\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2014 (2014), No. 208, pp. 1--10.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2014 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2014/208\hfil Feynman-Kac theorem in Hilbert spaces]
{Feynman-Kac theorem in Hilbert spaces}

\author[I. V. Melnikova, V. S. Parfenenkova \hfil EJDE-2014/208\hfilneg]
{Irina V. Melnikova, Valentina S. Parfenenkova}  % in alphabetical order

\address{Irina V. Melnikova \newline
Ural Federal University, Lenin av. 51, 620083 Ekaterinburg, Russia}
\email{Irina.Melnikova@usu.ru}

\address{Valentina S. Parfenenkova \newline
Ural Federal University, Lenin av. 51, 620083 Ekaterinburg, Russia}
\email{vika8887@e1.ru}

\thanks{Submitted February 20, 2014. Published October 7, 2014.}
\subjclass[2000]{47D06, 60G15, 60H30}
\keywords{Semigroups of operators; infinite dimensional stochastic equations; \hfill\break\indent diffusion processes; Kolmogorov equations}

\begin{abstract}
 In this article we study the relationship between solutions to Cauchy problems
 for the abstract stochastic differential equation $dX(t)=AX(t)dt + BdW(t)$
 and solutions to Cauchy problems (backward and forward)
 for the infinite dimensional deterministic partial differential equation
 $$
 \pm\frac{\partial g}{\partial t}(t,x) + \frac{\partial g}{\partial x}(t,x)Ax
 + \frac{1}{2}\operatorname{Tr}[(BQ^{1/2})^* \frac{\partial^2 g}{\partial x^2}(t,x)
 (BQ^{1/2})] = 0,
 $$
 where $g$ is the probability characteristic  $g=\mathbb{E}^{t,x}[h(X(T))]$
 in the backward case and $g=\mathbb{E}^{0,x}[h(X(t))]$ in the forward case.
 This relationship, that is the inifinite dimensional Feynman-Kac theorem,
 is proved in both directions: from stochastic to deterministic and from
 deterministic to stochastic. Special attention is given to the definition and
 interpretation of objects in the equations.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{corollary}[theorem]{Corollary}
\allowdisplaybreaks

\section{Introduction}

Many practical problems lead to stochastic equations and it is not always possible
or necessary to find their solutions, sometimes is sufficient to have only
probability characteristics of the solutions.

The famous Feynman-Kac theorem allows to pass from solving stochastic
differential equations to deterministic partial differential equations for
probability characteristics in the finite dimensional case.
It relates solutions of the Cauchy problem for stochastic differential equations
with Brownian motion $\beta(t)$, $t\geq 0$:
\begin{equation}\label{mp1}
dX(t) = a(t,X(t))dt + b(t,X(t))d\beta(t), \quad t\in [0,T], \; X(0)=\xi,
\end{equation}
and solutions of the Cauchy problem for deterministic partial differential equations
\begin{equation}\label{mp2}
g_t(t,x) + a(t,x)g_x(t,x) + \frac{1}{2} b^2 (t,x)
g_{xx}(t,x) = 0, \quad g(T,x) = h(x),
\end{equation}
for the probability characteristic $g(t,x)=\mathbb{E}^{t,x}[h(X(T))]$
with an arbitrary Borel function $h$. Here $\mathbb{E}^{t,x}$ means
the mathematical expectation of a solution to the equation
\eqref{mp1} under the condition $X(t) = x$,
$0\leq t\leq T$.

The study of the relationship between problems
\eqref{mp1}--\eqref{mp2}
 was initially caused by needs from physics. For example, the
process $\{X(t), t\in[0,T]\}$ describes the random motion of particles in a liquid
or gas and $g(t,x)$ is a probability characteristic such as
temperature, determined by the Kolmogorov equation.
Recent years the importance of the relationship between stochastic and deterministic
problems became more acute with the development of numerical methods
(see, e.g. \cite{Milstein}) and applications in financial
mathematics. For example, $X(t)$ is a stock price at time $t$, then
$g(t,x)$ is the value of stock options, determined by the famous
Black-Scholes equation \cite{Bjork,Emamirad_Goldstein,Shreve}.

Along with applications to mathematical physics and financial mathematics
in finite dimensional case (see, e.g.
\cite{Allen,Bjork,Emamirad_Goldstein,Oksendal,Shreve}) there exist recent
infinite dimensional applications, for instance, stochastic equations in
financial mathematics \cite{Ekeland_Taflin,Filipovic}.
As an example, let $P(t,T)$
be a price at time $t\leq T$ of a coupon bond with maturity date
$T$ parametrized as $P(t,t) = 1$ for all $t$ and let
${f(t,T), t\leq T}$ be the forward curve, i.e. $P(t,T) = \exp
\left(-\int_t^T f(t,s)ds \right)$. Then the Musiela
reparametrization $r(t,\cdot):=f(t,\cdot+t)$ in the special case of zero
HJM-shift satisfies the following equation in Hilbert space
$H=L_2(\mathbb{R}_{+})$
\begin{equation}\label{mpBondEquation}
dr(t) = Ar(t) dt + b(t,r(t))dW(t), \quad t\geq 0, \quad r(0) = \xi,
\end{equation}
where $A$ is the generator of the right-shifts semigroup in $H$, $W$ is an
$\mathbb{H}$-valued $Q$-Wiener process, and $b$ is a random mapping from a
Hilbert space $\mathbb{H}$ to $H$.
The value of bond options may be calculated, at least numerically,
via $g(t,x)$ defined for $X(t)=r(t)$.
So the relationship between infinite dimensional stochastic and deterministic
problems is important both in theory and applications.
An extension of the Feynman-Kac theorem to the
infinite dimensional case raises many questions related with the very
formulation of the problems in infinite dimensional spaces, the definition
of relevant objects and a rigorous rationale for the
relationship.

