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Bayes Rule The product rule gives us two ways to factor a joint probability: Therefore, Why is this useful? –Can get diagnostic probability P(Cavity | Toothache) from causal probability P(Toothache | Cavity) –Can update our beliefs based on evidence –Key tool for probabilistic inference Rev. Thomas Bayes ( )

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Bayes Rule example Marie is getting married tomorrow, at an outdoor ceremony in the desert. In recent years, it has rained only 5 days each year (5/365 = 0.014). Unfortunately, the weatherman has predicted rain for tomorrow. When it actually rains, the weatherman correctly forecasts rain 90% of the time. When it doesn't rain, he incorrectly forecasts rain 10% of the time. What is the probability that it will rain on Marie's wedding?

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Bayes Rule example Marie is getting married tomorrow, at an outdoor ceremony in the desert. In recent years, it has rained only 5 days each year (5/365 = 0.014). Unfortunately, the weatherman has predicted rain for tomorrow. When it actually rains, the weatherman correctly forecasts rain 90% of the time. When it doesn't rain, he incorrectly forecasts rain 10% of the time. What is the probability that it will rain on Marie's wedding?

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Bayes rule: Another example 1% of women at age forty who participate in routine screening have breast cancer. 80% of women with breast cancer will get positive mammographies. 9.6% of women without breast cancer will also get positive mammographies. A woman in this age group had a positive mammography in a routine screening. What is the probability that she actually has breast cancer?

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Independence Two events A and B are independent if and only if P(A B) = P(A) P(B) –In other words, P(A | B) = P(A) and P(B | A) = P(B) –This is an important simplifying assumption for modeling, e.g., Toothache and Weather can be assumed to be independent Are two mutually exclusive events independent? –No, but for mutually exclusive events we have P(A B) = P(A) + P(B) Conditional independence: A and B are conditionally independent given C iff P(A B | C) = P(A | C) P(B | C)

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Conditional independence: Example Toothache: boolean variable indicating whether the patient has a toothache Cavity: boolean variable indicating whether the patient has a cavity Catch: whether the dentist’s probe catches in the cavity If the patient has a cavity, the probability that the probe catches in it doesn't depend on whether he/she has a toothache P(Catch | Toothache, Cavity) = P(Catch | Cavity) Therefore, Catch is conditionally independent of Toothache given Cavity Likewise, Toothache is conditionally independent of Catch given Cavity P(Toothache | Catch, Cavity) = P(Toothache | Cavity) Equivalent statement: P(Toothache, Catch | Cavity) = P(Toothache | Cavity) P(Catch | Cavity)

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Conditional independence: Example How many numbers do we need to represent the joint probability table P(Toothache, Cavity, Catch)? 2 3 – 1 = 7 independent entries Write out the joint distribution using chain rule: P(Toothache, Catch, Cavity) = P(Cavity) P(Catch | Cavity) P(Toothache | Catch, Cavity) = P(Cavity) P(Catch | Cavity) P(Toothache | Cavity) How many numbers do we need to represent these distributions? = 5 independent numbers In most cases, the use of conditional independence reduces the size of the representation of the joint distribution from exponential in n to linear in n

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Probabilistic inference Suppose the agent has to make a decision about the value of an unobserved query variable X given some observed evidence variable(s) E = e –Partially observable, stochastic, episodic environment –Examples: X = {spam, not spam}, e = message X = {zebra, giraffe, hippo}, e = image features Bayes decision theory: –The agent has a loss function, which is 0 if the value of X is guessed correctly and 1 otherwise –The estimate of X that minimizes expected loss is the one that has the greatest posterior probability P(X = x | e) –This is the Maximum a Posteriori (MAP) decision

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MAP decision Value x of X that has the highest posterior probability given the evidence E = e: Maximum likelihood (ML) decision: likelihood prior posterior

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Naïve Bayes model Suppose we have many different types of observations (symptoms, features) E 1, …, E n that we want to use to obtain evidence about an underlying hypothesis X MAP decision involves estimating –If each feature F i can take on k values, how many entries are in the (conditional) joint probability table P(E 1, …, E n |X = x)?

