State Space Models Let { xt:t T} and { yt:t T} denote two vector valued time series that satisfy the system of equations: yt = Atxt + vt (The observation equation) xt = Btxt-1 + ut (The state equation) The time series { yt:t T} is said to have state-space representation. Note: { ut:t T} and { vt:t T} denote two vector valued time series that satisfying: 1. 2. 3. 4. E(ut) = E(vt) = 0. E(utusˊ) = E(vtvsˊ) = 0 if t ≠ s. E(ututˊ) = Su and E(vtvtˊ) = Sv. E(utvsˊ) = E(vtusˊ) = 0 for all t and s. Example: One might be tracking an object with several radar stations. The process {xt:t T} gives the position of the object at time t. The process { yt:t T} denotes the observations at time t made by the several radar stations. As in the Hidden Markov Model we will be interested in determining position of the object, {xt:t T}, from the observations, {yt:t T} , made by the several radar stations Example: Many of the models we have considered to date can be thought of a StateSpace models Autoregressive model of order p: yt 1 yt 1 2 yt 2 p yt p ut yt p 1 xt yt 1 yt Define Then yt 0 0 1xt and xt Bxt 1 ut 0 0 1 1 0 2 Observation equation State equation 0 0 x t 1 ut 1 0 p 1 Hidden Markov Model: Assume that there are m states. Also that there the observations Yt are discreet and take on n possible values. Suppose that the m states are denoted by the vectors: 1 0 0 0 1 0 e1 , e 2 , , e m 0 0 1 Suppose that the n possible observations taken at each state are 1 0 0 0 1 0 f1 , f 2 , , f n 0 0 1 Let ij PX t e j X t 1 ei , ij mm and ij PYt f j X t ei , Β ij mn i1 i2 E X t X t 1 e i Πe i im Note Let ut X t EX t X t 1 ei X t Πei X t ΠX t 1 So that X t ΠX t 1 ut with The State Equation Eut X t 1 , X t 2 Eut X t 1 0 Also X t X t Π X t 1 ut Π X t 1 ut Π X t 1 Π X t 1 Π X t 1ut ut Π X t 1 ut ut Hence ut ut X t X t ΠX t 1 ΠX t 1 ΠX t 1ut ut ΠX t 1 and Σu Eut ut X t 1 EX t X t X t 1 t ΠX t 1 ΠX t 1 Ediag X t X t 1 Πdiag X t 1 Π where diag(v) = the diagonal matrix with the components of the vector v along the diagonal Since X t ΠX t 1 ut then diag X t diag ΠX t 1 diag ut and Ediag X t X t 1 diag ΠX t 1 Thus Σu diag ΠX t 1 Πdiag X t 1 Π We have defined ij PYt f j X t ei , Β ij mn Hence Let i1 E Yt X t e i i 2 Βe i in vt Yt EYt X t Yt ΒX t Then Yt ΒX t v t with The Observation Equation Evt X t 0 and Σ v Evt vt X t diag ΒX t Βdiag X t Β Hence with these definitions the state sequence of a Hidden Markov Model satisfies: X t ΠX t 1 ut The State Equation with Eut X t 0 and Σu Eut ut X t 1 diag ΠX t 1 Πdiag X t 1 Π The observation sequence satisfies: Yt ΒX t v t The Observation Equation with Evt X t 0 and Σ v Evt vt X t diag ΒX t Βdiag X t Β Kalman Filtering We are now interested in determining the state vector xt in terms of some or all of the observation vectors y1, y2, y3, … , yT. We will consider finding the “best” linear predictor. We can include a constant term if in addition one of the observations (y0 