Score Evaluation within the Extended Square-root
Score Evaluation within the Extended Square-root Information Filter Maria V. Kulikova
and
Innokenti V. Semoushin
University of the Witwatersrand, Johannesburg, South Africa Ulyanovsk State University, 42 Leo Tolstoy Str., 432970 Ulyanovsk, Russia E-mail:
[email protected] [email protected]
http://staff.ulsu.ru/semoushin/ Score Evaluation within the Extended Square-root Information Filter – p. 1/??
Introduction The method of maximum likelihood . . .
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Introduction The method of maximum likelihood . . . √ Gradient-search optimization algorithms;
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Introduction The method of maximum likelihood . . . √ Gradient-search optimization algorithms; √ Gradient of the negative Log Likelihood Function;
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Introduction The method of maximum likelihood . . . √ Gradient-search optimization algorithms; √ Gradient of the negative Log Likelihood Function; √ Kalman Filter
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Introduction The method of maximum likelihood . . . √ Gradient-search optimization algorithms; √ Gradient of the negative Log Likelihood Function; √ Kalman Filter =⇒ is known to be unstable;
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Introduction The method of maximum likelihood . . . √ Gradient-search optimization algorithms; √ Gradient of the negative Log Likelihood Function; √ Kalman Filter =⇒ is known to be unstable; [P. Park&T. Kailath, 1995] The extended Square Root Information Filter (eSRIF):
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Introduction The method of maximum likelihood . . . √ Gradient-search optimization algorithms; √ Gradient of the negative Log Likelihood Function; √ Kalman Filter =⇒ is known to be unstable; [P. Park&T. Kailath, 1995] The extended Square Root Information Filter (eSRIF): √ avoids numerical instabilities arising from computational errors;
Score Evaluation within the Extended Square-root Information Filter – p. 2/??
Introduction The method of maximum likelihood . . . √ Gradient-search optimization algorithms; √ Gradient of the negative Log Likelihood Function; √ Kalman Filter =⇒ is known to be unstable; [P. Park&T. Kailath, 1995] The extended Square Root Information Filter (eSRIF): √ avoids numerical instabilities arising from computational errors; √ appears to be better suited to parallel implementation and to very large scale integration (VLSI) implementation. Score Evaluation within the Extended Square-root Information Filter – p. 2/??
Problem Statement Consider the discrete-time linear dynamic stochastic system xt+1 = Ft xt + Gt wt , z t = H t xt + v t ,
t = 0, 1, . . . , N, t = 1, 2, . . . , N,
(1) (2)
with the system state xt ∈ Rn , the state disturbance wt ∈ Rq , the observed vector zt ∈ Rm , and the measurement error vt ∈ Rm ,
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Problem Statement Consider the discrete-time linear dynamic stochastic system xt+1 = Ft xt + Gt wt , z t = H t xt + v t ,
t = 0, 1, . . . , N, t = 1, 2, . . . , N,
(1) (2)
with the system state xt ∈ Rn , the state disturbance wt ∈ Rq , the observed vector zt ∈ Rm , and the measurement error vt ∈ Rm , such that the initial state x0 and each wt , vt of {wt : t = 0, 1, . . .}, {vt : t = 1, 2, . . .} are taken from mutually independent Gaussian distributions with the following expectations: io h nh i = x¯0 0 0 E and x 0 w t vt Score Evaluation within the Extended Square-root Information Filter – p. 3/??
Problem Statement ¯0 ) (x0 − x E wt vt
and E
wt wtT0
T (x0 − x ¯0 ) P0 wt = 0 vt 0
= 0, E
vt vtT0
= 0 if t 6= t0 .
0 Qt 0
0
0 Rt
Score Evaluation within the Extended Square-root Information Filter – p. 4/??
Problem Statement ¯0 ) (x0 − x E wt vt
wt wtT0
T (x0 − x ¯0 ) P0 wt = 0 vt 0
vt vtT0
0 Qt 0
0
0 Rt
= 0, E = 0 if t 6= t0 . Assume the and E system is parameterized by a vector θ ∈ Rp of unknown system parameters. This means that all the above characteristics, namely Ft , Gt , Ht , P0 ≥ 0, Qt ≥ 0 and Rt > 0 can depend upon θ (the corresponding notations Ft (θ), Gt (θ) and so on, are suppressed for the sake of simplicity).
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Problem Statement For example, (1), (2) may describe a discrete autoregressive (AR) process observed in the presence of additive noise 0 1 0 ... 0 0 0 0 1 ... 0 0 . . . . . . . . .. xt + .. wt , .. .. .. xt+1 = 0 0 0 ... 1 0 θ1 θ2 θ3 . . . θp
zt =
h
0 0 ... 1
γ
i
xt + v t .
In this case θ are the unknown AR parameters which need to be estimated. Score Evaluation within the Extended Square-root Information Filter – p. 5/??
Problem Statement The negative Log Likelihood Function (LLF) for system (1), (2) is given by N
Lθ Z1
N
o 1 X nm −1 = et ln(2π) + ln(det(Re,t )) + eTt Re,t 2 t=1 2
def
with et = zt − Ht xˆt being the zero-mean innovations whose def
covariance is determined as Re,t = E{et eTt } = Ht Pt HtT + Rt through matrix Pt , the error covariance of the time updated estimate xˆt generated by the Kalman Filter.
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Problem Statement Let lθ (zt ) denote the negative LLF for the t-th measurement zt in def
system (1), (2), given measurements Z1t−1 = {z1 , . . . , zt−1 }, then o 1 nm −1 ln(2π) + ln(det(Re,t )) + eTt Re,t lθ (zt ) = et . (3) 2 2
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Problem Statement Let lθ (zt ) denote the negative LLF for the t-th measurement zt in def
system (1), (2), given measurements Z1t−1 = {z1 , . . . , zt−1 }, then o 1 nm −1 ln(2π) + ln(det(Re,t )) + eTt Re,t lθ (zt ) = et . (3) 2 2 By differentiating (3) we obtain
1 ∂ 1 ∂ T −1 ∂lθ (zt ) et Re,t et , i = 1, p. = [ln(det(Re,t ))] + ∂θi 2 ∂θi 2 ∂θi (4) As can be seen the computation of (3) and (4) leads to implementation of a Kalman Filter (and its derivative with respect to each parameter) which is known to be unstable. Score Evaluation within the Extended Square-root Information Filter – p. 7/??
The eSRIF [P. Park&T. Kailath, 1995]: assume that Π0 > 0, Rt > 0 and Ft −T /2 −T /2 −T /2 −T /2 are invertible. Given P0 = Π0 and P0 xˆ0 = Π0 x¯0 , then
Ot
−T /2 Rt
0 0
−T /2 −Rt Ht Ft−1 −T /2
Pt
Ft−1
0
−T /2 T /2 Rt Ht Ft−1 Gt Qt −T /2
−Pt
T /2
Ft−1 Gt Qt
−T /2 = −P t+1 Kp,t ∗
0 −T /2
Pt+1 ∗
−T /2
Pt
x ˆt
0
Iq −T /2 Re,t
−T /2 −Rt zt
0
−¯ et −T /2
0
Pt+1 x ˆt+1
∗
∗
where Ot is any orthogonal transformation such that the matrix on the right-hand side of the formula is block lower triangular.
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The eSRIF The eSRIF is a modification of the conventional one:
Ot
−T /2 Rt
0 0
−T /2 −Rt Ht Ft−1 −T /2
Pt
Ft−1
−T /2 T /2 Rt Ht Ft−1 Gt Qt −T /2
−Pt
0
T /2
Ft−1 Gt Qt Iq
−T /2 Re,t
0
−T /2 = −P t+1 Kp,t ∗
−T /2
Pt+1 ∗
0
0 ∗
where Ot is any orthogonal transformation such that the matrix on the right-hand side of the formula is lower triangular.
