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1 Naïve Bayes A probabilistic ML algorithm. 2 Axioms of Probability Theory All probabilities between 0 and 1 True proposition has probability 1, false.

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Presentation on theme: "1 Naïve Bayes A probabilistic ML algorithm. 2 Axioms of Probability Theory All probabilities between 0 and 1 True proposition has probability 1, false."— Presentation transcript:

1 1 Naïve Bayes A probabilistic ML algorithm

2 2 Axioms of Probability Theory All probabilities between 0 and 1 True proposition has probability 1, false has probability 0. P(true) = 1 P(false) = 0. The probability of disjunction is: A B

3 3 Conditional Probability P(A | B) is the probability of A given B Assumes that B is all and only information known. Defined by: A B

4 4 Independence A and B are independent iff: Therefore, if A and B are independent: These two constraints are logically equivalent

5 5 Joint Distribution The joint probability distribution for a set of random INDEPENDENT variables, X 1,…,X n gives the probability of every combination of values (an n-dimensional array with v n values if all variables are discrete with v values, all v n values must sum to 1): P(X 1,…,X n ) The probability of all possible conjunctions (assignments of values to some subset of variables) can be calculated by summing the appropriate subset of values from the joint distribution. Therefore, all conditional probabilities can also be calculated. circlesquare red0.200.02 blue0.020.01 circlesquare red0.050.30 blue0.20 Class=positive Class=negative

6 6 Probabilistic Classification Let Y be the random variable for the class which takes values {y 1,y 2,…y m } (m possible classifications for our instances). Let X be the random variable describing an instance consisting of a vector of values for n features, let x k be a possible value for X and x ik a possible value for X i. For classification, we need to compute: P(Y=y i | X=x k ) for i=1…m E.g. the objective is to classify a new unseen x k by estimating the probability of each possible classification y i, given the feature values of the instance to be classified x k :

7 Probabilistic Classification (2) However, given no other assumptions, this requires a table giving the probability of each category for each possible instance in the instance space, which is impossible to accurately estimate from a reasonably-sized training set. –Assuming that Y and all X i are binary, we need 2 n entries to specify P(Y=pos | X=x k ) for each of the 2 n possible x k since: –P(Y=neg | X=x k ) = 1 – P(Y=pos | X=x k ) –Compared to 2 n+1 – 1 entries for the joint distribution P(Y,X 1,X 2 …X n ) 7

8 8 Bayes Theorem Simple proof from definition of conditional probability: QED: (Def. cond. prob.)

9 9 Bayesian Categorization For each classification value y i we have (applying Bayes): P(Y=yi) and P(X=x k ) are called priors and can be estimated from learning set D since categories are complete and disjoint

10 Complete and Disjoint Complete: Y can only assume values in Disjoint: If a set of categories is complete and disjoint, Z is a random variable, and z is any of its possible values, then: 10

11 11 Bayesian Categorization (cont.) To know P(Y=y i |X=x k ) need to know: –Priors: P(Y=y i ) –Conditionals: P(X=x k | Y=y i ) P(Y=y i ) are easily estimated from data. –If n i of the examples in D have value Y=y i then P(Y=y i ) = n i / |D|

12 Bayesian Categorization (cont.) To summarize: We need to estimate Which is equivalent to estimate (Bayes) We know that: Therefore the problem is to estimate But: So we really need to estimate only 12 P(Y=y i ) = n i / |D|

13 Bayesian Categorization (cont.) In other terms, to estimate the probability of a category given an instance P(Y=y i /X=x k ), we need to estimate the probability of seeing that instance given the category P(X=x k /Y=y i ). IS THIS SIMPLER THAN THE ORIGINAL PROBLEM? Too many possible instances (e.g. 2 n for binary features) to estimate all P(X=x k | Y=y i ) for all i. Still need to make some sort of independence assumptions about the features to make learning tractable 13

14 14 Generative Probabilistic Models Assume a simple (usually unrealistic) probabilistic method by which the data was generated. For categorization, each category has a different parameterized generative model that characterizes that category. Training: Use the data for each category to estimate the parameters of the generative model for that category. –Maximum Likelihood Estimation (MLE): Set parameters to maximize the probability that the model produced the given training data. –If M λ denotes a model with parameter values λ and D i is the training data for the i-th class, find model parameters for class i (λ i ) that maximize the likelihood of D k : Testing: Use Bayesian analysis to determine the category model that most likely generated a specific test instance. A generative model is a model for randomly generating observable data, typically given some hidden parameters. It specifies a joint probability distribution over observation and label sequences. A generative model is a model for randomly generating observable data, typically given some hidden parameters. It specifies a joint probability distribution over observation and label sequences.

15 Model parameters 15 M λ is estimated on the learning set D Our (hidden) model parameters are the conditional probabilities, those generating our observable data

16 16 Naïve Bayes Generative Model Size Color Shape Positive Negative pos neg pos neg sm med lg med sm med lg red blue grn circ sqr tri circ sqr tri sm lg med sm lg med lg sm blue red grn blue grn red grn blue circ sqr tri circ sqr circ tri Category

17 17 Naïve Bayes Inference Problem Size Color Shape Positive Negative pos neg pos neg sm med lg med sm med lg red blue grn circ sqr tri circ sqr tri sm lg med sm lg med lg sm blue red grn blue grn red grn blue circ sqr tri circ sqr circ tri Category lg red circ ?? I estimate on the learning set the probability of extracting lg, red, circ from the red or blue urns. I estimate on the learning set the probability of extracting lg, red, circ from the red or blue urns.

