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\usepackage{amssymb}
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\divide\pst@dimz\tw@
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% Definition of the mixed color MixedColor
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\LogoPosition{10.3, 8.1}
\Logo{\includegraphics[width=1.2cm]{iccs06.eps}}

\title{A Log Likelihood Gradient Evaluation by Using the Extended
Square-Root Information Filter}
\subtitle{}
\author{
\href{mkulikova@cam.wits.ac.za }{M.V. Kulikova, I.V. Semushin}} \institution{School
of Computational and Applied Mathematics, University of the
Witwatersrand, Private Bag~3, Wits~2050, Johannesburg, South Africa\\
E-mail: \quad {\LightBlue mkulikova@cam.wits.ac.za}}

%\DefaultTransition{Glitter}
%\usepackage[pdftex]{graphicx}
%\usepackage{pdfslide}

\begin{document}


\maketitle



%------------------------------------------------------------------------------

%---------------------------------------------------------------------- SLIDE -
\overlays{8}{%
\begin{slide}[Glitter]{Introduction}
\small The method of maximum likelihood $\ldots$ \FromSlide{2}
\begin{itemstep}
\item Gradient-search optimization algorithms;
\FromSlide{3}\item Gradient of the negative Log Likelihood Function;
\FromSlide{4}\item Kalman Filter \FromSlide{5}  $\Longrightarrow$ is
known to be {\green unstable};
\end{itemstep}
\FromSlide{6} [P.~Park\&T.~Kailath, 1995] The extended Square Root
Information Filter (eSRIF):
\begin{itemstep}
\FromSlide{7}\item avoids numerical instabilities arising from
computational errors; \FromSlide{8}\item appears to be better suited
to parallel implementation and to very large scale integration
(VLSI) implementation.
\end{itemstep}
\end{slide}
}
%------------------------------------------------------------------------------


%---------------------------------------------------------------------- SLIDE -

%------------------------------------------------------------------------------
\overlays{6}{%
\begin{slide}[Glitter]{Problem Statement}
\small
Consider the discrete-time linear dynamic stochastic system
  $$
    x_{t+1} = F_t x_t+G_t w_t,  \qquad t=0,1, \ldots, N, %\phantom{\quad t=0,1, \ldots, N,}
    \eqno{(1)}
  $$
  $$
    z_t = H_t x_t+v_t, \qquad t=1,2, \ldots, N,  \eqno{(2)}
  $$
%\begin{figure}%[h]
%\input{fig1-1-1.txt}\vspace{-1cm}
%\caption{}\label{f:1-1-1}
%\end{figure}
\FromSlide{2} \hspace{-1.2ex}  with the system state {\green $x_t \in \Rset^n$},
\FromSlide{3} the state disturbance {\green $w_t \in  \Rset^q$},
\FromSlide{4} the observed vector {\green $z_t \in \Rset^m$},
\FromSlide{5} and  the measurement error {\green $v_t \in \Rset^m$,}
\FromSlide{6} such that the initial state {\green $x_0$} and each
{\green $w_t$}, {\green $v_{t}$} of
$\{w_t : t=0,1, \ldots\}$,   % and
$\{v_t : t=1,2, \ldots\}$ are taken from mutually independent
Gaussian  distributions with the following expectations:
\end{slide}
}

%---------------------------------------------------------------------- SLIDE -

%------------------------------------------------------------------------------
\overlays{2}{%
\begin{slide}[Glitter]{Problem Statement}
\small
{\scriptsize
\[
\E{\left[\begin{array}{c}x_0\\w_t\\v_{t}\end{array}\right]}=
\left[\begin{array}{c}\bar x_0\\0\\0\end{array}\right]\]\[
\E{\left[\begin{array}{c}(x_0-\bar
x_0)\\w_t\\v_{t}\end{array}\right] \left[\begin{array}{c}(x_0-\bar
x_0)\\w_t\\v_{t}\end{array}\right]^T }= \left[\begin{array}{ccc}
P_0&0&0\\
0&Q_t&0\\
0&0&R_{t}
\end{array}\right]
\]
}\hspace{-0.6ex}  and $\E{w_t w_{t^{\prime}}^T}=0$, $\E{v_t v_{t^{\prime}}^T}=0$ if
$t\ne t^{\prime}$. \FromSlide{2} Assume the system is parameterized
by a vector {\green $\theta \in \Rset^p$} of unknown system
parameters. This means that all the above characteristics, namely
\FromSlide{2} {\green $F_t$,  $G_t$, $H_t$, $P_0\ge 0$,  $Q_t\ge 0$
and $R_t
>0$} can depend upon $\theta$ (the corresponding notations
$F_t(\theta)$, $G_t(\theta)$ and so on, are suppressed  for the sake
of simplicity).
\end{slide}
}

%---------------------------------------------------------------------- SLIDE -
\begin{slide}{Problem Statement}
\small {\LightBlue For example},  (1), (2) may describe a discrete
autoregressive (AR) process observed in the presence of additive
noise
$$
x_{t+1}=
\left[
\begin{array}{ccccc}
0 & 1 & 0 & \ldots & 0 \\
0 & 0 & 1 & \ldots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \ldots & 1 \\
{\green \theta_1} & {\green \theta_2} & {\green \theta_3} & \ldots &
{\green \theta_p}
\end{array}
\right]
x_t+
\left[
\begin{array}{c}
0  \\
0 \\
\vdots \\
0 \\
\gamma
\end{array}
\right]
w_t,
$$
$$
 z_t=
\left[
\begin{array}{cccc}
0 & 0 & \ldots & 1 \\
\end{array}
\right]
x_t+v_t.
$$
In this case {\green $\theta$} are the unknown AR parameters which
need to be estimated.
\end{slide}

