\fProblem 4 This problem is recursive MMSE estimation. We are given: The two independent Gaussian processes are: i). {Un : n >=2} ii). Vn : n >= 1 Where: Un ~ i.i.d N(0, 1) Vn ~ i.i.d N(0, 1) We also know that: X1 = U1 X2 = U1 + U2 Xn = PnXn-1 + (1 - Pn)Xn-2 + Un, n >= 3 Part a Pn = {1 n is even = {0 n is odd The observation sequences are: Yn = Xn + Vn, n>=1 Therefore, we proceed to derive the Kalman filter as the best linear estimator for our linear, nonGaussian model. We slightly generalize the model that was treated so far. Hence we get: Yn = 1/2 (x0 - x'0)TP0-1(x0 - x'0) Kalman filter: Initialization: Update equations: For i = 0,1,2,...,n, the estimate is: Where the filter again is: The MSE is: This can be solved explicitly as: The gain is: The recursive estimate is: Part a We are given: Pn = for n >= 1 We know the sequence: Zn = Xn - Xn-1 + Vn ; n >= 2 Since: Zn = A + X(t), the mean of Y(t) is: E[Y(t)] = E[A] + E[X(t)] = A[A] + ux .............(i) Then: RY (t, ) = E [(A + X(t)) (A + X(t + ))] .............(ii) = E[A2] + 2E[A]ux + Rx( ) + AE [X(t + )] + E [X(t)X(t + )] ............(iii) = E[A2] + 2E [A] X + RX( )..................(iv) Where the last equality is justified by the fact that we are given that X(t) is wide sense stationary. Therefore the expectation becomes: = E[A] + E[X(t)] Problem 4 This problem is recursive MMSE estimation. We are given: The two independent Gaussian processes are: i). {Un : n >=2} ii). Vn : n >= 1 Where: Un ~ i.i.d N(0, 1) Vn ~ i.i.d N(0, 1) We also know that: X1 = U1 X2 = U1 + U2 Xn = PnXn-1 + (1 - Pn)Xn-2 + Un, n >= 3 Part a Pn = {1 n is even = {0 n is odd The observation sequences are: Yn = Xn + Vn, n>=1 Therefore, we proceed to derive the Kalman filter as the best linear estimator for our linear, nonGaussian model. We slightly generalize the model that was treated so far. Hence we get: Yn = 1/2 (x0 - x'0)TP0-1(x0 - x'0) Kalman filter: Initialization: Update equations: For i = 0,1,2,...,n, the estimate is: Where the filter again is: The MSE is: This can be solved explicitly as: The gain is: The recursive estimate is: Part a We are given: Pn = for n >= 1 We know the sequence: Zn = Xn - Xn-1 + Vn ; n >= 2 Since: Zn = A + X(t), the mean of Y(t) is: E[Y(t)] = E[A] + E[X(t)] = A[A] + ux .............(i) Then: RY (t, ) = E [(A + X(t)) (A + X(t + ))] .............(ii) = E[A2] + 2E[A]ux + Rx( ) + AE [X(t + )] + E [X(t)X(t + )] ............(iii) = E[A2] + 2E [A] X + RX( )..................(iv) Where the last equality is justified by the fact that we are given that X(t) is wide sense stationary. Therefore the expectation becomes: = E[A] + E[X(t)]