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The eSRIF The eSRIF is a modification of the conventional one:
Ot
−T /2 Rt
0 0
−T /2 −Rt Ht Ft−1 −T /2
Pt
Ft−1
−T /2 T /2 Rt Ht Ft−1 Gt Qt −T /2
−Pt
0
T /2
Ft−1 Gt Qt Iq
−T /2 Re,t
0
−T /2 = −P t+1 Kp,t ∗
−T /2
Pt+1 ∗
0 0 ∗
−T /2 −Rt zt −T /2
Pt
x ˆt
0 −¯ et −T /2
Pt+1 x ˆt+1 ∗
where Ot is any orthogonal transformation such that the matrix on the right-hand side of the formula is lower triangular.
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The eSRIF Remark. The predicted estimate now can be found from the entries of the post-array by solving the triangular system −1/2 −1/2 − ˜t+1 P xˆ− = xˆt+1 . P˜t+1 (5) t+1
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The LLG in terms of the eSRIF The Log Likelihood Gradient (LLG) in terms of the eSRIF is given by i ∂ h et ∂lθ (zt ) 1/2 T ∂¯ = , i = 1, . . . p, ln(det(Re,t )) + e¯t ∂θi ∂θi ∂θi 1/2
where Re,t is a square-root factor of the matrix Re,t , i. e. T /2
1/2
Re,t = Re,t Re,t , and e¯t are the normalized innovations, i. e. −T /2
e¯t = Re,t
et .
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The LLG in terms of the eSRIF 1/2
Taking into account that matrix Re,t is upper triangular, we can show
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The LLG in terms of the eSRIF 1/2
Taking into account that matrix Re,t is upper triangular, we can show 1/2 i h ∂ R e,t ∂ 1/2 −1/2 , i = 1, . . . , p, ln det(Re,t ) = tr Re,t ∂θi ∂θi where tr [ · ] is a trace of matrix.
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The LLG in terms of the eSRIF 1/2
Taking into account that matrix Re,t is upper triangular, we can show 1/2 i h ∂ R e,t ∂ 1/2 −1/2 , i = 1, . . . , p, ln det(Re,t ) = tr Re,t ∂θi ∂θi
where tr [ · ] is a trace of matrix. Finally, we obtain the expression for the LLG in terms of the eSRIF: −1/2 ∂ Re,t et ∂lθ (zt ) 1/2 T ∂¯ +¯ et = − tr Re,t , i = 1, . . . , p. (6) ∂θi ∂θi ∂θi Score Evaluation within the Extended Square-root Information Filter – p. 12/??
LLG Evaluation
−1/2 Re,t
∂ ∂l(zt ) 1/2 = −tr Re,t ∂θi ∂θi
et + e¯Tt ∂¯ , i = 1, . . . , p. ∂θi
(6)
Score Evaluation within the Extended Square-root Information Filter – p. 13/??
LLG Evaluation
−1/2 Re,t
∂ ∂l(zt ) 1/2 = −tr Re,t ∂θi ∂θi
et + e¯Tt ∂¯ , i = 1, . . . , p. ∂θi
(6)
Score Evaluation within the Extended Square-root Information Filter – p. 13/??
LLG Evaluation
∂ ∂l(zt ) 1/2 = −tr Re,t ∂θi ∂θi
The eSRIF: −T /2 Rt Ot 0 0
−1/2 Re,t
−T /2 −Rt Ht Ft−1 −T /2 −1 Pt Ft
0
et + e¯Tt ∂¯ , i = 1, . . . , p. ∂θi
−T /2 T /2 Rt Ht Ft−1 Gt Qt −T /2 −1 T /2 −Pt F t Gt Q t
−T /2 R e,t = −P −T /2 Kp,t t+1 ∗
Iq 0 −T /2
Pt+1 ∗
−T /2 −Rt zt −T /2 Pt x ˆt
0
0
−¯ et
0
Pt+1 x ˆt+1
∗
−T /2
∗
(6)
Score Evaluation within the Extended Square-root Information Filter – p. 13/??
LLG Evaluation
∂ ∂l(zt ) 1/2 = −tr Re,t ∂θi ∂θi
The eSRIF: −T /2 Rt Ot 0 0
−1/2 Re,t
−T /2 −Rt Ht Ft−1 −T /2 −1 Pt Ft
0
et + e¯Tt ∂¯ , i = 1, . . . , p. ∂θi
−T /2 T /2 Rt Ht Ft−1 Gt Qt −T /2 −1 T /2 −Pt F t Gt Q t
−T /2 R e,t = −P −T /2 Kp,t t+1 ∗
Iq 0 −T /2
Pt+1 ∗
−T /2 −Rt zt −T /2 Pt x ˆt
0
0
−¯ et
0
Pt+1 x ˆt+1
∗
−T /2
∗
(6)
Score Evaluation within the Extended Square-root Information Filter – p. 13/??
LLG Evaluation
∂ ∂l(zt ) 1/2 = −tr Re,t ∂θi ∂θi
The eSRIF: −T /2 Rt Ot 0 0
−1/2 Re,t
−T /2 −Rt Ht Ft−1 −T /2 −1 Pt Ft
0
et + e¯Tt ∂¯ , i = 1, . . . , p. ∂θi
−T /2 T /2 Rt Ht Ft−1 Gt Qt −T /2 −1 T /2 −Pt F t Gt Q t
−T /2 R e,t = −P −T /2 Kp,t t+1 ∗
Iq 0 −T /2
Pt+1 ∗
−T /2 −Rt zt −T /2 Pt x ˆt
0
0
−¯ et
0
Pt+1 x ˆt+1
∗
−T /2
∗
(6)
Score Evaluation within the Extended Square-root Information Filter – p. 13/??
LLG Evaluation Lemma 1. Let QA = L (7) where Q is any orthogonal transformation such that the matrix on the right-hand side of formula (7) is lower triangular and A is a nonsingular matrix. If the elements of A are differentiable functions of a parameter θ then the upper triangular matrix U in Q0θ QT = U¯ T − U¯ is, in fact, the strictly upper triangular part of the matrix QA0θ L−1 : ¯ + D + U¯ QA0θ L−1 = L
(8)
(9)
¯ D and U¯ are, respectively, strictly lower triangular, where L, diagonal and strictly upper triangular.Score Evaluation within the Extended Square-root Information Filter – p. 14/??
Algorithm LLG-eSRIF I. For each θi , i = 1, . . . , p, apply the eSRIF
Ot
−T /2 Rt
0 0
−T /2 −Rt Ht Ft−1 −T /2
Pt
Ft−1
0
−T /2 T /2 Rt Ht Ft−1 Gt Qt −T /2
−Pt
T /2
Ft−1 Gt Qt
−T /2 = −P t+1 Kp,t ∗
0 −T /2
Pt+1 ∗
−T /2
Pt
x ˆt
0
Iq −T /2 Re,t
−T /2 −Rt zt
0
−¯ et −T /2
0
Pt+1 x ˆt+1
∗
∗
where Ot is any orthogonal transformation such that the matrix on the right-hand side of the formula is block lower triangular.
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Algorithm LLG-eSRIF II. For each θi , i = 1, . . . , p, calculate
∂ −T /2 ∂θi Rt Ot 0 0
∂ ∂θi ∂ ∂θi
(1)
St
(4)
St
0
∂ ∂θi ∂ ∂θi
(2)
St
(5)
St
0
Xi
= Ni ∗
∂ ∂θi ∂ ∂θi
(3)
St
(6) St 0 M i Li Wi K i , ∗ ∗
Yi Vi ∗
Ot is the same orthogonal transformation as in the eSRIF and (1)
St
(4)
St
−T /2
= −Rt
−T /2
= Pt
(2)
= Rt
(5)
= −Pt
Ht Ft−1 , St
Ft−1 ,
St
−T /2
T /2
Ht Ft−1 Gt Qt
−T /2
T /2
Ft−1 Gt Qt
,
(3)
, St
(6)
St
−T /2
= −Rt
−T /2
= Pt
zt ,
x ˆt .
