Naïve Bayes Classification
CS 6243 Machine Learning
Modified from the slides by Dr. Raymond J. Mooney
http://www.cs.utexas.edu/~mooney/cs391L/
1
Probability Basics
• Definition (informal)
– Probabilities are numbers assigned to events
that indicate “how likely” it is that the event
will occur when a random experiment is
performed
– A probability law for a random experiment is a
rule that assigns probabilities to the events in
the experiment
– The sample space S of a random experiment is
the set of all possible outcomes
2
Probabilistic Calculus
• All probabilities between 0 and 1
0  P( A)  1
• If A, B are mutually exclusive:
– P(A  B) = P(A) + P(B)
• Thus: P(not(A)) = P(Ac) = 1 – P(A)
S
A
B
3
Conditional probability
• The joint probability of two events A and B P(AB), or
simply P(A, B) is the probability that event A and B occur
at the same time.
• The conditional probability of P(A|B) is the probability
that A occurs given B occurred.
P(A | B) = P(A  B) / P(B)
<=> P(A  B) = P(A | B) P(B)
<=> P(A  B) = P(B|A) P(A)
4
Example
• Roll a die
– If I tell you the number is less than 4
– What is the probability of an even number?
• P(d = even | d < 4) = P(d = even  d < 4) / P(d < 4)
• P(d = 2) / P(d = 1, 2, or 3) = (1/6) / (3/6) = 1/3
5
Independence
• A and B are independent iff:
P( A | B)  P( A)
P( B | A)  P( B)
These two constraints are logically equivalent
• Therefore, if A and B are independent:
P( A  B)
P( A | B) 
 P( A)
P( B)
P( A  B)  P( A) P( B)
6
Examples
• Are P(d = even) and P(d < 4) independent?
–
–
–
–
P(d = even and d < 4) = 1/6
P(d = even) = ½
P(d < 4) = ½
½ * ½ > 1/6
• If your die actually has 8 faces, will P(d = even)
and P(d < 5) be independent?
• Are P(even in first roll) and P(even in second roll)
independent?
• Playing card, are the suit and rank independent?
7
Theorem of total probability
• Let B1, B2, …, BN be mutually exclusive events whose union equals
the sample space S. We refer to these sets as a partition of S.
• An event A can be represented as:
•
•
Since B1, B2, …, BN are mutually exclusive, then
P(A) = P(A  B1) + P(A  B2) + … + P(A  BN) Marginalization
And therefore
P(A) = P(A|B1)*P(B1) + P(A|B2)*P(B2) + … + P(A|BN)*P(BN)
Exhaustive conditionalization
= i P(A | Bi) * P(Bi)
8
Example
• A loaded die:
– P(6) = 0.5
– P(1) = … = P(5) = 0.1
• Prob of even number?
P(even)
= P(even | d < 6) * P (d<6) +
P(even | d = 6) * P (d=6)
= 2/5 * 0.5 + 1 * 0.5
= 0.7
9
Another example
• A box of dice:
– 99% fair
– 1% loaded
• P(6) = 0.5.
• P(1) = … = P(5) = 0.1
– Randomly pick a die and roll, P(6)?
• P(6) = P(6 | F) * P(F) + P(6 | L) * P(L)
– 1/6 * 0.99 + 0.5 * 0.01 = 0.17
10
Bayes theorem
• P(A  B) = P(B) * P(A | B) = P(A) * P(B | A)
Conditional probability
(likelihood)
=> P(B | A) =
Posterior probability
P ( A | B ) P (B )
Prior of B
P(A )
Prior of A (Normalizing constant)
This is known as Bayes Theorem or Bayes Rule, and is (one of) the
most useful relations in probability and statistics
Bayes Theorem is definitely the fundamental relation in Statistical Pattern
Recognition
11
Bayes theorem (cont’d)
• Given B1, B2, …, BN, a partition of the sample
space S. Suppose that event A occurs; what is the
probability of event Bj?
Posterior probability
Likelihood
Prior of Bj
• P(Bj | A) = P(A | Bj) * P(Bj) / P(A)
Normalizing constant
= P(A | Bj) * P(Bj) / jP(A | Bj)*P(Bj)
Bj: different models / hypotheses
(theorem of total probabilities)
In the observation of A, should you choose a model that maximizes
P(Bj | A) or P(A | Bj)? Depending on how much you know about Bj !
12
Example
• A test for a rare disease claims that it will report
positive for 99.5% of people with disease, and
negative 99.9% of time for those without.
• The disease is present in the population at 1 in
100,000
• What is P(disease | positive test)?
– P(D|+) = P(+|D)P(D)/P(+) = 0.01
• What is P(disease | negative test)?
– P(D|-) = P(-|D)P(D)/P(-) = 5e-8
13
Joint Distribution
• The joint probability distribution for a set of random variables,
X1,…,Xn gives the probability of every combination of values (an ndimensional array with vn values if all variables are discrete with v
values, all vn values must sum to 1): P(X1,…,Xn)
negative
positive
circle
square
red
0.20
0.02
blue
0.02
0.01
circle
square
red
0.05
0.30
blue
0.20
0.20
• The probability of all possible conjunctions (assignments of values to
some subset of variables) can be calculated by summing the
appropriate subset of values from the joint distribution.
P(red  circle )  0.20  0.05  0.25
P(red )  0.20  0.02  0.05  0.3  0.57
• Therefore, all conditional probabilities can also be calculated.
P( positive | red  circle ) 
P( positive  red  circle ) 0.20

