1. Bayesian Learning No reading assignment for this topic

2. 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?

3. Event space = all possible pairs of cardsX Y

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

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

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

7. Deriving Bayes Rule

8. Bayesian Learning

9. Application to Machine Learning
In machine learning we have a space H of hypotheses:
h1 , h2 , ... , hn
We also have a set D of data
We want to calculate P(h | D)

10. Terminology 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.

11. Machine Learning Formulation
Bayes Rule:
Machine Learning Formulation

12. Example

13. 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?

14. Bayesian probability formulation
Hypothesis space H: h1 = Car is behind door A h2 = Car is behind door B h3 = Car is behind door C Data D = Monty opened B What is P(h1 | D)? What is P(h2 | D)? What is P(h3 | D)?

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

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

17. Using Bayes’ Rule to solve the Monty Hall problem
You pick door A.
Data D = Monty opened door B
Hypothesis space H:
h1 = Car is behind door A
h2 = Car is behind door C
h3 = Car is behind door B
What is P(h1 | D)?
What is P(h2 | D)?
What is P(h3 | D)?
Prior probability:
P(h1) = 1/3 P(h2) =1/3 P(h3) =1/3
Likelihood:
P(D | h1) = 1/2
P(D | h2) = 1
P(D | h3) = 0
P(D) = p(D|h1)p(h1) + p(D|h2)p(h2) +
p(D|h3)p(h3) = 1/6 + 1/3 + 0 = 1/2
By Bayes rule:
P(h1|D) = P(D|h1)p(h1) / P(D) = ½ 1/3 / ½ = 1/3
P(h2|D) = P(D|h2)p(h2) / P(D) = 1 1/3 / ½ = 2/3
So you should switch!

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

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

20. 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 of the population has leukemia.
Toby’s test is positive.
Which is more likely: Toby has leukemia or Toby does not have leukemia?

21. Hypothesis space:
h1 = T. has leukemia
h2 = T. does not have leukemia
Prior: of the population has leukemia. Thus
P(h1) = 0.008
P(h2) = 0.992
Likelihood:
P(+ | h1) = 0.98, P(− | h1) = 0.02
P(+ | h2) = 0.03, P(− | h2) = 0.97
Posterior knowledge:
Blood test is + for this patient.

22. In summary
P(h1) = 0.008, P(h2) = 0.992
P(+ | h1) = 0.98, P(− | h1) = 0.02
P(+ | h2) = 0.03, P(− | h2) = 0.97
Thus:

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

24. In-Class Exercise
Suppose you receive an message with the subject “Hi”. You have been keeping statistics on your , and have found that while only 10% of the total 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?

25. 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.

26. 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?

27. Naive Bayes Classifier
Let f (x) be a target function for classification: f (x)  {+1, −1}.
Let x = <x1, x2, ..., xn>
We want to find the most probable class value, hMAP,
given the data x:

28. 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(x1, x2, ..., xn | class). Why not?

29. Naive Bayes classifier: Assume
Is this a good assumption?
Given this assumption, here’s how to classify an instance
x = <x1, x2, ...,xn>:
Naive Bayes classifier:
Estimate the values of these various probabilities over the training set.

30. Training data: Test data: Day Outlook Temp Humidity Wind PlayTennis
D1 Sunny Hot High Weak No
D2 Sunny Hot High Strong No
D3 Overcast Hot High Weak Yes
D4 Rain Mild High Weak Yes
D5 Rain Cool Normal Weak Yes
D6 Rain Cool Normal Strong No
D7 Overcast Cool Normal Strong Yes
D8 Sunny Mild High Weak No
D9 Sunny Cool Normal Weak Yes
D10 Rain Mild Normal Weak Yes
D11 Sunny Mild Normal Strong Yes
D12 Overcast Mild High Strong Yes
D13 Overcast Hot Normal Weak Yes
D14 Rain Mild High Strong No
Test data:
D15 Sunny Cool High Strong ?

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

32. Estimating probabilities
Recap: In previous example, we had a training set and a new example,
<Outlook=sunny, Temperature=cool, Humidity=high, Wind=strong>
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(xi = ai , c) be the number of training instances with attribute value xi=ai and class c. Then

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

34. Now suppose we want to classify a new instance:
<Outlook=overcast, Temperature=cool, Humidity=high, Wind=strong>. Then:
This incorrectly gives us zero probability due to small sample.

