ECE531 Lecture 6: Detection of Discrete-Time Signals with Random Parameters

Transcription

1 ECE531 Lecture 6: Detection of Discrete-Time Signals with Random Parameters D. Richard Brown III Worcester Polytechnic Institute 26-February-2009 Worcester Polytechnic Institute D. Richard Brown III 26-February / 27

2 Introduction Today we consider parametric detection problems for signals with one or more unknown parameters. We focus on the case of an N-sample discrete time observation Y Y with binary hypotheses H 0 : Y p x (y) for x X 0 H 1 : Y p x (y) for x X 1 Our approach: Absorb the unknown parameters into our notion of the state x and the associated state space X. In many textbooks, it is common to denote the random parameter(s) as Θ and a realization of these parameters as θ. We will stay consistent with our earlier notation, however, and use x here. Worcester Polytechnic Institute D. Richard Brown III 26-February / 27

3 Example Suppose we have a known signal s R n and we have a communication system that transmits either s 0 = s or s 1 = s. The signal arrives with unknown amplitude a > 0 and is corrupted by zero-mean AWGN with variance σ 2. The hypotheses can be written as H 0 : Y N(as 0,σ 2 I) H 1 : Y N(as 1,σ 2 I) for unknown a > 0. What is the state space X and what are the sets X 0 and X 1? An equivalent, but more clever way to write these hypotheses is H 0 : Y N(as,σ 2 I) with a < 0 H 1 : Y N(as,σ 2 I) with a > 0 for unknown a R\0. What is the state space X and what are the sets X 0 and X 1? Worcester Polytechnic Institute D. Richard Brown III 26-February / 27

4 Types of Detectors: What we Already Know Problem: Find a decision rule to decide between for unknown a R\0 H 0 : Y N(as,σ 2 I) with a < 0 H 1 : Y N(as,σ 2 I) with a > 0 We know that these types of problems are composite binary hypothesis testing problems. Bayes decision rules involve computation of two discriminant functions (or commodity costs) g 0 (y,π) and g 1 (y,π) and subsequent comparison. Under the N-P criterion with significance level (or size) α: Our strategy: reduce the composite HT problem to a simple one. Check for the existence of a UMP decision rule (check the critical region and/or monotone likelihood ratio) that maximizes P D,x for all x X 1 subject to P fp,x α for all x X 0. If a UMP rule doesn t exist, we can usually find an LMP decision rule. Worcester Polytechnic Institute D. Richard Brown III 26-February / 27

5 Example: Known Signal with Unknown Amplitude s k a + Y k n-sample detector decisions W k We would like to decide on the presence of a known signal s R n (H 1 ) versus the absence of the signal (H 0 ). The known signal s arrives at the detector corrupted by noise W and with unknown amplitude a. We observe a realization y R n of the random variable Y = as + W. a = 0 when no signal is present (H 0 ) a > 0 when a signal is present (H 1 ) What kind of hypothesis testing problem is this? Binary, composite, one-sided. What is the state space here? x = a X = [0, ). Worcester Polytechnic Institute D. Richard Brown III 26-February / 27

6 Bayes Detector for this Example To develop a Bayes detector for this problem, we need 3 things: 1. We need a conditional pdf or pmf statistically describing the observations y Y for each state x X: p x (y). 2. We need a cost assignment for each state x X and each hypothesis H i : C 0 (x) and C 1 (x). 3. We need a prior (pdf) on the states: π(x). The Bayes risk of decision rule ρ in this case can be written as r(ρ, π) = π(x) ρ (y)c(x)p x (y)dy dx X Y ( ) = ρ (y) C(x)π(x)p x (y)dx dy Y } X {{ } g(y,π) We know how to solve this problem: For each y Y, the Bayes decision rule just compares g 0 (y,π) to g 1 (y,π) and sets δ Bπ equal to the index of the smaller discriminant function (or commodity cost). Worcester Polytechnic Institute D. Richard Brown III 26-February / 27

7 Bayes Detector for this Example Suppose the noise is distributed as W N(0,Σ). Conditioned on x = a, we can write the density of the vector observation as ( 1 p x=a (y) = (2π) n/2 exp (y as) Σ 1 ) (y as) Σ 1/2 2 Suppose also that we have the UCA and the prior π(x) = π 0 δ(x) + (1 π 0 )(I x (0) I x (1)) where I x (t) = 1 for all x > t and I x (t) = 0 otherwise. Note that the state space here is X = [0,1]. Worcester Polytechnic Institute D. Richard Brown III 26-February / 27

8 Bayes Decision Rule: π 0 = 0.5 and s = [1.5, 0.5] ybar ybar0 Worcester Polytechnic Institute D. Richard Brown III 26-February / 27

9 Bayes Decision Rule: π 0 = 0.4 and s = [1.5, 0.5] ybar ybar0 Worcester Polytechnic Institute D. Richard Brown III 26-February / 27

10 Bayes Decision Rule: π 0 = 0.75 and s = [1.5, 0.5] ybar ybar0 Worcester Polytechnic Institute D. Richard Brown III 26-February / 27

