2
3
5.5.1


        Theorem)                  n
        X = (T1 = x1 )∧(T2 = x2 )∧· · ·∧(Tn = xn )
   CH X               C = {...
P (CH ∩ X)   P (X | CH )P (CH )
P (CH   | X) =            =
                  P (X)           P (X)
P (C =   | X) > P (C = × | X)




P (C =   | X) < P (C = × | X)




                                6
P (X | C )P (C )
   P (C   | X) =
                          P (X)
        P (C ) = N /N

           X   = (T1 = x1 ) ∧ (T2...
P (C ) = 4/10 = 0.4, P (C× ) = 6/10 = 0.6
X = (T1 = No) ∧ (T2 = No) ∧ (T3 = Yes) ∧ (T4 = Yes)
                            ...
X = (T1 = No) ∧ (T2 = No) ∧ (T3 = Yes) ∧ (T4 = Yes)
   P (X | C )   = P (T1 = No | C ) × P (T2 = No | C )
                ...
X = (T1 = No) ∧ (T2 = No) ∧ (T3 = Yes) ∧ (T4 = Yes)
P (X | C× )   =   P (T1 = No | C× ) × P (T2 = No | C× )×
             ...
P (X | C ) · P (C ) = 0.0
  P (X | C× ) · P (C× ) = 0.0198

P (X | C ) · P (C ) < (X | C× ) · P (C× )




                ...
12
Datamining 3rd Naivebayes
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Datamining 3rd Naivebayes

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Datamining 3rd Naivebayes

  1. 1. 2
  2. 2. 3
  3. 3. 5.5.1 Theorem) n X = (T1 = x1 )∧(T2 = x2 )∧· · ·∧(Tn = xn ) CH X C = {C1 , C2 , ...} X CH X CH 5.3 P (CH ∩ X) P (X | CH )P (CH ) P (CH | X) = CH = (C = ) P (X) P (X) P (CH |A) A P (CH |X) 5.4 P (CH ∩ X) = P (CH | X)P (X) = PP (C| CHX) (CH ) 4 H ) (X H ∩ )P P (X|C
  4. 4. P (CH ∩ X) P (X | CH )P (CH ) P (CH | X) = = P (X) P (X)
  5. 5. P (C = | X) > P (C = × | X) P (C = | X) < P (C = × | X) 6
  6. 6. P (X | C )P (C ) P (C | X) = P (X) P (C ) = N /N X = (T1 = x1 ) ∧ (T2 = x2 ) ∧ · · · ∧ (Tn = xn ) = x1 ∧ x2 ∧ · · · ∧ x3 P (X | C ) = P (x1 ∧ x2 ∧ · · · ∧ xn | C ) = P (x1 | C )P (x2 | C ) · · · P (xn | C ) n = P (xk | C ) k=1 P (X) CH
  7. 7. P (C ) = 4/10 = 0.4, P (C× ) = 6/10 = 0.6 X = (T1 = No) ∧ (T2 = No) ∧ (T3 = Yes) ∧ (T4 = Yes) 8
  8. 8. X = (T1 = No) ∧ (T2 = No) ∧ (T3 = Yes) ∧ (T4 = Yes) P (X | C ) = P (T1 = No | C ) × P (T2 = No | C ) ×P (T3 = Yes) | C ) × P (T4 = Yes | C ) P (T1 = No | C ) = 2/4 = 0.5 P (T2 = No | C ) = 0/4 = 0.0 P (T3 = Yes | C ) = 2/4 = 0.5 P (T4 = Yes | C ) = 0/4 = 0.0 P (X|C ) = 0.5 × 0.0 × 0.5 × 0.0 = 0.0 P (X | C ) · P (C ) = 0.0 × 0.4 = 0.0
  9. 9. X = (T1 = No) ∧ (T2 = No) ∧ (T3 = Yes) ∧ (T4 = Yes) P (X | C× ) = P (T1 = No | C× ) × P (T2 = No | C× )× P (T3 = Yes | C× ) × P (T4 = Yes | C× ) P (T1 = No | C× ) = 4/6 = 0.667 P (T2 = No | C× ) = 4/6 = 0.667 P (T3 = Yes | C× ) = 1/6 = 0.167 P (T4 = Yes | C× ) = 4/6 = 0.667 P (X|C× ) = 0.667 × 0.667 × 0.167 × 0.667 = 0.0494 P (X | C× ) · P (C× ) = 0.0494 × 0.4 = 0.0198
  10. 10. P (X | C ) · P (C ) = 0.0 P (X | C× ) · P (C× ) = 0.0198 P (X | C ) · P (C ) < (X | C× ) · P (C× ) 11
  11. 11. 12

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