%---------------------------------------------------------------------- SLIDE -
\begin{slide}{Problem Statement}
\small The negative Log Likelihood Function (LLF) for system~(1),
(2) is given by
$$
L_{\theta}\left( Z_1^N \right)= \frac{1}{2} \sum \limits_{t=1}^N  \left\{
 \frac{m}{2}\ln(2\pi)
+ \ln ( \det (R_{e,t})) + e_t^T R_{e,t}^{-1} e_t
\right\}
$$
with ${\green e_t} \eqdef z_t-H_t \hat x_t$  being the zero-mean
innovations whose covariance is determined as ${\green R_{e,t}}
\eqdef E \{e_t e_t^T\}=H_tP_tH_t^T+R_t$ through matrix {\green
$P_t$}, the error covariance of the time updated estimate {\green
$\hat x_t$} generated by the {\green Kalman Filter}.
\end{slide}

%---------------------------------------------------------------------- SLIDE -
\overlays{4}{%
\begin{slide}{Problem Statement}
\small Let {\green $l_{\theta}(z_t)$} denote the negative  LLF for
the $t$-th measurement $z_t$ in system~(1), (2), given measurements
$Z_1^{t-1}\eqdef\{ z_1, \ldots, z_{t-1} \}$, \FromSlide{2} then
$$
{\green l_{\theta}(z_t)}= \frac{1}{2} \left\{ \frac{m}{2}\ln(2\pi)+
\ln ( \det (R_{e,t}))+ e_t^T R_{e,t}^{-1} e_t \right\}. \eqno{(3)}
$$

\FromSlide{3} By differentiating~(3) we obtain
$$
{\green \displaystyle \frac{\partial{l_{\theta}(z_t)
}}{\partial{\theta_i}}} = \frac{1}{2} \displaystyle
\frac{\partial}{\partial{\theta_i}} \left[ \ln ( \det (R_{e,t}))
\right] + \frac{1}{2} \displaystyle
\frac{\partial}{\partial{\theta_i}} \left[ e_t^T R_{e,t}^{-1} e_t
\right] i=1,\ldots p. \eqno{(4)}
$$
 \FromSlide{4} As can be seen the
computation of~(3) and (4) {\green leads to} implementation of a
{\green Kalman Filter} (and its derivative with respect to each
parameter) which is known to be {\green unstable}.
\end{slide}
}

%---------------------------------------------------------------------- SLIDE -
\begin{slide}{The eSRIF}
\small [P.~Park\&T.~Kailath, 1995]: assume that ${\green \Pi_0}>0$,
${\green R_t}>0$ and ${\green F_t}$ are invertible. Given ${\green
P_0^{-T/2}}=\Pi_0^{-T/2}$ and ${\green P_0^{-T/2}\hat x_0}=
\Pi_0^{-T/2}\bar x_0$, then
$$
{\scriptsize
\begin{array}{r}
O_t \left[
\begin{array}{ccc|c}
R_t^{-T/2} & -R_t^{-T/2}H_tF_t^{-1} &
R_t^{-T/2}H_tF_t^{-1}G_tQ_t^{T/2} & -R_t^{-T/2}z_t   \\
[5pt] 0 & P_t^{-T/2}F_t^{-1}  & -P_t^{-T/2}F_t^{-1}G_tQ_t^{T/2} &
P_t^{-T/2}\hat x_t \  \\ [8pt]
0 & 0 & I_q  & 0 \\
\end{array}
\right]  \\[5pt]
 = \left[
\begin{array}{ccc|c}
R_{e,t}^{-T/2} & 0 & 0\quad   &- \bar e_t  \\[5pt]
-P_{t+1}^{-T/2} K_{p,t} &
P_{t+1}^{-T/2} &  0\quad  &P_{t+1}^{-T/2} \hat x_{t+1}  \\[5pt]
* & * &  * \quad & *\\
\end{array}
\right]
\end{array}
}
 $$
where {\green $O_t$} is  any orthogonal transformation such that the
matrix on the right-hand side of the formula is block lower
triangular.
\end{slide}

%---------------------------------------------------------------------- SLIDE -
\overlays{2}{%
\begin{slide}{The eSRIF}
\small
The {\green eSRIF} is a modification of the conventional one:
$$
\begin{array}{rr}
O_t \left. \left[
\begin{array}{ccc}
R_t^{-T/2} & -R_t^{-T/2}H_tF_t^{-1} &
R_t^{-T/2}H_tF_t^{-1}G_tQ_t^{T/2} \\ [5pt]
 0 & P_t^{-T/2}F_t^{-1}  &
-P_t^{-T/2}F_t^{-1}G_tQ_t^{T/2}  \\ [5pt]
0 & 0 & I_q   \\
\end{array}
 \OnlySlide{1}\right]
 \FromSlide{2}\right| &
\left.
\begin{array}{c}
\FromSlide{2} {\green -R_t^{-T/2}z_t} \\ [5pt] \FromSlide{2} {\green
P_t^{-T/2}\hat x_t}  \\ [5pt]
\FromSlide{2} {\green 0} \\
\end{array}\FromSlide{2} \right] \\[8pt]
= \left.\left[
\begin{array}{ccc}
R_{e,t}^{-T/2} & 0 & 0   \\[5pt]
-P_{t+1}^{-T/2} K_{p,t} &
P_{t+1}^{-T/2} &  0  \\[5pt]
* & * &  *  \\
\end{array}
\OnlySlide{1} \right] \FromSlide{2}  \right|&
 \left.
\begin{array}{c}
\FromSlide{2} {\green - \bar e_t}  \\[5pt]
\FromSlide{2} {\green P_{t+1}^{-T/2} \hat x_{t+1}}  \\[5pt]
\FromSlide{2} {\green *}\\
\end{array}
\FromSlide{2} \right]
\end{array}
$$
where $O_t$ is  any orthogonal transformation  such that the matrix
on the right-hand side of the formula is \OnlySlide{1} lower
triangular.
\FromSlide{2} is block lower
triangular.
\end{slide}
}

