Sequential Probability Ratio Test for Sparse Signals
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- The document discusses using a sequential probability ratio test (SPRT) for sparse signal detection from compressed measurements. SPRT allows observations to continue until sufficient evidence is accumulated to decide between two hypotheses. - It outlines how to apply SPRT to the problem by deriving the likelihood ratio test statistic and decision thresholds based on desired false alarm and detection probabilities. Performance is characterized by bounding the probability of error based on properties of compressed sensing measurements. - Simulation results show decreasing decision time with higher compression ratios and numbers of observations, consistent with theoretical predictions. Future work includes refining the analysis and extending it to random sparse signals.
Sequential Probability Ratio Test for Sparse Signals
- 1. Sequential Probability Ratio Test for Sparse Signals Mohamed Seif, PhD student ECE Department October 2017The University of Arizona
- 2. Outline November 2017The University of Arizona • Motivation • Preliminaries on Compressive Sensing • Sequential Probability Ratio Test: Recap • Work Discussion 2
- 3. Compressive Sensing November 2017The University of Arizona 3 s sparse signalN ⇥ 1 K (=3) non-zero elements Sparsity of order K M ⇥ N random matrix Gaussian distribution = y Dimensionality Reduction M ⇥ 1 measurements K < M << N Scanning the support of the signal has exponential complexity!
- 4. How to recover the sparse signal? November 2017The University of Arizona 4 RN dimensional space s {s0 : y = s0 } (1 )ksk2 2 k sk2 2 (1 + )ksk2 2 Restricted Isometry Property: Gaussian distribution works! M 2K log ✓ N M ◆ Required number of measurements: RM s
- 5. How to recover the sparse signal? (Cont’d) November 2017The University of Arizona 5 RN dimensional space s {s0 : y = s0 } Question: How to estimate the signal? l2 norm recovery ˆs Not accurate! ˆs = arg min y= s0 ksk2 ksk2 2 = |s1|2 + |s2|2 + · · · + |sN |2 Where, RM
- 6. How to recover the sparse signal? (Cont’d) November 2017The University of Arizona 6 RN dimensional space s {s0 : y = s0 } Question: How to estimate the signal? ˆsl1 norm recovery ˆs = arg min y= s0 ksk1 ksk1 = |s1| + |s2| + · · · + |sN | Where, RM
- 7. Stopping Problem: Recap November 2017The University of Arizona • Stopping problems are a simple but important class of learning problems. •In this problem class, information arrives over time, and we have to choose whether to view the information or stop and make a decision. 7 Event Detector t=0 t=1 t=2 t=3 . . . t=10 Input Signal Sudden Change in observation t=0 t=1 t=2 t=3 . . . t=10 Ideal case: Stop observing and detect the change accurately! Output Signal
- 8. Solution of the problem November 2017The University of Arizona 8 ⇢0 0 Risk 0.5 1 ⇢0 0 ⇢0 0 Risk is a concave function ⇢L ⇢U Stop and decide H1Stop and decide H0 Continue updating priors
- 9. Solution of the problem (Cont’d) November 2017The University of Arizona 9 • Now we are interested to obtain ⇢L , ⇢U • The updated prior is ⇢n+1 0 = Ln (Sn ) Ln(Sn) + ⇢0/(1 ⇢0) Ln (Sn ) = nY k=1 P1(Wk ) P0(Wk) • The likelihood ratio is defined as
- 10. Solution of the problem (Cont’d) November 2017The University of Arizona 10 • Determining if ⇢n+1 0 = Ln (Sn ) Ln(Sn) + ⇢0/(1 ⇢0) ⇢n+1 0 (Sn ) ⇢L or ⇢n+1 0 (Sn ) ⇢U is the same as testing Ln (Sn ) Aor Ln (Sn ) B • Why? Quasi linear function 0 Increasing function for Ln (Sn ) 0
