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

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1 An Introduction to Signal Detection and Estimation - Second Edition Chapter III: Selected Solutions H. V. Poor Princeton University March 17, 5 Exercise 1: Let {h k,l } denote the impulse response of a general discrete-time linear filter. The output at time n due to the input signal is n l=1 h n,l s l, and that due to noise is n l=1 h n,l N l.thus the output SNR at time n is SNR = n l=1 h n,l s l E{( n l=1 h n,l N l ) = ht ns h T nσ N h n where h n =(h n,1,h n,,...,h n,n ) T. Since Σ N >, we can write Σ N = Σ 1/ N Σ 1/ N when Σ 1/ N is invertible and symmetric. Thus, SNR = (Σ1/ N h n ) T Σ 1/ N s Σ 1/ N h n By the Schwarz Inequality ( x T y x y, we have SNR Σ 1/ N s with equality if and only if Σ 1/ N h n = λσ 1/ N s for a constant λ. Thus, max SNR occurs when h n = λσ 1 N s. The constant λ is arbitrary (it does not affect SNR), so we can take λ = 1, which gives the desired result. Exercise 3: a. From Exercise 15 of Chapter II, the optimum test here has critical regions: Γ k = {y R n p k (y) = 1 max p l(y)}. l M 1

2 Since p l is the N(s l,σ I) density, this reduces to Γ k = {y R n y s k = min y s l } l M 1 = {y R n s T k y = max l M 1 st l y}. b. We have P e = 1 M P k (Γ c M k), k= and P k (Γ c k)=1 P k (Γ k )=1 P k ( max l k M 1 st l Y <s T k Y ) Due to the assumed orthogonality of s 1,,s n, it is straighforward to show that, under H k, s T 1 Y,s T Y,,s T ny, are independent Gaussian random variable with variances σ s 1, and with means zero for l k and mean s 1 for l = k. Thus Now = P k ( max l k M 1 st l Y <s T k Y ) 1 P k ( max πσ s1 l k M 1 st LY <z) e (z s 1 )/σ s 1 P k ( max l k M 1 st LY <z)=p k ( {s T l Y <z}) l k M 1 = P k (s T l Y <z) l k M 1 [ ] M 1 z = Φ( σ s 1 ). Combining the above and setting x = z/σ s 1 yields dz 1 P k (Γ k )= 1 [Φ(x)] M 1 e (x d) dx, k =,,M 1, π and the desired expression for P e follows. Exercise 6: Since Y N(µ, Σ), it follows that Ŷk is linear in Y 1,,Y k 1, and that ˆσ Y k does not depend on Y 1,,Y k 1. Thus, I is a linear transformation of Y and is Gaussian. We need only show that E{I} and cov(i) =I. = We have E{I k } = E{Y k} E{Ŷk}. ˆσ k

3 Since Ŷk = E{Y k Y 1,,Y k 1 }, E{Ŷk} is an iterated expectation of Y k ; hence E{Y k } = E{Ŷk} and E{I k } =,k =1,,n. To see that cov(i) =I, note first that Var (I k )=E{I k} = E{(Y k Ŷk) } ˆσ Y k = ˆσ Y k ˆσ Y k =1. Now, for l<k, we have cov (I k,i l ) = E{I k I l } = E{(Y k Ŷk)(Y l Ŷl)} ˆσ Yk ˆσ Yl. Noting that E{(Y k Ŷk)(Y l Ŷl)} = E{E{(Y k Ŷk)(Y l Ŷl) Y 1,,Y k 1 }} = E{(E{Y k Y 1,,Y k 1 } Ŷk)(Y l Ŷl)} = E{(Ŷk Ŷk)(Y l Ŷl)} =, we have cov(i k,i l ) = for l<k. By symmetry we also have cov(i k,i l ) = for l>k, and the desired result follows. Exercise 7: a. The likelihood ratio is L(y) = 1 est Σ 1 y d / + 1 e st Σ 1 y d / = e d / cosh s T Σ 1 y, which is monotone increasing in the statistic T (y) s T Σ 1 y. (Here, as usual, d = s T Σ 1 s.) Thus, the Neyman-Pearson test is of the form To set the threshold η, we consider 1 if T (y) >η δ NP (y) = γ, if T (y) =η if T (y) <η. P (T (Y ) >η)=1 P ( η s T Σ 1 N η ) =1 Φ(η/d)+Φ( η/d) = [1 Φ(η/d)], 3