%---------------------------------------------------------------------- SLIDE -
\begin{slide}{The eSRIF}
\small {\LightBlue Remark.} The predicted estimate now can be found
from the entries of the post-array by solving the triangular system
$$ \left(\tilde
P^{-1/2}_{t+1}\right)\left( \hat x_{t+1}^- \right)=\left( \tilde
P^{-1/2}_{t+1} \hat x_{t+1}^- \right). \eqno{(5)}
$$
\end{slide}


%---------------------------------------------------------------------- SLIDE -
\overlays{3}{%
\begin{slide}{The LLG in terms of the eSRIF}
\small The Log Likelihood Gradient (LLG):
$$
\displaystyle \frac{\partial{l_{\theta}(z_t) }}{\partial{\theta_i}}
=
\frac{1}{2} \displaystyle \frac{\partial}{\partial{\theta_i}}
\left[ \ln ( \det (R_{e,t})) \right]
+
\frac{1}{2}
\displaystyle \frac{\partial}{\partial{\theta_i}}
\left[ e_t^T R_{e,t}^{-1} e_t \right], \quad i=1,2, \ldots p,
$$
\FromSlide{2} in terms of the eSRIF is given by
$$
\displaystyle \frac{\partial{l_{\theta}(z_t) }}{\partial{\theta_i}}
=
\displaystyle \frac{\partial}{\partial{\theta_i}}
\left[ \ln ( \det ({\green R_{e,t}^{1/2}})) \right]
+
{{\green \bar e_t}}^T
\displaystyle \frac{\partial{{\green \bar e_t} }}{\partial{\theta_i}},
\quad i=1,2, \ldots p,
$$
\FromSlide{3} where {\green $R_{e,t}^{1/2}$} is a square-root factor
of the matrix $R_{e,t}$, i.~e. $R_{e,t}=R_{e,t}^{T/2}
R_{e,t}^{1/2}$, and {\green $\bar e_t$} are the normalized
innovations, i.~e. $\bar e_t=R_{e,t}^{-T/2} e_t$.
\end{slide}
}

%---------------------------------------------------------------------- SLIDE -
\overlays{3}{%
\begin{slide}{The LLG in terms of the eSRIF}
\small
Taking into account that matrix $R_{e,t}^{1/2}$ is  upper
triangular, we  can  show \FromSlide{2}
   $$
   \displaystyle \frac{\partial}{\partial{\theta_i}}
   \left[ \ln \left( \det (R_{e,t}^{1/2}) \right) \right]
   =
   \tr{ R_{e,t}^{-1/2} \ \displaystyle \frac{\partial{\left(
   R_{e,t}^{1/2}\right) }}
   {\partial{\theta_i}}},\quad i=1, 2, \ldots, p
   $$
   where $\tr{\ \cdot \ }$ is a trace of matrix.

\FromSlide{3} Finally, we obtain  the expression  for the LLG
{\green in terms of the eSRIF}:
$$ \displaystyle \frac{\partial{l_{\theta}(z_t)
}}{\partial{\theta_i}} = - \tr{ R_{e,t}^{1/2} \ \displaystyle
\frac{\partial{\left( R_{e,t}^{-1/2}\right) }}{\displaystyle\partial{\theta_i}} }
+ {\bar e_t}^T \displaystyle \frac{\partial{\bar e_t
}}{\partial{\theta_i}}, \ i=1,2, \ldots, p. \eqno{(6)}
$$
\end{slide}
}