say) is the vector of 1’s. We will consider estimation of xt in terms of 1. y1, y2, y3, … , yt-1 (the prediction problem) 2. y1, y2, y3, … , yt (the filtering problem) 3. y1, y2, y3, … , yT (t < T, the smoothing problem) For any vector x define: xˆ s xˆ 1 s xˆ 2 s xˆ p s where xˆ i s is the best linear predictor of x(i), the ith component of x, based on y0, y1, y2, … , ys. The best linear predictor of x(i) is the linear function that of x, based on y0, y1, y2, … , ys that minimizes i E x xˆ i s 2 Remark: The best predictor is the unique vector of the form: xˆ s C0 y 0 C1y1 Cs y s Where C0, C1, C2, … ,Cs, are selected so that: x xˆ s y i i 0,1,2,, s i.e. Ex xˆ s yi 0 i 0,1,2,, s Remark: If x, y1, y2, … ,ys are normally distributed then: xˆ s Ex y1 , y 2 ,, y s Remark Let u and v, be two random vectors than uˆ v Cv is the optimal linear predictor of u based on v if C EuvEvv 1 State Space Models Let { xt:t T} and { yt:t T} denote two vector valued time series that satisfy the system of equations: yt = Atxt + vt (The observation equation) xt = Btxt-1 + ut (The state equation) The time series { yt:t T} is said to have state-space representation. Note: { ut:t T} and { vt:t T} denote two vector valued time series that satisfying: 1. 2. 3. 4. E(ut) = E(vt) = 0. E(utusˊ) = E(vtvsˊ) = 0 if t ≠ s. E(ututˊ) = Su and E(vtvtˊ) = Sv. E(utvsˊ) = E(vtusˊ) = 0 for all t and s. Kalman Filtering: Let { xt:t T} and { yt:t T} denote two vector valued time series that satisfy the system of equations: yt = Atxt + vt xt = Bxt-1 + ut xˆ t s Ext y1 , y 2 ,, y s Let and s Σtu E xt xˆ t s xu xˆ u s y1 , y 2 ,, y s Then xˆ t t 1 Βxˆ t 1 t 1 xˆ t t xˆ t t 1 K t y t At xˆ t t 1 where t 1 K t Σ tt t 1 At A t Σ tt At Σ v 1 One also assumes that the initial vector x0 has mean m and covariance matrix S an that xˆ 0 0 μ The covariance matrices are updated Σttt 1 B Σtt11,t1 B Σu Σ ttt Σ ttt 1 K t AΣ ttt 1 with 0 Σ 00 Σ Summary: The Kalman equations 1. Σttt 1 B Σtt11,t1 B Σu t 1 t 1 At A t Σ tt At Σ v 1 2. K t Σ tt 3. Σ ttt Σ ttt 1 K t AΣ ttt 1 4. xˆ t t 1 Βxˆ t 1 t 1 5. xˆ t t xˆ t t 1 K t y t At xˆ t t 1 with xˆ 0 0 μ and 0 Σ 00 Σ Proof: Now xˆ t s Ext y1 , y 2 ,, y s hence xˆ t t 1 Ext y1 , y 2 ,, y t 1 EΒxt 1 ut y1 , y 2 ,, y t 1 ΒExt 1 y1 , y 2 ,, y t 1 Βxˆ t 1 t 1 proving (4) Note yˆ t t 1 Ey t y1 , y 2 ,, y t 1 EAt xt vt y1 , y 2 ,, y t 1 At Ext y1 , y 2 ,, y t 1 At xˆ t t 1 Let et y t yˆ t t 1 y t At xˆ t t 1 At xt v t At xˆ t t 1 At xt xˆ t t 1 v t Let dt xt xˆ t t 1 Given y0, y1, y2, … , yt-1 the best linear predictor of dt using et is: 1 Edt et Eet et et Edt y 0 , y1 ,, y t 1 , et Edt y 0 , y1 ,, y t 1 , y t xˆ t t xˆ t t 1 Hence xˆ t t xˆ t t 1 K t et where et y t At xˆ t t 1 and K t Edt et Eet et (5) 1 Ex xˆ t 1x xˆ t 1 A Now Edt et E xt xˆ t t 1At xt xˆ t t 1 vt t t 1 Σ tt t At t t t Also Eet et EAt xt xˆ t t 1 v t At xt xˆ t t 1 vt At E xt xˆ t t 1xt xˆ t t 1 At At Ext xˆ t t 1vt E vt xt xˆ t