# On Function of Evolution of Distribution for Time Homogeneous Markov
Processes

Tomasz R. Bielecki111 Department of Applied Mathematics, Illinois Institute of
Technology, Chicago, IL 60616, USA e-mail<EMAIL_ADDRESS>Jacek
Jakubowski222 Institute of Mathematics, University of Warsaw, Banacha 2,
02-097 Warszawa, Poland<EMAIL_ADDRESS>and Maciej Wiśniewolski333
corresponding author: Institute of Mathematics, University of Warsaw, Banacha
2, 02-097 Warszawa, Poland<EMAIL_ADDRESS>
###### Abstract

A study of time homogeneous, real valued Markov processes with a special
property and a non-atomic initial distribution is provided. The new notion of
a function of evolution of distribution which determines the dependency
between one dimensional distributions of a process is introduced. This, along
with the notion of bridge operators which determine the backward structure, as
opposed to the forward structure determined by the usual semi-group operators,
paves a way to the new approach for dealing with finite-dimensional
distributions of Markov processes. This, in particular, produces explicit
formulas which effectively simplify the computations of finite-dimensional
distributions, giving an alternative to the standard approach based on
computations using the chain rule of transition densities. Various examples
illustrating the new approach are presented.

> Key words: Markov processes, function of evolution of distribution, Feller
> processes, bridge operator, conditional dependence, Kolmogorov distance
>
>
>
>
> AMS Subject Classification: 60J35, 60H30, 91G80.

##  1 Introduction

We study a class of Markov processes with a non-atomic initial distribution
and provide their special property. We introduce a new notion of a function of
evolution of distribution (in short FOED), which determines the dependency
between one dimensional distributions of the process. We develop an original
analysis based on the so called bridge operators which determine the backward
structure which is opposite to the forward structure determined by the usual
semi-group operators.

The usual distributional description of a Markov process relies on determining
the form of its transition density which by the classical chain rule gives the
formula for finite dimensional distributions of the process (see [25]). If the
transition density can not be determined explicitly the usual tool to find its
approximate form involves Fokker-Planck equations. Some relatively new studies
in this area can be find in [27], [19] and in case of diffusions the older
results are in [29].

Development of various techniques to determine finite dimensional
distributions of Markov processes has been motivated by their role in various
applications, in particular in finance. Surprisingly, the list of Markovian
models for which such distributions have not yet been described explicitly is
still quite vast. These include the so called stochastic volatility models,
which are systems of stochastic differential equations of various types ([7],
[8] [22], [2]). While studying a class of stochastic volatility models
Jakubowski and Wiśniewolski ([16], [17]) discovered phenomena which we present
and study in this work. Specifically, for Markov process with some non-trivial
initial distribution we present a new way to describe the relations between
its finite dimensional distributions. This is done by introduction of a
function, which we call the function of evolution of distribution, and a
related property which we call the FOED property. We derive several important
consequences stemming from the FOED property. It turns out that for the class
of Feller processes having the FOED property a one-dimensional distribution
can be determined explicitly in terms of the initial distribution and the
generator. There is no need to look for the transition density.

The FOED property for a stochastic process is strictly connected to the Markov
property, and in general to properties of Markov families. The notion of FOED
leads to a new backward approach to analysis of a Markov process in terms of
so called bridge operators. Instead of looking at the starting point and
continue with classical chain rule, the bridge operators pave a way to new
iteration-type formulas describing the finite dimensional distributions of
Markov processes. It turns out that under some assumptions about the
underlying Markov process the computations of finite dimensional distributions
of the process can be reduced to dealing with independent copies of the
initial distribution of this process. Last but not least, the bridge operators
and the FOED property deliver a new description of conditional distributions
of Markov processes.

The paper is organized as follows. In Section 2 we define the FOED property,
we present properties of processes with the FOED property and we give
illustrating examples. In Section 3 we investigate Feller processes with the
FOED property. In particular, we prove that a one-dimensional distribution can
be determined explicitly in terms of the initial distribution and the
infinitesimal generator of the process. In Section 4 we define a bridge
operator and we observe that the finite dimensional distribution of a Markov
process $X$, enjoying the FOED property, may be determined in terms of bridge
operators and the FOED property. This gives a new algorithmic approach to the
computational problems related to finite dimensional distributions of a Markov
process. The algorithm uses the initial distribution of $X$ and a form of
bridge operator. Using our methodology, in Section 5 we compute the
expectation
$\mathbb{E}\big{(}\Psi\big{(}X_{0},\ldots,X_{t_{n}}\big{)}\big{|}X_{s+t_{1}+\ldots
t_{n}}=z\big{)}$ for a Markov process $X$ with the FOED property, a bounded
Borel function $\Psi$, the discrete time tenor $0<t_{1}<\ldots<t_{n}$ and
$s\geq 0$. In Section 6 we present an application of the FOED property to
finding the Kolmogorov distance between $B_{t}$ and $X_{t}$, where $B$ is a
Brownian motion with the initial distribution $\gamma$ and $X$ is a Markov
process having the FOED property and the initial distribution $\gamma$.

##  2 Function of Evolution of Distribution

We consider a complete probability space $(\Omega,\mathcal{F},\mathbb{P})$. On
this probability space we consider a cádlág process $X$ with values in a
Polish space $(E,\mathcal{E})$ and with the initial non-atomic distribution
$\gamma$. In examples we usually have $E\subset\mathbb{R}^{d}$. The assumption
about non-atomic $\gamma$ is not restrictive since for a time-homogeneous
Markov process $X$ the process defined by $(Y_{t})=(X_{\epsilon+t})$,
$\epsilon>0$ suits our setup. By $\mathbb{F}^{X}=(\mathcal{F}^{X}_{t})_{t\geq
0}$ we denote the filtration generated by $X$ and augmented to satisfy usual
conditions. When we say that $X$ is a Markov process, we understand that $X$
is a Markov process with respect to $\mathbb{F}^{X}$. We assume that the shift
operator $\theta$ is well defined, so $X_{s}\circ\theta_{t}=X_{t+s}$.We start
with the notion of a function of evolution of distribution.

######  Definition 2.1.

We say that process $X$ admits the FOED property if there exists a Borel
measurable function $F:R_{+}\times E\mapsto R$ such that for every $y\in E$
the function $u\mapsto F(u,y)$ is locally integrable and

$\displaystyle\mathbb{E}f(X_{t})=\mathbb{E}\big{(}f(X_{0})\,e^{\int_{0}^{t}F(u,X_{0})du}\big{)},$
(1)

for every bounded Borel function $f$ and every $t\geq 0$. We call such $F$ the
function of evolution of distribution for $X$.

Please note that distribution of the random variable $X_{t}$ is determined by
values of $\mathbb{E}f(X_{t})$ for bounded Borel functions $f$, or by values
of $\mathbb{E}f(X_{t})$ for positive Borel functions $f$. Thus, we may replace
in the above definition the class of bounded Borel functions $f$ with the
class of positive Borel functions $f$.

Moreover, if such $F$ exists then it is uniquely determined on the support of
distribution of $X_{0}$. Also, it should be noted that in general $F$ depends
on the initial distribution $\gamma$, but for simplicity of notation we do not
explicitly exhibit this dependence. Let observe however that the notion of the
function of evolution of distribution for $X$ is trivial, ergo not useful, in
case of some initial distributions of $X$. For example, if $X_{0}=x$ and $X$
possesses the FOED property, then $X_{t}=x$ a.s. for every $t$. Indeed, taking
$f\equiv 1$ shows that $e^{\int_{0}^{t}F(u,x)}=1$ for every $t$ and (1)
implies $\mathbb{E}f(X_{t})=f(x)$ for every bounded Borel function $f$. For a
nontrivial example, where the FOED property is not useful, it is enough to
consider a $X_{t}=X_{0}+B_{t}$, where $B$ is a standard Brownian motion and
$X_{0}$ is a random variable independent of $B$, uniformly distributed on
$[0,1]$.

As usual, given a process $X$, then by $\mathbb{P}_{x}$ for $x\in E$ we denote
the probability measure defined on $\mathcal{F}$ by

$\displaystyle\mathbb{P}_{x}(A)=\mathbb{P}(A|X_{0}=x),\quad A\in\mathcal{F},$

where the right hand side should be understood in the sense of the regular
conditional probabilities.

The following example is a result of an earlier study, done by Jakubowski and
Wiśniewolski in [17], of a class of SDEs that are widely used in finance. It
turns out that processes given in terms of these SDEs posses the FOED
property. Actually, the paper [17] initiated the study of the FOED property,
motivation for which was the desire to simplify complexity of calculations
that traditionally plagued applications of the stochastic volatility models
(as documented in, e.g., [9], [11], [12], [18], [24], [28]).

