Lecture 3: Functions of Symmetric Matrices

Transcription

1 Lecture 3: Functions of Symmetric Matrices Yilin Mo July 2, Recap 1 Bayes Estimator: (a Initialization: (b Correction: f(x 0 Y 1 = f(x 0 f(x k Y k = αf(y k x k f(x k Y k 1, where ( 1 α = f(y k x k f(x k Y k 1 d x k The MMSE estimation can be derived as ˆx = E(x k Y k = x k f(x k Y k d x k (c Prediction: f(x k+1 Y k = f(x k+1 x k f(x k Y k d x k 2 Kalman Filter: (a Initialization: (b Prediction: ˆx 0 1 = 0, P 0 1 = Σ (1 ˆx k+1 k = Aˆx k k, P k+1 k = AP k k A T + Q (2 (c Correction: ˆx k+1 k+1 = ˆx k+1 k + P k+1 k C T (CP k+1 k C T + R 1 (y k+1 C ˆx k+1 k, (3 P k+1 k+1 = P k+1 k P k+1 k C T (CP k+1 k C T + R 1 CP k+1 k (4 1

2 3 Linear Estimator: (a Initialization: (b Prediction: ˆx 0 1 = 0 ˆx k+1 k = Aˆx k k (c Correction: ˆx k+1 k+1 = ˆx k+1 k + K k+1 ( yk+1 C ˆx k+1 k Estimation error covariance of the linear filter satisfies: P 0 1 = Σ, P k+1 k = AP k k A T + Q, P k+1 k+1 = (I K k+1 CP k+1 k (I K k+1 C T + K k+1 RK k+1 2 Kalman Filtering with Intermittent Observations: Problem Formulation Suppose the sensor send its measurements through an erasure channel: Sensor γ k KF Figure 1: Kalman Filtering with Intermittent Observations Let γ k be a binary variable, such that γ k = 0 implies that the KF does not receive y k and γ k = 1 implies that the KF receives y k We assume that γ k is an iid Bernoulli random variable with P (γ k = 1 = λ, which is independent from x 0, {w k }, {v k } Hence, the information that the KF has at time k is γ 0,, γ k, γ 0 y 0,, γ k y k The optimal estimator is a time varying KF: 1 Initialization: ˆx 0 1 = 0, P 0 1 = Σ (5 2 Prediction: ˆx k+1 k = Aˆx k k, P k+1 k = AP k k A T + Q (6 2

3 3 Correction: ˆx k+1 k+1 = ˆx k+1 k + γ k+1 P k+1 k C T (CP k+1 k C T + R 1 (y k+1 C ˆx k+1 k, (7 P k+1 k+1 = P k+1 k γ k+1 P k+1 k C T (CP k+1 k C T + R 1 CP k+1 k (8 To simplify notations, we define Furthermore, define P k P k k h(x AXA T + Q, g(x h(x h(xc T (Ch(XC T + R 1 Ch(X As a result, P k = { h(p k 1 if γ k = 0 g(p k 1 if γ k = 1 h is called a Lyapunov equation and g is called a discrete-time algebraic Riccati equation 3 Properties of Discrete-time Algebraic Riccati Equation 31 Symmetric Matrix Let S n be the space of real symmetric n by n matrices S n is a linear space with dimension n(n + 1/2 Definition 1 S n + S n is the set of all positive semidefinite matrices S n ++ S n is the set of all positive definite matrices 1 For any X, Y S n +, α, β 0, αx + βy S n + S n + is a convex cone 2 S n + ( S n + = {0} S n + induces a partial order on S n : 1 0 S n + = X X X Y = X Y S n + 2 S n + ( S n + = {0} implies that if X Y and Y X, then X = Y 3 Convexity implies that if X Y and Y Z, then X Z 3

4 However, it is not a total order: Neither X Y nor Y X X = 0, Y = [ ] Theorem 1 If the sequence {X k } is monotonically increasing, ie, X k+1 X k, and there exists an M, such that for all k, X k M, then the following entrywise limit is well-defined Proof Diagonal Elements: lim X k = X k X k+1 (i, i X k (i, i implies that the diagonal element X k+1 (i, i X k (i, i Hence, X k (i, i is increasing and is bounded by M(i, i Therefore X k (i, i converges Off-diagonal Elements: Consider k 1 k 2, then X k1 X k2, which implies that all principal minor is non-negative, ie, X k1 (i, j X k2 (i, j 2 X k1 (i, i X k2 (i, i X k1 (j, j X k2 (j, j Use Cauchy Criterion to prove that the off-diagonal elements also converge 32 Functions on S n Definition 2 A function f : S n S n is monotonically increasing if for any X Y, f(x f(y A function f is decreasing if f is increasing Definition 3 A function f : S n S n is convex if for any X, Y and α, β > 0, α + β = 1, the following inequality holds αf(x + βf(y f(αx + βy A function f is concave if f is convex Some functions: 1 Affine function: h(x = AXA T + Q h(x is increasing, convex and concave 2 Inverse function: f(x = X 1 f(x is decreasing and convex on S n ++ 4

5 Proof Consider X, Y S n ++ There exists an orthogonal matrix Q 1, such that Q 1 XQ T 1 = Λ X, where Λ X is a diagonal matrix Define Λ 1/2 X as the square root of Λ X Hence, Q 1 Λ X Q T 1 Q 1 Λ X Q T 1 = X Let X 1/2 = Q 1 Λ 1/2 X QT 1 Then there exists another orthogonal matrix Q 2, such that Q 2 X 1/2 Y X 1/2 Q T 2 = Λ Y, On the other hand Q 2 X 1/2 XX 1/2 Q T 2 = I The proof can be done by using the matrix Q 2 X 1/2 to diagonalize both X and Y and use the fact that 1/x is decreasing and concave on R + 3 Discrete-time algebraic Riccati equation: Matrix Inversion Lemma: Therefore, (A + UCV 1 = A 1 A 1 U(C 1 + V A 1 U 1 V A 1 (9 g(x = [ (h(x 1 + C T R 1 C ] 1 g(x is increasing, concave and non-negative on S n + (why? Another way of thinking: Consider the update equation of a linear filter: ϕ(x, K = (I KCh(X(I KC T + KRK T = K(Ch(XC T + RK T KCh(X h(xc T K T + h(x Define K = h(xc T (Ch(XC T + R 1, then Thus ϕ(x, K = g(x + (K K (Ch(XC T + R(K K T g(x = min ϕ(x, K K Fix K, ϕ(x, K is increasing and affine Thus, g(x is increasing, concave and non-negative on S n + (why? 5