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Naïve Bayes model Suppose we have many different types of observations (symptoms, features) E 1, …, E n that we want to use to obtain evidence about an underlying hypothesis X MAP decision involves estimating We can make the simplifying assumption that the different features are conditionally independent given the hypothesis: –If each feature can take on k values, what is the complexity of storing the resulting distributions?

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Naïve Bayes Spam Filter MAP decision: to minimize the probability of error, we should classify a message as spam if P(spam | message) > P(¬spam | message)

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Naïve Bayes Spam Filter MAP decision: to minimize the probability of error, we should classify a message as spam if P(spam | message) > P(¬spam | message) We have P(spam | message) P(message | spam)P(spam) and ¬P(spam | message) P(message | ¬spam)P(¬spam)

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Naïve Bayes Spam Filter We need to find P(message | spam) P(spam) and P(message | ¬spam) P(¬spam) The message is a sequence of words (w 1, …, w n ) Bag of words representation –The order of the words in the message is not important –Each word is conditionally independent of the others given message class (spam or not spam)

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Naïve Bayes Spam Filter We need to find P(message | spam) P(spam) and P(message | ¬spam) P(¬spam) The message is a sequence of words (w 1, …, w n ) Bag of words representation –The order of the words in the message is not important –Each word is conditionally independent of the others given message class (spam or not spam) Our filter will classify the message as spam if

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Bag of words illustration US Presidential Speeches Tag Cloud

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Bag of words illustration US Presidential Speeches Tag Cloud

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Bag of words illustration US Presidential Speeches Tag Cloud

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Naïve Bayes Spam Filter prior likelihood posterior

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Parameter estimation In order to classify a message, we need to know the prior P(spam) and the likelihoods P(word | spam) and P(word | ¬spam) –These are the parameters of the probabilistic model –How do we obtain the values of these parameters? spam: 0.33 ¬spam: 0.67 P(word | ¬spam)P(word | spam)prior

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Parameter estimation How do we obtain the prior P(spam) and the likelihoods P(word | spam) and P(word | ¬spam)? –Empirically: use training data –This is the maximum likelihood (ML) estimate, or estimate that maximizes the likelihood of the training data: P(word | spam) = # of word occurrences in spam messages total # of words in spam messages d: index of training document, i: index of a word

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Parameter estimation How do we obtain the prior P(spam) and the likelihoods P(word | spam) and P(word | ¬spam)? –Empirically: use training data Parameter smoothing: dealing with words that were never seen or seen too few times –Laplacian smoothing: pretend you have seen every vocabulary word one more time than you actually did P(word | spam) = # of word occurrences in spam messages total # of words in spam messages P(word | spam) = # of word occurrences in spam messages + 1 total # of words in spam messages + V (V: total number of unique words)

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Summary of model and parameters Naïve Bayes model: Model parameters: P(spam) P(¬spam) P(w 1 | spam) P(w 2 | spam) … P(w n | spam) P(w 1 | ¬spam) P(w 2 | ¬spam) … P(w n | ¬spam) Likelihood of spam prior Likelihood of ¬ spam

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Bag-of-word models for images Csurka et al. (2004), Willamowski et al. (2005), Grauman & Darrell (2005), Sivic et al. (2003, 2005)

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Bag-of-word models for images 1.Extract image features

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Bag-of-word models for images 1.Extract image features

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2.Learn “visual vocabulary” Bag-of-word models for images

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1.Extract image features 2.Learn “visual vocabulary” 3.Map image features to visual words Bag-of-word models for images

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Bayesian decision making: Summary Suppose the agent has to make decisions about the value of an unobserved query variable X based on the values of an observed evidence variable E Inference problem: given some evidence E = e, what is P(X | e)? Learning problem: estimate the parameters of the probabilistic model P(X | E) given a training sample {(x 1,e 1 ), …, (x n,e n )}

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