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Algorithm LLG-eSRIF III. For each θi , i = 1, . . . , p, compute
Ji =
Xi Y i M i Ni V i W i
−T /2 Re,t −T /2 −Pt+1 Kp,t
∗
0 −T /2 Pt+1
∗
0
−1
0 ∗
.
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Algorithm LLG-eSRIF IV. For each θi , i = 1, . . . , p, split the matrices m+n+q
Ji =
zh
}|
Li + Di + Ui | {z } m+n
∗∗∗
i{ o
m+n
where Li , Di and Ui are the strictly lower triangular, diagonal and strictly upper triangular parts of Ji , respectively.
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Algorithm LLG-eSRIF V. For each θi , i = 1, . . . , p, compute the following quantities: −T /2 ∂Re,t 0 ∂θi = Li + Di + UiT −T /2 ˜ −T /2 P K ∂ ˜ p,t t+1 ∂ Pt+1 − (12) ∂θi ∂θi −T /2 Re,t 0 , × −T /2 −T /2 −P˜t+1 Kp,t P˜t+1
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Algorithm LLG-eSRIF ∂¯ et = ∂θi
"
−T /2 ∂Re,t
∂θi
#
T /2
− Xi Re,t e¯t + Yi Ft xˆt − Li ,
(13)
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Algorithm LLG-eSRIF ∂¯ et = ∂θi (6)
∂St+1 ∂θi
"
−T /2 ∂Re,t
#
T /2
− Xi Re,t e¯t + Yi Ft xˆt − Li ,
∂θi −T /2 ∂ P˜t+1 Kp,t T /2 = + Ni Re,t e¯t ∂θi " −T /2 # ∂ P˜t+1 + − Vi Ft xˆt + Ki . ∂θi
(13)
(14)
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Algorithm LLG-eSRIF VI. Finally, we have all values to compute the LLG according to (6): −1/2 ∂ R e,t ∂lθ (zt ) ∂¯ et 1/2 T = − tr Re,t , i = 1, p. + e¯t ∂θi ∂θi ∂θi
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Algorithm LLG-eSRIF VI. Finally, we have all values to compute the LLG according to (6): −1/2 ∂ R e,t ∂lθ (zt ) ∂¯ et 1/2 T = − tr Re,t , i = 1, p. + e¯t ∂θi ∂θi ∂θi The LLG-eSRIF:
The LLG-eSRIF:
Stage I
Stages II —V
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Algorithm LLG-eSRIF VI. Finally, we have all values to compute the LLG according to (6): −1/2 ∂ R e,t ∂lθ (zt ) ∂¯ et 1/2 T = − tr Re,t , i = 1, p. + e¯t ∂θi ∂θi ∂θi The LLG-eSRIF: Stage
I
The LLG-eSRIF: Stages
II —V
Score Evaluation within the Extended Square-root Information Filter – p. 21/??
Algorithm LLG-eSRIF VI. Finally, we have all values to compute the LLG according to (6): −1/2 ∂ R e,t ∂lθ (zt ) ∂¯ et 1/2 T = − tr Re,t , i = 1, p. + e¯t ∂θi ∂θi ∂θi The LLG-eSRIF: Stage
I
The LLG-eSRIF: Stages II —V
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Algorithm LLG-eSRIF The LLG-eSRIF consists of two parts:
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Algorithm LLG-eSRIF The LLG-eSRIF consists of two parts: √ The source filtering algorithm, i.e. eSRIF.
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Algorithm LLG-eSRIF The LLG-eSRIF consists of two parts: √ The source filtering algorithm, i.e. eSRIF. √ The "differentiated" part: Stages II — V.
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Algorithm LLG-eSRIF The LLG-eSRIF consists of two parts: √ The source filtering algorithm, i.e. eSRIF. √ The "differentiated" part: Stages II — V. Thus, the Algorithm LLG-eSRIF is ideal for simultaneous
Score Evaluation within the Extended Square-root Information Filter – p. 22/??
Algorithm LLG-eSRIF The LLG-eSRIF consists of two parts: √ The source filtering algorithm, i.e. eSRIF. √ The "differentiated" part: Stages II — V. Thus, the Algorithm LLG-eSRIF is ideal for simultaneous √ state estimation
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Algorithm LLG-eSRIF The LLG-eSRIF consists of two parts: √ The source filtering algorithm, i.e. eSRIF. √ The "differentiated" part: Stages II — V. Thus, the Algorithm LLG-eSRIF is ideal for simultaneous √ state estimation √ parameter identification.
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Algorithm LLG-eSRIF Remark. Since the matrices in LLG (6) are triangular, only the
diagonal elements of
1/2 Re,t
∂
and
−1/2 Re,t
need to be computed.
∂θi Hence, the Algorithm LLG-eSRIF allows the m × m-matrix inversion of Re,t to be avoided in the evaluation of LLG.
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Ill-Conditioned Example Problems Problem 1 Given: h i θ 0 , H = 1, 0 , R = e2 θ, F = I2 , Q = 0, P0 = 0 θ h iT G = 0, 0 , 0< e « 1
where θ is an unknown parameter, I2 is an identity 2 × 2 matrix; to simulate roundoff we assume
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Ill-Conditioned Example Problems Problem 1 Given: h i θ 0 , H = 1, 0 , R = e2 θ, F = I2 , Q = 0, P0 = 0 θ h iT G = 0, 0 , 0< e « 1
where θ is an unknown parameter, I2 is an identity 2 × 2 matrix; r 2 to simulate roundoff we assume e + 1 6= 1 but e + 1 = 1.
Score Evaluation within the Extended Square-root Information Filter – p. 24/??
Ill-Conditioned Example Problems Problem 1 Given: h i θ 0 , H = 1, 0 , R = e2 θ, F = I2 , Q = 0, P0 = 0 θ h iT G = 0, 0 , 0< e « 1
where θ is an unknown parameter, I2 is an identity 2 × 2 matrix; r 2 to simulate roundoff we assume e + 1 6= 1 but e + 1 = 1. Calculate: (P1 )0θ at the point θ = 1.
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Ill-Conditioned Example Problems Problem 1 Given: h i θ 0 , H = 1, 0 , R = e2 θ, F = I2 , Q = 0, P0 = 0 θ h iT G = 0, 0 , 0< e « 1
where θ is an unknown parameter, I2 is an identity 2 × 2 matrix; r 2 to simulate roundoff we assume e + 1 6= 1 but e + 1 = 1. Calculate: (P1 )0θ at the point θ = 1. For the θ = 1, this example illustrates the initialization problems [Kaminski P.G., Bryson A.E., Schmidt S.F., 1971] that result when H1 Π0 H1T + R1 is rounded to H1 Π0 H1T . Score Evaluation within the Extended Square-root Information Filter – p. 24/??
Comparison (Problem 1) Filter
Exact Answer
Rounded Answer
Score Evaluation within the Extended Square-root Information Filter – p. 25/??
Comparison (Problem 1) Filter
’diff’ KF
Exact Answer 2 e 1 + e2 0 0 (P1 )θ = 0 1
Rounded Answer 0 0 r = 0 1
Score Evaluation within the Extended Square-root Information Filter – p. 25/??
