Example 2: Finite dimensional realization of a nonlinear filter

The aim here is to construct a finite dimensional realization of a nonlinear filter for the stochastic system described by (see [28,19]):

$$dx(t) = dv(t),\ \ x(0)=x_0$$ (29)

$$dy(t) = x(t)dt+dw(t)$$

where $v$ and $w$ are independent Brownian motions. As suggested in [5] such a realization can be derived by applying Lie algebra techniques to the DMZ equation for the unnormalized conditional density $\sigma(t,x)$, given the observation process $\{y(s); 0\leq s\leq t\}$ for system (29). The DMZ equation here is

$$d\sigma(t,x)=L_0\sigma(t,x)+L_1\sigma(t,x)\circ dy(t),\ \sigma(0,x)=\sigma_0(x),\ \sigma_0\in \mathbb{L}_2(\mathbb{R})$$

where the differential operators $L_0, L_1:D(\mathbb{R})\rightarrow \mathbb{L}_2(\mathbb{R})$ are defined by the following expressions on their common invariant domain $D(\mathbb{R})$ which is dense in $\mathbb{L}_2(\mathbb{R})$ (see [28]):

$$L_0=\frac{1}{2}\frac{\partial^2}{\partial{x}^2}-\frac{x^2}{2},\hspace{5mm} L_1=x$$

It will first be shown that the estimation Lie algebra $L_E\stackrel{def}{=} L(L_0,L_1)$ for the above problem is finite dimensional and solvable. Then, the solution of the Cauchy problem for any given $\sigma_0\in \mathbb{L}_2(\mathbb{R})$, representing the conditional density of $x(0)$, can be written in the form of a product of exponentials, see [23]:

$$\sigma(t,x)=\prod_{i=0}^{r} e^{\gamma_i(t) L_i}\sigma_0(x)$$ (30)

where $L_i$, $i=0,\ldots,r$ is a basis for the Lie algebra $L_E$. The exponential $e^{t L_i}$ represents here a strongly continuous one-parameter semi-group operator defined on a Banach space $\mathbb{L}_2(\mathbb{R})$ and corresponding to the infinitesimal generator $L_i$. The last representation is only valid if the Baker-Campbell-Hausdorff-Zassenhaus formula:

$$e^{t L_i}L_j = \left (\sum_{m=0}^{\infty} \frac{t^m}{m!} (ad L_i)^m L_j\right ) e^{t L_i}$$ (31)

holds for all the $L_i$, $L_j$, $i,j=0,\ldots,r$. As pointed out in [28] the validity of (31) is guaranteed if there exists a common, dense (in $\mathbb{L}_2(\mathbb{R})$), invariant under $L_E$, set of analytic vectors for the estimation Lie algebra spanned by $L_i$, $i=0,\ldots,r$. Such a set can be constructed as the linear span of eigenvectors of the operator $L_0$. To check the solvability of $L_E$, the differential operators $L_0$ and $L_1$ are first defined in Maple as follows:

> L0:=xi->(1/2)*diff(xi,x$2)-(1/2)*x^2*xi;
  L1:=xi->x*xi;

$$\begin{eqnarray*} L0&:=& \xi\rightarrow \frac{1}{2}\frac{\partial^2\xi}{\partial{x}^2}-\frac{x^2}{2}\,\xi \\ L1&:=& \xi\rightarrow x\,\xi \end{eqnarray*}$$

A basis for the Lie algebra of operators $L(L_0,L_1)$ is obtained by considering a free nilpotent Lie algebra $L_n(X_0,X_1)$, where $n$ is a sufficiently high order and calculating its P. Hall basis. For example, for $n=7$, the P. Hall basis for $L_7(X_0,X_1)$, $B=\{B_1,\ldots,B_{41}\}$ counts 41 elements and is constructed by invoking B:=phb(2,7). In this process, the package also delivers explicit bracket expressions for the basis elements in $B$ which are omitted here for reason of brevity. Identifying $L_{i-1}=Ev(B_i)$, $i=1,2$, the basis elements in $B$ can thus be evaluated next by executing the LTP function calcLBdiffop(B[i],B[1..2],[L0,L1],[x]), for i $=3,4,\ldots,41$, yielding:

$$\begin{eqnarray*} L_2\stackrel{def}{=}Ev(B_3)=[L_0,L_1] = \frac{\partial}{\part... ...v(B_2)=L_1\\ L_4\stackrel{def}{=}Ev(B_5)=[L_1,[L_0,L_1]] = -1 \end{eqnarray*}$$