%---------------------------------------------------------------------- SLIDE -
\overlays{5}{%
\begin{slide}{LLG Evaluation}
\scriptsize
$$
\displaystyle \frac{ \partial{ l(z_t)}}{\partial{\theta_i}}  = -
\tr{
\fromSlide*{2}{\rnode{NA1}{\pscirclebox[linecolor=green]{R_{e,t}^{1/2}}}}
\onlySlide*{1}{\rnode{NA1}{\pscirclebox[linecolor=green,linestyle=none]{R_{e,t}^{1/2}}}}
\fromSlide*{5}{\red \displaystyle \frac{\partial{\left(
R_{e,t}^{-1/2}\right)}}{\partial{\theta_i}} } \onlySlide*{1}
{\displaystyle \frac{\partial{\left(
R_{e,t}^{-1/2}\right)}}{\partial{\theta_i}}} \onlySlide*{2}
{\displaystyle \frac{\partial{\left(
R_{e,t}^{-1/2}\right)}}{\partial{\theta_i}}} \onlySlide*{3}
{\displaystyle \frac{\partial{\left(
R_{e,t}^{-1/2}\right)}}{\partial{\theta_i}}} \onlySlide*{4}
{\displaystyle \frac{\partial{\left(
R_{e,t}^{-1/2}\right)}}{\partial{\theta_i}}} }+
\fromSlide*{2}{\rnode{NA2}{\pscirclebox[linecolor=green]{{\bar
e_t}^T}}}
\onlySlide*{1}{\rnode{NA2}{\pscirclebox[linecolor=green,linestyle=none]{{\bar
e_t}^T}}} \onlySlide*{5}{{\red \displaystyle\frac{\partial{\bar e_t
}}{\partial{\theta_i}} }, \  i=1, \ldots, p. \eqno{(6)}}
\onlySlide*{1} {\displaystyle\frac{\partial{\bar e_t
}}{\partial{\theta_i}}, \ i=1, \ldots, p. \eqno{(6)}} \onlySlide*{2}
{\displaystyle\frac{\partial{\bar e_t }}{\partial{\theta_i}}, \ i=1,
\ldots, p. \eqno{(6)}} \onlySlide*{3}
{\displaystyle\frac{\partial{\bar e_t }}{\partial{\theta_i}}, \ i=1,
\ldots, p. \eqno{(6)}} \onlySlide*{4}{
\displaystyle\frac{\partial{\bar e_t }}{\partial{\theta_i}}, \  i=1,
\ldots, p. \eqno{(6)}},
$$
\FromSlide{3} The eSRIF:
$$
\begin{array}{r}
O_t \left[
\begin{array}{ccc|c}
R_t^{-T/2} & -R_t^{-T/2}H_tF_t^{-1} &
R_t^{-T/2}H_tF_t^{-1}G_tQ_t^{T/2} & -R_t^{-T/2}z_t  \\
0 &
P_t^{-T/2}F_t^{-1}  & -P_t^{-T/2}F_t^{-1}G_tQ_t^{T/2} &
P_t^{-T/2}\hat x_t \ \\
0 & 0 & I_q  & 0 \\
\end{array}
\right]   \\[3pt]
= \left[
\begin{array}{ccc|c}
\fromSlide*{4}{\rnode{NB1}{\pscirclebox[linecolor=green]{R_{e,t}^{-T/2}}}}
\onlySlide*{3}{\rnode{NB1}{\pscirclebox[linecolor=green,linestyle=none]{R_{e,t}^{-T/2}}}}
 & 0 & 0  \quad &
\fromSlide*{4}{\rnode{NB2}{\pscirclebox[linecolor=green]{ -\bar
e_t}}}
\onlySlide*{3}{\rnode{NB2}{\pscirclebox[linecolor=green,linestyle=none]{-\bar e_t}}}\\
-P_{t+1}^{-T/2} K_{p,t} &
P_{t+1}^{-T/2} &  0  \quad &P_{t+1}^{-T/2} \hat x_{t+1}  \\
* & * &  *  & *\\
\end{array}
\right]
\end{array}
$$
\fromSlide*{4}{\nccurve[linecolor=green,angleA=90,angleB=270]{->}{NB1}{NA1}}
\fromSlide*{4}{\nccurve[linecolor=green,angleA=90,angleB=270]{->}{NB2}{NA2}}
\end{slide}
}


%---------------------------------------------------------------------- SLIDE -
\begin{slide}{LLG Evaluation}
\small {\LightBlue Lemma 1.} Let $\qquad \qquad QA=L \qquad
\qquad\qquad \qquad\qquad \qquad (7)$ \\
 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 $\theta$ then the upper
triangular matrix $U$ in
$$
Q'_{\theta}Q^T=\bar U^T-\bar U \eqno{(8)}
$$
is, in fact, the strictly upper triangular part of the matrix
$QA'_{\theta}L^{-1}$:
$$
QA'_{\theta}L^{-1}=\bar L +D+\bar U        \eqno{(9)}
$$
where $\bar L$, $D$ and $\bar U$ are, respectively, strictly
lower triangular, diagonal and
strictly upper triangular.
\end{slide}

%---------------------------------------------------------------------- SLIDE -

\overlays{6}{%
\begin{slide}{LLG Evaluation}
\small
Lemma~1 easily leads to a method for computing {\green $L'_{\theta}$}.

\FromSlide{2} By differenting  {\green $QA=L$} we obtain
$$
L'_{\theta}=QA'_{\theta}+Q'_{\theta}A \eqno{(10)}
$$

\FromSlide{3} Multiply by $L^{-1}$:
$$
L'_{\theta}{\green L^{-1}}=QA'_{\theta}{\green
L^{-1}}+Q'_{\theta}A{\green L^{-1}}  \FromSlide{4} \Rightarrow
L'_{\theta}L^{-1}=QA'_{\theta}L^{-1}+{\green Q'_{\theta}Q^T}
\eqno{(11)}
$$

\FromSlide{5} By substituting~(8) and (9) into (11), we find that
$$
L'_{\theta}L^{-1}=\underbrace{(L+D+U)}_{QA'_{\theta}L^{-1}}+
\underbrace{(U^T-U)}_{Q'_{\theta}Q^T} \quad \FromSlide{6}
\Longrightarrow \quad {\green L'_{\theta}=(L+D+U^T)L^{-1}}.
$$
\end{slide}
}