Comparison (Problem 1) Filter
’diff’ KF
’diff’ IF
Exact Answer 2 e 1 + e2 0 0 (P1 )θ = 0 1 2 1+e 0 e2 −1 0 P1 θ = − 0 1
Rounded Answer 0 0 r = 0 1
1 r e2 0 = − 0 1
Score Evaluation within the Extended Square-root Information Filter – p. 25/??
Comparison (Problem 1) Filter
’diff’ KF
’diff’ IF
LLGeSRIF
Exact Answer Rounded Answer 2 e 0 0 r 1 + e2 0 0 (P1 )θ = = 0 1 0 1 2 1 1+e 0 0 2 0 2 r P1−1 θ = − e = − e 0 1 0 1 √ 2 1 1+e 0 0 0 r 1 1 −1/2 e − P1 =− e 2 2 θ 0 1 0 1 Score Evaluation within the Extended Square-root Information Filter – p. 25/??
Ill-Conditioned Example Problems Problem 2 Given: h i θ 0 , H = 1, 1 , R = e2 θ, F = I2 , Q = 0, P0 = 0 θ h iT G = 0, 0 , 0< e « 1
where θ is an unknown parameter, I2 is an identity 2 × 2 matrix; to simulate roundoff we assume
Score Evaluation within the Extended Square-root Information Filter – p. 26/??
Ill-Conditioned Example Problems Problem 2 Given: h i θ 0 , H = 1, 1 , R = e2 θ, F = I2 , Q = 0, P0 = 0 θ h iT G = 0, 0 , 0< e « 1
where θ is an unknown parameter, I2 is an identity 2 × 2 matrix; r 2 to simulate roundoff we assume e + 1 6= 1 but e + 1 = 1.
Score Evaluation within the Extended Square-root Information Filter – p. 26/??
Ill-Conditioned Example Problems Problem 2 Given: h i θ 0 , H = 1, 1 , R = e2 θ, F = I2 , Q = 0, P0 = 0 θ h iT G = 0, 0 , 0< e « 1
where θ is an unknown parameter, I2 is an identity 2 × 2 matrix; r 2 to simulate roundoff we assume e + 1 6= 1 but e + 1 = 1. Calculate: (P1 )0θ at the point θ = 1.
Score Evaluation within the Extended Square-root Information Filter – p. 26/??
Comparison (Problem 2) Filter
Exact Answer
Rounded Answer
Score Evaluation within the Extended Square-root Information Filter – p. 27/??
Comparison (Problem 2) Filter
’diff’ KF
Exact Answer 2 1 + e −1 1 2 + e2 −1 1 + e2
Rounded Answer 1 1 −1 2 −1 1
Score Evaluation within the Extended Square-root Information Filter – p. 27/??
Comparison (Problem 2) Filter
’diff’ KF ’diff’ IF
Exact Answer 2 1 + e −1 1 2 + e2 −1 1 + e2
1 − 2 e
1 + e2 1
1 1 + e2
Rounded Answer 1 1 −1 2 −1 1
1 1 1 − 2 e 1 1
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Comparison (Problem 2) Filter
’diff’ KF ’diff’ IF
LLGeSRIF
Exact Answer 2 1 + e −1 1 2 + e2 −1 1 + e2
1 + e2
1 − 2 e 1 r 2 + e2 1 1 + e2 − 2 0
1 1 + e2
Rounded Answer 1 1 −1 2 −1 1
1 √ e 1 + e2 √ 1 + e2 e
1 1 1 − 2 e 1 1 √ 1 2 − 2 0
1 e 1 e
Score Evaluation within the Extended Square-root Information Filter – p. 27/??
Numerical Results Example. Let the test system (1), (2) be defined as follows: 1 ∆t 0 xt + w t , xt+1 = 0 e−∆t/τ 1 zt =
h
1
0
i
xt + v t
where τ is an unknown parameter which needs to be estimated.
Score Evaluation within the Extended Square-root Information Filter – p. 28/??
Numerical Results Example. Let the test system (1), (2) be defined as follows: 1 ∆t 0 xt + w t , xt+1 = 0 e−∆t/τ 1 zt =
h
1
0
i
xt + v t
where τ is an unknown parameter which needs to be estimated. For the test problem, τ ∗ =15 was chosen as the true value of parameter τ .
Score Evaluation within the Extended Square-root Information Filter – p. 28/??
Numerical Results The negative Log Likelihood Function
The Log Likelihood Gradient
Score Evaluation within the Extended Square-root Information Filter – p. 29/??
Numerical Results The negative Log Likelihood Function
The Log Likelihood Gradient
500
The negative Log Likelihood Function
498
496
494
492
490
488
486 0
CKF CIF eSRIF 15
30
45
tau (sec.)
Score Evaluation within the Extended Square-root Information Filter – p. 29/??
Numerical Results The negative Log Likelihood Function
The Log Likelihood Gradient
500
The negative Log Likelihood Function
498
496
494
492
490
488
486 0
CKF CIF eSRIF 15
30
45
tau (sec.)
Score Evaluation within the Extended Square-root Information Filter – p. 29/??
Numerical Results The negative Log Likelihood Function
The Log Likelihood Gradient
500
0.5
0 The Log Likelihood Gradient
The negative Log Likelihood Function
498
496
494
492
490
−0.5
−1
−1.5
488
486 0
CKF CIF eSRIF 15
30 tau (sec.)
"differentiated" CKF "differentiated" CIF Algorithm LLG−eSRIF
45
−2 0
15
30 tau (sec.)
Score Evaluation within the Extended Square-root Information Filter – p. 29/??
45
Numerical Results The negative Log Likelihood Function
The Log Likelihood Gradient
500
0.5
0 The Log Likelihood Gradient
The negative Log Likelihood Function
498
496
494
492
490
−0.5
−1
−1.5
488
486 0
CKF CIF eSRIF 15
30 tau (sec.)
"differentiated" CKF "differentiated" CIF Algorithm LLG−eSRIF
45
−2 0
15
30 tau (sec.)
Score Evaluation within the Extended Square-root Information Filter – p. 29/??
45
Conclusion In this paper, the new algorithm for evaluating the Log Likelihood Gradient (score) of linear discrete-time dynamic systems has been developed. The necessary theory has been given and substantiated by the computational experiments. Two ill-conditioned example problems have been constructed to show the superior perfomance of the Algorithm LLG-eSRIF over the conventional approach. All of these are good reasons to use the presented algorithm in practice.
Score Evaluation within the Extended Square-root Information Filter – p. 30/??
Q&A Maria Kulikova University of the Witwatersrand, Johannesburg, South Africa
Score Evaluation within the Extended Square-root Information Filter – p. 31/??
and
Innokenti V. Semoushin
University of the Witwatersrand, Johannesburg, South Africa Ulyanovsk State University, 42 Leo Tolstoy Str., 432970 Ulyanovsk, Russia E-mail:
[email protected] [email protected]
http://staff.ulsu.ru/semoushin/ Score Evaluation within the Extended Square-root Information Filter – p. 1/??
Introduction The method of maximum likelihood . . .
Score Evaluation within the Extended Square-root Information Filter – p. 2/??
Introduction The method of maximum likelihood . . . √ Gradient-search optimization algorithms;
Score Evaluation within the Extended Square-root Information Filter – p. 2/??
Introduction The method of maximum likelihood . . . √ Gradient-search optimization algorithms; √ Gradient of the negative Log Likelihood Function;
Score Evaluation within the Extended Square-root Information Filter – p. 2/??
Introduction The method of maximum likelihood . . . √ Gradient-search optimization algorithms; √ Gradient of the negative Log Likelihood Function; √ Kalman Filter
Score Evaluation within the Extended Square-root Information Filter – p. 2/??