######  Example 2.2.

Consider a linear stochastic volatility model, where we assume that the price
$X_{t}$ at time $t$ of some underlying financial asset has a stochastic
volatility $Y_{t}$, which has the dynamics given by a time dependent SDE, so
the dynamics of the vector $(X,Y)$ is given by

$\displaystyle dX_{t}$ $\displaystyle=Y_{t}X_{t}dW_{t},$ (2) $\displaystyle
dY_{t}$ $\displaystyle=\mu(t,Y_{t})\ dt+\sigma(t,Y_{t})dZ_{t},$ (3)

with $X_{0}=x$, $Y_{0}=y$ which are positive constants. Here the processes
$W,Z$ are correlated Brownian motions, $d{\left\langle
W,Z\right\rangle}_{t}=\rho dt$ with $\rho\in(-1,1)$, and
$\mu:\mathbb{R}_{+}\times\mathbb{R}_{+}\rightarrow\mathbb{R}$,
$\sigma:\mathbb{R}_{+}\times\mathbb{R}_{+}\rightarrow\mathbb{R}$ are
continuous functions such that there exists a unique positive strong solution
of (3). It turns out that for such a process $X$ and independent of it gamma
random variable $\gamma$ with parameters $\alpha=1/2$ and $\beta=1$, i.e.
random variable with density $g(x)=\frac{1}{\sqrt{\pi x}}e^{-x}$ we have

$\displaystyle
d\Big{(}\frac{X_{t}^{2}}{4\gamma}\Big{)}=2\Big{(}\frac{X_{t}^{2}}{4\gamma}\Big{)}Y_{t}dW_{t}+\Big{(}\frac{X_{t}^{2}}{4\gamma}\Big{)}Y_{t}^{2}dt$

and

$\mathbb{P}\Big{(}\frac{X_{t}^{2}}{4\gamma}\in
dz\Big{)}=\mathbb{P}\Big{(}\frac{x^{2}}{4\gamma}\in
dz\Big{)}e^{\int_{0}^{t}F(u,z)du}$

([17, Thm. 3.5, Corr. 3.6]). This means that process
$\frac{X_{t}^{2}}{4\gamma}$ has the FOED property. The use of the FOED
property also gives

$\displaystyle\mathbb{E}\Big{(}X_{t}e^{-\lambda
X_{t}^{2}}\Big{)}=x\exp\Big{(}-\lambda
x^{2}+\int_{0}^{t}F(u,1/4\lambda)du\Big{)},\quad\lambda>0,$

where $F$ is the function of evolution of distribution of
$\frac{X_{t}^{2}}{4\gamma}$ ([17, Prop.3.7]). This shows that determining $F$
gives the explicit formula for a (slightly transformed) Laplace transform of
the random variable $X_{t}^{2}$.

Since we are interested in time homogenous Markov processes we start with a
proposition giving a broad class of such processes admitting the FOED property
for each initial distribution with strictly positive density. In particular,
the proposition gives a way to construct the FOED function $F$. By
$\mathcal{C}^{1}[0,\infty)$ we understand the class of functions defined on
$[0,\infty)$ which belong to $\mathcal{C}^{1}(0,\infty)$ and have the right-
continuous derivative at $0$.

######  Proposition 2.3.

Let $X$ be a Markov process with the transition density $p$ with respect to
the Lebesgue measure such that $t\mapsto
p_{t}(x,y)\in\mathcal{C}^{1}[0,\infty)$ for $x,y\in E$. Assume that $X$ has
the initial distribution with the strictly positive density $g$ (with respect
to the Lebesgue measure). Then $X$ has the FOED property and

$e^{\int_{0}^{t}F(u,x)du}=\frac{\int_{E}p_{t}(z,x)g(z)dz}{g(x)},\
\textrm{for}\ t\geq 0.$ (4)

###### Proof.

Let $\phi(t,x)=\int_{E}p_{t}(z,x)g(z)dz,\ t>0$, $\phi(0,x)=g(x)$. Put

$\displaystyle F(t,x)=\frac{\partial\phi(t,x)}{\partial t},\qquad t>0,$ (5)

and $F(0,x)=\frac{1}{g(x)}\frac{\partial\phi(t,x)}{\partial t}|_{t=0^{+}}$.
Then for any $t>0$ and $x\in E$ we have

$\displaystyle e^{\int_{0}^{t}F(u,x)du}=\frac{\phi(t,x)}{g(x)},$ (6)

which implies (1), so $X$ has the FOED property with function $F$ satisfying
(4). ∎

######  Example 2.4.

Let $X$ be as in Proposition 2.3 and $X_{0}=x\in E$. Let $s>0$ be such that
the distribution of $X_{s}$ has the strictly positive density $g_{s}$. For
this $s>0$, we define a new process $Y_{t}=X_{t+s}$, $t\geq 0$. Then the
initial distribution of $Y$ is $\gamma(dz)=g_{s}(z)dz$. It turns out that
under the above assumptions the process $Y$ has the FOED property. Indeed, for
any bounded Borel $f$ on $E$ and $t>0$

$\displaystyle\mathbb{E}f(Y_{t})$
$\displaystyle=\mathbb{E}\mathbb{E}_{X_{s}}f(X_{t})=\int_{E\times
E}f(z)p_{t}(v,z)g_{s}(v)dvdz=\int_{E}f(z)e^{\int_{0}^{t}F(u,z)du}g_{s}(z)dz,$

where, in line with the proof of Proposition 2.3, we take $F$ such that

$\displaystyle
e^{\int_{0}^{t}F(u,z)du}=\frac{\int_{E}p_{t}(v,z)g_{s}(v)dv}{g_{s}(z)}.$

Other examples of processes with the FOED property will be given later.

In the sequel we will use the following notation

$\displaystyle\mu_{t}(dz)=\mathbb{P}(X_{t}\in dz),\quad t>0.$

The next proposition gives two immediate consequences of the FOED property.

######  Proposition 2.5.

Suppose that $X$ has the FOED property. Then,

a) For $0\leq s<t$ and for every positive Borel function $f$

$\mathbb{E}f(X_{t})=\mathbb{E}\big{(}f(X_{0})\,e^{\int_{s}^{t}F(u,X_{0})du}e^{\int_{0}^{s}F(u,X_{0})du}\big{)}=\mathbb{E}\big{(}f(X_{s})\,e^{\int_{s}^{t}F(u,X_{s})du}\big{)}.$

b) The function $F$ determines uniquely the density of
$\frac{d\mu_{t}}{d\mu_{s}}$, namely

$\displaystyle\mu_{t}(dz)=\mu_{s}(dz)e^{\int_{s}^{t}F(u,z)du},$ (7)

in particular $\mu_{t}(dz)=\gamma(dz)e^{\int_{0}^{t}F(u,z)du}$, so $E$ is the
support of $\mu_{t}$ provided that $E$ is the support of $\mu_{0}$. This means
the FOED property implies absolute continuity of one dimensional distributions
of $X$.

Another immediate consequence of the FOED property is

$\displaystyle\mathbb{E}\big{(}f(X_{t})e^{-\int_{0}^{t}F(u,X_{t})du}\big{)}=\mathbb{E}f(X_{0}).$
(8)

Taking $f\equiv 1$ we obtain

$\displaystyle\mathbb{E}e^{-\int_{0}^{t}F(u,X_{t})du}=1.$

######  Theorem 2.6.

Assume that process $X$ on $(\Omega,\mathcal{F},\mathbb{P})$ has the FOED
property and the inital distribution with strictly positive density. The
following statements are equivalent:
(i) For each $\mathcal{F}^{X}_{\infty}$ measurable positive or bounded random
variable $Z$ and for each $t>0$ and $x,y\in E$ it holds

$\displaystyle\mathbb{E}(Z\circ\theta_{t})=\mathbb{E}Ze^{\int_{0}^{t}F(u,X_{0})du},\qquad\mathbb{E}(Z\circ\theta_{t}|X_{t}=x,X_{0}=y)=\mathbb{E}(Z\circ\theta_{t}|X_{t}=x).$

(ii) $X$ is a Markov process with respect to $\mathbb{F}^{X}$.