The present article concerns  the stochastic Cauchy problem in Hilbert spaces
\begin{equation}\label{mp3}
dX(t)=AX(t)dt + BdW(t), \quad t\in [0,T],\quad X(0)=\xi.
\end{equation}
We prove the infinite dimensional case of the Feynman-Kac theorem under the basic condition on $A$ to be the generator of a $C_0$-semigroup in a Hilbert space $H$. We suppose $B\in\mathcal{L}(\mathbb{H},H)$ in the case of an $\mathbb{H}$-valued $Q$-Wiener process $W$ and $B\in \mathcal{L}_{\mathtt{HS}}(\mathbb{H},H)$ in the case of a cylindrical Wiener process $W$. For the problem \eqref{mp3} we associate a
problem for a deterministic partial differential equation which is an extension of the problem \eqref{mp2} in the case of Hilbert spaces:
\begin{equation}\label{mp4}
\frac{\partial g}{\partial t}(t,x) + \frac{\partial g}{\partial
x}(t,x)Ax + \frac{1}{2}\operatorname{Tr} \big[(BQ^{1/2})^* \frac{\partial^2
g}{\partial x^2}(t,x)(BQ^{1/2})\big] = 0, g(T,x)=h(x),
\end{equation}
and show that $g$ satisfies the infinite dimensional deterministic
problem \eqref{mp4} with the trace class operator $Q$
in the case of $Q$-Wiener process and with $Q=I$ in the case of cylindrical
Wiener process.
We give the rigorous interpretation of objects included in stochastic and
deterministic equations and prove the connection between their solutions
in both directions: ``from stochastic to deterministic'' and
``from deterministic to stochastic''.

The implication ``from stochastic to deterministic'' consists of several steps.
The first step is the proof of the Markov property for the Cauchy problem
solution $X$, then the martingal property for the function $g(t,x)|_{x=X(t)}$
and the last step is the formal usage of infinite dimensional Ito's formula
for $g(t,X(t))$. Particular attention is paid to the subtle issue of transition
from zero expectation for a function of $g$ to equality for $g$ itself.
The implication ``from deterministic to stochastic'' also uses the infinite
dimensional Ito's formula for $g(t,X(t))$ too.

Comparing with our previous article \cite{Melnikova_Parfenenkova}
in the current paper we proved the implication ``from deterministic to stochastic''  and improved the proof of the implication ``from stochastic to deterministic'',
we proved the relationship for the stochastic problem
\eqref{mp3} with $Q$-Wiener and cylindrical Wiener processes,
and at last along with $g(t,x)=\mathbb{E}^{t,x}[h(X(T))]$ we introduce
a different probability characteristic $\tilde{g}(t,x)=\mathbb{E}^{0,x}[h(X(t))]$
and show that $g$ leads to the backward deterministic Cauchy problem and
$\tilde{g}$ leads to the forward deterministic Cauchy problem.

\section{Definitions and auxiliary statements}

We start with interpretation for objects of the stochastic
problem \eqref{mp3}. Let the operator $A$ be
the generator of a $C_0$-semigroup in Hilbert space $H$. This
ensures uniform well-posedness of the Cauchy problem for the
corresponding homogeneous equation $X'(t)=AX(t)$ and existence of strongly
continuous solution operators $U(t), t\geq 0$ to the homogeneous problem,
as well as existence and uniqueness of a weak solution to the stochastic
problem \eqref{mp3} with an $\mathbb{H}$-valued Wiener process $W$
(see, e.g. \cite{Alshanskiy_Melnikova,DaPrato_Zabczyk}):
$$
X(t) = U(t)\xi + W_A(t) = U(t)\xi + \int_0^t U(t-s)BdW(s), \quad t\in[0,T].
$$

The stochastic convolution $W_A(t)$ with respect to Wiener process
(as $Q$-Wiener as cylindrical Wiener) formally is defined under the
following condition
\begin{equation}\label{solution_via_stoch_convolution}
\mathbb{E}\Big[ \int_0^t \| U(s)B \|_{\mathtt{HS}}^2 ds \Big] < \infty,
\end{equation}
where $\|U(s)B\|_{\mathtt{HS}}^2 := \operatorname{Tr}
[U(s)BQ(U(s)B)^*]$. In the case of
a $Q$-Wiener process the operator $Q$ is a trace class operator and for the
validity of \eqref{solution_via_stoch_convolution} it is sufficient the
operators $U(s)B,\, s\in[0,T]$, hence $B$, to be bounded. In the case of
cylindrical Wiener process the operator $Q$ is only bounded operator
with $\operatorname{Tr} Q = \infty$ (for simplicity we will suppose $Q=I$)
 and for the validity of \eqref{solution_via_stoch_convolution} it is 
sufficient the operators
$U(s)B$, $s\in[0,T]$, hence $B$, to be Hilbert-Schmidt operator.