Introduction The method of maximum likelihood . . . √ Gradient-search optimization algorithms; √ Gradient of the negative Log Likelihood Function; √ Kalman Filter =⇒ is known to be unstable;
Score Evaluation within the Extended Square-root Information Filter – p. 2/??
Introduction The method of maximum likelihood . . . √ Gradient-search optimization algorithms; √ Gradient of the negative Log Likelihood Function; √ Kalman Filter =⇒ is known to be unstable; [P. Park&T. Kailath, 1995] The extended Square Root Information Filter (eSRIF):
Score Evaluation within the Extended Square-root Information Filter – p. 2/??
Introduction The method of maximum likelihood . . . √ Gradient-search optimization algorithms; √ Gradient of the negative Log Likelihood Function; √ Kalman Filter =⇒ is known to be unstable; [P. Park&T. Kailath, 1995] The extended Square Root Information Filter (eSRIF): √ avoids numerical instabilities arising from computational errors;
Score Evaluation within the Extended Square-root Information Filter – p. 2/??
Introduction The method of maximum likelihood . . . √ Gradient-search optimization algorithms; √ Gradient of the negative Log Likelihood Function; √ Kalman Filter =⇒ is known to be unstable; [P. Park&T. Kailath, 1995] The extended Square Root Information Filter (eSRIF): √ avoids numerical instabilities arising from computational errors; √ appears to be better suited to parallel implementation and to very large scale integration (VLSI) implementation. Score Evaluation within the Extended Square-root Information Filter – p. 2/??
Problem Statement Consider the discrete-time linear dynamic stochastic system xt+1 = Ft xt + Gt wt , z t = H t xt + v t ,
t = 0, 1, . . . , N, t = 1, 2, . . . , N,
(1) (2)
with the system state xt ∈ Rn , the state disturbance wt ∈ Rq , the observed vector zt ∈ Rm , and the measurement error vt ∈ Rm ,
Score Evaluation within the Extended Square-root Information Filter – p. 3/??
Problem Statement Consider the discrete-time linear dynamic stochastic system xt+1 = Ft xt + Gt wt , z t = H t xt + v t ,
t = 0, 1, . . . , N, t = 1, 2, . . . , N,
(1) (2)
with the system state xt ∈ Rn , the state disturbance wt ∈ Rq , the observed vector zt ∈ Rm , and the measurement error vt ∈ Rm , such that the initial state x0 and each wt , vt of {wt : t = 0, 1, . . .}, {vt : t = 1, 2, . . .} are taken from mutually independent Gaussian distributions with the following expectations: io h nh i = x¯0 0 0 E and x 0 w t vt Score Evaluation within the Extended Square-root Information Filter – p. 3/??
Problem Statement ¯0 ) (x0 − x E wt vt
and E
wt wtT0
T (x0 − x ¯0 ) P0 wt = 0 vt 0
= 0, E
vt vtT0
= 0 if t 6= t0 .
0 Qt 0
0
0 Rt
Score Evaluation within the Extended Square-root Information Filter – p. 4/??
Problem Statement ¯0 ) (x0 − x E wt vt
wt wtT0
T (x0 − x ¯0 ) P0 wt = 0 vt 0
vt vtT0
0 Qt 0
0
0 Rt
= 0, E = 0 if t 6= t0 . Assume the and E system is parameterized by a vector θ ∈ Rp of unknown system parameters. This means that all the above characteristics, namely Ft , Gt , Ht , P0 ≥ 0, Qt ≥ 0 and Rt > 0 can depend upon θ (the corresponding notations Ft (θ), Gt (θ) and so on, are suppressed for the sake of simplicity).
Score Evaluation within the Extended Square-root Information Filter – p. 4/??
Problem Statement For example, (1), (2) may describe a discrete autoregressive (AR) process observed in the presence of additive noise 0 1 0 ... 0 0 0 0 1 ... 0 0 . . . . . . . . .. xt + .. wt , .. .. .. xt+1 = 0 0 0 ... 1 0 θ1 θ2 θ3 . . . θp
zt =
h
0 0 ... 1
γ
i
xt + v t .
In this case θ are the unknown AR parameters which need to be estimated. Score Evaluation within the Extended Square-root Information Filter – p. 5/??
Problem Statement The negative Log Likelihood Function (LLF) for system (1), (2) is given by N
Lθ Z1
N
o 1 X nm −1 = et ln(2π) + ln(det(Re,t )) + eTt Re,t 2 t=1 2
def
with et = zt − Ht xˆt being the zero-mean innovations whose def
covariance is determined as Re,t = E{et eTt } = Ht Pt HtT + Rt through matrix Pt , the error covariance of the time updated estimate xˆt generated by the Kalman Filter.
Score Evaluation within the Extended Square-root Information Filter – p. 6/??
Problem Statement Let lθ (zt ) denote the negative LLF for the t-th measurement zt in def
system (1), (2), given measurements Z1t−1 = {z1 , . . . , zt−1 }, then o 1 nm −1 ln(2π) + ln(det(Re,t )) + eTt Re,t lθ (zt ) = et . (3) 2 2
Score Evaluation within the Extended Square-root Information Filter – p. 7/??
Problem Statement Let lθ (zt ) denote the negative LLF for the t-th measurement zt in def
system (1), (2), given measurements Z1t−1 = {z1 , . . . , zt−1 }, then o 1 nm −1 ln(2π) + ln(det(Re,t )) + eTt Re,t lθ (zt ) = et . (3) 2 2 By differentiating (3) we obtain
1 ∂ 1 ∂ T −1 ∂lθ (zt ) et Re,t et , i = 1, p. = [ln(det(Re,t ))] + ∂θi 2 ∂θi 2 ∂θi (4) As can be seen the computation of (3) and (4) leads to implementation of a Kalman Filter (and its derivative with respect to each parameter) which is known to be unstable. Score Evaluation within the Extended Square-root Information Filter – p. 7/??
The eSRIF [P. Park&T. Kailath, 1995]: assume that Π0 > 0, Rt > 0 and Ft −T /2 −T /2 −T /2 −T /2 are invertible. Given P0 = Π0 and P0 xˆ0 = Π0 x¯0 , then
Ot
−T /2 Rt
0 0
−T /2 −Rt Ht Ft−1 −T /2
Pt
Ft−1
0
−T /2 T /2 Rt Ht Ft−1 Gt Qt −T /2
−Pt
T /2
Ft−1 Gt Qt
−T /2 = −P t+1 Kp,t ∗
0 −T /2
Pt+1 ∗
−T /2
Pt
x ˆt
0
Iq −T /2 Re,t
−T /2 −Rt zt
0
−¯ et −T /2
0
Pt+1 x ˆt+1
∗
∗
where Ot is any orthogonal transformation such that the matrix on the right-hand side of the formula is block lower triangular.
Score Evaluation within the Extended Square-root Information Filter – p. 8/??
The eSRIF The eSRIF is a modification of the conventional one:
Ot
−T /2 Rt
0 0
−T /2 −Rt Ht Ft−1 −T /2
Pt
Ft−1
−T /2 T /2 Rt Ht Ft−1 Gt Qt −T /2
−Pt
0
T /2
Ft−1 Gt Qt Iq
−T /2 Re,t
0
−T /2 = −P t+1 Kp,t ∗
−T /2
Pt+1 ∗
0
0 ∗
where Ot is any orthogonal transformation such that the matrix on the right-hand side of the formula is lower triangular.
Score Evaluation within the Extended Square-root Information Filter – p. 9/??
The eSRIF The eSRIF is a modification of the conventional one:
Ot
−T /2 Rt
0 0
−T /2 −Rt Ht Ft−1 −T /2
Pt
Ft−1
−T /2 T /2 Rt Ht Ft−1 Gt Qt −T /2
−Pt
0
T /2
Ft−1 Gt Qt Iq
−T /2 Re,t
0
−T /2 = −P t+1 Kp,t ∗
−T /2
Pt+1 ∗
0 0 ∗
−T /2 −Rt zt −T /2
Pt
x ˆt
0 −¯ et −T /2
Pt+1 x ˆt+1 ∗
where Ot is any orthogonal transformation such that the matrix on the right-hand side of the formula is lower triangular.