- 11. Solution of the problem (Cont’d) November 2017The University of Arizona 11 • The new bounds A and B are obtained as A = ⇢n 0 ⇢L (1 ⇢n 0 )(1 ⇢L) B = ⇢n 0 ⇢U (1 ⇢n 0 )(1 ⇢U ) • Now the decision rule is Ln (Sn ) = 8 >< >: B stop and chooseY n = 1 A stop and choose Y n = 0 otherwise continue observing
- 12. Solution of the problem (Cont’d) November 2017The University of Arizona 12 • Since getting A, B exactly is difficult P⇡ F ⇡ 1 A B A P⇡ M ⇡ A(B 1) B A Wald’s approximation • Then for an acceptable P⇡ F , P⇡ M A = P⇡ M 1 P⇡ F B = 1 P⇡ M P⇡ F
- 13. November 2017The University of Arizona 13 Sequential Probability Ratio Test for Compressed Signal
- 14. Hypothesis Testing November 2017The University of Arizona 14 Random Deterministic (Today’s Talk) H1 : yi = s + ni H0 : yi = ni noise only signal + noise
- 15. Hypothesis Testing November 2017The University of Arizona 15 Likelihood functions: f(yi|H0) = exp( 1 2 yT i ( 1 t ) 1 yi) | 1 |1/2(2⇡)M/2 f(yi|H1) = exp( 1 2 (yi s)T ( 1 t ) 1 (yi s) | 1 |1/2(2⇡)M/2 = 1 2 (noise variance) Typo: T
- 16. Sequential Probability Ratio Test November 2017The University of Arizona 16 • We need to compute the likelihood function Ln (Sn ) = nY i=1 f1(yi|H1) f0(yi|H0) Ln (Sn ) = 8 >< >: B stop and chooseY n = 1 A stop and choose Y n = 0 otherwise continue observing • Remember the decision rule
- 17. Sequential Probability Ratio Test November 2017The University of Arizona 17 • After algebraic simplifications ⇤(y) 1 H1conclude ⇤(y) 2 conclude H0 2 ⇤(y) 1 continue observing Decision statistic • Where, ⇤(y) = nX i=1 yT i ( T ) 1 s 1 = 2 log(B) + n( s)T ( T ) 1 s 2 = 2 log(A) + n( s)T ( T ) 1 s
- 18. Performance Characterization November 2017The University of Arizona 18 • Average probability of error Pe = P(H0)Pfa + P(H1)(1 Pd) Pfa = P(⇤(y) 1|H0) Pd = P(⇤(y) 1|H1) • We assume, P(H0) = P(H1) = 1/2 • We get, ˆP = T ( T ) 1 , False alarm probability Detection probability Pe = Q ✓ 1 2 r n 1 kˆPsk ◆ A = B = 1, In SPRT, PfaPd and are predefined
- 19. Performance Characterization (Cont’d) November 2017The University of Arizona 19 Pe = Q ✓ 1 2 r n 1 kˆPsk ◆ This expression does not show the impact of compressed measurement! stable embedding Theorem Davenport, Mark A., et al. "Signal processing with compressive measurements." IEEE Journal of Selected Topics in Signal Processing 4.2 (2010): 445-460. Suppose that r N M ˆP provides a -stable embedding property. It satisfies! Then for any deterministic signal s, the probability of error has the following bounds: Q ✓ p 1 + p n 2 r M N ksk2 p 1 ◆ Pe Q ✓ p 1 p n 2 r M N ksk2 p 1 ◆ Theorem:
- 20. Performance Characterization (Cont’d) November 2017The University of Arizona 20 • Then, approximately Pe ⇡ Q ✓p n 2 r M N ksk2 p 1 ◆ Compression ratio < 1 Time p SNR @ SNR = 3 dB Comment: monotonically decreasing function of compression ratio and number of time.
- 21. Simulation Results November 2017The University of Arizona 21 Less measurements, more delay! Compression ratio (CR) = M N Sparsity order (K) = 5 Signal dimension (N) =100 Number of measurements Pd = 0.1, Pfa = 0.1 Monte-Carlo Simulation (number of iterations = 10000)
- 22. Simulation Results November 2017The University of Arizona 22 Compression ratio (CR) = M N Sparsity order (K) = 5 Signal dimension (N) =50 Number of measurements Pd = 0.1, Pfa = 0.1 Monte-Carlo Simulation (number of iterations = 100) (lack of time) Reconstruction using l1 norm algorithm min s ksk1 s.t. ky sk2 P(ˆs 6= s) M = 10
- 23. Future Work November 2017The University of Arizona • Refining the results • Studying the SPRT mechanism for random signals • Expecting a research paper
- 24. Thank You!