###### Proof.

$(i)\Rightarrow(ii)$ It is enough to prove for arbitrary $n\in\mathbb{N}$,
$0\leq t_{1}<t_{2}<\ldots<t_{n}<t$ and a positive, Borel function $f$ on $E$
that

$\displaystyle\mathbb{E}[f(X_{t})|X_{t_{1}},\ldots,X_{t_{n}}]=\mathbb{E}_{X_{t_{n}}}f(X_{t-t_{n}}).$
(9)

Indeed, from (9) using the monotone class theorem we see that

$\displaystyle\mathbb{E}_{X_{t_{n}}}[f(X_{t-t_{n}})]=\mathbb{E}[f(X_{t})|\mathcal{F}^{X}_{t_{n}}].$

Hence for any $A\in\mathcal{F}^{X}_{t_{n}}$

$\displaystyle\mathbb{E}_{X_{0}}[f(X_{t})1_{A}]$
$\displaystyle=\mathbb{E}[f(X_{t})1_{A}|X_{0}]=\mathbb{E}[\mathbb{E}[f(X_{t})1_{A}|X_{0}]|\mathcal{F}^{X}_{t_{n}}]=\mathbb{E}[1_{A}\mathbb{E}[f(X_{t})|\mathcal{F}^{X}_{t_{n}}]|X_{0}]$
$\displaystyle=\mathbb{E}[1_{A}\mathbb{E}_{X_{t_{n}}}[f(X_{t-t_{n}})]|X_{0}]=\mathbb{E}_{X_{0}}[1_{A}\mathbb{E}_{X_{t_{n}}}[f(X_{t-t_{n}})]],$

which yields for every $x\in E$

$\displaystyle\mathbb{E}_{x}[f(X_{t})1_{A}]=\mathbb{E}_{x}[1_{A}\mathbb{E}_{X_{t_{n}}}[f(X_{t-t_{n}})]].$
(10)

Hence we obtain for $s\leq t$ that

$\mathbb{P}_{x}(X_{t}\in A|\mathcal{F}_{s})=\mathbb{P}_{X_{s}}(X_{t-s}\in
A)\quad\mathbb{P}_{x}\ a.s.$

Integrating both sides with respect to the initial distribution yields

$\mathbb{P}(X_{t}\in A|\mathcal{F}_{s})=\mathbb{P}_{X_{s}}(X_{t-s}\in
A)\quad\mathbb{P}\ a.s.$

showing that $X$ is a Markov process.

So it is enough to prove (9). We start with showing that for any
$\mathcal{F}^{X}_{\infty}$ measurable, positive or bounded random variable $Z$
we have

$\displaystyle\mathbb{E}_{x}Z=\mathbb{E}[Z\circ\theta_{v}|X_{v}=x],\qquad x\in
E,\ v\geq 0.$ (11)

Let $f$ and $g$ be positive, Borel functions on $E$ and $t,v\in[0,\infty)$.
Define $h$ by formula

$h(x,v,t+v)=\mathbb{E}(f(X_{t+v})|X_{v}=x).$

The function $h$ is well defined since $E$ is, by assumption, the support of
distribution of $X_{u}$ for each $u\geq 0$. From the definition of FOED we
have

$\displaystyle\mathbb{E}[f(X_{t+v})g(X_{v})]=\mathbb{E}[h(X_{v},v,t+v)g(X_{v})]=\mathbb{E}\big{(}h(X_{0},v,t+v)g(X_{0})e^{\int_{0}^{v}F(u,X_{0})du}\big{)}.$
(12)

Note that for $Z=f(X_{t})g(X_{0})$ we have that
$Z\circ\theta_{v}=f(X_{t+v})g(X_{v})$. Thus, if $(i)$ holds, then

$\displaystyle\mathbb{E}[f(X_{t+v})g(X_{v})]$
$\displaystyle=\mathbb{E}\big{(}f(X_{t})g(X_{0})e^{\int_{0}^{v}F(u,X_{0})du}\big{)}.$
(13)

Comparing (12) and (13) we see that (almost surely) $f(X_{t})=h(X_{0},v,t+v)$,
so that

$\displaystyle\mathbb{E}[f(X_{t})|X_{0}=x]=h(x,v,t+v),$

and so, actually, $h(x,0,t)=h(x,v,t+v)$. Therefore

$\displaystyle\mathbb{E}_{x}f(X_{t})=\mathbb{E}[f(X_{t+v})|X_{v}=x].$ (14)

Proceeding in the similar way we show (11). It is enough to prove (11) for
$Z=\Pi_{i=1}^{n}f_{i}(X_{t_{i}})$, where $n\in\mathbb{N}$, $0\leq
v<t_{1}<t_{2}<\ldots<t_{n}$ and $f_{i}$, $i=1,\ldots,n$, are positive Borel
functions, and then use the monotone class theorem. Let $\widehat{h}$ be
defined by

$\displaystyle\widehat{h}(x,v,t_{1},\ldots,t_{n})=\mathbb{E}\Big{[}\Pi_{i=1}^{n}f_{i}(X_{t_{i}})|X_{v}=x\Big{]}.$

For a positive, Borel function $g$ we have

$\displaystyle\mathbb{E}\Big{[}g(X_{v})\Pi_{i=1}^{n}f_{i}(X_{t_{i}})\Big{]}$
$\displaystyle=\mathbb{E}\Big{[}g(X_{v})\widehat{h}(X_{v},v,t_{1},\ldots,t_{n})\Big{]}$
(15)
$\displaystyle=\mathbb{E}\Big{[}g(X_{0})\widehat{h}(X_{0},v,t_{1},\ldots,t_{n})e^{\int_{0}^{v}F(u,X_{0})du}\Big{]},$

where in the last equality we used the definition of FOED. On the other hand
by $(i)$

$\displaystyle\mathbb{E}\Big{[}g(X_{v})\Pi_{i=1}^{n}f_{i}(X_{t_{i}})\Big{]}$
$\displaystyle=\mathbb{E}\Big{[}\Big{(}g(X_{0})\Pi_{i=1}^{n}f_{i}(X_{t_{i}-v})\Big{)}\circ\theta_{v}\Big{]}$
(16)
$\displaystyle=\mathbb{E}\Big{[}g(X_{0})\mathbb{E}\Big{[}\Pi_{i=1}^{n}f_{i}(X_{t_{i}-v})|X_{0}\Big{]}e^{\int_{0}^{v}F(u,X_{0})du}\Big{]}.$

So comparing (15) and (16) we have for any $x\in E$

$\displaystyle\widehat{h}(x,v,t_{1},\ldots,t_{n})=\mathbb{E}\Big{[}\Pi_{i=1}^{n}f_{i}(X_{t_{i}-v})|X_{0}=x\Big{]}=\mathbb{E}_{x}\Big{[}\Pi_{i=1}^{n}f_{i}(X_{t_{i}-v})\Big{]}.$

As a result, recalling the definition of $\widehat{h}$ and using the last
equality we obtain

$\displaystyle\mathbb{E}\Big{[}\Pi_{i=1}^{n}f_{i}(X_{t_{i}})|X_{v}=x\Big{]}=\mathbb{E}_{x}\Big{[}\Pi_{i=1}^{n}f_{i}(X_{t_{i}-v})\Big{]},$

and (11) follows for $Z=\Pi_{i=1}^{n}f_{i}(X_{t_{i}})$.

We now show (9). Let

$\displaystyle
h_{1}(t,t_{1},\ldots,t_{n},x_{1},\ldots,x_{n})=\mathbb{E}[f(X_{t})|X_{t_{1}}=x_{1},\ldots,X_{t_{n}}=x_{n}].$

We see from $(i)$ that for arbitrary positive Borel functions
$f_{1},\ldots,f_{n}$

$\displaystyle\mathbb{E}$ $\displaystyle[f(X_{t})f_{1}(X_{t_{1}})\ldots
f_{n}(X_{t_{n}})]=\mathbb{E}[h_{1}(t,t_{1},\ldots,t_{n},X_{t_{1}},\ldots,X_{t_{n}})f_{1}(X_{t_{1}})\ldots
f_{n}(X_{t_{n}})]$ (17)
$\displaystyle=\mathbb{E}f_{1}(X_{0})e^{\int_{0}^{t_{1}}F(X_{0},u)du}h_{1}(t,t_{1},\ldots,t_{n},X_{0},X_{t_{2}-t_{1}},\ldots,X_{t_{n}-t_{1}})\Pi_{i=2}^{n}f_{i}(X_{t_{i}-t_{1}})$
$\displaystyle=\mathbb{E}\Big{[}f_{1}(X_{0})e^{\int_{0}^{t_{1}}F(X_{0},u)du}\mathbb{E}_{X_{0}}\Big{(}h_{1}(t,t_{1},\ldots,t_{n},X_{0},X_{t_{2}-t_{1}},\ldots,X_{t_{n}-t_{1}})\Pi_{i=2}^{n}f_{i}(X_{t_{i}-t_{1}})\Big{)}\Big{]},$

where in the second equality we used identity (11). On the other hand defining
$\widetilde{h}$ by