Now we give the interpretation for the objects of the deterministic partial
differential equation \eqref{mp4}. Define the function
$g(t,x) := \mathbb{E}^{t,x}[h(X(T))]$ that
transforms $[0,T] \times H$ into $\mathbb{R}$, supposing $h$ a
measurable function from $H$ to $\mathbb{R}$. We show that $g$
satisfies the infinite dimensional deterministic Cauchy problem \eqref{mp4}
corresponding to the stochastic one \eqref{mp3}.

As we mentioned above, in the case of cylindrical Wiener process $W$
the operator $Q$ is equal to $I$ and the problem \eqref{mp4} becomes as follows
\begin{equation}\label{mpcylindrical}
\frac{\partial g}{\partial t}(t,x) + \frac{\partial g}{\partial
x}(t,x)Ax + \frac{1}{2}\operatorname{Tr}\big[B^* \frac{\partial^2
g}{\partial x^2}(t,x)B\big] = 0, \quad g(T,x)=h(x).
\end{equation}
In this case condition \eqref{solution_via_stoch_convolution}
on operator $B$ guarantees that operator under the trace sign,
$B^* \frac{\partial^2 g}{\partial x^2}(t,x)B$, is really of trace class.

First we make sense to terms in  \eqref{mp4}.
The derivatives $\frac{\partial g}{\partial x}$ and
$\frac{\partial^2 g}{\partial x^2}$ are understood in the sense of Frechet; that
means $\frac{\partial g}{\partial x}: [0,T] \times H \to
H^*$ and $\frac{\partial^2 g}{\partial x^2}: [0,T] \times H
\to \mathcal{L}(H, H^*)$. More precisely
\begin{gather*}
\frac{\partial g}{\partial x}(t,x)(\cdot): H \to \mathbb{R}, \quad
\frac{\partial^2 g}{\partial x^2}(t,x)(\cdot): H\to H^*,
\quad \text{for any fixed $t\in [0,T]$, $x \in H$}, \\
BQ^{1/2}:\mathbb{H}\to H, \quad (BQ^{1/2})^*:H^*\to \mathbb{H}^*.
\end{gather*}
The term $\operatorname{Tr}[(BQ^{1/2})^* \frac{\partial^2 g}{\partial
x^2}(BQ^{1/2})]$ requires special attention. Expression $\operatorname{Tr}$ is
usually defined as the trace of an operator acting in the same
Hilbert space. The operator under the trace sign in equation \eqref{mp4}
maps Hilbert space $\mathbb{H}$ to its adjoint $\mathbb{H}^*$.

Using the traditional definition of the trace and the Riesz theorem on the
isomorphism $\mathbb{H}$ and $\mathbb{H}^*$; that is identifying
$\mathbb{H}^*$ with $\mathbb{H}$, we can make sense to the trace sign.
Note that the isomorphism allows us to consider operators $BQ^{1/2}$,
$(BQ^{1/2})^*$, and $\frac{\partial^2 g}{\partial x^2}$ as mappings
from $\mathbb{H}$ to $H$, from $H$ to $\mathbb{H}$, and $H$ to $H$, respectively.
Then operator $(BQ^{1/2})^* \frac{\partial^2 g}{\partial x^2}(BQ^{1/2})$
transfers the Hilbert space $\mathbb{H}$ to $\mathbb{H}$ and trace of this
operator can be understood in the usual sense:
in the case of a $Q$-Wiener process with a trace class
operator $Q$, bounded operators $\frac{\partial^2 g}{\partial^2 x}$ and $B$
we have
\begin{align*}
\big|\operatorname{Tr}\big[(BQ^{1/2})^* \frac{\partial^2 g}{\partial
x^2}(BQ^{1/2})\big]\big|
&\leq \sum_{j=1}^\infty\big|\big\langle \frac{\partial^2 g}{\partial x^2}
(BQ^{1/2})e_j, (BQ^{1/2})e_j\big\rangle\big|\\
& \leq \sum_{j=1}^\infty\sigma_j^2\|B\|^2
\| \frac{\partial^2 g}{\partial x^2}\|^2<\infty.
\end{align*}
In the case of a cylindrical Wiener process the estimate for
$\operatorname{Tr}[B^* \frac{\partial^2 g}{\partial
x^2}B]$ takes place for the bounded operator
 $\frac{\partial^2 g}{\partial^2 x}$ and a Hilbert-Schmidt operator $B$.