%---------------------------------------------------------------------- SLIDE -
\begin{slide}{Algorithm~LLG-eSRIF}
\small
\begin{enumerate}
\item[{\LightBlue I.}]
For each $\theta_i$, $i=1,2, \ldots, p$, apply the eSRIF
$$
{\scriptsize
\begin{array}{r}
O_t \left[
\begin{array}{ccc|c}
R_t^{-T/2} & -R_t^{-T/2}H_tF_t^{-1} &
R_t^{-T/2}H_tF_t^{-1}G_tQ_t^{T/2} & -R_t^{-T/2}z_t   \\
[5pt] 0 & P_t^{-T/2}F_t^{-1}  & -P_t^{-T/2}F_t^{-1}G_tQ_t^{T/2} &
P_t^{-T/2}\hat x_t \  \\ [8pt]
0 & 0 & I_q  & 0 \\
\end{array}
\right]  \\[5pt]
 = \left[
\begin{array}{ccc|c}
R_{e,t}^{-T/2} & 0 & 0\quad   &- \bar e_t  \\[5pt]
-P_{t+1}^{-T/2} K_{p,t} &
P_{t+1}^{-T/2} &  0\quad  &P_{t+1}^{-T/2} \hat x_{t+1}  \\[5pt]
* & * &  * \quad & *\\
\end{array}
\right]
\end{array}
}
 $$
where {\green $O_t$} is  any orthogonal transformation such that the
matrix on the right-hand side of the formula is block lower
triangular.
\end{enumerate}
\end{slide}

%---------------------------------------------------------------------- SLIDE -
\begin{slide}{Algorithm~LLG-eSRIF}
\small
\begin{enumerate}
\item[{\LightBlue II.}] For each $\theta_i$, $i=1,2, \ldots,
p$, calculate
$$
{\scriptsize
\begin{array}{r}
O_t \left[
\begin{array}{ccc|c}
\displaystyle \frac{\partial}{\partial{\theta_i}}
\left( R_t^{-T/2} \right)
& \displaystyle \frac{\partial}{\partial{\theta_i}}
\left( S_t^{(1)} \right)
& \displaystyle \frac{\partial}{\partial{\theta_i}}
\left( S_t^{(2)} \right)
& \displaystyle \frac{\partial}{\partial{\theta_i}}
\left( S_t^{(3)} \right)\\
 0
& \displaystyle \frac{\partial}{\partial{\theta_i}} \left( S_t^{(4)} \right)
& \displaystyle \frac{\partial}{\partial{\theta_i}} \left( S_t^{(5)} \right)
&  \displaystyle \frac{\partial}{\partial{\theta_i}}\left( S_t^{(6)}\right) \\
  0 & 0 & 0 & 0\\
\end{array}
\right] \phantom{,}\\
= \left[
\begin{array}{ccc|c}
X_i & Y_i & M_i & L_i\\
N_i & V_i & W_i & K_i \\
 * & * & * & * \\
\end{array}
\right],
\end{array}
}
$$
{\green $O_t$} is the same orthogonal transformation as in the eSRIF
and
$$
{\scriptsize
\begin{array}{lll}
S_t^{(1)}=-R_t^{-T/2}H_tF_t^{-1}, &
S_t^{(2)}=R_t^{-T/2}H_tF_t^{-1}G_tQ_t^{T/2}, &
S_t^{(3)}=-R_t^{-T/2}z_t, \\
 S_t^{(4)}=P_t^{-T/2}F_t^{-1}, &
S_t^{(5)}=-P_t^{-T/2}F_t^{-1}G_tQ_t^{T/2}, &
S_t^{(6)}=P_t^{-T/2}\hat x_t.
\end{array}
}
$$
\end{enumerate}
\end{slide}


%---------------------------------------------------------------------- SLIDE -
\begin{slide}{Algorithm~LLG-eSRIF}
\small
\begin{enumerate}
\item[{\LightBlue III.}]
For each  $\theta_i$, $i=1,2, \ldots, p$, compute
$$
J_i=
\left[
\begin{array}{ccc}
X_i & Y_i & M_i \\
N_i & V_i & W_i \\
\end{array}
\right]
\left[
\begin{array}{ccc}
R_{e,t}^{-T/2} & 0 & 0 \\
-P_{t+1}^{-T/2}K_{p,t}\ &  \
P_{t+1}^{-T/2} &   0   \\
* & * & * \\
\end{array}
\right]^{-1}.
$$
\end{enumerate}
\end{slide}

%---------------------------------------------------------------------- SLIDE -
\begin{slide}{Algorithm~LLG-eSRIF}
\small
\begin{enumerate}
\item[{\LightBlue IV.}] For each $\theta_i$, $i=1,2, \ldots, p$,
split the matrices
$$
J_i=
\lefteqn{
 \left.
\phantom{
 \begin{array}{c}
    L_i^{(1)}+D_i^{(1)}+U_i^{(1)} \\
 \end{array}
 \begin{array}{c}
  |\qquad *** \qquad\\
 \end{array}
}
 \right\}  \scriptstyle{m+n}
}
 \overbrace{
  \rule{0pt}{5mm}
  \underbrace{
    \left[
      \begin{array}{c|}
      L_i+D_i+U_i \\
     \end{array}
    \right.
  }_{m+n}
   \left.
    \begin{array}{c}
     \qquad *** \qquad\\
     \end{array}
    \right]
  }^{m+n+q} \quad  \qquad
$$
where {\green $L_i$}, {\green $D_i$} and {\green $U_i$} are the
strictly lower triangular, diagonal and
strictly upper triangular parts of {\green $J_i$}, respectively.
\end{enumerate}
\end{slide}