$\displaystyle\widetilde{h}(t,t_{1},t_{2},\ldots,t_{n},x)=\mathbb{E}\Big{[}f(X_{t})\Pi_{i=2}^{n}f_{i}(X_{t_{i}})|X_{t_{1}}=x\Big{]}$

we have

$\displaystyle\mathbb{E}[f(X_{t})f_{1}(X_{t_{1}})\ldots f_{n}(X_{t_{n}})]$
$\displaystyle=\mathbb{E}\Big{[}f_{1}(X_{t_{1}})\widetilde{h}(t,t_{1},t_{2},\ldots,t_{n},X_{t_{1}})\Big{]}$
(18)
$\displaystyle=\mathbb{E}\Big{[}f_{1}(X_{0})\widetilde{h}(t,t_{1},t_{2},\ldots,t_{n},X_{0})e^{\int_{0}^{t_{1}}F(X_{0},u)du}\Big{]},$

where in the last equality we used the definition of FOED. As a result
comparing (17) and (18) we obtain for any $x_{1}\in E$

$\displaystyle\widetilde{h}(t,t_{1},t_{2},\ldots,t_{n},x_{1})=\mathbb{E}_{x_{1}}\Big{(}h_{1}(t,t_{1},\ldots,t_{n},x_{1},X_{t_{2}-t_{1}},\ldots,X_{t_{n}-t_{1}})\Pi_{i=2}^{n}f_{i}(X_{t_{i}-t_{1}})\Big{)}.$
(19)

Observe now that by (11), definition of $\widetilde{h}$, and then (19) we have

$\displaystyle\mathbb{E}_{x_{1}}\Big{[}f(X_{t-t_{1}})\Pi_{i=2}^{n}f_{i}(X_{t_{i}-t_{1}})\Big{]}=\mathbb{E}\Big{[}f(X_{t})\Pi_{i=2}^{n}f_{i}(X_{t_{i}})|X_{t_{1}}=x_{1}\Big{]}=\widetilde{h}(t,t_{1},t_{2},\ldots,t_{n},x_{1})$
$\displaystyle=\mathbb{E}_{x_{1}}\Big{(}h_{1}(t,t_{1},\ldots,t_{n},x_{1},X_{t_{2}-t_{1}},\ldots,X_{t_{n}-t_{1}})\Pi_{i=2}^{n}f_{i}(X_{t_{i}-t_{1}})\Big{)}.$
(20)

We write the left-hand side of (20) as follows:

$\displaystyle\mathbb{E}_{x_{1}}\Big{[}f(X_{t-t_{1}})\Pi_{i=2}^{n}f_{i}(X_{t_{i}-t_{1}})\Big{]}=\mathbb{E}_{x_{1}}\Big{(}f_{2}(X_{t_{2}-t_{1}})\mathbb{E}_{x_{1}}\Big{[}f(X_{t-t_{1}})\Pi_{i=3}^{n}f_{i}(X_{t_{i}-t_{1}})|X_{t_{2}-t_{1}}\Big{]}\Big{)}.$

On the other hand, the right-hand side of (20) is

$\displaystyle\mathbb{E}_{x_{1}}$
$\displaystyle\Big{(}h_{1}(t,t_{1},\ldots,t_{n},x_{1},X_{t_{2}-t_{1}},\ldots,X_{t_{n}-t_{1}})\Pi_{i=2}^{n}f_{i}(X_{t_{i}-t_{1}})\Big{)}$
$\displaystyle=\mathbb{E}_{x_{1}}\Big{(}f_{2}(X_{t_{2}-t_{1}})\mathbb{E}_{x_{1}}\Big{[}h_{1}(t,t_{1},\ldots,t_{n},x_{1},X_{t_{2}-t_{1}},\ldots,X_{t_{n}-t_{1}})\Pi_{i=3}^{n}f_{i}(X_{t_{i}-t_{1}})|X_{t_{2}-t_{1}}\Big{]}\Big{)}.$

So from (20) we have for any $x_{2}\in E$

$\displaystyle\mathbb{E}_{x_{1}}$
$\displaystyle\Big{[}f(X_{t-t_{1}})\Pi_{i=3}^{n}f_{i}(X_{t_{i}-t_{1}})|X_{t_{2}-t_{1}}=x_{2}\Big{]}$
(21)
$\displaystyle=\mathbb{E}_{x_{1}}\Big{[}h_{1}(t,t_{1},\ldots,t_{n},x_{1},x_{2},\ldots,X_{t_{n}-t_{1}})\Pi_{i=3}^{n}f_{i}(X_{t_{i}-t_{1}})|X_{t_{2}-t_{1}}=x_{2}\Big{]}.$

Observe that the last equality may be rewritten as

$\displaystyle\mathbb{E}$
$\displaystyle\Big{[}\Big{(}f(X_{t-t_{2}})\Pi_{i=3}^{n}f_{i}(X_{t_{i}-t_{2}})\Big{)}\circ\theta_{t_{2}-t_{1}}\big{|}X_{t_{2}-t_{1}}=x_{2},X_{0}=x_{1}\Big{]}$
$\displaystyle=\mathbb{E}\Big{[}\Big{(}h_{1}(t,t_{1},\ldots,t_{n},x_{1},x_{2},X_{t_{3}-t_{2}},\ldots,X_{t_{n}-t_{2}})\Pi_{i=3}^{n}f_{i}(X_{t_{i}-t_{2}})\Big{)}\circ\theta_{t_{2}-t_{1}}\big{|}X_{t_{2}-t_{1}}=x_{2},X_{0}=x_{1}\Big{]},$

so by the second condition in (i) we get

$\displaystyle\mathbb{E}$
$\displaystyle\Big{[}\Big{(}f(X_{t-t_{2}})\Pi_{i=3}^{n}f_{i}(X_{t_{i}-t_{2}})\circ\theta_{t_{2}-t_{1}}\Big{)}\big{|}X_{t_{2}-t_{1}}=x_{2}\Big{]}$
$\displaystyle=\mathbb{E}\Big{[}\Big{(}h_{1}(t,t_{1},\ldots,t_{n},x_{1},x_{2},X_{t_{3}-t_{2}},\ldots,X_{t_{n}-t_{2}})\Pi_{i=3}^{n}f_{i}(X_{t_{i}-t_{2}})\Big{)}\circ\theta_{t_{2}-t_{1}}\big{|}X_{t_{2}-t_{1}}=x_{2}\Big{]}.$

Using (11) we obtain

$\displaystyle\mathbb{E}_{x_{2}}\Big{[}$ $\displaystyle
f(X_{t-t_{2}})\Pi_{i=3}^{n}f_{i}(X_{t_{i}-t_{2}})\Big{]}$
$\displaystyle=\mathbb{E}_{x_{2}}\Big{[}h_{1}(t,t_{1},\ldots,t_{n},x_{1},x_{2},X_{t_{3}-t_{2}},\ldots,X_{t_{n}-t_{2}})\Pi_{i=3}^{n}f_{i}(X_{t_{i}-t_{2}})\Big{]}.$

Repeating the last procedure $(n-3)$ times, we obtain

$\displaystyle\mathbb{E}_{x_{n}}\Big{[}f(X_{t-t_{n}})\Big{]}=h_{1}(t,t_{1},\ldots,t_{n},x_{1},x_{2},\ldots,x_{n}),$

which after recalling the definition of $h$ yields

$\displaystyle\mathbb{E}_{X_{t_{n}}}\Big{[}f(X_{t-t_{n}})\Big{]}=\mathbb{E}[f(X_{t})|X_{t_{1}},\ldots,X_{t_{n}}].$

This proves (9). The proof of implication $(i)\Rightarrow(ii)$ is finished.

$(ii)\Rightarrow(i)$ The second condition in (i) follows directly from the
Markov property, so we only need to show the first one. Using the argument of
monotone-class theorem it is enough to prove that

$\displaystyle\mathbb{E}\big{(}\Pi_{i=1}^{n}f_{i}(X_{t_{i}+t})\big{)}=\mathbb{E}\Big{(}\big{(}\Pi_{i=1}^{n}f_{i}(X_{t_{i}})\big{)}\circ\theta_{t}\Big{)}=\mathbb{E}\big{(}\Pi_{i=1}^{n}f_{i}(X_{t_{i}})e^{\int_{0}^{t}F(X_{0},u)du}\big{)}$
(22)

for arbitrary $0<t_{1}<\ldots<t_{n}$ and $f_{1},\ldots,f_{n}$ positive Borel
functions on $E$.