t 1 At Evt vt A t Σ ttt 1 At Σ v hence t 1 t 1 K t Σ tu At A t Σ tt At Σ v 1 (2) Thus xˆ t t 1 Βxˆ t 1 t 1 (4) xˆ t t xˆ t t 1 K t y t At xˆ t t 1 where t 1 K t Σ tt t 1 At A t Σ tt At Σ v 1 (5) (2) Also t 1 ˆ ˆ Σtt E xt xt t 1xt xt t 1 y 0 ,, y t 1 EΒxt 1 ut Βxˆ t 1 t 1 Βxt ut Βxˆ t t 1 y 0 ,, y t 1 Hence Σttt 1 B Σtt11,t1 B Σu (3) The proof that Σ ttt Σ ttt 1 K t AΣ ttt 1 will be left as an exercise. (1) Example: Suppose we have an AR(2) time series xt 1 xt 1 2 xt 2 ut What is observe is the time series yt xt vt {ut|t T} and {vt|t T} are white noise time series with standard deviations su and sv. This model can be expressed as a state-space model by defining: xt 1 x t , Β 1 xt 1 then xt 1 x 1 t 1 2 ut , ut 0 0 2 xt 1 ut 0 xt 2 0 or xt Βxt 1 ut The equation: yt xt vt can be written xt yt 1,0 vt Ax t vt xt 1 Note: Σ v s v2 s u2 Σu 0 0 0 The Kalman equations 1. Σttt 1 B Σtt11,t1 B Σu t 1 t 1 At A t Σ tt At Σ v 1 2. K t Σ tt 3. Σ ttt Σ ttt 1 K t AΣ ttt 1 4. xˆ t t 1 Βxˆ t 1 t 1 5. xˆ t t xˆ t t 1 K t y t At xˆ t t 1 Let t 1 Σtt s s t 11 t 12 s s t 12 t 22 t t r r t 11 12 Σtt t t r12 r22 The Kalman equations Σttt 1 B Σtt11,t1 B Σu 1. t s11 t s12 s 1 s 1 t 12 t 22 2 r11t 1 r12t 1 1 1 s u2 0 t 1 t 1 0 r12 r22 2 0 0 0 s r 2r 2 r s t 11 t 1 2 11 1 t 1 12 1 s r r 2 t 12 t 1 11 1 t 1 11 s r t 22 t 1 12 t 1 22 2 2 2 u 2. t 1 K t Σ tt s Kt s t 11 t 12 t 1 At A t Σ tt s s 1 1 0 s 0 s t 12 t 22 t 11 t 12 At Σ v s 1 s v s 0 t 12 t 22 t s11 t t s11 t s s 1 11 v t s11 s v t s 12 s12 st s v 11 1 1 t t 1 Σ tt Σ tt 3. t r11t r12t s11 t t t r12 r22 s12 r s t 11 s s s r s t 22 t 22 2 v 2 s t 12 s s t 11 t t s s11 s12 K t 1 0 t t s s12 s22 t s11 t t 2 s11 s12 s t v t s11 s12 t t s22 s12 st s 2 v 11 t t s11s12 t t r12 s12 t s11 s v2 t 2 11 t 11 K t AΣ tt t 12 t 22 s11t t s12 t 11 t 1 2 v 4. xˆ t t 1 Βxˆ t 1 t 1 xˆt t 1 1 xˆ t 1 1 t 1 2 xˆt 1 t 1 ˆ 0 xt 2 t 1 xˆt t 1 1 xˆt 1 t 1 2 xˆt 2 t 1 5. xˆ t t xˆ t t 1 K t y t At xˆ t t 1 xˆt t xˆt t 1 xˆt t 1 xˆ t xˆ t 1 K t yt 1 0 xˆ t 1 t 1 t 1 t 1 t s11 t 2 xˆt t xˆt t 1 s11 s v yt xˆt t 1 t xˆ t xˆ t 1 t 1 t 1 s12 st s 2 v 11 t 11 s yt xˆt t 1 xˆt t xˆt t 1 t 2 s11 s v t 12 s yt xˆt t 1 xˆt 1 t xˆt 1 t 1 t 2 s11 s v Kalman Filtering (smoothing): Now consider finding xˆ t T Ext y1 , y 2 ,, yT These can be found by successive backward recursions for t = T, T – 1, … , 2, 1 xˆ t 1 T xˆ t 1 t 1 J t 1 xˆ t T xˆ t t 1 where s Σtu t 1 t 1 J t 1 Σt 1,t 1Β Σtt 1 E xt xˆ t s xu xˆ u s y1 , y 2 ,, y s The covariance matrices satisfy the recursions T t 1 T t 1 Σt 1,t 1 Σt 1,t 1 J t 1 Σtt Σtt J t 1 The backward recursions 1. 2. 3. t 1 t 1 J t 1 Σt 1,t 1Β Σtt 1 xˆ t 1 T xˆ t 1 t 1 J t 1 xˆ t T xˆ t t 1 T t 1 T t 1 Σt 1,t 1 Σt 1,t 1 J t 1 Σtt Σtt J t 1 In the example: t 1 t r11t r12t s12 Β Σ 1 tt t t t s22 r12 r22 t 1 t 1 Σ tt , Σ tt , xˆt t 1 and xˆt t t s t 1 11 Σtt t s12 - calculated in forward recursion 2 0
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