%---------------------------------------------------------------------- SLIDE -
\begin{slide}{Algorithm~LLG-eSRIF}
\small
{\LightBlue V.}
For each $\theta_i$, $i=1,2, \ldots, p$, compute the following
quantities:
$$
\begin{array}{r}
\left[
\begin{array}{cc}
{\green \displaystyle
\frac{\partial{R_{e,t}^{-T/2}}}{\partial{\theta_i}} }
 & 0 \\[8pt]
-\displaystyle \frac{\partial{\left( \tilde P_{t+1}^{-T/2}K_{p,t}
\right)}} {\partial{\theta_i}} & {\LightBlue \displaystyle
\frac{\partial{\tilde P_{t+1}^{-T/2}}}{\partial{\theta_i}}}
\end{array}
\right]
 =
\left[ L_i+D_i+U_i^T \right]\phantom{,}\\[40pt]
  \times  \left[
\begin{array}{cc}
R_{e,t}^{-T/2} & 0 \\
-\tilde P_{t+1}^{-T/2}K_{p,t}  & \tilde P_{t+1}^{-T/2} \\
\end{array}
\right],
\end{array} \eqno{(12)}
$$
\end{slide}

%---------------------------------------------------------------------- SLIDE -
\overlays{2}{%
\begin{slide}{Algorithm~LLG-eSRIF}
\small
$$
{\green \displaystyle \frac{\partial{\bar e_t}}{\partial{\theta_i}}
} = \left[ \displaystyle
\frac{\partial{R_{e,t}^{-T/2}}}{\partial{\theta_i}} - X_i
\right]R_{e,t}^{T/2}\bar e_t +Y_iF_t\hat x_t -L_i,  \eqno{(13)}
$$
\FromSlide{2}
$$
\begin{array}{r}
{\LightBlue \displaystyle
\frac{\partial{S_{t+1}^{(6)}}}{\partial{\theta_i}} } = \left[
\displaystyle \frac{\partial{\left( \tilde
P_{t+1}^{-T/2}K_{p,t}\right)}} {\partial{\theta_i}}   + N_i \right]
R_{e,t}^{T/2}\bar e_t \qquad \quad \\[15pt]
+
 \left[ \displaystyle \frac{\partial{\tilde
P_{t+1}^{-T/2} }}{\partial{\theta_i}}
 - V_i
\right] F_t\hat x_t +K_i.
\end{array} \eqno{(14)}
$$
\end{slide}
}

%---------------------------------------------------------------------- SLIDE -
\overlays{4}{%
\begin{slide}{Algorithm~LLG-eSRIF}
\small
{\LightBlue VI.}
Finally, we have all values to compute the LLG according to (6):
$$
\displaystyle \frac{\partial{l_{\theta}(z_t) }}{\partial{\theta_i}}
= - \tr{
\fromSlide*{3}{\rnode{mah1+}{\pscirclebox[linecolor=LightBlue]{R_{e,t}^{1/2}}}}
\onlySlide*{1}{\rnode{mah1+}{\pscirclebox[linecolor=LightBlue,linestyle=none]{R_{e,t}^{1/2}}}}
\onlySlide*{2}{\rnode{mah1+}{\pscirclebox[linecolor=LightBlue,linestyle=none]{R_{e,t}^{1/2}}}}
\fromSlide*{4}{\rnode{mah2+}{\pscirclebox[linecolor=green]{
\displaystyle \frac{\partial{\left( R_{e,t}^{-1/2}\right) }}
{\displaystyle\partial{\theta_i}} }}}
\onlySlide*{1}{\rnode{mah2+}{\pscirclebox[linecolor=green,linestyle=none]{
\displaystyle \frac{\partial{\left( R_{e,t}^{-1/2}\right) }}
{\displaystyle\partial{\theta_i}} }}}
\onlySlide*{2}{\rnode{mah2+}{\pscirclebox[linecolor=green,linestyle=none]{
\displaystyle \frac{\partial{\left( R_{e,t}^{-1/2}\right) }}
{\displaystyle\partial{\theta_i}} }}}
\onlySlide*{3}{\rnode{mah2+}{\pscirclebox[linecolor=green,linestyle=none]{
\displaystyle \frac{\partial{\left( R_{e,t}^{-1/2}\right) }}
{\displaystyle\partial{\theta_i}} }}} }+
\fromSlide*{3}{\rnode{mah1-}{\pscirclebox[linecolor=LightBlue]{{\bar
e_t}^T}}}
\onlySlide*{1}{\rnode{mah1-}{\pscirclebox[linecolor=LightBlue,linestyle=none]
{{\bar e_t}^T}}}
\onlySlide*{2}{\rnode{mah1-}{\pscirclebox[linecolor=LightBlue,linestyle=none]
{{\bar e_t}^T}}}
\fromSlide*{4}{\rnode{mah3+}{\pscirclebox[linecolor=green]{
\displaystyle \frac{\partial{\bar e_t}}{\partial{\theta_i}} }}}
\onlySlide*{1}{\rnode{mah3+}{\pscirclebox[linecolor=green,linestyle=none]{
\displaystyle \frac{\partial{\bar e_t}}{\partial{\theta_i}} }}}
\onlySlide*{2}{\rnode{mah3+}{\pscirclebox[linecolor=green,linestyle=none]{
\displaystyle \frac{\partial{\bar e_t}}{\partial{\theta_i}} }}}
\onlySlide*{3}{\rnode{mah3+}{\pscirclebox[linecolor=green,linestyle=none]{
\displaystyle \frac{\partial{\bar e_t}}{\partial{\theta_i}} }}}, \
i=1,\ldots, p.
$$
\vspace{4cm}
 \FromSlide{2}
\begin{tabular}{lcl}
The LLG-eSRIF:  & \qquad \qquad \qquad \qquad & The LLG-eSRIF:  \\
Stage \onlySlide*{2}{
\rnode{mah1}{\pscirclebox[linecolor=LightBlue,linestyle=none]{I}}}
\fromSlide*{3}{\rnode{mah1}{\pscirclebox[linecolor=LightBlue]{I}}} &
\qquad & Stages \fromSlide*{4}
{\rnode{mah2}{\pscirclebox[linecolor=green]{II~--- IV}}}
\onlySlide*{2}{\rnode{mah2}{\pscirclebox[linecolor=green,linestyle=none]
{II~--- IV}}}
\onlySlide*{3}{\rnode{mah2}{\pscirclebox[linecolor=green,linestyle=none]
{II~--- IV}}}
\end{tabular}
\fromSlide*{3}{\nccurve[linecolor=LightBlue,angleA=90,angleB=270]{->}{mah1}{mah1+}}
\fromSlide*{3}{\nccurve[linecolor=LightBlue,angleA=90,angleB=270]{->}{mah1}{mah1-}}
\fromSlide*{4}{\nccurve[linecolor=green,angleA=90,angleB=270]{->}{mah2}{mah2+}}
\fromSlide*{4}{\nccurve[linecolor=green,angleA=90,angleB=270]{->}{mah2}{mah3+}}
\end{slide}
}