18 HOW? 18

19 19 Naïve Bayesian Categorization Assume features of an instance are independent given the category (conditionally independent). Therefore, we only need to know P(X i | Y) for each possible pair of a feature-value and a category. If Y and all X i and binary, this requires specifying only 2n parameters: –P(X i =true | Y=true) and P(X i =true | Y=false) for each X i –P(X i =false | Y) = 1 – P(X i =true | Y) Compared to specifying 2 n parameters without any independence assumptions.

20 20 Naïve Bayes Example ProbabilityY=positiveY=negative P(Y)0.5 P(small | Y)3/8 P(medium | Y)3/82/8 P(large | Y)2/83/8 P(red | Y)5/82/8 P(blue | Y)2/83/8 P(green | Y)1/83/8 P(square | Y)1/83/8 P(triangle | Y)2/83/8 P(circle | Y)5/82/8 Test Instance: Training set Have 3 sm out of 8 instances in red “Size” urn then P(small/pos)=3/8=0,375 (round 4) Have 3 sm out of 8 instances in red “Size” urn then P(small/pos)=3/8=0,375 (round 4)

21 21 Naïve Bayes Example Probabilitypositivenegative P(Y)0.5 P(medium | Y)3/82/8 P(red | Y)5/82/8 P(circle | Y)5/82/8 P(positive | X) = P(positive)*P(medium | positive)*P(red | positive)*P(circle | positive) / P(X) 0.5 * 3/8 * 5/8 * 5/8 P(negative | X) = P(negative)*P(medium | negative)*P(red | negative)*P(circle | negative) / P(X) 0.5 * 2/8 * 2/8 * 2/8 = 0,073 =0.0078 Test Instance: P(positive/X)>P(negative/X)  positive

22 Naive summary Classify any new datum instance x k =(x 1,…x n ) as: To do this based on training examples, estimate the parameters from the training examples in D: –For each target value of the classification variable (hypothesis) y i –For each attribute value a t of each datum instance

23 23 Estimating Probabilities Normally, as in previous example, probabilities are estimated based on observed frequencies in the training data. If D contains n k examples in category y k, and n ijk of these n k examples have the jth value for feature X i, x ij, then: However, estimating such probabilities from small training sets is error-prone. If due only to chance, a rare feature, X i, is always false in the training data,  y k :P(X i =true | Y=y k ) = 0. If X i =true then occurs in a test example, X, the result is that  y k : P(X | Y=y k ) = 0 and  y k : P(Y=y k | X) = 0

24 24 Probability Estimation Example Probabilitypositivenegative P(Y)0.5 P(small | Y)0.5 P(medium | Y)0.0 P(large | Y)0.5 P(red | Y)1.00.5 P(blue | Y)0.00.5 P(green | Y)0.0 P(square | Y)0.0 P(triangle | Y)0.00.5 P(circle | Y)1.00.5 ExSizeColorShapeCategory 1smallredcirclepositive 2largeredcirclepositive 3smallredtrianglenegitive 4largebluecirclenegitive Test Instance X: P(positive | X) = 0.5 * 0.0 * 1.0 * 1.0 / P(X) = 0 P(negative | X) = 0.5 * 0.0 * 0.5 * 0.5 / P(X) = 0

25 25 Smoothing To account for estimation from small samples, probability estimates are adjusted or smoothed. Laplace smoothing using an m-estimate assumes that each feature is given a prior probability, p, that is assumed to have been previously observed in a “ virtual ” sample of size m. For binary features, p is simply assumed to be 0.5.

26 26 Laplace Smothing Example Assume training set contains 10 positive examples: –4: small –0: medium –6: large Estimate parameters as follows (if m=1, p=1/3) –P(small | positive) = (4 + 1/3) / (10 + 1) = 0.394 –P(medium | positive) = (0 + 1/3) / (10 + 1) = 0.03 –P(large | positive) = (6 + 1/3) / (10 + 1) = 0.576 –P(small or medium or large | positive) = 1.0

27 27 Continuous Attributes If X i is a continuous feature rather than a discrete one, need another way to calculate P(X i | Y). Assume that X i has a Gaussian distribution whose mean and variance depends on Y. During training, for each combination of a continuous feature X i and a class value for Y, y k, estimate a mean, μ ik, and standard deviation σ ik based on the values of feature X i in class y k in the training data. μ ik is the mean value of Xi observed over instances for which Y= y k in D During testing, estimate P(X i | Y=y k ) for a given example, using the Gaussian distribution defined by μ ik and σ ik.

28 28 Comments on Naïve Bayes Tends to work well despite strong assumption of conditional independence. Experiments show it to be quite competitive with other classification methods on standard UCI datasets. Although it does not produce accurate probability estimates when its independence assumptions are violated, it may still pick the correct maximum-probability class in many cases. –Able to learn conjunctive concepts in any case Does not perform any search of the hypothesis space. Directly constructs a hypothesis from parameter estimates that are easily calculated from the training data. –Strong bias Not guarantee consistency with training data. Typically handles noise well since it does not even focus on completely fitting the training data.


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