35. One solution: Laplace smoothing (also called “add-one” smoothing)
For each class cj and attribute xi with value ai, add one “virtual” instance.
That is, recalculate:
where k is the number of possible values of attribute a.

36. Training data: Day Outlook Temp Humidity Wind PlayTennis
D1 Sunny Hot High Weak No
D2 Sunny Hot High Strong No
D3 Overcast Hot High Weak Yes
D4 Rain Mild High Weak Yes
D5 Rain Cool Normal Weak Yes
D6 Rain Cool Normal Strong No
D7 Overcast Cool Normal Strong Yes
D8 Sunny Mild High Weak No
D9 Sunny Cool Normal Weak Yes
D10 Rain Mild Normal Weak Yes
D11 Sunny Mild Normal Strong Yes
D12 Overcast Mild High Strong Yes
D13 Overcast Hot Normal Weak Yes
D14 Rain Mild High Strong No
Add virtual instances for Outlook:
Outlook=Sunny: Yes Outlook=Overcast: Yes Outlook=Rain: Yes
Outlook=Sunny: No Outlook=Overcast: No Outlook=Rain: No
P(Outlook=Overcast| No) = 0 / / = 1/8

37. Etc.

38. In-class exercises

39. 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)

40. Simplest discretization method
For each attribute xi , create k equal-size bins in interval from min(xi ) to max(xi).
Choose thresholds in between bins.
P(Humidity < 40 | yes) P(40<=Humidity < 80 | yes) P(80<=Humidity < 120 | yes)
P(Humidity < 40 | no) P(40<=Humidity < 80 | no) P(80<=Humidity < 120 | no)
Humidity: 25, 38, 50, 80, 90, 92, 96, 99
Threshold: 40
Threshold: 80
Threshold: 120

41. Questions: What should k be? What if some bins have very few instances?
Problem with balance between discretization bias and variance.
The more bins, the lower the bias, but the higher the variance, due to small sample size.

42. Alternative simple (but effective) discretization method (Yang & Webb, 2001)
Let n = number of training examples. For each attribute Ai , create bins. Sort values of Ai in ascending order, and put of them in each bin.
Don’t need add-one smoothing of probabilities
This gives good balance between discretization bias and variance.

43. Alternative simple (but effective) discretization method (Yang & Webb, 2001)
Let n = number of training examples. For each attribute Ai , create bins. Sort values of Ai in ascending order, and put of them in each bin.
Don’t need add-one smoothing of probabilities
This gives good balance between discretization bias and variance.
Humidity: 25, 38, 50, 80, 90, 92, 96, 99

44. Beyond Independence: Conditions for the Optimality of the Simple Bayesian Classifer (P. Domingos and M. Pazzani)
Naive Bayes classifier is called “naive” because it assumes attributes are independent of one another.

45. This paper asks: why does the naive (“simple”) Bayes classifier, SBC, do so well in domains with clearly dependent attributes?

46. Experiments
Compare five classification methods on 30 data sets from the UCI ML database.
SBC = Simple Bayesian Classifier
Default = “Choose class with most representatives in data”
C4.5 = Quinlan’s decision tree induction system
PEBLS = An instance-based learning system
CN2 = A rule-induction system

47. For SBC, numeric values were discretized into ten equal-length intervals.

49. Number of domains in which SBC was more accurate versus less accurate than corresponding classifier
Same as line 1, but significant at 95% confidence
Average rank over all domains (1 is best in each domain)

50. Measuring Attribute Dependence
They used a simple, pairwise mutual information measure:
For attributes Am and An , dependence is defined as
where AmAn is a “derived attribute”, whose values consist of the possible combinations of values of Am and An
Note: If Am and An are independent, then D(Am, An | C) = 0.

51. Results:
(1) SBC is more
successful than more complex
methods, even when there is substantial dependence among attributes.
(2) No correlation between degree
of attribute dependence and
SBC’s rank.
But why????

52. Explanation:
Suppose C = {+1,−1} are the possible classes. Let x be a new example with attributes <a1, a2, ..., an>..
What the naive Bayes classifier does is calculates two probabilities,
and returns the class that has the maximum probability given x.

53. The probability calculations are correct only if the independence assumption is correct.
However, the classification is correct in all cases in which the relative ranking of the two probabilities, as calculated by the SBC, is correct!
The latter covers a lot more cases than the former.
Thus, the SBC is effective in many cases in which the independence assumption does not hold.