Score Evaluation within the Extended Square-root Information Filter – p. 9/??
The eSRIF Remark. The predicted estimate now can be found from the entries of the post-array by solving the triangular system −1/2 −1/2 − ˜t+1 P xˆ− = xˆt+1 . P˜t+1 (5) t+1
Score Evaluation within the Extended Square-root Information Filter – p. 10/??
The LLG in terms of the eSRIF The Log Likelihood Gradient (LLG) in terms of the eSRIF is given by i ∂ h et ∂lθ (zt ) 1/2 T ∂¯ = , i = 1, . . . p, ln(det(Re,t )) + e¯t ∂θi ∂θi ∂θi 1/2
where Re,t is a square-root factor of the matrix Re,t , i. e. T /2
1/2
Re,t = Re,t Re,t , and e¯t are the normalized innovations, i. e. −T /2
e¯t = Re,t
et .
Score Evaluation within the Extended Square-root Information Filter – p. 11/??
The LLG in terms of the eSRIF 1/2
Taking into account that matrix Re,t is upper triangular, we can show
Score Evaluation within the Extended Square-root Information Filter – p. 12/??
The LLG in terms of the eSRIF 1/2
Taking into account that matrix Re,t is upper triangular, we can show 1/2 i h ∂ R e,t ∂ 1/2 −1/2 , i = 1, . . . , p, ln det(Re,t ) = tr Re,t ∂θi ∂θi where tr [ · ] is a trace of matrix.
Score Evaluation within the Extended Square-root Information Filter – p. 12/??
The LLG in terms of the eSRIF 1/2
Taking into account that matrix Re,t is upper triangular, we can show 1/2 i h ∂ R e,t ∂ 1/2 −1/2 , i = 1, . . . , p, ln det(Re,t ) = tr Re,t ∂θi ∂θi
where tr [ · ] is a trace of matrix. Finally, we obtain the expression for the LLG in terms of the eSRIF: −1/2 ∂ Re,t et ∂lθ (zt ) 1/2 T ∂¯ +¯ et = − tr Re,t , i = 1, . . . , p. (6) ∂θi ∂θi ∂θi Score Evaluation within the Extended Square-root Information Filter – p. 12/??
LLG Evaluation
−1/2 Re,t
∂ ∂l(zt ) 1/2 = −tr Re,t ∂θi ∂θi
et + e¯Tt ∂¯ , i = 1, . . . , p. ∂θi
(6)
Score Evaluation within the Extended Square-root Information Filter – p. 13/??
LLG Evaluation
−1/2 Re,t
∂ ∂l(zt ) 1/2 = −tr Re,t ∂θi ∂θi
et + e¯Tt ∂¯ , i = 1, . . . , p. ∂θi
(6)
Score Evaluation within the Extended Square-root Information Filter – p. 13/??
LLG Evaluation
∂ ∂l(zt ) 1/2 = −tr Re,t ∂θi ∂θi
The eSRIF: −T /2 Rt Ot 0 0
−1/2 Re,t
−T /2 −Rt Ht Ft−1 −T /2 −1 Pt Ft
0
et + e¯Tt ∂¯ , i = 1, . . . , p. ∂θi
−T /2 T /2 Rt Ht Ft−1 Gt Qt −T /2 −1 T /2 −Pt F t Gt Q t
−T /2 R e,t = −P −T /2 Kp,t t+1 ∗
Iq 0 −T /2
Pt+1 ∗
−T /2 −Rt zt −T /2 Pt x ˆt
0
0
−¯ et
0
Pt+1 x ˆt+1
∗
−T /2
∗
(6)
Score Evaluation within the Extended Square-root Information Filter – p. 13/??
LLG Evaluation
∂ ∂l(zt ) 1/2 = −tr Re,t ∂θi ∂θi
The eSRIF: −T /2 Rt Ot 0 0
−1/2 Re,t
−T /2 −Rt Ht Ft−1 −T /2 −1 Pt Ft
0
et + e¯Tt ∂¯ , i = 1, . . . , p. ∂θi
−T /2 T /2 Rt Ht Ft−1 Gt Qt −T /2 −1 T /2 −Pt F t Gt Q t
−T /2 R e,t = −P −T /2 Kp,t t+1 ∗
Iq 0 −T /2
Pt+1 ∗
−T /2 −Rt zt −T /2 Pt x ˆt
0
0
−¯ et
0
Pt+1 x ˆt+1
∗
−T /2
∗
(6)
Score Evaluation within the Extended Square-root Information Filter – p. 13/??
LLG Evaluation
∂ ∂l(zt ) 1/2 = −tr Re,t ∂θi ∂θi
The eSRIF: −T /2 Rt Ot 0 0
−1/2 Re,t
−T /2 −Rt Ht Ft−1 −T /2 −1 Pt Ft
0
et + e¯Tt ∂¯ , i = 1, . . . , p. ∂θi
−T /2 T /2 Rt Ht Ft−1 Gt Qt −T /2 −1 T /2 −Pt F t Gt Q t
−T /2 R e,t = −P −T /2 Kp,t t+1 ∗
Iq 0 −T /2
Pt+1 ∗
−T /2 −Rt zt −T /2 Pt x ˆt
0
0
−¯ et
0
Pt+1 x ˆt+1
∗
−T /2
∗
(6)
Score Evaluation within the Extended Square-root Information Filter – p. 13/??
LLG Evaluation Lemma 1. Let QA = L (7) where Q is any orthogonal transformation such that the matrix on the right-hand side of formula (7) is lower triangular and A is a nonsingular matrix. If the elements of A are differentiable functions of a parameter θ then the upper triangular matrix U in Q0θ QT = U¯ T − U¯ is, in fact, the strictly upper triangular part of the matrix QA0θ L−1 : ¯ + D + U¯ QA0θ L−1 = L
(8)
(9)
¯ D and U¯ are, respectively, strictly lower triangular, where L, diagonal and strictly upper triangular.Score Evaluation within the Extended Square-root Information Filter – p. 14/??
Algorithm LLG-eSRIF I. For each θi , i = 1, . . . , p, apply the eSRIF
Ot
−T /2 Rt
0 0
−T /2 −Rt Ht Ft−1 −T /2
Pt
Ft−1
0
−T /2 T /2 Rt Ht Ft−1 Gt Qt −T /2
−Pt
T /2
Ft−1 Gt Qt
−T /2 = −P t+1 Kp,t ∗
0 −T /2
Pt+1 ∗
−T /2
Pt
x ˆt
0
Iq −T /2 Re,t
−T /2 −Rt zt
0
−¯ et −T /2
0
Pt+1 x ˆt+1
∗
∗
where Ot is any orthogonal transformation such that the matrix on the right-hand side of the formula is block lower triangular.
Score Evaluation within the Extended Square-root Information Filter – p. 15/??
Algorithm LLG-eSRIF II. For each θi , i = 1, . . . , p, calculate
∂ −T /2 ∂θi Rt Ot 0 0
∂ ∂θi ∂ ∂θi
(1)
St
(4)
St
0
∂ ∂θi ∂ ∂θi
(2)
St
(5)
St
0
Xi
= Ni ∗
∂ ∂θi ∂ ∂θi
(3)
St
(6) St 0 M i Li Wi K i , ∗ ∗
Yi Vi ∗
Ot is the same orthogonal transformation as in the eSRIF and (1)
St
(4)
St
−T /2
= −Rt
−T /2
= Pt
(2)
= Rt
(5)
= −Pt
Ht Ft−1 , St
Ft−1 ,
St
−T /2
T /2
Ht Ft−1 Gt Qt
−T /2
T /2
Ft−1 Gt Qt
,
(3)
, St
(6)
St
−T /2
= −Rt
−T /2
= Pt
zt ,
x ˆt .