\section{From stochastic to deterministic}

At first we prove necessary properties of the process $X$ that is a solution
of \eqref{mp3}, and the function $g(t,x)$ that determines the relationship
between solutions of problems \eqref{mp3} and
\eqref{mp4}. We obtain required properties for the case of more general processes,
 diffusion processes, to which the solution of the equation \eqref{mp3}
is a special case.

An $H$-valued Ito process $\{X(t),t \geq 0 \}$ is called diffusion if it can be
written in the form
\begin{equation} \label{mpDiff}
dX(t) = a(X(t))dt + b(X(t))dW(t),
\end{equation}
where $a$ and $b$ are some measurable mappings.
Consider the Cauchy problem for the equation \eqref{mpDiff},
\begin{equation}\label{mpDiff_CP}
dX(t) = a(X(t))dt + b(X(t))dW(t), \quad t\in [0,T], \quad X(0)=\xi.
\end{equation}
In this article we consider only diffusion processes such that
existence and uniqueness of a solution to the stochastic Cauchy
problem \eqref{mpDiff_CP} take place.
It may be reached by different ways. For example, it is guaranteed by
the estimate to coefficients $a$ and $b$:
 $\|a(z_1) - a(z_2)\| + \|b(z_1)- b(z_2)\| \leq c\|z_1 - z_2\|$,
$z_1,z_2\in H$, $c\in\mathbb{R}$ (see \cite[Theorem 2.1, ch. VII]{Dalecky_Fomin}.
In this article we are not interested in a concrete form of conditions,
we only suppose that existence and uniqueness of the Cauchy problem solution hold.

As pointed out in previous section this unique solution can be written
as a sum of the term depending on the initial data and the stochastic
convolution term.
To prove the infinite dimensional Feynman-Kac theorem it is important
to establish the Markov property for the solution of the Cauchy problem
\eqref{mpDiff_CP}. The following statement is an extension of the
finite dimensional case result (see \cite[Theorem 7.1.2]{Oksendal})
 to the case of Hilbert spaces.

\begin{proposition} \label{prop1}
Let $h=h(z)$, $z\in H$ be Borel-measurable and $X=X(t),\, t\in [0,T]$,
be a unique solution of \eqref{mpDiff_CP}. Then $X$
satisfies the Markov property with respect to the filter
$\mathfrak{F}_t$, $t\geq 0$ defined by Wiener process $W$:
\begin{equation}\label{mpMarkov}
\mathbb{E}\left[ h(X(t+s))| \mathfrak{F}_t \right] =
\mathbb{E}^{0,X(t)}[h(X(s))].
\end{equation}
\end{proposition}

\begin{proof}
Let $X^{t,x}$, $t\in [0,T]$, $x\in H$ be the solution of the Cauchy problem
for the equation \eqref{mpDiff} with the initial data $X(t)=x$.
In this notation $X$, which is the solution of the Cauchy problem \eqref{mpDiff_CP},
 may be written as $X^{0,\xi}$.
 It is easy to see that a constriction of $X(\tau)$ to the segment $\tau\in [t,T]$
is also a solution of the Cauchy problem for the equation \eqref{mpDiff} with the
initial data $x=X(t)$.
By the uniqueness of a solution to the Cauchy problem
\eqref{mpDiff_CP} we have $X(\tau) = X^{t,X(t)}(\tau)$,
$\tau\geq t$, almost surely. So we have
\begin{equation}\label{prop_equalities_0}
\mathbb{E}[h(X(\tau))|\mathfrak{F}_t] =
\mathbb{E}[h(X^{t,X(t)}(\tau))|\mathfrak{F}_t]=
\mathbb{E}[h(X^{t,X(t)}(\tau))]=
\mathbb{E}[h(X(\tau))].
\end{equation}
The first and the last equalities are consequences of the uniqueness
of a solution \eqref{mpDiff_CP}. The validity of the second equality
follows from the independence $X^{t,x}(\tau)$ from $\mathfrak{F}_t$
because $t$ is a starting point for the solution $X^{t,x}$.

Suppose $s$ is defined as $\tau = t + s$. Then
\begin{equation}\label{prop_equalities}
\mathbb{E}[h(X(t+s))|\mathfrak{F}_t] =
\mathbb{E}[h(X(t+s))] = \mathbb{E}[h(X^{t,X(t)}(t+s))]=
\mathbb{E}[h(X^{t,z}(t+s))]_{z=X(t)}.
\end{equation}
Here the first equality follows from \eqref{prop_equalities_0}.
The second equality is a consequence of solution's uniqueness.