 0.80
P(red  circle )
0.25
14
Probabilistic Classification
• Let Y be the random variable for the class which takes values
{y1,y2,…ym}.
• Let X be the random variable describing an instance consisting
of a vector of values for n features <X1,X2…Xn>, let xk be a
possible value for X and xij a possible value for Xi.
• For classification, we need to compute P(Y=yi | X=xk) for i=1…m
• However, given no other assumptions, this requires a table
giving the probability of each category for each possible instance
in the instance space, which is impossible to accurately estimate
from a reasonably-sized training set.
– Assuming Y and all Xi are binary, we need 2n entries to specify
P(Y=pos | X=xk) for each of the 2n possible xk’s since
P(Y=neg | X=xk) = 1 – P(Y=pos | X=xk)
– Compared to 2n+1 – 1 entries for the joint distribution P(Y,X1,X2…Xn)
15
Bayesian Categorization
• Determine category of xk by determining for each yi
P(Y  yi | X  xk ) 
P( X  xk | Y  yi ) P(Y  yi )
P ( X  xk )
• P(X=xk) can be determined since categories are
complete and disjoint.
m
m
i 1
i 1
 P(Y  yi | X  xk )  
P( X  xk | Y  yi ) P(Y  yi )
1
P( X  xk )
m
P( X  xk )   P( X  xk | Y  yi ) P(Y  yi )
i 1
16
Bayesian Categorization (cont.)
• Need to know:
– Priors: P(Y=yi)
– Conditionals: P(X=xk | Y=yi)
• P(Y=yi) are easily estimated from data.
– If ni of the examples in D are in yi then P(Y=yi) = ni / |D|
• Too many possible instances (e.g. 2n for binary
features) to estimate all P(X=xk | Y=yi).
• Need to make some sort of independence
assumptions about the features to make learning
tractable.
17
Conditional independence
• Chain rule:
P(x1, x2, x3) = P(x1, x2, x3) / P(x2, x3)
* P(x2, x3) / P(x3)
* P(x3)
= P(x1 | x2, x3) P(x2 | x3) P(x3)
• P(x1, x2, x3 | Y)
= P(x1 | x2, x3, Y) P(x2 | x3 , Y) P(x3 | Y)
= P(x1 | Y) P(x2 | Y) P(x3 , Y) assuming conditional independence
18
Naïve Bayes Generative Model
neg
pos pos
pos neg
pos neg
Category
med
sm lg
med
lg lg sm
sm med
red
blue
red grn red
red blue
red
circ
tri tricirc
circ circ
circ sqr
lg
sm
med med
sm lglg
sm
red
blue
grn grn
red blue
blue grn
circ
sqr
tri
circ
circ tri sqr
sqr tri
Size
Color
Shape
Size
Color
Shape
Positive
Negative
20
Naïve Bayes Inference Problem
lg red circ
??
??
neg
pos pos
pos neg
pos neg
Category
med
sm lg
med
lg lg sm
sm med
red
blue
red grn red
red blue
red
circ
tri tricirc
circ circ
circ sqr
lg
sm
med med
sm lglg
sm
red
blue
grn grn
red blue
blue grn
circ
sqr
tri
circ
circ tri sqr
sqr tri
Size
Color
Shape
Size
Color
Shape
Positive
Negative
21
Naïve Bayesian Categorization
• If we assume features of an instance are independent given
the category (conditionally independent).
n
P( X | Y )  P( X 1 , X 2 , X n | Y )   P( X i | Y )
i 1
• Therefore, we then only need to know P(Xi | Y) for each
possible pair of a feature-value and a category.
• If Y and all Xi and binary, this requires specifying only 2n
parameters:
– P(Xi=true | Y=true) and P(Xi=true | Y=false) for each Xi
– P(Xi=false | Y) = 1 – P(Xi=true | Y)
• Compared to specifying 2n parameters without any
independence assumptions.
22
Naïve Bayes Example
Probability
positive
negative
P(Y)
0.5
0.5
P(small | Y)
0.4
0.4
P(medium | Y)
0.1
0.2
P(large | Y)
0.5
0.4
P(red | Y)
0.9
0.3
P(blue | Y)
0.05
0.3
P(green | Y)
0.05
0.4
P(square | Y)
0.05
0.4
P(triangle | Y)
0.05
0.3
P(circle | Y)
0.9
0.3
Test Instance:
<medium ,red, circle>
23
Naïve Bayes Example
Probability
positive
negative
P(Y)
0.5
0.5
P(medium | Y)
0.1
