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Unit 5 Probabilistic Model

Transcript of Unit 5 - khyatirnirmal.files.wordpress.com · Unit 5 Probabilistic Model •Machine Learning Task...

Page 1: Unit 5 - khyatirnirmal.files.wordpress.com · Unit 5 Probabilistic Model •Machine Learning Task Given X and Y X is Feature Vector Y is target value associated with X •For given

Unit 5

Probabilistic Model

Page 2: Unit 5 - khyatirnirmal.files.wordpress.com · Unit 5 Probabilistic Model •Machine Learning Task Given X and Y X is Feature Vector Y is target value associated with X •For given

• Machine Learning Task

Given X and Y

X is Feature Vector

Y is target value associated with X

• For given training data as (X,Y) pair the objective is:

i. To construct a learning Model

ii. To use the learning model to predict value of Y associated with unseen X

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• X can be x1,x2… xi Y can be y1,y2…yi

• For binary Classification problem j=2 therefor y = y1 and y2

• Let N number of training pairs, (X,Y) are given as i/p.

• Yj= all feature vectors or instance belongs to Yj

• Rj = number of feature vectors belongs to class Yj

• Xi= set of feature value or instances of type Xi

• Ci= total number of instances of type xi

• Nij= number of features of type Xi belonging class Yj

xi

Yj

Ci

rj

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Joint Probability Example

P(Red and Ace)

Black

Color Type Red Total

Ace 2 2 4

Non-Ace 24 24 48

Total 26 26 52

52

2

cards of number total

ace and red are that cards of number

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Join Probability

• What is the probability that selected sample of type Xi and also belongs to class Yj

P(X=xi, Y=yj)

= P(X=xi ^ Y=yj)

=

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Marginal Probability Example

P(Ace)

Black

Color Type Red Total

Ace 2 2 4

Non-Ace 24 24 48

Total 26 26 52

52

4

52

2

52

2)BlackandAce(P)dReandAce(P

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Marginal Probability

• Let an instance is selected from input data

• What is the probability that selected sample is of type Xi

• Required Probability

• P(X=xi)=

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Conditional Probabilities

Liberal Moderate Conservative Total

Gender Male 17 29 14 60

Female 30 24 23 77

Total 47 53 37 137

Political views

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Here is a contingency table that gives the counts of ECO 138 students by their gender and political views. (Data are from Fall 2005 Class Survey)

P(Female) = 77/137 = 0.562

P(Female and Liberal) = 30/137 = 0.219

What is the probability that a selected student has moderate political views given that we have selected a female?

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Conditional Probabilities (continued)

Liberal Moderate Conservative Total

Gender Male 17 29 14 60

Female 30 24 23 77

Total 47 53 37 137

Political views

9

What is the probability that a selected student has moderate political views given that we have selected a female?

P(Moderate | Female) = 24/77 = 0.311

Conditional probability, P (B|A) – the probability of event B given event A.

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Conditional Probabilities (continued)

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Formal Definition:

P(B|A) = P(A and B)

P(A)

Example: P(Moderate and Female)

P(Female)

=(24/137) / (77/137)

= 0.175 / 0.562

= 0.311

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Conditional Probability

• Let an instance is selected from input data

• What is the probability that selected sample is of type Xi

• What is the probability that selected input belongs to Yj

• P(Y=yj | X= xi) =

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• Conditional probability:

• Bayes theorem:

Bayesian classification

)(

)()|()|(

x

xx

p

CpCpCp

)(

),()|(

)(

),()|(

Cp

CpCp

p

CpCp

xx

x

xx

posterior

probability

likelihood prior probability

evidence

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Naïve Bayes Classification

• Based on Bayes Rule

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Finally, we classify X as RED since its class membership achieves the largest posterior probability.

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Naïve Bayes Solved Example

OUTLOOK TEMPERATURE HUMIDITY WINDY PLAY GOLF

0 Rainy Hot High False No

1 Rainy Hot High True No

2 Overcast Hot High False Yes

3 Sunny Mild High False Yes

4 Sunny Cool Normal False Yes

5 Sunny Cool Normal True No

6 Overcast Cool Normal True Yes

7 Rainy Mild High False No

8 Rainy Cool Normal False Yes

9 Sunny Mild Normal False Yes

10 Rainy Mild Normal True Yes

11 Overcast Mild High True Yes

12 Overcast Hot Normal False Yes

13 Sunny Mild High True No

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Example

In this example we have 4 inputs (predictors). The final posterior probabilities can be standardized between 0 and 1.

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P (N0 | Today) > P (Yes | Today)

So, prediction that golf would be played is ‘No’.