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12 Neyman-Pearson Detector To develop a N-P detector for this problem, we need 2 things: 1. We need a conditional pdf or pmf statistically describing the observations y Y for each state x X: p x (y). 2. We need a significance-level for the test: P fp α. How should we approach the problem? Check to see if a UMP detector exists. If not, derive an LMP detector with good performance for a near zero. A new option: If we can specify a prior on the states, denoted as π(x), we can reduce each composite hypothesis to a simple one by taking a weighted average of the density with respect to the prior, i.e. H j : Y p j (y) = p x (y)π j (x)dx X j π(x) where π j (x) = Prob(x X j ) when x X j and is equal to zero otherwise. Then testing H 0 versus H 1 is a simple hypothesis testing problem for which the N-P lemma gives a unique optimal solution. Worcester Polytechnic Institute D. Richard Brown III 26-February / 27

13 Neyman-Pearson Detector Suppose the noise is distributed as W N(0,σ 2 I). Conditioned on x = a, we can write the density of the vector observation as ( 1 p x=a (y) = (2πσ 2 exp (y ) as) (y as) ) n/2 2σ 2 = = n 1 k=0 n 1 k=0 p x=a (y k ) 1 2πσ exp ( (y k as k ) 2 ) Note that any results we derive for this case can also be applied to the case when W N(0,Σ) by using our coordinate-transformation (decorrelation) trick. 2σ 2 Worcester Polytechnic Institute D. Richard Brown III 26-February / 27

14 Neyman-Pearson Detector: UMP Decision Rule Let s set up a simple HT test: H 0 : a = 0 versus H 1 : a = a 1 > 0. The N-P lemma says that, for this simple HT problem, we can find an α-level decision rule of the form 1 ln (L(y)) > ln v ρ(y) = γ ln (L(y)) = ln v 0 ln (L(y)) < ln v ( ) where ln (L(y)) := ln p1 (y) p 0 (y) with v 0 and γ [0,1] chosen such that P fp = α. Worcester Polytechnic Institute D. Richard Brown III 26-February / 27

15 Log-Likelihood Ratio The log-likelihood ratio for this simple HT test: ( ) p1 (y) ln (L(y)) = ln p 0 (y) ( n 1 k=0 = ln exp ( ) n 1 k=0 exp y2 k 2σ ( 2 n 1 = ln = k=0 n 1 a 1 σ 2 k=0 = a 1 σ 2 ) (y k a 1 s k ) 2 2σ 2 exp ( 2yk a 1 s k a 2 1 s2 k 2σ 2 ( y k s k a 1s 2 ) k 2 (s y a 1 s 2 ) 2 ) ) Worcester Polytechnic Institute D. Richard Brown III 26-February / 27

16 False Positive Probability P fp = Prob (ln (L(Y )) > lnv a = 0) + γprob (ln(l(y )) = ln v a = 0) = Prob (s Y > σ2 ln v + a 1 s 2 ) a = 0 a 1 2 How is s Y distributed when a = 0? Hence, v must be selected such that σ 2 P fp = Q a 1 ln v + a 1 s 2 2 = α s σ Worcester Polytechnic Institute D. Richard Brown III 26-February / 27

17 Neyman-Pearson Detector: UMP Decision Rule Does a UMP decision rule exist? To answer this question, we need to determine if the critical region Γ 1 = {y Y : ρ(y) decides H 1 } depends on our choice of a 1. We decide H 1 for sure when s y > σ2 ln v + a 1 s 2 a 1 2 and we decide H 1 with probability γ when s y = σ2 ln v + a 1 s 2 a 1 2 (which happens with probability zero). Critical region depends on a 1 no UMP decision rule. What should we do now? Worcester Polytechnic Institute D. Richard Brown III 26-February / 27

18 Neyman-Pearson Detector: UMP Decision Rule Hold on. Does the critical region Γ 1 = {y R n : s y > σ2 a 1 ln v + a 1 s 2 really depend on a 1? Given 0 α 1 and t > 0, the unique solution to Q(z/t) = α is z = tq 1 (α). Hence, the unique solution to σ 2 Q a 1 ln v + a 1 s 2 2 = α s σ is σ 2 2 } ln v + a 1 s 2 = s σq 1 (α). a 1 2 Hence the critical region for a significance-level α N-P decision rule can be written as Γ 1 = {y R n : s y > s σq 1 (α)} Does this depend on a 1? Does a UMP decision rule exist? Worcester Polytechnic Institute D. Richard Brown III 26-February / 27

19 Neyman-Pearson Detector: UMP Decision Rule 1 s y > s σq 1 (α) ρ UMP (y) = γ s y = s σq 1 (α) 0 s y < s σq 1 (α) How would this change if the noise was distributed as W N(0,Σ)? Worcester Polytechnic Institute D. Richard Brown III 26-February / 27