Using the diffusion property of process $X$, we obtain
$$
h(X^{t,z}(t+s))=h(X^{0,z}(s))
$$
and hence the equality for mathematical expectations
\begin{equation}\label{mp5}
\mathbb{E}[h(X^{t,z}(t+s))]_{z=X(t)}=\mathbb{E}[h(X^{0,z}(s))]_{z=X(t)}
\end{equation}
is also valid. Equalities \eqref{prop_equalities} and \eqref{mp5}
imply the desired relation
\eqref{mpMarkov}.
\end{proof}

\begin{corollary} \label{coro1}
Suppose a process $X$ is a solution of
\eqref{mp3}. Then it is unique, diffusive, and by Proposition \ref{prop1} has
 the Markov property.
\end{corollary}

By the homogeneity in time of diffusion processes, equation \eqref{mp5}
can be written in the ensuing form:
\begin{equation}\label{prop_homog}
\mathbb{E}^{0,X(t)}[h(X(s))] = \mathbb{E}^{t,X(t)}[h(X(t+s))].
\end{equation}
As a consequence of Proposition \ref{prop1} and \eqref{prop_homog} we obtain
the following result.

\begin{corollary} \label{coro2}
Let $h(z), z\in H$ be Borel-measurable and $X(t)$, $t\geq 0$, be a unique
solution of \eqref{mpDiff_CP}. Then the Markov property for $X$ may be
rewritten as follows:
$$
\mathbb{E}[ h(X(t+s))| \mathfrak{F}_t] =
\mathbb{E}^{t,X(t)}[h(X(t+s))].
$$
\end{corollary}

On the basis of Markov property for $X$ we prove the martingale property for $g$,
 useful for the proof of Feynman-Kac theorem.
The following statement generalizes \cite[Theorem 5.50]{Shreve-Lectures}.

\begin{proposition} \label{prop2}
Suppose a process $X$ satisfies conditions of Proposition \ref{prop1}.
Then the process $g(t,X(t)) := \mathbb{E}^{t,x}[h(X(T))]|_{x=X(t)}$
is martingale; i.e.,
$$
\mathbb{E}\left[g(\tau,X(\tau))|\mathfrak{F}_t\right] = g(t, X(t)),
\quad 0 \leq t \leq \tau \leq T.
$$
\end{proposition}

\begin{proof}
According to Proposition \ref{prop1}, $X$ has the Markov property.
Therefore
$$
\mathbb{E}\big[ h(X(T)) | \mathfrak{F}_\tau \big]
= \mathbb{E}^{\tau,X(\tau)}[h(X(T))]
= g(\tau,X(\tau)),
$$
and we obtain
\begin{align*}
\mathbb{E}[g(\tau,X(\tau))|\mathfrak{F}_t]
&= \mathbb{E}[\mathbb{E}[ h(X(T)) | \mathfrak{F}_\tau]|\mathfrak{F}_t]
= \mathbb{E}[h(X(T)) | \mathfrak{F}_t] \\
&= \mathbb{E}^{t,X(t)}[h(X(T))] = g(t,X(t)).
\end{align*}
The first equality implies the obtained representation for the
process $g(\tau,X(\tau))$ via the conditional expectation. The second
equality follows from properties of conditional expectation. The
third one is the direct consequence of the Markov property for
$X$. The last equality follows from the definition of the
process $g(t,X(t))$ and completes the proof.
\end{proof}

Now we can  prove the connection between problems \eqref{mp3} and \eqref{mp4}.

\begin{theorem} \label{thm1}
Consider the stochastic Cauchy problem \eqref{mp3}
where $A$ is the generator of a $C_0$-semigroup in a Hilbert space $H$,
$B\in \mathcal{L}(\mathbb{H},H)$ in the case of a $Q$-Wiener process $W$
and $B\in \mathcal{L}_{\mathtt{HS}}(\mathbb{H},H)$ in the case of a
 cylindrical Wiener process $W$.
Define $g=g(t,x) := \mathbb{E}^{t,x}[h(X(T))]:  [0,T] \times H \to \mathbb{R}$,
$h$ is a measurable function from $H$ to $\mathbb{R}$.
Suppose $\mathbb{E}^{t,x}|h(X(T))| < \infty$ and derivatives
$\frac{\partial g}{\partial t}$, $\frac{\partial g}{\partial x}$,
 $\frac{\partial^2 g}{\partial x^2}$ exist for all pairs $(t,x)$.
Then $g$ is a solution of the infinite dimensional backward Cauchy problem
\eqref{mp4} in the case of $Q$-Wiener process and the problem
\eqref{mpcylindrical} in the case of cylindrical Wiener process.
\end{theorem}

\begin{proof}
Let $X$ be a solution of \eqref{mp3}. Fix some $\tau\in [0,T]$ and consider
the stochastic Cauchy problem with the initial data $x=X(\tau)$:
\begin{equation}\label{Stochastic_CP_common}
dX^{\tau,x}(t)=AX^{\tau,x}(t)dt + BdW(t), \quad t\in [\tau,T],\quad
X^{\tau,x}(\tau)=x.
\end{equation}
Then $X^{\tau,x}$ is a constriction of $X$ on the segment $[\tau,T]$; i.e.,
they coincide on $[\tau,T]$. It means that $g$ also coincide for Cauchy
problems \eqref{mp3} and \eqref{Stochastic_CP_common}. So further we will
simply write $X$ instead of $X^{\tau,x}$.

