Bayesian Learning No reading assignment for this topic.

Bayesian Learning No reading assignment for this topic

Conditional Probability Probability of an event given the occurrence of some other event. E.g., Consider choosing a card from a well-shuffled standard deck of 52 playing cards. Given that the first card chosen is an ace, what is the probability that the second card chosen will be an ace?

Event space = all possible pairs of cards Y X

Y First card is Ace Event space = all possible pairs of cards

Y = First card is Ace X = Second card is Ace Event space = all possible pairs of cards

P(Y) = 4 / 52 P(X,Y) = # possible pairs of aces / total # of pairs = 43/5251 = 12/2652. P(X | Y) = (12/2652) / (4 / 52) = 3/51.

Deriving Bayes Rule

Bayesian Learning

Application to Machine Learning In machine learning we have a space H of hypotheses: h 1, h 2,..., h n We also have a set D of data We want to calculate P(h | D)

Prior probability of h: P(h): Probability that hypothesis h is true given our prior knowledge If no prior knowledge, all h H are equally probable Posterior probability of h: P(h | D): Probability that hypothesis h is true, given the data D. Likelihood of D: P(D | h): Probability that we will see data D, given hypothesis h is true. Terminology

Bayes Rule: Machine Learning Formulation

Example

The Monty Hall Problem You are a contestant on a game show. There are 3 doors, A, B, and C. There is a new car behind one of them and goats behind the other two. Monty Hall, the host, asks you to pick a door, any door. You pick door A. Monty tells you he will open a door, different from A, that has a goat behind it. He opens door B: behind it there is a goat. Monty now gives you a choice: Stick with your original choice A or switch to C. Should you switch? http://math.ucsd.edu/~crypto/Monty/monty.html

Bayesian probability formulation Hypothesis space H: h 1 = Car is behind door A h 2 = Car is behind door B h 3 = Car is behind door C Data D = Monty opened B What is P(h 1 | D)? What is P(h 2 | D)? What is P(h 3 | D)?

Event space Event space = All possible configurations of cars and goats behind doors A, B, C Y = Goat behind door B X = Car behind door A

Y = Goat behind door B X = Car behind door A Bayes Rule: Event space

MAP (maximum a posteriori) Learning Bayes rule: Goal of learning: Find maximum a posteriori hypothesis h MAP : because P(D) is a constant independent of h.

Note: If every h H is equally probable, then This is called the maximum likelihood hypothesis.

A Medical Example Toby takes a test for leukemia. The test has two outcomes: positive and negative. It is known that if the patient has leukemia, the test is positive 98% of the time. If the patient does not have leukemia, the test is positive 3% of the time. It is also known that 0.008 of the population has leukemia. Tobys test is positive. Which is more likely: Toby has leukemia or Toby does not have leukemia?

Hypothesis space: h 1 = T. has leukemia h 2 = T. does not have leukemia Prior: 0.008 of the population has leukemia. Thus P(h 1 ) = 0.008 P(h 2 ) = 0.992 Likelihood: P(+ | h 1 ) = 0.98, P( | h 1 ) = 0.02 P(+ | h 2 ) = 0.03, P( | h 2 ) = 0.97 Posterior knowledge: Blood test is + for this patient.

In summary P(h 1 ) = 0.008, P(h 2 ) = 0.992 P(+ | h 1 ) = 0.98, P( | h 1 ) = 0.02 P(+ | h 2 ) = 0.03, P( | h 2 ) = 0.97 Thus:

What is P(leukemia|+)? So, These are called the posterior probabilities.

In-Class Exercise Suppose you receive an e-mail message with the subject Hi. You have been keeping statistics on your e-mail, and have found that while only 10% of the total e-mail messages you receive are spam, 50% of the spam messages have the subject Hi and 2% of the non-spam messages have the subject Hi. What is the probability that the message is spam?

Bayesianism vs. Frequentism Classical probability: Frequentists Probability of a particular event is defined relative to its frequency in a sample space of events. E.g., probability of the coin will come up heads on the next trial is defined relative to the frequency of heads in a sample space of coin tosses. Bayesian probability: Combine measure of prior belief you have in a proposition with your subsequent observations of events. Example: Bayesian can assign probability to statement There was life on Mars a billion years ago but frequentist cannot.

Independence and Conditional Independence Two random variables, X and Y, are independent if Two random variables, X and Y, are independent given Z if Examples?

Naive Bayes Classifier Let f (x) be a target function for classification: f (x) {+1, 1}. Let x = We want to find the most probable class value, h MAP, given the data x:

By Bayes Theorem: P(class) can be estimated from the training data. How? However, in general, not practical to use training data to estimate P(x 1, x 2,..., x n | class). Why not?

Naive Bayes classifier: Assume Is this a good assumption? Given this assumption, heres how to classify an instance x = : Naive Bayes classifier: Estimate the values of these various probabilities over the training set.

DayOutlook Temp Humidity Wind PlayTennis D1Sunny HotHigh Weak No D2Sunny HotHigh Strong No D3Overcast HotHigh Weak Yes D4Rain MildHigh Weak Yes D5Rain CoolNormal Weak Yes D6Rain CoolNormal Strong No D7Overcast CoolNormal StrongYes D8Sunny MildHigh Weak No D9Sunny CoolNormal Weak Yes D10Rain MildNormal Weak Yes D11Sunny MildNormal Strong Yes D12Overcast MildHigh Strong Yes D13Overcast HotNormal Weak Yes D14Rain MildHigh Strong No Training data: D15Sunny CoolHigh Strong ? Test data:

In practice, use training data to compute a probablistic model:

Estimating probabilities Recap: In previous example, we had a training set and a new example, We asked: What classification is given by a naive Bayes classifier? Let n(c) be the number of training instances with class c, and n(x i = a i, c) be the number of training instances with attribute value x i =a i and class c. Then

Problem with this method: If n(c) is very small, gives a poor estimate. E.g., P(Outlook = Overcast | no) = 0.

Now suppose we want to classify a new instance:. Then: This incorrectly gives us zero probability due to small sample.

One solution: Laplace smoothing (also called add-one smoothing) For each class c j and attribute x i with value a i, add one virtual instance. That is, recalculate: where k is the number of possible values of attribute a.

DayOutlook Temp Humidity Wind PlayTennis D1Sunny HotHigh Weak No D2Sunny HotHigh Strong No D3Overcast HotHigh Weak Yes D4Rain MildHigh Weak Yes D5Rain CoolNormal Weak Yes D6Rain CoolNormal Strong No D7Overcast CoolNormal StrongYes D8Sunny MildHigh Weak No D9Sunny CoolNormal Weak Yes D10Rain MildNormal Weak Yes D11Sunny MildNormal Strong Yes D12Overcast MildHigh Strong Yes D13Overcast HotNormal Weak Yes D14Rain MildHigh Strong No Training data: Add virtual instances for Outlook: Outlook=Sunny: YesOutlook=Overcast: YesOutlook=Rain: Yes Outlook=Sunny: NoOutlook=Overcast: NoOutlook=Rain: No P(Outlook=Overcast| No) = 0 / 5 0 + 1 / 5 + 3 = 1/8

In-class exercises

Naive Bayes on continuous-valued attributes How to deal with continuous-valued attributes? Two possible solutions: Discretize Assume particular probability distribution of classes over values (estimate parameters from training data)

Simplest discretization method For each attribute x i, create k equal-size bins in interval from min(x i ) to max(x i ). Choose thresholds in between bins. P(Humidity < 40 | yes) P(40

Bayesian Learning No reading assignment for this topic.

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