%---------------------------------------------------------------------- SLIDE -
\overlays{6}{%
\begin{slide}{Algorithm~LLG-eSRIF}
\small The LLG-eSRIF consists of two parts:
\begin{itemstep}
\FromSlide{2}
\item The source filtering algorithm, i.e. {\green eSRIF}.
\FromSlide{3}
\item  The "differentiated" part: {\green Stages II~--- IV}.
\end{itemstep}

\FromSlide{4} Thus the {\LightBlue Algorithm LLG-eSRIF} is ideal for
simultaneous
\begin{itemstep}
\FromSlide{5} \item state estimation
 \FromSlide{6}
\item parameter identification.
\end{itemstep}
\end{slide}
}

%---------------------------------------------------------------------- SLIDE -

\begin{slide}{Algorithm~LLG-eSRIF}
\small {\LightBlue Remark.} Since, the matrices in LLG~(6) are
triangular, only the diagonal elements of {\green $R_{e,t}^{1/2}$}
and {\green $\displaystyle \frac{\partial{\left(
R_{e,t}^{-1/2}\right) }}{\partial{\theta_i}}$} need to be computed.
Hence, the Algorithm~LLG-eSRIF allows the $m\times m$-matrix
inversion of $R_{e,t}$ {\LightBlue to be avoided} in the evaluation
of LLG.
\end{slide}


%---------------------------------------------------------------------- SLIDE -
\overlays{4}{%
\begin{slide}{Ill-Conditioned  Example Problems}

\small
{\LightBlue Problem 1}
{\green Given:}
$$
\begin{array}{r}
P_0=\left[
    \begin{array}{cc}
      \theta & 0  \\
      0 & \theta  \\
    \end{array}
    \right],
H=\left[
    \begin{array}{cc}
      1, & 0  \\
    \end{array}
    \right],
R=e^2\theta,\   F=I_2, \  Q=0,  \\
 \ G=\left[
    \begin{array}{cc}
      0, & 0  \\
    \end{array}
    \right]^T,\   $0< e << 1$
    \end{array}
$$
where $\theta$ is an unknown parameter, $I_2$ is an identity
$2\times 2$ matrix; to simulate roundoff we assume
\FromSlide{2}{\LightBlue $e+1\ne1$ but $e^2+1\stackrel{r}{=}1$}.

\FromSlide{3}{\green Calculate:} $\left( P_1\right)'_{\theta}$ at the point $\theta=1$.

\FromSlide{4} For the $\theta=1$, this example illustrates  the
initialization problems [Kaminski~P.G., Bryson~A.E., Schmidt~S.F., 1971]
that result when  $H_1\Pi_0H_1^T+R_1$ is rounded to $H_1\Pi_0H_1^T$.
\end{slide}
}


%---------------------------------------------------------------------- SLIDE -
\overlays{4}{%
\begin{slide}{Comparison (Problem~1)}
\small
\begin{tabular}{p{1.5cm}|c|c}
\hline
Filter &  Exact Answer & Rounded  Answer\\
\hline
\FromSlide{2} {\green 'diff' KF} & \FromSlide{2}
    $\left.\left(P_1\right)'_{\theta} \right|_{\theta=1}
          =
          \left[
          \begin{array}{cc}
          \displaystyle\frac{e^2}{1+e^2}
           & 0\\[8pt]
          0   & 1 \\
          \end{array}
          \right]$
        & \FromSlide{2}  $\stackrel{r}{=}{\red
          \left[
          \begin{array}{rr}
          0 & 0\\[4pt]
          0 & 1 \\
          \end{array}
          \right]}$  \\
 \hline
\FromSlide{3} {\green 'diff' IF} & \FromSlide{3}
    $ \left.\left(P_1^{-1}\right)'_{\theta} \right|_{\theta=1}
          =
          -\left[
          \begin{array}{cc}
          \displaystyle\frac{1+e^2}{e^2}
           & 0\\[8pt]
          0   & 1 \\
          \end{array}
          \right]$
        & \FromSlide{3} $\stackrel{r}{=}-
          \left[
          \begin{array}{rr}
          \displaystyle\frac{1}{e^2} & 0\\[8pt]
          0 & 1 \\
          \end{array}
          \right]$  \\
\hline
 \FromSlide{4}  {\green LLG-eSRIF}  & \FromSlide{4}
$\left.\left(P_1^{-1/2}\right)'_{\theta}\right|_{\theta=1}=
         -\displaystyle\frac{1}{2}
         \left[
          \begin{array}{cc}
          \displaystyle\frac{\sqrt{1+e^2}}{e}
           & 0\\[7pt]
          0   & 1 \\
          \end{array}
         \right]$  &
         \FromSlide{4} $ \stackrel{r}{=}
         -\displaystyle\frac{1}{2}
          \left[
         \begin{array}{cc}
          \displaystyle\frac{1}{e} & 0\\[7pt]
          0   & 1 \\
          \end{array}
         \right]$ \\
\hline
\end{tabular}
\end{slide}
}