The proof goes by induction. For $n=1$ the assertion follows directly by the
definition of FOED. Indeed, using Markov and FOED properties we obtain that

$\displaystyle\mathbb{E}$ $\displaystyle
f_{1}(X_{t_{1}+t})=\mathbb{E}\mathbb{E}\big{(}f_{1}(X_{t_{1}+t})|\mathcal{F}^{X}_{t}\big{)}=\mathbb{E}\mathbb{E}_{X_{t}}f_{1}(X_{t_{1}})$
$\displaystyle=\mathbb{E}\big{(}e^{\int_{0}^{t}F(X_{0},u)du}\mathbb{E}_{X_{0}}f_{1}(X_{t_{1}})\big{)}=\mathbb{E}\mathbb{E}\big{(}e^{\int_{0}^{t}F(u,X_{0})du}f_{1}(X_{t_{1}})|X_{0}\big{)}=\mathbb{E}e^{\int_{0}^{t}F(u,X_{0})du}f_{1}(X_{t_{1}}).$

Assume that the assertion holds for $0<t_{1}<\ldots<t_{n-1}$ and $(n-1)$
positive, Borel functions. We conclude from the Markov property that

$\displaystyle\mathbb{E}\Pi_{i=1}^{n}f_{i}(X_{t_{i}+t})$
$\displaystyle=\mathbb{E}\mathbb{E}\Big{(}\Pi_{i=1}^{n}f_{i}(X_{t_{i}+t})|\mathcal{F}^{X}_{t_{n-1}+t}\Big{)}=\mathbb{E}\Big{(}\Pi_{i=1}^{n-1}f_{i}(X_{t_{i}+t})\mathbb{E}_{X_{t_{n-1}+t}}f_{n}(X_{t_{n}-t_{n-1}})\Big{)}.$

From the induction assumption the last expression is equal to

$\displaystyle\mathbb{E}\Big{(}\Pi_{i=1}^{n-1}f_{i}(X_{t_{i}})\Big{(}\mathbb{E}_{X_{t_{n-1}}}f_{n}(X_{t_{n}-t_{n-1}})\Big{)}e^{\int_{0}^{t}F(u,X_{0})du}\Big{)},$

which from the Markov property is equal to

$\displaystyle\mathbb{E}\Big{(}\Pi_{i=1}^{n-1}f_{i}(X_{t_{i}})\mathbb{E}\big{(}f_{n}(X_{t_{n}-t_{n-1}})|\mathcal{F}^{X}_{t_{n-1}}\big{)}e^{\int_{0}^{t}F(u,X_{0})du}\Big{)}=\mathbb{E}\Big{(}\Pi_{i=1}^{n}f_{i}(X_{t_{i}})e^{\int_{0}^{t}F(u,X_{0})du}\Big{)}.$

This gives (22). The proof is complete. ∎

For an $\mathcal{F}^{X}_{\infty}$ measurable positive or bounded random
variable $Z$ we have the following representation theorem.

######  Theorem 2.7.

Assume that process $X$ on $(\Omega,\mathcal{F},\mathbb{P})$ has the FOED
property and the inital distribution with strictly positive density such that
$\mathbb{E}|F(t,X_{0})|<\infty$ for every $t\geq 0$. If
$(X,\mathbb{P}_{x})_{x\in E}$ is a Markov family with respect to
$\mathbb{F}^{X}$, then for an $\mathcal{F}^{X}_{\infty}$-measurable positive
or bounded random variable $Z$ we have

$\displaystyle\mathbb{E}(Z\circ\theta_{t})=\mathbb{E}Z+\mathbb{E}\int_{0}^{t}(Z\circ\theta_{r})F(r,X_{r})dr.$

###### Proof.

Using the assumption $\mathbb{E}|F(t,X_{0})|<\infty$ we conclude from Theorem
2.6 for $r>0$ and $\widetilde{Z}(r)=F(r,X_{0})Z$ that

$\displaystyle\mathbb{E}\big{(}\widetilde{Z}(r)\circ\theta_{r}\big{)}=\mathbb{E}\widetilde{Z}(r)e^{\int_{0}^{r}F(X_{0},u)du}.$

Since $F(\cdot,y)$ is locally integrable, integrating both sides with respect
to $r$ and using Fubini’s theorem yield for any $t>0$

$\displaystyle\mathbb{E}\int_{0}^{t}\widetilde{Z}(r)\circ\theta_{r}dr=\mathbb{E}\int_{0}^{t}\widetilde{Z}(r)e^{\int_{0}^{r}F(X_{0},u)du}dr.$

Hence, upon noting that
$(\widetilde{Z}(r)\circ\theta_{r})(\omega)=\widetilde{Z}(r)(\theta_{r}(\omega))=F(r,X_{0}(\theta_{r}(\omega)))Z(\theta_{r}(\omega))=(Z\circ\theta_{r})(\omega)F(r,X_{r})(\omega)$),
we obtain

$\displaystyle\mathbb{E}\int_{0}^{t}$
$\displaystyle(Z\circ\theta_{r})F(r,X_{r})dr=\mathbb{E}\int_{0}^{t}F(r,X_{0})Ze^{\int_{0}^{r}F(u,X_{0})du}dr$
$\displaystyle=\mathbb{E}Z\Big{[}e^{\int_{0}^{t}F(u,X_{0})du}-1\Big{]}=\mathbb{E}(Z\circ\theta_{t})-\mathbb{E}Z,$

where in the last equality we used again Theorem 2.6, this time for $Z$. ∎

We end this section with a few examples of processes which are Markov and have
the FOED property.

######  Example 2.8.

Let $B$ be the (real valued) standard Brownian motion process.

a) (A Gauss-Gauss process with a parameter $a>0$)
Consider $X_{t}=B_{t+a}$. Here we have

$\mu_{0}(dz)=\gamma(dz)=\frac{1}{\sqrt{2\pi a}}e^{-\frac{z^{2}}{2a}}dz\
\textrm{and}\
\mu_{t}(dz)=\frac{1}{\sqrt{2\pi(t+a)}}e^{-\frac{z^{2}}{2(t+a)}}dz.$

Thus, using Propositions 2.3 and 2.5 we conclude that $X$ has the FOED
property and the function of evolution of distribution, say $F_{a}(u,z)$,
satisfies

$\displaystyle
e^{\int_{0}^{t}F_{a}(u,z)du}=\sqrt{\frac{a}{a+t}}e^{\frac{z^{2}t}{2a(a+t)}}.$
(23)

and thus it has the form

$\displaystyle
F_{a}(t,z)=-\frac{1}{2(a+t)}+\Big{(}\frac{1}{2a}+\frac{1}{2(a+t)^{2}}\Big{)}z^{2}.$
(24)

b) Consider a model analogous to Gauss-Gauss, this time built on Ornstein-
Uhlenbeck process. Namely, let $y\in\mathbb{R}$, $\lambda\geq 0$ and $Y$ be
the unique strong solution of

$\displaystyle Y_{t}=y+B_{t}-\lambda\int_{0}^{t}Y_{s}ds,\quad t\geq 0.$

For a brief characterization of such a process see [3, App. I, point 24]. It
is well known that

$Y_{t}=ye^{-\lambda t}+e^{-\lambda t}\int_{0}^{t}e^{\lambda s}dB_{s},\quad
t\geq 0.$

Thus, $Y_{t}\sim N(\nu_{t},\sigma^{2}_{t})$, where $\nu_{t}=ye^{-\lambda t}$
and $\sigma^{2}_{t}=e^{-2\lambda t}\int_{0}^{t}e^{2\lambda s}ds$. Now, let
$X_{t}=Y_{t+a}$ for $a>0.$ Then,

$\mu_{0}(dz)=\gamma(dz)=\frac{1}{\sqrt{2\pi\sigma^{2}_{a}}}e^{-\frac{(z-\nu_{a})^{2}}{2\sigma^{2}_{a}}}dz\
\textrm{and}\
\mu_{t}(dz)=\frac{1}{\sqrt{2\pi\sigma^{2}_{t+a}}}e^{-\frac{(z-\nu_{t+a})^{2}}{2\sigma^{2}_{t+a}}}dz.$

Consequently, using Propositions 2.3 and 2.5 we conclude that $X$ has the FOED
property and the function of evolution of distribution, say $F_{a}(u,z)$,
satisfies

$\displaystyle
e^{\int_{0}^{t}F_{a}(u,z)du}=\sqrt{\frac{\sigma^{2}_{a}}{\sigma^{2}_{t+a}}}e^{-\frac{(z-\nu_{t+a})^{2}}{2\sigma^{2}_{t+a}}+\frac{(z-\nu_{a})^{2}}{2\sigma^{2}_{a}}}.$
(25)

Clearly, with $y=\lambda=0$ the result (25) agrees with (23).
In [3, App. I, point 24] the transition density for $X$ is given relative to
the speed measure of $X$. Accordingly, it can be shown that

