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Washington University in St. Louis - ESE 524HW 5 Solutions. ESE 524 - Detection and Estimation Theory

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ESE 524 - Detection and Estimation Theory Homework 5 Assigned March 31, Due: April 14 100 points + 40 EC 1) Gaussian linear model (20 pts) We are interested in estimating the physical health of a ... group of candidates for an astronaut position. In order to do so, we evaluate the heart rate (θ1) and respiration rate after physical exercise (θ2). Assume that the prior distribution of these parameters based on studies from the general population is given by a jointly Gausssian distribution: θ = θ θ1 2 , with mean µθ = 100 30 and covariance matrix Cθ = 1 3 3 1. Since we are interested in a single score to compare the candidates, instead of measuring the heart rate and the respiration rate, we measure the score of the first principal component (PCA) of these measurements. Derive the posterior distribution of the measurements according to the following steps: a) (10 pts)Find the largest eigenvalue (λ) and it’s corresponding eigenvector (v 2 R2×1) of Cθ. Normalize the eigenvector so that it has unit norm. b) (10 pts) Now our system is given by: x = vT · θ + w, where w ∼ N(0; 1) is the measurement noise. Find the posterior p.d.f. p(θjx) Solution: a) the largest eigenvalue of Cθ is 4, and its’s corresponding eigenvalue is vT = [p12 p12] b) Here we use the simplified expresions from theorem 3 (L4: page 54), with: H = v = v v1 1, Cθ = 1 3 3 1, µθ = 100 30 , and Cw = 1. Then we get Cθjx = (HT Cw−1H + Cθ−1)−1 = −10:4 0 :6 1: :4 6 And E(θjx) = (HT Cw−1H + Cθ−1)−1(HT Cw−1x + Cθ−1µθ) = 0567 :567 · x + −48 22 Then the posterior pdf is given by fθjx(θjx) ∼ N(E(θjx); Cθjx) 2) Soft Thresholding (20 pts): We collect conditionally i.i.d. measurements (zn)N n=0 −1 given the signal (sn)N n=0 −1, following p(znjsn) = N(znjsn; 1) where zn and sn are real-valued. (a) (10 pts.)If the signal (sn)N n=0 −1 is an i.i.d sequence following the Laplace pdf π(s) = λ 2 e−λjsj; λ > 0 (1) find the MAP estimates of (sn)N n=0 −1. Solution: Because the measurements are i.i.d. and (sn)N n=0 −1 is an i.i.d. sequence, p(snjzn) / p(znjsn)π(sn) / exp −1 2(zn − sn)2 exp(−λjsnj) = exp − 1 2zn2 + 1 2s2 n − znsn + λjsnj [Show More]

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