Applying the Ito formula in Hilbert spaces
\cite{DaPrato_Zabczyk} to $g$ as a function from the
solution of the problem \eqref{Stochastic_CP_common} we obtain
\begin{align*}
dg(t,X(t))
&=\frac{\partial g}{\partial x}(t,X(t))BdW(t)
+ \Big( \frac{\partial g}{\partial t}(t,X(t))
+ \frac{\partial g}{\partial x}(t,X(t))AX(t) \\
&\quad +\frac{1}{2}\operatorname{Tr}
\big[(BQ^{1/2})^* \frac{\partial^2 g}{\partial
x^2}(t,X(t))(BQ^{1/2}) \big] \Big) dt.
\end{align*}
This equality is written in the form of differentials
(increments). In the integral form it can be written as
\begin{align*}
g(t,X(t))
&= g(\tau,x) + \int_\tau^t \frac{\partial g}{\partial
x}(s,X(s))BdW(s) + \int_\tau^t \Big( \frac{\partial g}{\partial s}(s,X(s))\\
&\quad + \frac{\partial g}{\partial x}(s,X(s))AX(s) + \frac{1}{2}\operatorname{Tr}
\big[(BQ^{1/2})^* \frac{\partial^2 g}{\partial
x^2}(s,X(s))(BQ^{1/2}) \big] \Big) ds.
\end{align*}
Apply the expectation to both sides of the equation. From the
definition of an Ito integral (via the approximation in the mean square
by step processes) and the properties of the $Q$-Wiener process, we
obtain $\mathbb{E}[ \int_\tau^t \frac{\partial g}{\partial
x}(s,X(s))BdW(s)] = 0$. Further, since the process
$g$ is martingale, we have
\begin{equation}\label{123}
\mathbb{E}[g(t,X(t))] =
\mathbb{E}[g(t,X(t))|\mathfrak{F}_\tau] = g(\tau,x).
\end{equation}
Hence using the theorem of Tonelli-Fubbini in Hilbert spaces and
equalities \eqref{123} we conclude that
\begin{align*}
0 &= \mathbb{E}\Big[ \int_{\tau}^t \Big( \frac{\partial g}{\partial
s}(s,X(s)) + \frac{\partial g}{\partial x}(s,X(s))AX(s) \\
&\quad  + \frac{1}{2}\operatorname{Tr} \big[(BQ^{1/2})^* \frac{\partial^2 g}{\partial
x^2}(s,X(s))(BQ^{1/2}) \big] \Big) ds \Big] \\
&=\int_{\tau}^t \mathbb{E} \Big[ \frac{\partial g}{\partial s}(s,X(s))
+ \frac{\partial g}{\partial x}(s,X(s))AX(s) \\
&\quad  + \frac{1}{2}\operatorname{Tr}
\big[(BQ^{1/2})^* \frac{\partial^2 g}{\partial
x^2}(s,X(s))(BQ^{1/2}) \big] \Big] ds.
\end{align*}
The above equality is valid for all $t\in [\tau,T]$. Therefore
$$
\mathbb{E} \Big[ \frac{\partial g}{\partial t}(t,X(t)) +
\frac{\partial g}{\partial x}(t,X(t))AX(t) + \frac{1}{2}\operatorname{Tr}
\big[(BQ^{1/2})^* \frac{\partial^2 g}{\partial
x^2}(t,X(t))(BQ^{1/2}) \big] \Big] = 0.
$$
Rewrite this equality at the origin point $(\tau,x)$, as
$$
\mathbb{E} \Big[ \frac{\partial g}{\partial t}(\tau,x) +
\frac{\partial g}{\partial x}(\tau,x)Ax + \frac{1}{2}\operatorname{Tr}
\big[(BQ^{1/2})^* \frac{\partial^2 g}{\partial x^2}(\tau,x)(BQ^{1/2})
\big] \Big] = 0;
$$
that is,
$$
\mathbb{E} [ \frac{\partial g}{\partial t}(\tau,x)] +
\mathbb{E} [\frac{\partial g}{\partial x}(\tau,x)Ax ] +
\frac{1}{2}\mathbb{E} \big[\operatorname{Tr} [(BQ^{1/2})^* \frac{\partial^2
g}{\partial x^2}(\tau,x)(BQ^{1/2})] \big] = 0.
$$
Note that $Ax$ does not depend on $\omega$, thus
\[
\mathbb{E} \big[\frac{\partial g}{\partial x}(\tau,x)Ax \big]
= \mathbb{E} \big[\frac{\partial g}{\partial x}(\tau,x) \big] Ax.
\]
 Using fact that mappings
$\frac{\partial}{\partial t}$, $\frac{\partial}{\partial x}$ and
$\frac{\partial^2}{\partial x^2}$ are independent of the variable
$\omega$, and that the expectation $\mathbb{E}$ is an integral of
the variable $\omega$, we conclude that all these operators commute
with the operator $\mathbb{E}$. Furthermore, according to the
interpretation of trace given in section 2, it also commutes with
the operator $\mathbb{E}$ by following arguments:
\begin{align*}
\mathbb{E} \big[\operatorname{Tr} \big[(BQ^{1/2})^* \frac{\partial^2 g}{\partial
x^2}(\tau,x)(BQ^{1/2}) \big] \big]
&=\mathbb{E} \Big[
 \sum_{k=1}^\infty \langle (BQ^{1/2})^* \frac{\partial^2 g}{\partial
 x^2}(\tau,x)(BQ^{1/2})e_j, e_j \rangle \Big]
 \\
&=\big\langle \mathbb{E} \big[ (BQ^{1/2})^* \frac{\partial^2
g}{\partial x^2}(\tau,x)(BQ^{1/2})e_j \big], e_j \big\rangle
\\
&= \operatorname{Tr} \mathbb{E}
\big[ (BQ^{1/2})^* \frac{\partial^2 g}{\partial
x^2}(\tau,x)(BQ^{1/2})\big].
\end{align*}
Note
$$
\mathbb{E}[g(\tau,x)]
= \mathbb{E}[\mathbb{E}^{\tau,x}[h(X(T))]]
= \mathbb{E}\big[\mathbb{E}[h(X(T))]\big]
= \mathbb{E}[h(X(T))] = g(\tau,x).
$$
Hence we obtain
\begin{equation}\label{mpMain}
\frac{\partial g}{\partial \tau}(\tau,x) + \frac{\partial g}{\partial
x}(\tau,x)Ax + \frac{1}{2}\operatorname{Tr}
\big[(BQ^{1/2})^* \frac{\partial^2
g}{\partial x^2}(\tau,x)(BQ^{1/2}) \big] = 0.
\end{equation}
Varying $\tau \in [0,T]$ we obtain \eqref{mpMain} for all pairs
$\{(\tau,x): x=X(\tau)\}$. It remains to see that
$$
g(T,x) := \mathbb{E}^{T,x}[h(X(T))] = h(X(T))|_{x = X(T)} = h(x),
$$
which completes the proof.
\end{proof}

