regression Sometimes we can replace the loss function we care about with one which is easier to optimize. This is known as a surrogate loss function. We already know how to fit a linear regression model. Can we use this instead? y = w x + b L SE (y, t) = 1 (y t)2 2 Doesn t matter that the targets are actually binary. Threshold predictions at y = 1/2. UofT CSC 411: 07-Linear Classification 15 / 23
16 Attempt 2: Linear Regression The problem: The loss function hates when you make correct predictions with high confidence! If t = 1, it s more unhappy about y = 10 than y = 0. UofT CSC 411: 07-Linear Classification 16 / 23
17 Attempt 3: Logistic Activation Function There s obviously no reason to predict values outside [0, 1]. Let s squash y into this interval. The logistic function is a kind of sigmoidal, or S-shaped, function: σ(z) = e z A linear model with a logistic nonlinearity is known as log-linear: z = w x + b y = σ(z) L SE (y, t) = 1 2 (y t)2. Used in this way, σ is called an activation function, and z is called the logit. UofT CSC 411: 07-Linear Classification 17 / 23
18 Attempt 3: Logistic Activation Function The problem: (plot of L SE as a function of z) L = L z w j z w j w j w j α L w j In gradient descent, a small gradient (in magnitude) implies a small step. If the prediction is really wrong, shouldn t you take a large step? UofT CSC 411: 07-Linear Classification 18 / 23
19 Logistic Regression Because y [0, 1], we can interpret it as the estimated probability that t = 1. The pundits who were 99% confident Clinton would win were much more wrong than the ones who were only 90% confident. Cross-entropy loss captures this intuition: L CE (y, t) = { log y if t = 1 log(1 y) if t = 0 = t log y (1 t) log(1 y) UofT CSC 411: 07-Linear Classification 19 / 23
20 Logistic Regression Logistic Regression: z = w x + b y = σ(z) 1 = 1 + e z L CE = t log y (1 t) log(1 y) [[gradient derivation in the notes]] UofT CSC 411: 07-Linear Classification 20 / 23
21 Logistic Regression Problem: what if t = 1 but you re really confident it s a negative example (z 0)? If y is small enough, it may be numerically zero. This can cause very subtle and hard-to-find bugs. y = σ(z) y 0 L CE = t log y (1 t) log(1 y) computes log 0 Instead, we combine the activation function and the loss into a single logistic-cross-entropy function. L LCE (z, t) = L CE (σ(z), t) = t log(1 + e z ) + (1 t) log(1 + e z ) Numerically stable computation: E = t * np.logaddexp(0, -z) + (1-t) * np.logaddexp(0, z) UofT CSC 411: 07-Linear Classification 21 / 23
22 Logistic Regression Comparison of loss functions: UofT CSC 411: 07-Linear Classification 22 / 23
23 Logistic Regression Comparison of gradient descent updates: Linear regression: w w α N N (y (i) t (i) ) x (i) i=1 Logistic regression: w w α N N (y (i) t (i) ) x (i) i=1 Not a coincidence! These are both examples of matching loss functions, but that s beyond the scope of this course. UofT CSC 411: 07-Linear Classification 23 / 23
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