0.2
P(red | Y)
0.9
0.3
P(circle | Y)
0.9
0.3
Test Instance:
<medium ,red, circle>
P(positive | X) = P(positive)*P(medium | positive)*P(red | positive)*P(circle | positive) / P(X)
0.5
*
0.1
*
0.9
*
0.9
= 0.0405 / P(X) = 0.0405 / 0.0495 = 0.8181
P(negative | X) = P(negative)*P(medium | negative)*P(red | negative)*P(circle | negative) / P(X)
0.5
*
0.2
*
0.3
* 0.3
= 0.009 / P(X) = 0.009 / 0.0495 = 0.1818
P(positive | X) + P(negative | X) = 0.0405 / P(X) + 0.009 / P(X) = 1
P(X) = (0.0405 + 0.009) = 0.0495
24
Estimating Probabilities
• Normally, probabilities are estimated based on observed
frequencies in the training data.
• If D contains nk examples in category yk, and nijk of these nk
examples have the jth value for feature Xi, xij, then:
P( X i  xij | Y  yk ) 
nijk
nk
• However, estimating such probabilities from small training
sets is error-prone.
• If due only to chance, a rare feature, Xi, is always false in
the training data, yk :P(Xi=true | Y=yk) = 0.
• If Xi=true then occurs in a test example, X, the result is that
yk: P(X | Y=yk) = 0 and yk: P(Y=yk | X) = 0
25
Probability Estimation Example
Ex
Size
Color
Shape
Category
1
small
red
circle
positive
2
large
red
circle
positive
3
small
red
triangle
negitive
4
large
blue
circle
negitive
Test Instance X:
<medium, red, circle>
Probability
positive
negative
P(Y)
0.5
0.5
P(small | Y)
0.5
0.5
P(medium | Y)
0.0
0.0
P(large | Y)
0.5
0.5
P(red | Y)
1.0
0.5
P(blue | Y)
0.0
0.5
P(green | Y)
0.0
0.0
P(square | Y)
0.0
0.0
P(triangle | Y)
0.0
0.5
P(circle | Y)
1.0
0.5
P(positive | X) = 0.5 * 0.0 * 1.0 * 1.0 / P(X) = 0
P(negative | X) = 0.5 * 0.0 * 0.5 * 0.5 / P(X) = 0
26
Smoothing
• To account for estimation from small samples,
probability estimates are adjusted or smoothed.
• Laplace smoothing using an m-estimate assumes that
each feature is given a prior probability, p, that is
assumed to have been previously observed in a
“virtual” sample of size m.
P( X i  xij | Y  yk ) 
nijk  mp
nk  m
• For binary features, p is simply assumed to be 0.5.
27
Laplace Smothing Example
• Assume training set contains 10 positive examples:
– 4: small
– 0: medium
– 6: large
• Estimate parameters as follows (if m=1, p=1/3)
–
–
–
–
P(small | positive) = (4 + 1/3) / (10 + 1) = 0.394
P(medium | positive) = (0 + 1/3) / (10 + 1) = 0.03
P(large | positive) = (6 + 1/3) / (10 + 1) = 0.576
P(small or medium or large | positive) =
1.0
28
Missing values
• Training: instance is not included in
frequency count for attribute value-class
combination
• Classification: attribute will be omitted
from calculation
• Example: Outlook Temp. Humidity Windy Play
?
Cool
High
True
?
Likelihood of “yes” = 3/9  3/9  3/9  9/14 = 0.0238
Likelihood of “no” = 1/5  4/5  3/5  5/14 = 0.0343
P(“yes”) = 0.0238 / (0.0238 + 0.0343) = 41%
P(“no”) = 0.0343 / (0.0238 + 0.0343) = 59%
29
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Continuous Attributes
• If Xi is a continuous feature rather than a discrete one, need
another way to calculate P(Xi | Y).
• Assume that Xi has a Gaussian distribution whose mean
and variance depends on Y.
• During training, for each combination of a continuous
feature Xi and a class value for Y, yk, estimate a mean, μik ,
and standard deviation σik based on the values of feature Xi
in class yk in the training data.
• During testing, estimate P(Xi | Y=yk) for a given example,
using the Gaussian distribution defined by μik and σik .
P ( X i | Y  yk ) 
1
 ik
  ( X i  ik ) 2 

