An Introduction to Signal Detection and Estimation - Second Edition Chapter II: Selected Solutions

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1 An Introduction to Signal Detection Estimation - Second Edition Chapter II: Selected Solutions H V Poor Princeton University March 16, 5 Exercise : The likelihood ratio is given by L(y) (y +1), y 1 a With uniform costs equal priors, the critical region for minimum Bayes error is given by y [, 1] L(y) 1} y [, 1] (y +1)} [, 1/] Thus the Bayes rule is given by 1 if y 1/ δ B (y) if 1/ <y 1 The corresponding minimum Bayes risk is r(δ B ) 1 1/ (y +1)dy + 1 1/ dy 11 4 b With uniform costs, the least-favorable prior will be interior to (, 1), so we examine the conditional risks of Bayes rules for an equalizer condition The critical region for the Bayes rule δ is given by Γ 1 y [, 1] L(y) } [,τ ], 1 where τ 1 ( if 7 1 5) if < 7 < 5 if 7 1 1

2 Thus, the conditional risks are: τ R (δ ) (y +1)dy 1 if 7 τ ( τ +1) if < 7 < 5 if if 7 R 1 (δ ) dy 1 τ if τ 7 < < 5 1 if 7 1 By inspection, a minimax threshold τ L is the solution to the equation τ L ( ) τ L +1 1 τ L, which yields τ L ( 7 5)/ The minimax risk is the value of the equalized conditional risk; ie, V ( L )1 τ L c The Neyman-Pearson test is given by 1 if >η (y+1) δ NP (y) γ if η (y+1), if <η (y+1) where η γ are chosen to give false-alarm probability α Since L(y) is monotone decreasing in y, the above test is equivalent to 1 if y<η δ NP (y) γ if y η, if y>η where η 1 Since Y is a continuous rom variable, we can ignore the rom- η ization Thus, the false-alarm probability is: P F (δ NP )P (Y<η ) η (y +1)dy if η The threshold for P F (δ NP )α is the solution to η ( ) η +1 α, which is η 1+α 1 So, an α-level Neyman-Pearson test is 1 if y 1+α 1 δ NP (y) if y> 1+α 1 The detection probability is P D (δ NP ) η η dy η 1+α 1, ( η +1) if <η < 1 1 if η 1 <α<1,

3 Exercise 4: Here the likelihood ratio is given by y L(y) ey a Thus, Bayes critical regions are of the form e (y 1) e, y Γ 1 y (y 1) τ }, where τ e log ( 1 ) There are three cases: φ [ if τ < Γ 1 1 τ, 1+ ] τ if τ 1 [, 1+ τ ] if τ > 1 The condition τ < is equivalent to < 1, where e 1+ e ; the condition τ 1 is equivalent to, where 1+ ; the condition τ > 1is equivalent to < The minimum Bayes risk V ( ) can be calculated for the three regions: V ( )1, < 1, [ 1+ τ 1 τ ] V ( ) 1 e y dy +(1 ) e y dy + τ 1+ e y dy, τ, 1+ τ V ( ) e y dy +(1 ) 1+ e y dy, < τ b The minimax rule can be found by equating conditional risks Investigation of the above shows that this equality occurs in the intermediate region, thus corresponds to a threshold τ L (, 1) solving e τ L e τ L e(1 + Φ(1 + τ L) Φ(1 τ L)) The minimax risk is then either of the equal conditional risks; eg, V ( L )e 1+ τ L e 1 τ L c Here, romization is unnecessary, the Neyman-Pearson critical regions are of the form Γ 1 y (y 1) η },

4 where η log(η) There are three cases: e φ [ if η < Γ 1 1 η, 1+ ] η if η 1 [, 1+ η ] if η > 1 The false-alarm probability is thus: P F (δ NP ) 1+ η 1 η P F (δ NP ) P F (δ NP ), η < e y dy e 1+ η e 1 η e sinh ( η ), η 1, 1+ η e y dy 1 e 1 η, η > 1 From this we see that the threshold for α level NP testing is [ ( )] η sinh 1 αe if <α 1 e [1 + log(1 α)] if 1 e <α<1 The detection probability is thus [ ( ) ( )] P D (δ NP ) Φ 1+ η Φ 1 η [ ( ( )) αe Φ 1 + sinh 1 P D (δ NP ) Exercise 6a&b: ( Φ 1 sinh 1 ( αe ))], <α 1 e, [ ( ) Φ 1+ η 1 ] [ Φ ( + log(1 α)) 1 ], 1 e <α 1 Here we have p (y) p N (y + s) p 1 (y) p N (y s), which gives L(y) 1+(y + s) 1+(y s) a With equal priors uniform costs, the critical region for Bayes testing is Γ 1 L(y) 1} 1+(y + s) 1+(y s) } sy sy} [, ) Thus, the Bayes test is 1 if y δ B (y) if y< 4

5 The minimum Bayes risk is then r(δ B ) 1 1 [1 + (y + s) ] dy [1 + (y s) ] dy 1 tan 1 (s) b Because of the symmetry of this problem with uniform costs, we can guess that 1/ is the least-favorable prior To confirm this, we can check that this answer from Part a gives an equalizer rule: 1 R (δ 1/ ) [1 + (y + s) ] dy 1 1 [1 + (y s) ] dy R 1(δ 1/ ) Exercise 7: a The densities under the two hypotheses are: p 1 (y) Thus, the likelihood ratio is p (y) p(y) e y, y >, p(y s)p(y)ds L(y) p 1(y) p (y) y e s y e s ds ye y, y > y, y > b Romization is irrelevant here, so the false-alrm probability for threshold η is P F (δ NP )P (Y>η)e η, which gives the threshold η log α, for α level Neyman-Pearson testing The corresponding detection probability is P D (δ NP )P 1 (Y>η) η ye y dy (η +1)e η α(1 log α), <α<1 c Here the densities under the two hypotheses become: n n p (y) p(y k ) e y k, < miny 1,y,,y n }, [ n ] [ miny1,y,,y n} n ] p 1 (y) p(y k s) p(s)ds e s y k e s ds p [ (y) e (n 1) miny 1,y,,y n} 1 ], < miny 1,y,,y n } 5

