Recap
•We have studied the Bayes Classifier last class.
PR NPTEL course – p.1/128
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Deriving Bayes Classifier: An Example from NPTEL Course, Slides of Design and Analysis of Algorithms
An example of deriving bayes classifier from nptel course. It explains the concept of bayes classifier, its optimization for minimizing risk, and the method to derive it when class conditional densities are known. The document also covers related topics such as neyman-pearson classifier, minmax classifier, and receiver operating characteristic (roc).
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2012/2013
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Recap^ •
We have studied the Bayes Classifier last class. PR NPTEL course – p.1/
Recap^ •
We have studied the Bayes Classifier last class.
- We have seen the Bayes classifier for general lossfunction and proved its optimality. PR NPTEL course – p.2/
An Example
Consider another example of deriving Bayes classifier.
- Suppose we have
K
classes. The classifier is allowed the option to ‘reject’ a pattern and this is done by theclassifier assigning class
K
to the pattern. PR NPTEL course – p.4/
An Example
Consider another example of deriving Bayes classifier.
- Suppose we have
K
classes. The classifier is allowed the option to ‘reject’ a pattern and this is done by theclassifier assigning class
K
to the pattern. Define the loss function by
L
i, j
if i
j and i, j
, K
ρ m if i
, K,
and i
j
ρ r if i
K
PR NPTEL course – p.5/
Example Contd.^ •
Recall that the Bayes classifier is h B
X
α i if R
α i
X
R
α j
X
j. where
R
α i
X
K ∑^ j =
L
α i
, C
j
q j
X
) PR NPTEL course – p.7/
Example Contd.^ •
Recall that the Bayes classifier is h B
X
α i if R
α i
X
R
α j
X
j. where
R
α i
X
K ∑^ j =
L
α i
, C
j
q j
X
- So, we now need to calculate
R
α i
X
for different actions, α i available to the classifier. PR NPTEL course – p.8/
- For α i
, K
, we have
L
α i
, C
j
ρ m if α i
C
j and it is zero otherwise.
- Hence,
R
i
X
j 6 = i ρ m q j
X
ρ m
q i
X
. PR NPTEL course – p.10/
- For α i
, K
, we have
L
α i
, C
j
ρ m if α i
C
j and it is zero otherwise.
- Hence,
R
i
X
j 6 = i ρ m q j
X
ρ m
q i
X
.
- Also,
R
K
X
j ρ r q j
X
ρ r PR NPTEL course – p.11/
- For α i
, K
, we have
L
α i
, C
j
ρ m if α i
C
j and it is zero otherwise.
- Hence,
R
i
X
j 6 = i ρ m q j
X
ρ m
q i
X
.
- Also,
R
K
X
j ρ r q j
X
ρ r
- Hence, h B
X
i ,
i
K
, if ρ m
q i
X
ρ m
q j
X
j and ρ m
q i
X
ρ r PR NPTEL course – p.13/
- We have, h B
X
i ,
i
K
, if ρ m
q i
X
ρ m
q j
X
j and ρ m
q i
X
ρ r PR NPTEL course – p.14/
- We saw, h B
X
i ,
i
K
, if (i). q i
X
q j
X
j , and (ii). q i
X
ρ r ρ m ; else h B
X
K
. PR NPTEL course – p.16/
- We saw, h B
X
i ,
i
K
, if (i). q i
X
q j
X
j , and (ii). q i
X
ρ r ρ m ; else h B
X
K
.
- If ρ r
ρ m PR NPTEL course – p.17/
- We saw, h B
X
i ,
i
K
, if (i). q i
X
q j
X
j , and (ii). q i
X
ρ r ρ m ; else h B
X
K
.
- If ρ r
ρ m
Never reject a patter!
If ρ r
PR NPTEL course – p.19/
- We saw, h B
X
i ,
i
K
, if (i). q i
X
q j
X
j , and (ii). q i
X
ρ r ρ m ; else h B
X
K
.
- If ρ r
ρ m
Never reject a patter!
If ρ rAlways reject the pattern (unless you are absolutely sure) PR NPTEL course – p.20/