exp 
2

2

2
ik


30
Statistics for
weather data
Outlook
Temperature
Humidity
Windy
Yes
No
Yes
No
Yes
No
Sunny
2
3
64, 68,
65, 71,
65, 70,
70, 85,
Overcast
4
0
69, 70,
72, 80,
70, 75,
90, 91,
Rainy
3
2
72, …
85, …
80, …
95, …
Sunny
2/9
3/5
 =73
 =75
 =79
Overcast
4/9
0/5
 =6.2
 =7.9
 =10.2
Rainy
3/9
2/5
Play
Yes
No
Yes
No
False
6
2
9
5
True
3
3
 =86
False
6/9
2/5
 =9.7
True
3/9
3/5
9/14 5/14
• Example density value:
f (temperature  66 | yes ) 
1
2 6.2
e
( 66 73 )2

26.22
 0.0340
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Classifying a new day
• A new day:
Outlook
Temp.
Humidity
Windy
Play
Sunny
66
90
true
?
Likelihood of “yes” = 2/9  0.0340  0.0221  3/9  9/14 = 0.000036
Likelihood of “no” = 3/5  0.0291  0.0380  3/5  5/14 = 0.000136
P(“yes”) = 0.000036 / (0.000036 + 0. 000136) = 20.9%
P(“no”) = 0.000136 / (0.000036 + 0. 000136) = 79.1%
• Missing values during training are not
included in calculation of mean and standard
deviation
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*Probability densities
• Relationship between probability and density:


Pr[c   x  c  ]    f (c)
2
2
• But: this doesn’t change calculation of a
posteriori probabilities because  cancels out
• Exact relationship:
b
Pr[ a  x  b] 
 f (t )dt
a
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Naïve Bayes: discussion
•
•
•
•
•
•
Naïve Bayes works surprisingly well (even if independence
assumption is clearly violated)
– Experiments show it to be quite competitive with other
classification methods on standard UCI datasets.
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 (
kernel density estimators)
Does not perform any search of the hypothesis space. Directly
constructs a hypothesis from parameter estimates that are easily
calculated from the training data.
Typically handles noise well since it does not even focus on
completely fitting the training data.
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Naïve Bayes Extensions
• Improvements:
– select best attributes (e.g. with greedy search)
– often works as well or better with just a
fraction of all attributes
• Bayesian Networks
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