4400_645_Fall04_Lecture6

4400_645_Fall04_Lecture6 - Detection and Estimation Theory...

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Unformatted text preview: Detection and Estimation Theory Electrical & Computer Engineering 4400:693 (645) Fall 2004 Lecture 6 Monday September 20, 2004 © Dr. O. C. Ugweje, University of Akron Page 1 Review of Previous Lectures Electrical & Computer Engineering General Introduction – course mechanics √ Introduction to Detection Theory √ Types of Detection Problems Detection of known signals in noise Detection of signals with unknown parameters in noise Detection of random signals in noise Simplest Detection Problem Issues in Detection Introduction to Estimation Theory √ Tasks of Estimation Types of Estimation Problems Parameter Estimation Signal Estimation © Dr. O. C. Ugweje, University of Akron Page 2 Review (cont’d) Electrical & Computer Engineering Review of Important PDFs – Chapter 2 of textbook Detection Theory Binary Hypothesis Testing Problems (I) Bayesian Approach (II) Minimax Approach (III) Neyman-Parson Approach Bayesian Minimax Neyman-Parson Cost (Cij) Known Known Unknown Prior Probability Known Unknown Unknown Note: All approaches are determined by a likelihood ratio test. Bayes Hypothesis Testing Minimax Hypothesis Testing © Dr. O. C. Ugweje, University of Akron Page 3 Examples Electrical & Computer Engineering Example 1 (Bayes Approach) Consider the hypothesis testing problem H1 : X = a + n binary "1" transmitted H0 : X = n binary "0" transmitted where n is a Gaussian random variable with zero-mean and unit variance. We want to determine the decision rule and the decision space. Example 2 (Bayes Approach) For the binary channel shown below, find the likelihood ratio test and the test statistics. 1− λ 0 λ1 λ0 1 − λ0 © Dr. O. C. Ugweje, University of Akron Page 4 Examples (cont’d) Electrical & Computer Engineering Example 3 (Bayes Approach) Consider the hypothesis testing problem H1 : Y ~ P0 = N ( µ0 , σ 2 ) H 0 : Y ~ P = N ( µ1 , σ 2 ) 1 Determine the decision rule and the decision space. Example 4 (Bayes & Minimax Approach) Suppose Y is a random variable the following hyposeses 2 ( y + 1) , 0 ≤ y ≤ 1 H 0 : Y ~ P0 ( y ) = 3 0, else 1, 0 ≤ y ≤ 1 H1 : Y ~ P ( y ) = 1 0, else © Dr. O. C. Ugweje, University of Akron Page 5 Example 4 (cont’d) Electrical & Computer Engineering A. Find the Bayes Rule and the minimum Bayes risk for testing H0 vs. H1 with uniform cost and equal priors. B. Find the Minimax Rule and Minimax Risk for uniform cost. Solution: C00 = C11 = 0 For uniform cost Cij = 1-dij ⇒ C01 = C10 = 1 C - C00 π 0 P(H0)=P(H1) ⇒ 10 ⋅ = 1=γ C01 - C11 π 1 For (A) 3 , 0 ≤ y ≤1 P ( y | H1 ) P ( y ) L( y ) = 1 =1 = 2 ( y + 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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