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Bayesian Classification

Bayesian Classification. Instructor: Qiang Yang Hong Kong University of Science and Technology Qyang@cs.ust.hk Thanks: Dan Weld, Eibe Frank. Weather data set. Windy=True. Windy=False. Play=yes. Play=no. Basics. Unconditional or Prior Probability Pr(Play=yes) + Pr(Play=no)=1

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Bayesian Classification

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  1. Bayesian Classification Instructor: Qiang Yang Hong Kong University of Science and Technology Qyang@cs.ust.hk Thanks: Dan Weld, Eibe Frank

  2. Weather data set

  3. Windy=True Windy=False Play=yes Play=no Basics • Unconditional or Prior Probability • Pr(Play=yes) + Pr(Play=no)=1 • Pr(Play=yes) is sometimes written as Pr(Play) • Table has 9 yes, 5 no • Pr(Play=yes)=9/(9+5)=9/14 • Thus, Pr(Play=no)=5/14 • Joint Probability of Play and Windy: • Pr(Play=x,Windy=y) for all values x and y, should be 1 3/14 6/14 3/14 ?

  4. Probability Basics • Conditional Probability • Pr(A|B) • # (Windy=False)=8 • Within the 8, • #(Play=yes)=6 • Pr(Play=yes | Windy=False) =6/8 • Pr(Windy=False)=8/14 • Pr(Play=Yes)=9/14 • Applying Bayes Rule • Pr(B|A) = Pr(A|B)Pr(B) / Pr(A) • Pr(Windy=False|Play=yes)= 6/8*8/14/(9/14)=6/9

  5. Conditional Independence • “A and P are independent given C” • Pr(A | P,C) = Pr(A | C) C A P Probability F F F 0.534 F F T 0.356 F T F 0.006 F T T 0.004 T F F 0.048 T F T 0.012 T T F 0.032 T T T 0.008 Ache Cavity Probe Catches

  6. C A P Probability F F F 0.534 F F T 0.356 F T F 0.006 F T T 0.004 T F F 0.012 T F T 0.048 T T F 0.008 T T T 0.032 Suppose C=True Pr(A|P,C) = 0.032/(0.032+0.048) = 0.032/0.080 = 0.4 Pr(A|C) = 0.032+0.008/ (0.048+0.012+0.032+0.008) = 0.04 / 0.1 = 0.4 Conditional Independence • “A and P are independent given C” • Pr(A | P,C) = Pr(A | C) and also Pr(P | A,C) = Pr(P | C)

  7. C P(A) T 0.4 F 0.011 P(C) .1 C P(P) T 0.8 F 0.4 Conditional Independence Conditional probability table (CPT) • Can encode joint probability distribution in compact form C A P Probability F F F 0.534 F F T 0.356 F T F 0.006 F T T 0.004 T F F 0.012 T F T 0.048 T T F 0.008 T T T 0.032 Ache Cavity Probe Catches

  8. Creating a Network • 1: Bayes net = representation of a JPD • 2: Bayes net = set of cond. independence statements • If create correct structure that represents causality • Then get a good network • i.e. one that’s small = easy to compute with • One that is easy to fill in numbers

  9. Example • My house alarm system just sounded (A). • Both an earthquake (E) and a burglary (B) could set it off. • John will probably hear the alarm; if so he’ll call (J). • But sometimes John calls even when the alarm is silent • Mary might hear the alarm and call too (M), but not as reliably • We could be assured a complete and consistent model by fully specifying the joint distribution: • Pr(A, E, B, J, M) • Pr(A, E, B, J, ~M) • etc.

  10. Structural Models (HK book 7.4.3) Instead of starting with numbers, we will start with structural relationships among the variables There is a direct causal relationship from Earthquake to Alarm There is a direct causal relationship from Burglar to Alarm There is a direct causal relationship from Alarm to JohnCall Earthquake and Burglar tend to occur independently etc.