20 Neyman-Pearson Detector: LMP Decision Rule If the UMP detector did not exist or was too complicated, we could find an LMP detector for this example by comparing d da L a(y) a=0 to a threshold. When the noise samples are i.i.d., the likelihood ratio can be written as L a (y) = = p x=a(y) p x=0 (y) n 1 k=0 q(y k as k ) q(y k ) where q(x) is the pmf/pdf of the kth noise sample. In our example, q(x) = 1 2πσ e x2 /σ 2. We ll continue our analysis here for i.i.d. noise with general distribution q(x)... Worcester Polytechnic Institute D. Richard Brown III 26-February / 27

21 Neyman-Pearson Detector: LMP Decision Rule Taking the derivative of L a (y) with respect to a yields d da L n 1 q (y k as k ) a(y) = s k q(y k ) k=0 where q (x) = d dxq(x). Setting a = 0 yields j k q(y j as j ) q(y j ) d da L a(y) a=0 n 1 q (y k ) = s k q(y k ) k=0 n 1 q (y k ) = s k q(y k ) = k=0 n 1 s k h LMP (y k ) k=0 j k q(y j ) q(y j ) Worcester Polytechnic Institute D. Richard Brown III 26-February / 27

22 Neyman-Pearson Detector: LMP Decision Rule The locally most powerful decision rule then takes the form 1 n 1 k=0 s kh LMP (y k ) > τ ρ LMP (y) = γ n 1 k=0 s kh LMP (y k ) = τ 0 n 1 k=0 s kh LMP (y k ) < τ where γ and τ are selected such that P fp = α. In our example, h LMP (y k ) := q (y k ) q(y k ) = 2y k. Hence, if τ = τσ 2 /2, the σ 2 LMP decision rule then takes the form 1 s y > τ ρ LMP (y) = γ s y = τ 0 s y < τ with τ and γ selected so that P fp = α. How does this compare to the UMP decision rule? ρ LMP (y) = ρ UMP (y) here. Worcester Polytechnic Institute D. Richard Brown III 26-February / 27

23 LMP Decision Rule for i.i.d. Laplacian Noise For b > 0, the Laplacian density is given as q(x) = b 2 e b x. For Laplacian noise, h LMP (y k ) = q (y k ) q(y k ) = bsgn(x). Hence, the locally most powerful decision rule can be written as 1 n 1 k=0 s ksgn(y k ) > τ ρ LMP (y) = γ n 1 k=0 s ksgn(y k ) = τ 0 n 1 k=0 s ksgn(y k ) < τ In this case, ρ LMP (y) ρ UMP (y). Worcester Polytechnic Institute D. Richard Brown III 26-February / 27

24 The Third Alternative: Average Density Over a Prior The approach followed in pages of your textbook (as well as Example III.B.5) is to reduce the composite HT problem to a simple HT problem by using the a prior π(x) to compute a single density as a weighted average of the family of densities associated with H j, i.e. p j (y) = p x (y)π j (x)dx X j π(x) where π j (x) = Prob(x X j ) when x X j and is equal to zero otherwise. In this case, given a significance level α, the N-P lemma tells us that a unique optimal solution must exist based on a threshold test of the likelihood ratio L(y) = = p 1(y) p 0 (y) X 1 p x (y)π 1 (x)dx X 0 p x (y)π 0 (x)dx. Worcester Polytechnic Institute D. Richard Brown III 26-February / 27

25 Remarks on the Third Alternative If there is some uncertainty as to the prior, the usual approach is to chose π(x) such that it gives as little information about x as possible, i.e. such that the simple HT problem has maximum difficulty. Example III.B.5: Noncoherent Detection of a Modulated Sinusoidal Carrier. In this example, the unknown parameter is the phase of the received signal. What prior on this phase would give the least information? π(x) U(0, 2π) This is exactly the prior used in the example in your textbook. Please read this example and, in particular, note the development of the catalyst for deciding between two different (but known) signals with unknown phase on page 71. Worcester Polytechnic Institute D. Richard Brown III 26-February / 27

26 Conclusions Detection of known signals in noise: simple hypothesis testing. Detection of signals with one or more unknown parameters in noise: composite hypothesis testing. You are encouraged to at least skim the section on Stochastic Signals in section III.B (up to page 81). I also plan to discuss section III.D Sequential Detection in the optional lecture after the final exam. Worcester Polytechnic Institute D. Richard Brown III 26-February / 27

27 Midterm Exam: What You Need to Know Different types of hypothesis testing problems (binary, M-ary, simple, composite) Mathematical model of hypothesis testing problems. Intuition about good and bad decision rules. Poor textbook chapter II (whole chapter): Bayesian hypothesis testing (binary and M-ary, simple and composite) Minimax hypothesis testing (binary, simple) Neyman-Pearson hypothesis testing (binary, simple and composite) Poor textbook chapter III (up to page 72): Decorrelation of signals observed in correlated Gaussian noise. Detection of known discrete-time signals (binary and M-ary) Detection of discrete-time signals with random parameters (binary) Worcester Polytechnic Institute D. Richard Brown III 26-February / 27