It can further be verified that the application of the evaluation map $Ev$ to the remaining brackets in the basis B reveals several linear dependencies between $L_{i-1}=Ev(B_i)$, $i=1,\ldots,41$: $L_5=L_{26}=L_2$, $L_{11}=L_{35}=L_1$, $L_8=-L_{12}=-L_{20}=L_{31}=-L_{36}=-L_4$, and $L_i=0$, for the remaining Lie products. From this calculation it follows that a basis for $L(L_0,L_1)$ can be defined as $\{L_0,L_1,L_2,L_4\}\stackrel{def}{=}\{L_0,L_1,[L_0,L_1],[L_1,[L_0,L_1]]\}$. These calculations also show that the derived Lie algebra $[L_E,L_E]$ is spanned by $L_2$ and $L_4$, and is nilpotent because $[L_2,L_4]=Ev(B_{10})=0$. Hence, the Lie algebra $L_E$ is solvable, by Corollary 5.3 in [36]. The representation (30) now becomes:

$$\sigma(t,\cdot)=e^{\gamma_0(t) L_0} e^{\gamma_1(t) x} e^{\gamma_2(t) \frac{\partial }{\partial x}} e^{ -\gamma_3(t)} \sigma_0$$ (32)

is hence valid globally, see [45], and the functions $\gamma_i$, $i=0,\ldots,3$ can be computed by quadrature of the Wei-Norman equations. The analytic expression for the Wei-Norman equations can be derived by executing the sequence of commands:

r:=4;                 # Basis dimension.
max_bracket_order:=6; # Degree of nilpotency minus one.
wn:=wner(r,max_bracke_order,BB,B,[SR]):
wnfe:=wnde(wn,r,{},BB,{}):
F_g:=eval(wnfe[1]):

The symbol [SR] is a Maple list containing the dependencies between the members of the basis $B$ after application of the evaluation map $Ev$ as derived above. The symbol F_g assumes value of the matrix $\xi(\gamma)$ of equation (12) and is here:

$$F\!\_\,g:=\left [ \begin{array}{cccc} 1 & 0 & 0 & 0 \\ ... ...amma_0^6 & 0 \\ 0 & 0 & \gamma_1 & 1 \end{array} \right ]$$

The entries $(2,2)$ and $(3,3)$ of F_g can be clearly be recognized as the first few terms in the Taylor series expansion for $\cosh$. Similarly, the entries $(2,3)$ and $(3,2)$ are recognized as the first few terms of the Taylor series for $\sinh$. Now, it can be verified that if the above calculations are repeated using a Hall basis of order $n>7$, then the entries of F_g will contain higher order terms of these Taylor series. Thus, by induction, it can be shown that these entries truly are the $\cosh$ and $\sinh$ functions, so that the Wei-Norman equations for (32) in the form (14) are given by:

$$\dot{\gamma}_0=1,\ \ \dot{\gamma}_1=\cosh(\gamma_0)dy(t),\ ... ...}_3= \sinh(\gamma_0)\gamma_1 dy(t) = -\dot{\gamma}_2\gamma_1$$

where $u=[1\ dy(t)\ 0\ 0]^T$. The solution of these Wei-Norman equations constitutes the joint-sufficient statistics for the linear filtering problem of (29). Now, Mehler's formula (see [28]) allows to obtain the explicit expression for the one parameter semi-group $e^{\gamma_0(t)L_0}$ in the form of an integral operator as follows:

$$e^{\gamma_0(t)L_0}\phi(x) = e^{t L_0}\phi(x) = \int_{-\infty}^{\i... ...\frac{1}{2\pi\sinh(t)} e^{-1/2\coth(x^2+y^2)}e^{\frac{xy}{\sinh(t)}} \phi(y) dy$$ (33)

for any $\phi\in \mathcal{D}(\mathbb{R})$. Since $(e^{tx}\phi)(x) =e^{tx}\phi(x)$ and $(e^{t\frac{\partial}{\partial x}}\phi)(x)=\phi(x+t)$, then, finally, (32) and (33) combine into:

$$\sigma(t,x)=\int_{-\infty}^{\infty} \frac{1}{2\pi\sinh(t)} ... ...(t)}}e^{\gamma_3(t)}e^{\gamma_1(t)y}\sigma_0(\gamma_2(t)+y)dy$$

which is an explicit formula for the nonlinear filter for (29).