$\mu_{t}(dz)=\frac{1}{2}\frac{\sqrt{\lambda}}{\sqrt{2\pi\sinh(\lambda
t)}}\exp\Big{\\{}\frac{\lambda}{2}\Big{[}t+y^{2}+z^{2}-(y^{2}+z^{2})\coth(\lambda
t)+\frac{2zy}{\sinh(\lambda t)}\Big{]}\Big{\\}}m(dz),$

where $m(dz)=2e^{-\lambda z^{2}}dz$ is the speed measure for $X$. Given that,
the ”speed measure” representation for the function of evolution of
distribution takes the form

$\displaystyle e^{\int_{0}^{t}F(u,z)du}=\sqrt{\frac{\sinh(\lambda
a)}{\sinh(\lambda(a+t))}}\exp\Big{\\{}$
$\displaystyle\frac{\lambda}{2}\Big{(}t+(y^{2}+z^{2})(\coth(\lambda
a)-\coth(\lambda(a+t)))$ $\displaystyle+2yz\big{(}\frac{1}{\sinh(\lambda
a)}-\frac{1}{\sinh(\lambda(a+t))}\big{)}\Big{)}\Big{\\}}.$

The ”speed measure” representation approach will be taken in part c) below as
well.

c) Consider a model built on a squared Bessel process. Namely, let $y\geq 0$,
$\delta\geq 0$ and $Y$ be the unique strong solution of

$\displaystyle Y_{t}=y+2\int_{0}^{t}\sqrt{Y_{s}}dB_{s}+\delta t,\quad t\geq
0.$

For a brief characterization of such a process see [3, App. I, point 23]. It
is well known that

$\displaystyle\mu_{t}(dz)=\frac{1}{t}\big{(}{yz}\big{)}^{\frac{1}{2}(1-\frac{\delta}{2})}\exp\Big{(}-\frac{y+z}{2t}\Big{)}I_{\frac{\delta}{2}-1}\Big{(}\frac{\sqrt{yz}}{t}\Big{)}m(dz),$

where $I_{c}$ denotes the modified Bessel function with index
$c\in\mathbb{R}$. Then for fixed $a>0$ and $X_{t}=Y_{t+a}$, the process $X$ is
Markov with the FOED property and again by (7)

$\displaystyle
e^{\int_{0}^{t}F(u,z)du}=\frac{a}{a+t}\exp\Big{\\{}\frac{(y+z)t}{2a(a+t)}\Big{\\}}\frac{I_{\frac{\delta}{2}-1}\Big{(}\frac{\sqrt{yz}}{t+a}\Big{)}}{I_{\frac{\delta}{2}-1}\Big{(}\frac{\sqrt{yz}}{a}\Big{)}}.$

##  3 Feller processes with the FOED property

When we deal with Feller processes we will make standard assumption that $E$
is a LCCB space. Let $X$ is be a cádlág Feller process (in the sense of [25,
Ch. III]) admitting function $F$ of evolution of distribution and with the
infinite lifetime. By $\mathcal{C}_{0}(E)$ we denote the space of bounced and
continuous functions on $E$ vanishing at infinity, and by
$\mathbb{D}_{A}\subset\mathcal{C}_{0}(E)$ we denote the domain of
infinitesimal generator of $X$. By $C^{2}_{c}(E)$ we denote the space of
continuous functions on $E$ with compact support and both first and second
derivatives continuous.

The next result shows how one-dimensional distributions of $X$ depend on $A$
and $F.$

######  Proposition 3.1.

Let $f\in\mathbb{D}_{A}$. Then

$\displaystyle\mathbb{E}f(X_{t})=\mathbb{E}\Big{(}f(X_{0})+Af(X_{0})\int_{0}^{t}e^{\int_{0}^{s}F(u,X_{0})du}ds\Big{)},\quad
t\geq 0.$ (26)

###### Proof.

Since $f\in\mathbb{D}_{A}$ the process
$f(X_{t})-f(X_{0})-\int_{0}^{t}Af(X_{s})ds$ is a martingale, so

$\displaystyle\mathbb{E}f(X_{t})=\mathbb{E}\Big{(}f(X_{0})+\int_{0}^{t}Af(X_{s})ds\Big{)}.$

Because $Af\in\mathcal{C}_{0}(E)$ we conclude from Fubini’s theorem and the
FOED property that

$\displaystyle\mathbb{E}f(X_{t})$
$\displaystyle=\mathbb{E}f(X_{0})+\int_{0}^{t}\mathbb{E}Af(X_{s})ds=\mathbb{E}f(X_{0})+\int_{0}^{t}\mathbb{E}Af(X_{0})e^{\int_{0}^{s}F(u,X_{0})du}ds$
$\displaystyle=\mathbb{E}\Big{(}f(X_{0})+Af(X_{0})\int_{0}^{t}e^{\int_{0}^{s}F(u,X_{0})du}ds\Big{)}.$

The proof is complete. ∎

If $t\mapsto F(t,z)\in\mathcal{C}^{n}(\mathbb{R}_{+})$ for each $z\in E$, then
we define a sequence of functions $\mathcal{L}^{m}F:[0,\infty)\times
E\rightarrow\mathbb{R}$ by

$\displaystyle\mathcal{L}^{0}F\equiv 1,\ \mathcal{L}^{1}F\equiv F,\
\mathcal{L}^{m+1}F=\frac{d}{dt}\mathcal{L}^{m}F+F\mathcal{L}^{m}F,\quad m\leq
n.$ (27)

Next, define

$\displaystyle\mathbb{D}^{(n)}_{A}=\\{f\in\mathbb{D}_{A}:A^{i}f\in\mathbb{D}_{A}\
{\rm for}\ i\leq n\\},$

and $A^{n+1}f=(A\circ A^{n})f$ for $f\in\mathbb{D}^{(n)}_{A}$,
$n\in\mathbb{N}$, with $A^{1}=A$.

######  Proposition 3.2.

Fix $n\in\mathbb{N}$. Assume that the function $t\mapsto
F(t,z)\in\mathcal{C}^{n}(\mathbb{R}_{+})$ for each $z\in E$ and
$\mathbb{E}|F(t,X_{t})|<\infty$ for each $t\geq 0$. Let
$f\in\mathbb{D}^{(n)}_{A}$. Then

$\displaystyle\mathbb{E}(A^{n}f(X_{t}))=\mathbb{E}\Big{(}f(X_{t})\mathcal{L}^{n}F(t,X_{t})\Big{)}$
(28)

for every $t>0$.

###### Proof.

The proof goes by induction. For $n=1$ using Theorem 2.7 for $Z=f(X_{0})$
yields

$\displaystyle\mathbb{E}f(X_{t})=\mathbb{E}f(X_{0})+\mathbb{E}\int_{0}^{t}f(X_{s})F(s,X_{s})ds.$

On the other hand we have for $f\in C^{2}_{c}(E)$ (see [25, Remark p.295])

$\displaystyle\mathbb{E}f(X_{t})=\mathbb{E}f(X_{0})+\mathbb{E}\int_{0}^{t}Af(X_{s})ds.$

Comparing both equalities, using Fubini’s theorem and taking derivative with
respect to $t$ yield the assertion.

Assume now that identity (28) is true for any $k\leq n-1$. We have

$\displaystyle\mathbb{E}A^{n}f(X_{t})$
$\displaystyle=\mathbb{E}A(A^{n-1}f(X_{t}))=\mathbb{E}F(t,X_{t})A^{n-1}f(X_{t})$
$\displaystyle=\mathbb{E}F(t,X_{0})e^{\int_{0}^{t}F(u,X_{0})du}A^{n-1}f(X_{0}),$
(29)

where in the last equality we used the assumption that $X$ has the FOED
property.