  11. Earthquake Burglary Alarm MaryCalls JohnCalls Possible Bayesian Network

  12. P(E) .002 P(B) .001 B T T F F E T F T F P(A) .95 .94 .29 .01 A T F P(J) .90 .05 A T F P(M) .70 .01 Complete Bayesian Network Earthquake Burglary Alarm MaryCalls JohnCalls

  13. Microsoft Bayesian Belief Net • http://research.microsoft.com/adapt/MSBNx/ • Can be used to construct and reason with Bayesian Networks • Consider the example

  14. Mining for Structural Models • Difficult to mine • Some methods are proposed • Up to now, no good results yet • Often requires domain expert’s knowledge • Once set up, a Bayesian Network can be used to provide probabilistic queries • Microsoft Bayesian Network Software

  15. Use the Bayesian Net for Prediction • From a new day’s data we wish to predict the decision • New data: X • Class label: C • To predict the class of X, is the same as asking • Value of Pr(C|X)? • Pr(C=yes|X) • Pr(C=no|X) • Compare the two

  16. Naïve Bayesian Models • Two assumptions: Attributes are • equally important • statistically independent (given the class value) • This means that knowledge about the value of a particular attribute doesn’t tell us anything about the value of another attribute (if the class is known) • Although based on assumptions that are almost never correct, this scheme works well in practice!

  17. Why Naïve? • Assume the attributes are independent, given class • What does that mean? play outlook temp humidity windy Pr(outlook=sunny | windy=true, play=yes)= Pr(outlook=sunny|play=yes)

  18. Weather data set

  19. Is the assumption satisfied? • #yes=9 • #sunny=2 • #windy, yes=3 • #sunny|windy, yes=1 Pr(outlook=sunny|windy=true, play=yes)=1/3 Pr(outlook=sunny|play=yes)=2/9 Pr(windy|outlook=sunny,play=yes)=1/2 Pr(windy|play=yes)=3/9 Thus, the assumption is NOT satisfied. But, we can tolerate some errors (see later slides)

  20. Probabilities for the weather data • A new day:

  21. Bayes’ rule • Probability of event H given evidence E: • A priori probability of H: • Probability of event before evidence has been seen • A posteriori probability of H: • Probability of event after evidence has been seen

  22. Naïve Bayes for classification • Classification learning: what’s the probability of the class given an instance? • Evidence E = an instance • Event H = class value for instance (Play=yes, Play=no) • Naïve Bayes Assumption: evidence can be split into independent parts (i.e. attributes of instance are independent)

  23. The weather data example Evidence E Probability for class “yes”

  24. The “zero-frequency problem” • What if an attribute value doesn’t occur with every class value (e.g. “Humidity = high” for class “yes”)? • Probability will be zero! • A posteriori probability will also be zero! (No matter how likely the other values are!) • Remedy: add 1 to the count for every attribute value-class combination (Laplace estimator) • Result: probabilities will never be zero! (also: stabilizes probability estimates)

  25. Modified probability estimates • In some cases adding a constant different from 1 might be more appropriate • Example: attribute outlook for class yes • Weights don’t need to be equal (if they sum to 1) Sunny Overcast Rainy

  26. Missing values • Training: instance is not included in frequency count for attribute value-class combination • Classification: attribute will be omitted from calculation • Example:

  27. Dealing with numeric attributes • Usual assumption: attributes have a normal or Gaussian probability distribution (given the class) • The probability density function for the normal distribution is defined by two parameters: • The sample mean : • The standard deviation : • The density function f(x):

  28. Statistics for the weather data • Example density value:

  29. Classifying a new day • A new day: • Missing values during training: not included in calculation of mean and standard deviation

  30. Probability densities • Relationship between probability and density: • But: this doesn’t change calculation of a posteriori probabilities because  cancels out • Exact relationship:

  31. Example of Naïve Bayes in Weka • Use Weka Naïve Bayes Module to classify • Weather.nominal.arff

  32. Discussion of Naïve Bayes • Naïve Bayes works surprisingly well (even if independence assumption is clearly violated) • Why? Because classification doesn’t require accurate probability estimates as long as maximum probability is assigned to correct class • However: adding too many redundant attributes will cause problems (e.g. identical attributes) • Note also: many numeric attributes are not normally distributed

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