Score Evaluation within the Extended Square-root Information Filter – p. 16/??
Algorithm LLG-eSRIF III. For each θi , i = 1, . . . , p, compute
Ji =
Xi Y i M i Ni V i W i
−T /2 Re,t −T /2 −Pt+1 Kp,t
∗
0 −T /2 Pt+1
∗
0
−1
0 ∗
.
Score Evaluation within the Extended Square-root Information Filter – p. 17/??
Algorithm LLG-eSRIF IV. For each θi , i = 1, . . . , p, split the matrices m+n+q
Ji =
zh
}|
Li + Di + Ui | {z } m+n
∗∗∗
i{ o
m+n
where Li , Di and Ui are the strictly lower triangular, diagonal and strictly upper triangular parts of Ji , respectively.
Score Evaluation within the Extended Square-root Information Filter – p. 18/??
Algorithm LLG-eSRIF V. For each θi , i = 1, . . . , p, compute the following quantities: −T /2 ∂Re,t 0 ∂θi = Li + Di + UiT −T /2 ˜ −T /2 P K ∂ ˜ p,t t+1 ∂ Pt+1 − (12) ∂θi ∂θi −T /2 Re,t 0 , × −T /2 −T /2 −P˜t+1 Kp,t P˜t+1
Score Evaluation within the Extended Square-root Information Filter – p. 19/??
Algorithm LLG-eSRIF ∂¯ et = ∂θi
"
−T /2 ∂Re,t
∂θi
#
T /2
− Xi Re,t e¯t + Yi Ft xˆt − Li ,
(13)
Score Evaluation within the Extended Square-root Information Filter – p. 20/??
Algorithm LLG-eSRIF ∂¯ et = ∂θi (6)
∂St+1 ∂θi
"
−T /2 ∂Re,t
#
T /2
− Xi Re,t e¯t + Yi Ft xˆt − Li ,
∂θi −T /2 ∂ P˜t+1 Kp,t T /2 = + Ni Re,t e¯t ∂θi " −T /2 # ∂ P˜t+1 + − Vi Ft xˆt + Ki . ∂θi
(13)
(14)
Score Evaluation within the Extended Square-root Information Filter – p. 20/??
Algorithm LLG-eSRIF VI. Finally, we have all values to compute the LLG according to (6): −1/2 ∂ R e,t ∂lθ (zt ) ∂¯ et 1/2 T = − tr Re,t , i = 1, p. + e¯t ∂θi ∂θi ∂θi
Score Evaluation within the Extended Square-root Information Filter – p. 21/??
Algorithm LLG-eSRIF VI. Finally, we have all values to compute the LLG according to (6): −1/2 ∂ R e,t ∂lθ (zt ) ∂¯ et 1/2 T = − tr Re,t , i = 1, p. + e¯t ∂θi ∂θi ∂θi The LLG-eSRIF:
The LLG-eSRIF:
Stage I
Stages II —V
Score Evaluation within the Extended Square-root Information Filter – p. 21/??
Algorithm LLG-eSRIF VI. Finally, we have all values to compute the LLG according to (6): −1/2 ∂ R e,t ∂lθ (zt ) ∂¯ et 1/2 T = − tr Re,t , i = 1, p. + e¯t ∂θi ∂θi ∂θi The LLG-eSRIF: Stage
I
The LLG-eSRIF: Stages
II —V
Score Evaluation within the Extended Square-root Information Filter – p. 21/??
Algorithm LLG-eSRIF VI. Finally, we have all values to compute the LLG according to (6): −1/2 ∂ R e,t ∂lθ (zt ) ∂¯ et 1/2 T = − tr Re,t , i = 1, p. + e¯t ∂θi ∂θi ∂θi The LLG-eSRIF: Stage
I
The LLG-eSRIF: Stages II —V
Score Evaluation within the Extended Square-root Information Filter – p. 21/??
Algorithm LLG-eSRIF The LLG-eSRIF consists of two parts:
Score Evaluation within the Extended Square-root Information Filter – p. 22/??
Algorithm LLG-eSRIF The LLG-eSRIF consists of two parts: √ The source filtering algorithm, i.e. eSRIF.
Score Evaluation within the Extended Square-root Information Filter – p. 22/??
Algorithm LLG-eSRIF The LLG-eSRIF consists of two parts: √ The source filtering algorithm, i.e. eSRIF. √ The "differentiated" part: Stages II — V.
Score Evaluation within the Extended Square-root Information Filter – p. 22/??
Algorithm LLG-eSRIF The LLG-eSRIF consists of two parts: √ The source filtering algorithm, i.e. eSRIF. √ The "differentiated" part: Stages II — V. Thus, the Algorithm LLG-eSRIF is ideal for simultaneous
Score Evaluation within the Extended Square-root Information Filter – p. 22/??
Algorithm LLG-eSRIF The LLG-eSRIF consists of two parts: √ The source filtering algorithm, i.e. eSRIF. √ The "differentiated" part: Stages II — V. Thus, the Algorithm LLG-eSRIF is ideal for simultaneous √ state estimation
Score Evaluation within the Extended Square-root Information Filter – p. 22/??
Algorithm LLG-eSRIF The LLG-eSRIF consists of two parts: √ The source filtering algorithm, i.e. eSRIF. √ The "differentiated" part: Stages II — V. Thus, the Algorithm LLG-eSRIF is ideal for simultaneous √ state estimation √ parameter identification.
Score Evaluation within the Extended Square-root Information Filter – p. 22/??
Algorithm LLG-eSRIF Remark. Since the matrices in LLG (6) are triangular, only the
diagonal elements of
1/2 Re,t
∂
and
−1/2 Re,t
need to be computed.
∂θi Hence, the Algorithm LLG-eSRIF allows the m × m-matrix inversion of Re,t to be avoided in the evaluation of LLG.
Score Evaluation within the Extended Square-root Information Filter – p. 23/??
Ill-Conditioned Example Problems Problem 1 Given: h i θ 0 , H = 1, 0 , R = e2 θ, F = I2 , Q = 0, P0 = 0 θ h iT G = 0, 0 , 0< e « 1
where θ is an unknown parameter, I2 is an identity 2 × 2 matrix; to simulate roundoff we assume
Score Evaluation within the Extended Square-root Information Filter – p. 24/??
Ill-Conditioned Example Problems Problem 1 Given: h i θ 0 , H = 1, 0 , R = e2 θ, F = I2 , Q = 0, P0 = 0 θ h iT G = 0, 0 , 0< e « 1
where θ is an unknown parameter, I2 is an identity 2 × 2 matrix; r 2 to simulate roundoff we assume e + 1 6= 1 but e + 1 = 1.
Score Evaluation within the Extended Square-root Information Filter – p. 24/??
Ill-Conditioned Example Problems Problem 1 Given: h i θ 0 , H = 1, 0 , R = e2 θ, F = I2 , Q = 0, P0 = 0 θ h iT G = 0, 0 , 0< e « 1
where θ is an unknown parameter, I2 is an identity 2 × 2 matrix; r 2 to simulate roundoff we assume e + 1 6= 1 but e + 1 = 1. Calculate: (P1 )0θ at the point θ = 1.
Score Evaluation within the Extended Square-root Information Filter – p. 24/??