Next, by assumption $\mathbb{E}|F(0,X_{0})|<\infty$ and
$f\in\mathbb{D}^{(n)}_{A}$, so $z\mapsto A^{n-1}f(z)$ is bounded and we may
differentiate with respect to $t$ under expectation to conclude that RHS of
(3) is equal to

$\displaystyle\mathbb{E}F(t,X_{0})e^{\int_{0}^{t}F(u,X_{0})du}A^{n-1}f(X_{0})=\frac{d}{dt}\Big{(}\mathbb{E}e^{\int_{0}^{t}F(u,X_{0})du}A^{n-1}f(X_{0})\Big{)}.$

By the FOED property and the induction assumption we obtain

$\displaystyle\frac{d}{dt}\Big{(}\mathbb{E}e^{\int_{0}^{t}F(u,X_{0})du}A^{n-1}f(X_{0})\Big{)}=\frac{d}{dt}\Big{(}\mathbb{E}A^{n-1}f(X_{t})\Big{)}=\frac{d}{dt}\mathbb{E}\Big{(}f(X_{t})\mathcal{L}^{n-1}F(t,X_{t})\Big{)}$
$\displaystyle=\frac{d}{dt}\mathbb{E}\Big{(}f(X_{0})e^{\int_{0}^{t}F(u,X_{0})du}\mathcal{L}^{n-1}F(t,X_{0})\Big{)}=\mathbb{E}\Big{(}f(X_{0})\frac{d}{dt}\Big{(}e^{\int_{0}^{t}F(u,X_{0})du}\mathcal{L}^{n-1}F(t,X_{0})\Big{)}\Big{)},$

which is equal to

$\displaystyle\mathbb{E}\Big{(}f(X_{0})\Big{(}F(t,X_{0})e^{\int_{0}^{t}F(u,X_{0})du}\mathcal{L}^{n-1}F(t,X_{0})+e^{\int_{0}^{t}F(u,X_{0})du}\frac{d}{dt}\mathcal{L}^{n-1}F(t,X_{0})\Big{)}\Big{)}.$

Thus, using the above as well as (27) and the FOED property, we obtain

$\displaystyle\mathbb{E}$ $\displaystyle A^{n}f(X_{t})$
$\displaystyle=\mathbb{E}\Big{(}f(X_{0})\Big{(}F(t,X_{0})e^{\int_{0}^{t}F(u,X_{0})du}\mathcal{L}^{n-1}F(t,X_{0})+e^{\int_{0}^{t}F(u,X_{0})du}\frac{d}{dt}\mathcal{L}^{n-1}F(t,X_{0})\Big{)}\Big{)}$
$\displaystyle=\mathbb{E}f(X_{0})\mathcal{L}^{n}F(t,X_{0})e^{\int_{0}^{t}F(u,X_{0})du}=\mathbb{E}\Big{(}f(X_{t})\mathcal{L}^{n}F(t,X_{t})\Big{)}.$

The proof is complete. ∎

######  Example 3.3.

Consider the Gauss-Gauss process from Example 2.8 with parameter $a>0$. We
find a description of the functions $\mathcal{L}^{m}F$ in terms of the density
of $X_{t}$.

The assumptions of Proposition 3.2 are satisfied. Let $g_{t+a}$ denote the
density of $B_{t+a}$. Using formula (28) for $f(x)=e^{-\lambda x}$ with
$\lambda>0$ and $X_{t}=B_{t+a}$, we obtain

$\displaystyle\frac{\lambda^{2n}}{2^{n}}\mathbb{E}e^{-\lambda
B_{t+a}}=\mathbb{E}e^{-\lambda B_{t+a}}\mathcal{L}^{n}F(t,B_{t+a}),\qquad
t>0.$

Let now $f(s)=1_{\\{s>0\\}}\frac{s^{2n-1}}{(2n-1)!}$ and
$h(z)=\mathcal{L}^{n}F(t,z)g_{t+a}(z)$. Hence, since
$\lambda^{-2n}=\int_{0}^{\infty}e^{-\lambda
s}\frac{t^{2n-1}}{(2n-1)!}ds=\int_{-\infty}^{\infty}e^{-\lambda s}f(s)ds$,
then, using the convolution theorem for Laplace transforms, we obtain the
following chain of equalities for each $\lambda>0$,

$\displaystyle\mathbb{E}e^{-\lambda B_{t+a}}$
$\displaystyle=2^{n}\lambda^{-2n}\mathbb{E}e^{-\lambda
B_{t+a}}\mathcal{L}^{n}F(t,B_{t+a})$
$\displaystyle=2^{n}\Big{(}\int_{-\infty}^{\infty}e^{-\lambda
s}f(s)ds\Big{)}\Big{(}\int_{-\infty}^{\infty}e^{-\lambda z}g(z)dz\Big{)}$
$\displaystyle=2^{n}\int_{-\infty}^{\infty}e^{-\lambda
v}\left(\int_{-\infty}^{\infty}f(v-z)h(z)dz\right)dv$
$\displaystyle=2^{n}\int_{-\infty}^{\infty}e^{-\lambda
v}\left(\frac{1}{\Gamma(2n)}\int_{-\infty}^{v}(v-z)^{2n-1}g_{t+a}(z)\mathcal{L}^{n}F(t,z)dz\right)dv,$

so that we obtain an integral equation for $\mathcal{L}^{n}F$

$\displaystyle
g_{t+a}(v)=\frac{2^{n}}{\Gamma(2n)}\int_{-\infty}^{v}(v-z)^{2n-1}g_{t+a}(z)\mathcal{L}^{n}F(t,z)dz.$

Using the Leibnitz rule iteratively we can compute $\mathcal{L}^{n}F(t,z)$.
For example, for $n=1$ we get

$\displaystyle\mathcal{L}^{1}F(t,v)=\frac{1}{2}\frac{g^{\prime\prime}_{t+a}(v)}{g_{t+a}(v)}.$

∎

##  4 Bridge operators

In order to study the finite dimensional distributions of a Markov process $X$
with the initial distribution $\gamma$ having $E$ as the support and admitting
the FOED property, we introduce a bridge operator.

Towards this end we consider a Banach space
$L^{1}(\gamma):=L^{1}(E,\mathcal{B}(E),\gamma)$ with the norm given by
$\left\|h\right\|=\mathbb{E}|h(X_{0})|$. For $t\geq 0$ and for $f\in
L^{1}(\gamma)$ we denote

$\Lambda_{t}f=\mathbb{E}(f(X_{0})\big{|}X_{t}).$

For each $t\geq 0$ the operator $f\mapsto\Lambda_{t}f$ is a linear contraction
on $L^{1}(\gamma)$ as well as on $L^{2}(\gamma)$. We can treat $\Lambda_{t}f$
as a function on $E$, namely $\Lambda_{t}f(z)=h(z),z\in E$, where $h$ is a
Borel function such that $\mathbb{E}\big{(}f(X_{0})|X_{t}\big{)}=h(X_{t})$, so

$\Lambda_{t}f(z)=\mathbb{E}(f(X_{0})\big{|}X_{t}=z).$

We will always assume that $(t,z)\mapsto\Lambda_{t}(f)(z)$ is a measurable
function. Observe also that for a Markov process $X$ with the FOED property,
with the initial distribution $\gamma(dz)=g(z)dz$ and with the transition
density $p_{t}(\cdot,\cdot)$ we have a nice formula for $\Lambda_{t}f$

$\displaystyle\Lambda_{t}f(z)=\frac{\int_{E}f(v)p_{t}(v,z)\mu_{0}(dv)}{\mu_{t}(dz)/dz}=\frac{\int_{E}f(v)p_{t}(v,z)\gamma(dv)}{g(z)e^{\int_{0}^{t}F(u,z)du}},\quad
f\in L^{1}(\gamma),$ (30)

where for the second equality we used (7). So in this case clearly
$(t,z)\mapsto\Lambda_{t}f(z)$ is measurable. We will call the operator
$\Lambda_{t}$ the bridge operator.

We will demonstrate below that the finite dimensional distributions of $X$ can
be described in terms of the bridge operators $(\Lambda_{t})_{t>0}$ and the
FOED function. This gives a new algorithmic approach to the computational
problems related to finite dimensional distributions of a Markov process. The
algorithm uses the initial distribution $\gamma$ and the form of bridge
operator given by (30). The first step to achieve this goal is is the theorem
below.

######  Theorem 4.1.

Assume that $X$ is a Markov process having the FOED property. Let
$n\in\mathbb{N}$ and $0<t_{1}<\ldots<t_{n}$. Assume that
$f_{0},f_{1},\ldots,f_{n}$ are Borel and bounded functions on $E$. Then

$\displaystyle\mathbb{E}f_{0}(X_{0})f_{1}(X_{t_{1}})\ldots
f_{n}(X_{t_{n}})=\mathbb{E}\Big{(}f_{n}(X_{0})e^{\int_{0}^{t_{n}-t_{n-1}}F(u,X_{0})du}$
(31)
$\displaystyle\quad\times\Lambda_{t_{n}-t_{n-1}}\Big{(}f_{n-1}(\cdot)e^{\int_{0}^{t_{n-1}-t_{n-2}}F(u,\cdot)du}\Lambda_{t_{n-1}-t_{n-2}}\Big{(}\cdots\Lambda_{t_{1}}f_{0}\Big{)}(\cdot)\Big{)}(X_{0})\Big{)}.$

###### Proof.

Notice that each successive component of the RHS of (31) is well defined since
$f_{0},\ldots,f_{n}$ are Borel bounded functions on $E$. Let $c$ be such that
$\max_{i\leq n}\|f_{i}\|_{\infty}\leq c$.

