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Affine Constrained Kalman Bucy Smoother Newton Step

Syntax
[ds, dy, du] = ckbs_newton_step(
      mu, s, y, u, b, d, Bdia, Hdia, Hlow)

Purpose
This routine computes one step of Newton's method for solving the non-linear Kuhn-Tucker conditions for the  \mu -relaxed affine constrained Kalman-Bucy smoother problem.

Problem
Given  \mu \in \B{R}_+ ,  H \in \B{R}^{p \times p} ,  d \in \B{R}^p ,  b \in \B{R}^r , and  B \in \B{R}^{r \times p} , the  \mu -relaxed affine constrained Kalman-Bucy smoother problem is:  \[
\begin{array}{rl}
{\rm minimize} & \frac{1}{2} y^\R{T} H y + d^\R{T} y
- \mu \sum_{i=1}^r \log(s_i)
\; {\rm w.r.t} \; y \in \B{R}^p \; , \; s \in \B{R}_+^r
\\
{\rm subject \; to} & s + b + B y  = 0
\end{array}
\]  In addition,  H is symmetric block tri-diagonal with each block of size  n \times n and  B is block diagonal with each block of size  m \times n (there is an integer  N such that  p = n * N and  r = m * N ).

Lagrangian
We use  u \in \B{R}^r to denote the Lagrange multipliers corresponding to the constraint equation. The corresponding Lagrangian is  \[
L(y, s, u)  =
\frac{1}{2} y^\R{T} H y + d^\R{T} y
- \mu \sum_{i=1}^r \log(s_i)
+ u^\R{T} (s + b + B y)
\]  The partial gradients of the Lagrangian are given by  \[
\begin{array}{rcl}
\nabla_y L(y, s, u ) & = & H y + B^\R{T} u + d  \\
\nabla_s L(y, s, u ) & = & u - \mu / s \\
\nabla_u L(y, s, u ) & = & s + b + B y \\
\end{array}
\]  where  \mu / s  is the component by component division of  \mu  by the components of the  s . Note, from the second equation, that we only need consider  u \geq 0 because  s \geq 0 .

Kuhn-Tucker Conditions
We use  D(s) to denote the diagonal matrix with  s along its diagonal and  1_r to denote the vector, of length  r with all its components equal to one. We define  F : \B{R}^{r + p + r} \rightarrow \B{R}^{r + p + r} by  \[
F(s, u, y)
=
\left(
\begin{array}{c}
s + b + B y       \\
D(s) D(u) 1_r - \mu 1_r\\
H y + B^\R{T} u + d
\end{array}
\right)
\]  The Kuhn-Tucker conditions for a solution of the  \mu -relaxed constrained affine Kalman-Bucy smoother problem is  F(s, u, y) = 0  .

Newton Step
Given a value for  (s, u, y) , the Newton step  ( \Delta s^\R{T} , \Delta u^\R{T} , \Delta y^\R{T} )^\R{T} solves the problem:  \[
F^{(1)} (s, u, y)
\left( \begin{array}{c} \Delta s \\ \Delta u \\ \Delta y \end{array} \right)
=
- F(s, u, y)
\] 

mu
The argument mu is a positive scalar specifying the relaxation parameter  \mu .

s
The argument s is a column vector of length  r . All the elements of s are greater than zero.

y
The argument y is a column vector of length  p

u
The argument u is a column vector of length  r . All the elements of s are greater than zero.

b
The argument b is a column vector of length  r .

d
The argument d is a column vector of length  p

Bdia
The argument Bdia is an  m \times n \times N array. For  k = 1 , \ldots , N we define  B_k \in \B{R}^{m \times n} by  \[
      B_k = Bdia(:, :, k)
\] 

B
The matrix  B is defined by  \[
B
=
\left( \begin{array}{cccc}
B_1 & 0      & 0      &           \\
0   & B_2    & 0      & 0         \\
0   & 0      & \ddots & 0         \\
    & 0      & 0      & B_N
\end{array} \right)
\] 

Hdia
The argument Hdia is an  n \times n \times N array. For  k = 1 , \ldots , N we define  H_k \in \B{R}^{n \times n} by  \[
      H_k = Hdia(:, :, k)
\] 

Hlow
The argument Hlow is an  n \times n \times N array. For  k = 1 , \ldots , N we define  L_k \in \B{R}^{n \times n} by  \[
      L_k = Hlow(:, :, k)
\] 

H
The matrix  H is defined by  \[
H
=
\left( \begin{array}{cccc}
H_1 & L_2^\R{T} & 0      &           \\
L_2 & H_2    & L_3^\R{T} & 0         \\
0   & \ddots & \ddots & \ddots    \\
    & 0      & L_N    & H_N
\end{array} \right)
\] 

ds
The result ds is a column vector of length  r equal to the  \Delta s components of the Newton step.

dy
The result dy is a column vector of length  p equal to the  \Delta y components of the Newton step.

du
The result du is a column vector of length  r equal to the  \Delta u components of the Newton step.

Example
The file newton_step_ok.m contains an example and test of ckbs_newton_step. It returns true if ckbs_newton_step passes the test and false otherwise.

Method
The derivative of  F is given by  \[
F^{(1)} (s, y, u) =
\left(
\begin{array}{ccccc}
D( 1_r ) &  0  & B  \\
D( u )   & 0   & D(s) \\
0        & B^\R{T} & H
\end{array}
\right)
\]  It follows that  \[
\left(\begin{array}{ccc}
I & 0 & B \\
U & S & 0\\
0 & B^T & C \\
\end{array}\right)
\left(\begin{array}{ccc}
\Delta s \\ \Delta y \\ \Delta u
\end{array}\right)
=
-
\left(\begin{array}{ccc}
s + b + By \\
SU\B{1} - \mu\B{1}\\
Cy + B^Tu + d
\end{array}\right)\;.
\]  Below,  r_i refers to row  i . Applying the row operations  \[
\begin{array}{ccc}
r_2 &=& r_2 - U r_1 \\
r_3 &=& r_3 - B^T S^{-1} r_2
\end{array}\;,
\]  we obtain the equivalent system  \[
\left(\begin{array}{ccc}
I & 0 & B \\
0 & S & -UB\\
0 & 0 & C + B^T S^{-1} U B
\end{array}\right)
\left(\begin{array}{ccc}
\Delta s \\ \Delta y \\ \Delta u
\end{array}\right)
=
-
\left(\begin{array}{ccc}
s + b + B y \\
-U(b + B y) - \mu\B{1}\\
Cy + B^T u + d + B^T S^{-1}
\left(U(b + B y) + \mu \B{1}
\right)
\end{array}\right)\;.
\]   \Delta y is obtained from a single block tridiagonal solve (see third row of system). Then we immediately have  \[
\Delta u = US^{-1}(b + B(y + \Delta y)) + \frac{\mu}{s}
\]  and  \[
\Delta s = -s -b -B(y + \Delta y)\;.
\]  .
Input File: src/ckbs_newton_step.m