Ill-Conditioned Example Problems Problem 1 Given: h i θ 0 , H = 1, 0 , R = e2 θ, F = I2 , Q = 0, P0 = 0 θ h iT G = 0, 0 , 0< e « 1
where θ is an unknown parameter, I2 is an identity 2 × 2 matrix; r 2 to simulate roundoff we assume e + 1 6= 1 but e + 1 = 1. Calculate: (P1 )0θ at the point θ = 1. For the θ = 1, this example illustrates the initialization problems [Kaminski P.G., Bryson A.E., Schmidt S.F., 1971] that result when H1 Π0 H1T + R1 is rounded to H1 Π0 H1T . Score Evaluation within the Extended Square-root Information Filter – p. 24/??
Comparison (Problem 1) Filter
Exact Answer
Rounded Answer
Score Evaluation within the Extended Square-root Information Filter – p. 25/??
Comparison (Problem 1) Filter
’diff’ KF
Exact Answer 2 e 1 + e2 0 0 (P1 )θ = 0 1
Rounded Answer 0 0 r = 0 1
Score Evaluation within the Extended Square-root Information Filter – p. 25/??
Comparison (Problem 1) Filter
’diff’ KF
’diff’ IF
Exact Answer 2 e 1 + e2 0 0 (P1 )θ = 0 1 2 1+e 0 e2 −1 0 P1 θ = − 0 1
Rounded Answer 0 0 r = 0 1
1 r e2 0 = − 0 1
Score Evaluation within the Extended Square-root Information Filter – p. 25/??
Comparison (Problem 1) Filter
’diff’ KF
’diff’ IF
LLGeSRIF
Exact Answer Rounded Answer 2 e 0 0 r 1 + e2 0 0 (P1 )θ = = 0 1 0 1 2 1 1+e 0 0 2 0 2 r P1−1 θ = − e = − e 0 1 0 1 √ 2 1 1+e 0 0 0 r 1 1 −1/2 e − P1 =− e 2 2 θ 0 1 0 1 Score Evaluation within the Extended Square-root Information Filter – p. 25/??
Ill-Conditioned Example Problems Problem 2 Given: h i θ 0 , H = 1, 1 , R = e2 θ, F = I2 , Q = 0, P0 = 0 θ h iT G = 0, 0 , 0< e « 1
where θ is an unknown parameter, I2 is an identity 2 × 2 matrix; to simulate roundoff we assume
Score Evaluation within the Extended Square-root Information Filter – p. 26/??
Ill-Conditioned Example Problems Problem 2 Given: h i θ 0 , H = 1, 1 , R = e2 θ, F = I2 , Q = 0, P0 = 0 θ h iT G = 0, 0 , 0< e « 1
where θ is an unknown parameter, I2 is an identity 2 × 2 matrix; r 2 to simulate roundoff we assume e + 1 6= 1 but e + 1 = 1.
Score Evaluation within the Extended Square-root Information Filter – p. 26/??
Ill-Conditioned Example Problems Problem 2 Given: h i θ 0 , H = 1, 1 , R = e2 θ, F = I2 , Q = 0, P0 = 0 θ h iT G = 0, 0 , 0< e « 1
where θ is an unknown parameter, I2 is an identity 2 × 2 matrix; r 2 to simulate roundoff we assume e + 1 6= 1 but e + 1 = 1. Calculate: (P1 )0θ at the point θ = 1.
Score Evaluation within the Extended Square-root Information Filter – p. 26/??
Comparison (Problem 2) Filter
Exact Answer
Rounded Answer
Score Evaluation within the Extended Square-root Information Filter – p. 27/??
Comparison (Problem 2) Filter
’diff’ KF
Exact Answer 2 1 + e −1 1 2 + e2 −1 1 + e2
Rounded Answer 1 1 −1 2 −1 1
Score Evaluation within the Extended Square-root Information Filter – p. 27/??
Comparison (Problem 2) Filter
’diff’ KF ’diff’ IF
Exact Answer 2 1 + e −1 1 2 + e2 −1 1 + e2
1 − 2 e
1 + e2 1
1 1 + e2
Rounded Answer 1 1 −1 2 −1 1
1 1 1 − 2 e 1 1
Score Evaluation within the Extended Square-root Information Filter – p. 27/??
Comparison (Problem 2) Filter
’diff’ KF ’diff’ IF
LLGeSRIF
Exact Answer 2 1 + e −1 1 2 + e2 −1 1 + e2
1 + e2
1 − 2 e 1 r 2 + e2 1 1 + e2 − 2 0
1 1 + e2
Rounded Answer 1 1 −1 2 −1 1
1 √ e 1 + e2 √ 1 + e2 e
1 1 1 − 2 e 1 1 √ 1 2 − 2 0
1 e 1 e
Score Evaluation within the Extended Square-root Information Filter – p. 27/??
Numerical Results Example. Let the test system (1), (2) be defined as follows: 1 ∆t 0 xt + w t , xt+1 = 0 e−∆t/τ 1 zt =
h
1
0
i
xt + v t
where τ is an unknown parameter which needs to be estimated.
Score Evaluation within the Extended Square-root Information Filter – p. 28/??
Numerical Results Example. Let the test system (1), (2) be defined as follows: 1 ∆t 0 xt + w t , xt+1 = 0 e−∆t/τ 1 zt =
h
1
0
i
xt + v t
where τ is an unknown parameter which needs to be estimated. For the test problem, τ ∗ =15 was chosen as the true value of parameter τ .
Score Evaluation within the Extended Square-root Information Filter – p. 28/??
Numerical Results The negative Log Likelihood Function
The Log Likelihood Gradient
Score Evaluation within the Extended Square-root Information Filter – p. 29/??
Numerical Results The negative Log Likelihood Function
The Log Likelihood Gradient
500
The negative Log Likelihood Function
498
496
494
492
490
488
486 0
CKF CIF eSRIF 15
30
45
tau (sec.)
Score Evaluation within the Extended Square-root Information Filter – p. 29/??
Numerical Results The negative Log Likelihood Function
The Log Likelihood Gradient
500
The negative Log Likelihood Function
498
496
494
492
490
488
486 0
CKF CIF eSRIF 15
30
45
tau (sec.)
Score Evaluation within the Extended Square-root Information Filter – p. 29/??
Numerical Results The negative Log Likelihood Function
The Log Likelihood Gradient
500
0.5
0 The Log Likelihood Gradient
The negative Log Likelihood Function
498
496
494
492
490
−0.5
−1
−1.5
488
486 0
CKF CIF eSRIF 15
30 tau (sec.)
"differentiated" CKF "differentiated" CIF Algorithm LLG−eSRIF
45
−2 0
15
30 tau (sec.)
Score Evaluation within the Extended Square-root Information Filter – p. 29/??
45
Numerical Results The negative Log Likelihood Function
The Log Likelihood Gradient
500
0.5
0 The Log Likelihood Gradient
The negative Log Likelihood Function
498
496
494
492
490
−0.5
−1
−1.5
488
486 0
CKF CIF eSRIF 15
30 tau (sec.)
"differentiated" CKF "differentiated" CIF Algorithm LLG−eSRIF
45
−2 0
15
30 tau (sec.)
Score Evaluation within the Extended Square-root Information Filter – p. 29/??
45
Conclusion In this paper, the new algorithm for evaluating the Log Likelihood Gradient (score) of linear discrete-time dynamic systems has been developed. The necessary theory has been given and substantiated by the computational experiments. Two ill-conditioned example problems have been constructed to show the superior perfomance of the Algorithm LLG-eSRIF over the conventional approach. All of these are good reasons to use the presented algorithm in practice.
Score Evaluation within the Extended Square-root Information Filter – p. 30/??
Q&A Maria Kulikova University of the Witwatersrand, Johannesburg, South Africa
Score Evaluation within the Extended Square-root Information Filter – p. 31/??