4 where we have used the fact that s T Σ 1 N is Gaussian with zero mean and variance d. Thus, the threshold for size α is The randomization is unnecessary. The detection probability is = 1 η = dφ 1 (1 α/). P D ( δ NP )= 1 P 1(T (Y ) >η Θ = +1) + 1 P 1(T (Y ) >η Θ = 1) [ 1 P ( η d + s T Σ 1 N η )] + 1 [ 1 P ( η +d + s T Σ 1 N η )] = Φ ( Φ 1 (1 α/) + d ) Φ ( Φ 1 (1 α/) d ). b. Since the likelihood ratio is the average over the distribution of Θ of the likelihood ratio conditioned on Θ, we have L(y) = 1 πσθ e (θst y nθ s /)/σ e θ /σ θ dθ where and = k 1 e k s T y 1 πv e (θ µ) /v dθ = k 1 e k s T y, v = σ θn s σ θ σ + n s, µ = v s T y, k 1 = v, σ θ k = v 4. Exercise 13: a. In this situation, the problem is that of detecting a Gaussian signal with zero mean and covariance matrix Σ S = diag{as 1,As,...,As n}, in independent i.i.d. Gaussian noise with unit variance; and thus the Neyman-Pearson test is based on the quadratic statistic n As k T (y) = As k. k +1y 4

5 b. Assuming s k, for all k, a sufficient condition for a UMP test is that s k is constant. In this case, an equivalent test statistic is the radiometer n y k, which can be given size α without knowledge of A. c. From Eq. (III.B.11), we see that an LMP test can be based on the statistic Exercise 15: n T lo (y) = s kyk. Let L a denote the likelihood ratio conditioned on A = a. Then the undonditioned likelihood ratio is L(y) = L a (y)p A (a)da = e na /4σ I (a ˆr/σ )p A (a)da, with ˆr r/a, where r = y c + y s as in Example III.B.5. Note that ( n ˆr = ) ( n ) b k cos((k 1)ω c T s )y k + b k sin((k 1)ω c T s )y k, which can be computed without knowledge of A. Note further that L(y) = 1 e na /4σ a I ˆr σ (a ˆr/σ )p A (a)da >, where we have used the fact that I is monotone increasing in its argument. Thus, L(y) is monotone increasing in ˆr, and the Neyman-Pearson test is of the form 1 if ˆr>τ δ NP (y) = γ, if ˆr = τ. if ˆr<τ To get size α we choose τ so that P ( ˆR >τ )=α. From (III.B.7), we have that P ( ˆR >τ )=e (τ ) /nσ, from which the size-α desired threshold is τ = nσ log α. The detection probability can be found by first conditioning on A and then averaging the result over the distribution of A. (Note that we have not used the explicit form of the distribution of A to derive any of the above results.) It follows from (III.B.74) that P 1 ( ˆR >τ A = a) =Q(b, τ ) with b = na /σ and τ = /nτ /σ = log α. Thus, P D = Q( a σ n/,τ )p A (a)da = τ xe (x +na /σ )/ I (x a σ 5 n/) a A e a /A dxda

6 where a = = xe x / a e a /a τ A I (x a n/)dadx, σ A σ na +σ. On making the substitution y = a/a, this integral becomes P D = a xe x (1 b )/ ye (y +b A x )/ I (b xy)dydx = a xe x (1 b)/ Q(b τ A x, )dx, τ where b = na /σ. Since Q(b, ) = 1 for any value of b, and since 1 b = a /A, the detection probability becomes where x = 1 P D = a A τ. 1+ na σ Exercise 16: xe x (1 b )/ dx = e τ (1 b )/ = exp( τ ( 1+ na σ ))=α x, The right-hand side of the given equation is simply the likelihood ratio for detecting a N (, Σ S ) signal in independent N (,σ I) noise. From Eq. (III.B.84), this is given by exp( 1 σ yt Σ S (σ I + Σ S ) 1 y + 1 log( σ I / σ I + Σ S )). We thus are looking for a solution Ŝ to the equation ŜT y Ŝ = y T Σ S (σ I + Σ S ) 1 y + σ n σ log( ), σ + λ k where λ 1,λ,...,λ n are the eigenvalues of Σ S. On completing the square on the left-hand side of this equation, it can be rewritten as n Ŝ y = y y T Σ S (σ I + Σ S ) 1 y σ σ log( ) σ + λ k n σ [y T (σ I + Σ S ) 1 σ ] y log( ), σ + λ k which is solved by Ŝ = y ± for any nonzero vector v. σ [ y T (σ I + Σ S ) 1 y v 6 n σ ] 1/ log( ) v, σ + λ k