%---------------------------------------------------------------------- SLIDE -
\overlays{3}{%
\begin{slide}{Ill-Conditioned  Example Problems}
\small {\LightBlue Problem 2} {\green Given:}
$$
\begin{array}{r}
P_0=\left[
    \begin{array}{cc}
      \theta & 0  \\
      0 & \theta  \\
    \end{array}
    \right],
H=\left[
    \begin{array}{cc}
      1, & 1  \\
    \end{array}
    \right],
R=e^2\theta, \   F=I_2, \  Q=0, \\
 \ G=\left[
    \begin{array}{cc}
      0, & 0  \\
    \end{array}
    \right]^T,\   $0< e << 1$
    \end{array}
$$
where $\theta$ is an unknown parameter, $I_2$ is an identity
$2\times 2$ matrix; to simulate roundoff we assume
\FromSlide{2}{\LightBlue $e+1\ne1$ but $e^2+1\stackrel{r}{=}1$}.

\FromSlide{3}{\green Calculate:} $\left( P_1\right)'_{\theta}$ at
the point $\theta=1$.
\end{slide}
}

%---------------------------------------------------------------------- SLIDE -
\overlays{4}{%
\begin{slide}{Comparison (Problem~2)}
\small
\begin{tabular}{p{1.5cm}|c|c}
\hline
%\multicolumn{1}{c|}{Filter} &
%\multicolumn{2}{|c}{Solution} \\
% \cline{2-3}
Filter &  Exact Answer & Rounded  Answer\\
\hline \hline
\FromSlide{2} {\green 'diff' KF}& \FromSlide{2}
    $ \displaystyle\frac{1}{2+e^2}
          \left[
          \begin{array}{cc}
          1+e^2  & -1\\[5pt]
          -1   & 1+e^2 \\
          \end{array}
          \right]$
        & \FromSlide{2} $
          \stackrel{r}{=}
          \displaystyle\frac{1}{2}
          {\red \left[
          \begin{array}{rr}
          1 & -1\\[5pt]
          -1 & 1 \\
          \end{array}
          \right]}$  \\
\hline
\FromSlide{3}  {\green 'diff' IF} &
    \FromSlide{3} $  - \displaystyle\frac{1}{e^2}
          \left[
          \begin{array}{cc}
          1+e^2 & 1\\[5pt]
          1   & 1+e^2 \\
          \end{array}
          \right]$
        & \FromSlide{3} $
          \stackrel{r}{=}
          - \displaystyle\frac{1}{e^2}
          {\red \left[
          \begin{array}{rr}
          1 & 1\\[5pt]
          1 & 1 \\
          \end{array}
          \right]}$  \\
\hline  \FromSlide{4}  {\green LLG-eSRIF} & \FromSlide{4}
$-\displaystyle\frac{1}{2}
         \left[
          \begin{array}{cc}
          \displaystyle\sqrt{\frac{2+e^2}{1+e^2}} &
          \displaystyle\frac{1}{e\sqrt{1+e^2}}\\[10pt]
          0   &
          \displaystyle\frac{\sqrt{1+e^2}}{e}\\
          \end{array}
         \right]
          $  & \FromSlide{4}
          $ \stackrel{r}{=}
         -\displaystyle\frac{1}{2}
          \left[
         \begin{array}{cr}
          \sqrt{2}\quad & \quad\displaystyle\frac{1}{e}\\[7pt]
          0   & \displaystyle\frac{1}{e} \\
          \end{array}
         \right]$ \\
\hline
\end{tabular}
\end{slide}
}

%---------------------------------------------------------------------- SLIDE -

\overlays{2}{%
\begin{slide}{Numerical Results}
\small {\LightBlue Example.} Let the test system~(1), (2) be defined
as follows:
$$
x_{t+1}= \left [
\begin{array}{lcl}
1 & \Delta t  \\
0 &   e^{-\Delta t / {\green \tau}} \\
\end{array}
\right ] x_t+ \left [
\begin{array}{l}
0  \\
1\\
\end{array}
\right ] w_t,
$$
$$
z_t= \left [
\begin{array}{ll}
1 &\  \   0\\
\end{array}
\right ] x_t+v_t
$$
where {\green $\tau$}
 is a unknown parameter which needs to be estimated.
\FromSlide{2} For the test problem,  {\green $\tau^*$}={\red 10} was
chosen as the true value of  parameter $\tau$.
\end{slide}
}



%---------------------------------------------------------------------- SLIDE -
\begin{slide}{Conclusion}
\small
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.
\end{slide}


\end{document}


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