\begin{remark} \label{rmk1} \rm
Conditions on the function $g(t, x) = E^{t,x}[h(X(T))]$ for having  all
derivatives in the differential equation \eqref{mp4} are not usually
supplied by applications. In the case of function $h$ being not enough
smooth to ensure the conditions, some type of generalized problem \eqref{mp4}
have to be considered.

In \cite{Goldstein} in finite dimensional case and \cite{DaPrato}
in infinite dimensional case it was shown that $\{R_t, t\geq 0\}$
($R_t h(x) := \mathbb{E}^{0,x}[h(X(t))]$) forms a semigroup.
 According to this the Feynman-Kac theorem may be proved using the semigroup
techniques. Probably it will allow to weaken the conditions and it would
be the subject of our future research.
\end{remark}

\begin{remark} \label{rmk2} \rm
As we pointed out in the previous section $Q=I$ in the case of cylindrical
Wiener process. Further we will not focus on this fact and write equations
 with operator $Q$ for both types of Wiener processes.
\end{remark}

In the previous theorem we deduced backward Kolmogorov problem.
It is not some special feature of the relationship, we show that
the probability characteristic $\tilde{g}$ leads to the forward Kolmogorov problem.

\begin{theorem} \label{thm2}
Consider the stochastic Cauchy problem \eqref{mp3}
where $A$ is the generator of a $C_0$-semigroup in a Hilbert space $H$,
$B\in \mathcal{L}(\mathbb{H},H)$ in the case of $Q$-Wiener process $W$ and
$B\in \mathcal{L}_{\mathtt{HS}}(\mathbb{H},H)$ in the case of cylindrical
 Wiener process $W$.
Define $\tilde{g}(t,x) := \mathbb{E}^{0,x}[h(X(t))]: [0,T] \times H \to \mathbb{R}$,
$h$ is a measurable function from $H$ to $\mathbb{R}$. Suppose
$\mathbb{E}^{t,x}|h(X(T))| < \infty$, derivatives
$\frac{\partial \tilde{g}}{\partial t}$, $\frac{\partial \tilde{g}}{\partial x}$
and $\frac{\partial^2 \tilde{g}}{\partial x^2}$ exist for all pairs $(t,x)$.
Then $\tilde{g}(t,x)$ is a solution of the infinite dimensional forward Cauchy problem
\begin{equation}\label{mpforward}
\frac{\partial \tilde{g}}{\partial t}(t,x)
= \frac{\partial \tilde{g}}{\partial
x}(t,x)Ax + \frac{1}{2}\operatorname{Tr} [(BQ^{1/2})^* \frac{\partial^2
\tilde{g}}{\partial x^2}(t,x)(BQ^{1/2})] = 0, \quad \tilde{g}(0,x)=h(x).
\end{equation}
\end{theorem}

\begin{proof}
Let $s\in [0,T]$ be defined as $T=t+s$. Using \eqref{prop_homog},
homogeneity in time of diffusion process $X$, and definitions of $g$
and $\tilde{g}$ we have
$$
\tilde{g}(t,x)=\mathbb{E}^{0,x}[h(X(t))]=\mathbb{E}^{0,x}[h(X(T-s))]
=\mathbb{E}^{s,x}[h(X(T))] = g(s,x).
$$
So we have $\frac{\partial\tilde{g}}{\partial t}=-\frac{\partial g}{\partial s}$ and
$$
\frac{\partial \tilde{g}}{\partial s}(s,x)
= \frac{\partial \tilde{g}}{\partial x}(s,x)Ax
+ \frac{1}{2}\operatorname{Tr} \big[(BQ^{1/2})^* \frac{\partial^2
\tilde{g}}{\partial x^2}(s,x)(BQ^{1/2})\big] = 0.
$$
Boundary condition $g(T,x)=h(x)$ can be written in the form
$\tilde{g}(0,x)=h(x)$ that completes the proof.
\end{proof}

