HW4 - MS&E 226 Small Data Problem Set 4 Due 5:00 PM...

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MS&E 226 Problem Set 4 “Small” Data Due: November 18, 2016, 5:00 PM , submitted through Gradescope Problem 1. (Naive Bayes) In this problem, we study a particularly simple approach to classifica- tion: the naive Bayes classifier. This is one of the most widely used classification techniques, and despite its simplicity is known to work surprisingly well in a wide range of settings. We build the naive Bayes’ classifier for a spam detection problem. The data we work with is the dataset spam.csv , available from . csv . 1 This dataset has n = 4601 rows; each row represents information about a single e-mail mes- sage. There are 49 columns. The last column indicates whether the email was a spam or not ( Y = 1 means spam, Y = 0 means not spam). The remaining p = 48 columns are binary values. Each column j = 1 , ..., 48 represents a word, and the value in a particular row indicates whether that word was present in the email or not ( X j = 1 means the word was present, X j = 0 means that it was not present). As we discussed in class, the Bayes optimal classifier for 0-1 loss is the Bayes classifier : given a covariate vector ~ X , we compute the probabilities P ( Y = 0 | ~ X ) and P ( Y = 1 | ~ X ) . If the message is more likely to be spam than not, we predict 1; otherwise we predict 0. A limitation of Bayes classifiers is that you need to know the population model. In practice, we rarely know these probabilities, so they must be estimated from the data. That’s where the Naive Bayes approach comes in. (a) In general, we know that P ( Y = y | ~ X = ~x ) = P ( Y = y, ~ X = ~x ) P ( ~ X = ~x ) . That means a message with covariates ~x is more likely to be spam than not if P ( Y = 1 , ~ X = ~x ) > P ( Y = 0 , ~ X = ~x ) , as the denominator does not depend on Y . Show that ˆ Q EPM ~x,y = 1 n n X i =1 I { ~ X = ~x, Y i = y } is an unbiased estimate of P ( Y = y, ~ X = ~x ) . This is called the empirical population model (EPM). Consider the following classifier: if ˆ Q ~ X, 1 > ˆ Q ~ X, 0 , predict 1; otherwise predict 0. Explain why this is not a good classifier if n 2 p +1 . (b) (DELETED) 1 This dataset is derived from a spam dataset in the UCI ML repository; see . edu/ml/datasets/Spambase .
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The Naive Bayes classifier takes a different approach to reduce the “curse of dimensionality.” Note that by Bayes’ rule we have: P ( Y = y | ~ X = ~x ) = P ( ~ X = ~x | Y = y ) P ( Y = y ) P ( ~ X = ~x ) . (1) The Naive Bayes classifier assumes that the covariates are independent given Y = y ; that is: P ( ~ X = ~x | Y = y ) = p Y j =1 P ( X j = x j | Y = y ) . As before, we can drop the denominator P ( ~ X = ~x ) , so we want to classify Y = 1 if: P ( Y = 1) p Y j =1 P ( X j = x j | Y = 1) > P ( Y = 0) p Y j =1 P ( X j = x j | Y = 0) , and we want to classify Y = 0 otherwise. The Naive Bayes classifier uses an estimate for each of the quantities in the preceding expression.
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