6 Thus, the likelihood ratio is L(y) 1 [ e (n 1) miny 1,y,,y n} 1 ], < miny 1,y,,y n } d The false-alarm probability incurred by comparing L(y) from Part c to a threshold η is P F (δ NP )P (L(Y ) >η)p ( miny 1,Y,,Y n } >η n n n P ( (Y k >η )) P (Y k >η ) e η e nη, from which we have η 1 log α, or, equivalently, n ) log(()η +1) η e(n 1)η 1 α (n 1)/ Exercise 15 a The LMP test is we have thus 1 if p θ(y) θ θ >ηp (y) δ lo (y) γ, if p θ(y) θ θ ηp (y) if p θ(y) θ θ <ηp (y) p θ (y) θ θ p (y) sgn(y) ; 1 if sgn(y) >η δ lo (y) γ, if sgn(y) η if sgn(y) <η To set the threshold η, we consider if η 1 P (sgn(y ) >η) 1/ if 1 η<1 1 if η< 1 This implies that η 1 if <α<1/ 1 if 1/ α<1 6

7 The romization is γ α P (sgn(y ) >η) P (sgn(y ) η) α if <α<1/ α 1 if 1/ α<1 The LMP test is thus for <α<1/ ; it is δ lo (y) α if y> if y δ lo (y) for 1/ α<1 For fixed θ>, the detection probability is 1 if y α 1 if y< P D ( δ lo ; θ) P θ (sgn(y ) >η)+γp θ (sgn(y )η) α 1 e y θ dy if <α<1/ 1 e y θ dy +(α 1) 1 e y θ dy if 1/ α<1 α( e θ ) if <α<1/ 1+(α 1)e θ if 1/ α<1 b For fixed θ, the NP critical region is Γ θ y y θ >η } (, ) if η < θ (( η +θ ), ) if θ η θ φ if η >θ, from which 1 if η < θ 1 +θ)/ P (Γ θ ) e (η if θ η θ if η >θ Clearly, we must know θ to set η, thus the NP critical region depends on θ This implies that there is no UMP test The generalized likelihood ratio test uses this statistic: sup θ> e y y θ expsup( y y θ )} θ> 1 if y< e y if y 7

8 Exercise 16: We have M hypotheses H,H 1,H M 1, where Y has distribution P i density p i under hypothesis H i A decision rule δ is a partition of the observation set Γ into regions Γ, Γ 1,,Γ M 1, where δ chooses hypothesis H i when we observe y Γ i Equivalently, a decision rule can be viewed as a mapping from Γ to the set of decisions, 1,,M 1}, where δ(y) is the index of the hypothesis accepted when we observe Y y On assigning costs C ij to the acceptance of H i when H j is true, for i, j (M 1), we can define conditional risks, R j (δ),j, 1,,M 1, for a decision rule δ, where R j (δ) is the conditional expected cost given that H j is true We have R j (δ) i C ij P j (Γ i ) Assuming priors j P (H j occurs),j, 1,,M 1, we can define an overall average risk or Bayes risk as r(δ) A Bayes rule will minimize the Bayes risk We can write r(δ) j i j j C ij P j (Γ i ) j R j (δ) i j C ij P j (Γ i ) j j C ij p j (y)µ(dy) j C ij p j (y) µ(dy) i j Γ i i Γ i j Thus, by inspection, we see that the Bayes rule has decision regions given by Exercise 19: Γ i y Γ j C ij p j (y) j a The likelihood ratio is given by ( σ σ 1 ) n e n L(y) ( µ σ min k M 1 j n 1 σ1 e (y k µ 1 ) /σ 1 n 1 σ e (y k µ ) /σ ) ( µ 1 σ 1 e 1 σ ) n ( 1 σ 1 y k e µ 1 σ 1 j C kj p j (y) ) µ n σ y k, 8

9 which shows the structure indicated b If µ 1 µ µ σ 1 >σ, then the Neyman-Pearson test operates by comparing the quantity n (y k µ) to a threshold, choosing H 1 if the threshold is exceeded H otherwise Alternatively, if µ 1 >µ σ 1 σ, then the NP test compares n y k to a threshold, again choosing H 1 when the threshold is exceeded Note that, in the first case, the test statistic is quadratic in the observations, in the second case it is linear c For n 1,µ 1 µ µ σ 1 >σ,, the NP test is of the form δ NP (y) 1 if (y1 µ) η if (y 1 µ) <η, where η > is an appropriate threshold We have P F (δ NP )P ((Y 1 µ) >η )1 P ( η Y 1 µ η ) Thus, for size α we set ( ) ( ) [ ( )] η η η 1 Φ +Φ 1 Φ σ η σ [σ Φ 1 ( 1 α )], the detection probability is [ ( )] η P D (δ NP )1 P 1 ( η Y 1 µ η ) 1 Φ [ ( ( σ 1 Φ Φ 1 1 α ))], <α<1 σ 1 σ σ 1 9