\section{From deterministic to stochastic}

The statement in the opposite direction is also valid.

\begin{theorem} \label{thm3}
Let $g$ be the solution of the deterministic Cauchy problem \eqref{mp4}
$$
\frac{\partial g}{\partial t}(t,x) + \frac{\partial g}{\partial
x}(t,x)Ax + \frac{1}{2}\operatorname{Tr}
\big[(BQ^{1/2})^* \frac{\partial^2
g}{\partial x^2}(t,x)(BQ^{1/2})\big] = 0, \quad g(T,x)=h(x),
$$
where $A$ is the generator of a $C_0$-semigroup in a Hilbert space $H$,
$B\in \mathcal{L}(\mathbb{H},H)$ or $B\in \mathcal{L}_{\mathtt{HS}}(\mathbb{H},H)$.
 Suppose process $X$ be a weak solution of the stochastic Cauchy problem \eqref{mp3}
$$
dX(t)=AX(t)dt + BdW(t), \quad t\in [0,T],\quad X(0)=\xi.
$$
where $W$ is an $\mathbb{H}$-valued $Q$-Wiener process in the case of
$B\in \mathcal{L}(\mathbb{H},H)$ and $W$ is a cylindrical Wiener process
in the case of $B\in \mathcal{L}_{\mathtt{HS}}(\mathbb{H},H)$.
Then the equality $g(t,x)=\mathbb{E}^{t,x}[h(X(T))]$ between solutions of
Cauchy problems \eqref{mp4} and \eqref{mp3} takes place.
\end{theorem}

\begin{proof}
Let $X$ be a solution of \eqref{mp3}. Fix some $\tau\in [0,T]$ and consider
the stochastic Cauchy problem with the initial data $x=X(\tau)$:
$$
dX(t)=AX(t)dt + B dW(t),\quad t\in[\tau,T],\quad X(\tau)=x.
$$
As we pointed out earlier the solution of this problem is the constriction
of the solution \eqref{mp3} on the segment $[\tau, T]$.
Applying the Ito formula in a Hilbert space to $g(t,X(t)),\,t\in[\tau,T]$,
 where $g$ is the solution for the Cauchy problem \eqref{mp4}, we have
\begin{align*}
g(T,X(T))
&= g(\tau,X(\tau))+\int_\tau^T \frac{\partial g}{\partial x}(s,X(s))BdW(s)
+\int_\tau^T \Big( \frac{\partial g}{\partial s}(s,X(s)) \\
&\quad  + \frac{\partial g}{\partial x}(s,X(s))AX(s)+\frac{1}{2}
\operatorname{Tr} \big[ (BQ^{1/2})^* \frac{\partial^2 g}{\partial x^2}(s,X(s))
(BQ^{1/2}) \big] \Big) ds.
\end{align*}
Since $g$ is the solution for the Cauchy problem \eqref{mp4}, we have
$$
g(T,X(T))=g(\tau,X(\tau)) + \int_\tau^T \frac{\partial g}{\partial x}(s,X(s))BdW(s).
$$
Take the mathematical expectation both sides of this equation. From the
definition of Ito integral (via the approximation in the mean square
by step processes) and properties of the $Q$-Wiener process, we
obtain $\mathbb{E}[ \int_\tau^T \frac{\partial g}{\partial x}(s,X(s))BdW(s) ] = 0$.
 Therefore,
$$
\mathbb{E}^{\tau,x}[g(T,X(T))]=\mathbb{E}^{\tau,x}[g(\tau,X(\tau))].
$$
Rewrite both sides of this equation. According the fact $T$ is the
 end point of time for the Cauchy problem \eqref{mp4} we obtain the
 reformulation of the left side for this equation
$$
\mathbb{E}^{\tau,x}[g(T,X(T))]=\mathbb{E}^{\tau,x}[h(X(T))].
$$
On the other hand we may rewrite the right side of this equation
$$
\mathbb{E}^{\tau,x}[g(\tau,X(\tau))]=\mathbb{E}^{\tau,x}[g(\tau,x)]=g(\tau,x).
$$
As a result we have
$$
\mathbb{E}^{\tau,x}[h(X(T))] = g(\tau,x).
$$
Varying $\tau$ we have this equality on the segment $[0,T]$;
this completes the proof.
\end{proof}

\subsection*{Acknowledgments}
This research was supported by the Program for State Support of Leading
 Universities of the Russian Federation (agreement no. 02.A03.21.0006 of
 August 27, 2013) and grant RFBR no. 13-01-00090.

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\end{document}