4400_645_Fall04_Lecture6

O c ugweje university of akron page 6 example 4 contd

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Unformatted text preview: 1) P0 ( y | H 0 ) P0 ( y ) else 0, This means that 3 3 L( y ) = 1 ⇒ 3 2 ( y + 1) ⇒ − 1 y 2 ( y + 1) 2 © Dr. O. C. Ugweje, University of Akron Page 6 Example 4 (cont’d) Electrical & Computer Engineering This implies that 1 y H1 2 Thus the Bayes Rule becomes 1 1, 0 ≤ y ≤ 2 δB ( y) = 0, 1 < y ≤ 1 2 H0 Minimum Bayes Risk r (δ B ) = π 0 R0 (δ B ) + π 1 R1 (δ B ) = π 0 [C00 P0 ( Γ 0 ) + C10 P0 ( Γ1 )] + π 1 [C01 P ( Γ 0 ) + C11 P ( Γ1 )] 1 1 ↓ C00 = C11 = 0 = π 0C10 P0 ( Γ1 ) + π 1C01 P ( Γ 0 ) 1 © Dr. O. C. Ugweje, University of Akron Page 7 Example 4 (cont’d) Electrical & Computer Engineering But P0 [ Γ1 ] = 1/ 2 ∫ 0 1 2 5 ( y + 1) dy = 3 12 1 P [ Γ 0 ] = ∫ (1) dy = 1 2 1/ 2 Therefore 5 + π 1 = 1 5 + 1 1 = 11 r (δ B ) = π 0 1 12 2 2 12 2 2 24 For (B) L( y ) = C10 - C00 π 0 π π 3 = 0 = 0 =γ ⋅ 2 ( y + 1) C01 - C11 π 1 π1 1 − π 0 3 − 5π 0 ⇒y 2π 0 © Dr. O. C. Ugweje, University of Akron Page 8 Example 4 (cont’d) Electrical & Computer Engineering Letting 3 − 5π 0 γ '= 2π 0 we have Minimax Risk: 1, y <γ ' δ M ( y ) = 1 or 0, y = γ ', 0, y > γ ', 3 7 3 7 3 7 3 3 ≤ π 0 ≤ 5 or 0 ≤ π 0 ≤ 7 3 < π0 < 5 3 ≤ π 0 ≤ 5 or 3 5 ≤ π0 ≤ 1 V (π 0 ) = r (π 0 , δπ 0 ) = π 0 R0 (δ π 0 ) + (1 − π 0 ) R1 (δ π 0 ) where 0 R0 (δ π 0 ) = C00 P0 ( Γ 0 ) + C10 P0 ( Γ1 ) 0 R1 (δ π 0 ) = C11 P ( Γ1 ) + C01 P ( Γ 0 ) 1 1 © Dr. O. C. Ugweje, University of Akron Page 9 Example 4 (cont’d) Electrical & Computer Engineering and 3 1, 0 ≤ π0 ≤ 7 γ' 2 ( γ ')2 2 3 3 P0 ( Γ1 ) = ∫ 3 ( y + 1) dy = 3 , 7 < π0 < 5 +γ ' 2 0 3 0, ≤ π0 ≤ 1 5 ( ) 3 0, 0 ≤ π0 ≤ 7 1 3 3 P ( Γ 0 ) = ∫ (1) dy = (1 − γ ') , 7 < π 0 < 5 1 3 γ' 1, ≤ π0 ≤1 5 That is R0 (δ π 0 ) = C10 P0 ( Γ1 ) = 2 3 ( ( γ ')2 2 +γ ' R1 (δ π 0 ) = C01 P ( Γ 0 ) = (1 − γ ') 1 ) (**) Making these two equations equal, we can compute for the ' optimum γ L © Dr. O. C. Ugweje, University of Akron Page 10 Example 4 (cont’d) Electrical & Computer Engineering R0 (δ π 0 ) = R1 (δ π 0 ) ⇒2 3 ( ( γ ' )2 2 ) + γ ' = (1 − γ ') ⇒ ( γ ') + 2γ '+ 3γ '− 3 = 0 2 ⇒ ( γ ') + 5γ '− 3 = 0 2 ∴γ ' = −5 ± 37 37 − 5 ⇒ = 0.541 2 2 Thus the maximum risk is V (π 0 ) π ' 0 =γ L Therefore = π 0 R0 (δ π 0 ) + (1 − π 0 ) R1 (δ π 0 ) ' π 0 =γ L = π 0 { R0 (δ π 0 ) − R1 (δ π 0 )} + R1 (δ π 0 ) = R1 (δ π 0 ) π ' 0 =γ L = (1 − γ ' ) π ' π 0 =γ L ' 0 =γ L ' ' V ( γ L ) = (1 − γ L ) © Dr. O. C. Ugweje, University of Akron Page 11...
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This note was uploaded on 10/01/2010 for the course ELEC 6111 taught by Professor Brown during the Spring '10 term at E. Illinois.

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