The proof of (31) goes by induction. For $n=1$, we see that she function
$f_{1}(\cdot)e^{\int_{0}^{t_{1}}F(u,\cdot)du}\Lambda_{t_{1}}f_{0}(\cdot)\in
L^{1}(\gamma)$ since

$\displaystyle\mathbb{E}\Big{|}f_{1}(X_{0})e^{\int_{0}^{t_{1}}F(u,X_{0})du}\Lambda_{t_{1}}f_{0}(X_{0})\Big{|}\leq
c^{2}\mathbb{E}e^{\int_{0}^{t_{1}}F(u,X_{0})du}=c^{2},$

and then using the definitions of $\Lambda_{t}$ and having the FOED property
we obtain

$\displaystyle\mathbb{E}f_{0}(X_{0})f_{1}(X_{t_{1}})=\mathbb{E}f_{1}(X_{t_{1}})\Lambda_{t_{1}}f_{0}(X_{t_{1}})=\mathbb{E}f_{1}(X_{0})e^{\int_{0}^{t_{1}}F(u,X_{0})du}\Lambda_{t_{1}}f_{0}(X_{0}).$

For the induction step suppose that identity (31) holds for $n-1$ and for
bounded Borel functions $f_{0},\ldots,f_{n-1}$. Using the Markov property and
the induction assumption we obtain the following chain of equalities

$\displaystyle\mathbb{E}f_{0}(X_{0})f_{1}(X_{t_{1}})\ldots
f_{n}(X_{t_{n}})=\mathbb{E}\Big{(}f_{0}(X_{0})f_{1}(X_{t_{1}})\ldots\
\mathbb{E}_{X_{t_{n-1}}}f_{n}(X_{t_{n}-t_{n-1}})\Big{)}$
$\displaystyle=\mathbb{E}\Big{(}f_{n-1}(X_{0})\mathbb{E}_{X_{0}}f_{n}(X_{t_{n}-t_{n-1}})e^{\int_{0}^{t_{n-1}-t_{n-2}}F(u,X_{0})du}\Lambda_{t_{n-1}-t_{n-2}}\Big{(}\cdots\Lambda_{t_{1}}f_{0}\Big{)}(X_{0})\Big{)}$
$\displaystyle=\mathbb{E}\Big{(}f_{n-1}(X_{0})f_{n}(X_{t_{n}-t_{n-1}})e^{\int_{0}^{t_{n-1}-t_{n-2}}F(u,X_{0})du}\Lambda_{t_{n-1}-t_{n-2}}\Big{(}\cdots\Lambda_{t_{1}}f_{0}\Big{)}(X_{0})\Big{)}$
$\displaystyle=\mathbb{E}\Big{(}f_{n}(X_{t_{n}-t_{n-1}})$
$\displaystyle\qquad\times\mathbb{E}\Big{(}f_{n-1}(X_{0})e^{\int_{0}^{t_{n-1}-t_{n-2}}F(u,X_{0})du}\Lambda_{t_{n-1}-t_{n-2}}\Big{(}\cdots\Lambda_{t_{1}}f_{0}\Big{)}(X_{0})\Big{|}X_{t_{n}-t_{n-1}}\Big{)}\Big{)}$
$\displaystyle=\mathbb{E}\Big{(}f_{n}(X_{t_{n}-t_{n-1}})$
$\displaystyle\qquad\times\Lambda_{t_{n}-t_{n-1}}\Big{(}f_{n-1}(\cdot)e^{\int_{0}^{t_{n-1}-t_{n-2}}F(u,\cdot)du}\Lambda_{t_{n-1}-t_{n-2}}\Big{(}\cdots\Lambda_{t_{1}}f_{0}\Big{)}(\cdot)\Big{)}\Big{(}X_{t_{n}-t_{n-1}}\Big{)}\Big{)}.$

By the FOED property the last expression is equal to

$\displaystyle\mathbb{E}\Big{(}f_{n}(X_{0})$ $\displaystyle
e^{\int_{0}^{t_{n}-t_{n-1}}F(u,X_{0})du}$
$\displaystyle\quad\times\Lambda_{t_{n}-t_{n-1}}\Big{(}f_{n-1}(\cdot)e^{\int_{0}^{t_{n-1}-t_{n-2}}F(u,\cdot)du}\Lambda_{t_{n-1}-t_{n-2}}\Big{(}\cdots\Lambda_{t_{1}}f_{0}\Big{)}(\cdot)\Big{)}\big{(}X_{0}\big{)}\Big{)},$

and the assertion follows. ∎

For a fixed $t>0$ denote

$\displaystyle\operatorname{Ker}\Lambda_{t}=\\{h\in
L^{1}(\gamma):\Lambda_{t}h=0\quad\gamma-a.s.\\}.$

It follows from standard arguments of functional analysis that
$\operatorname{Ker}\Lambda_{t}$ is a closed subspace of $L^{1}(\gamma)$. Any
$f\in L^{1}(\gamma)$ decomposes

$\displaystyle f=\Lambda_{t}f+(f)^{\bot}_{t},\qquad t>0,$ (32)

where $(f)^{\bot}_{t}\in\operatorname{Ker}\Lambda_{t}$. We are ready to
formulate the result, which gives a simple recurrent formula for determining
the distribution of a vector $(X_{0},X_{t_{1}},\ldots,X_{t_{n}})$. The formula
is strictly connected with the notion of kernel of a bridge operator. We will
need the following definition: For a given non-negative integer $n$,
$0=t_{0}<t_{1}<\ldots<t_{n}$ and for bounded Borel functions
$f_{1},\ldots,f_{n}$ define

$\displaystyle\Xi^{(n)}_{[f_{1},..,f_{n}]}(t_{1},\ldots,t_{n},z)$
$\displaystyle=\prod_{i=1}^{n}f_{i}(z)e^{\int_{0}^{t_{i}-t_{i-1}}F(u,z)du},\quad
n\geq 1,\ z\in E.$ (33)

We will also need the following class of functions

$\displaystyle\mathcal{K}=\\{g\in
L^{1}(\gamma):\quad\alpha\in\operatorname{Ker}\Lambda_{s}\ {\rm for}\ {\rm
all}\ s>0\Rightarrow\ g\alpha\in\operatorname{Ker}\Lambda_{t}\ {\rm for}\ {\rm
all}\ t>0\\}.$ (34)

Actually, for a broad class of diffusions we may expect that
$\mathcal{K}=L^{1}(\gamma)$.

######  Example 4.2.

Consider a Gauss-Gauss process, that is $X_{t}=B_{t+a}$, where $B$ is a
standard Brownian motion and $a>0$ is a fixed number, so
$\gamma(dz)=\frac{1}{\sqrt{2\pi a}}e^{-\frac{z^{2}}{2a}}dz$. Then
$\mathcal{K}=L^{1}(\gamma)$. Indeed, first observe that if $h\in
L^{1}(\gamma)$ and

$\displaystyle\mathbb{E}\big{(}h(B_{a})\big{|}B_{t+a}=z\big{)}=0,$ (35)

for each $z\in\mathbb{R}$, then

$\displaystyle\mathbb{E}\big{(}h(B_{a})\big{|}B_{t+a}\big{)}=0.$

It follows that for every $\lambda\in\mathbb{R}$ we have

$\displaystyle 0$
$\displaystyle=\mathbb{E}\left(\mathbb{E}\big{(}h(B_{a})\big{|}B_{t+a}\big{)}e^{\lambda
B_{t+a}}\right)=\mathbb{E}\left(h(B_{a})e^{\lambda B_{t+a}}\right)$
$\displaystyle=\mathbb{E}\left(h(B_{a})e^{\lambda(B_{t+a}-B_{a})}e^{\lambda
B_{a}}\right)=\mathbb{E}\left(e^{\lambda(B_{t+a}-B_{a})}\right)\mathbb{E}\left(h(B_{a})e^{\lambda
B_{a}}\right).$

Consequently,

$\displaystyle\mathbb{E}\left(h(B_{a})e^{\lambda B_{a}}\right)=0$

for every $\lambda\in\mathbb{R}$, and so $h\equiv 0$. This yields that
$\operatorname{Ker}\Lambda_{t}=\\{0\\}$, so $\mathcal{K}=L^{1}(\gamma)$.
By a similar arguments we may conclude analogous results for an Ornstein-
Uhlenbeck process, a Bessel process and many other processes. We omit the
details.

Using the notion of class $\mathcal{K}$ we are able to give a description of
$\mathbb{E}\prod_{i=1}^{n}f_{i}(X_{t_{i}})$ in terms of distribution of
$X_{0}$, which allows to compute the distribution of vector
$(X_{t_{1}},\ldots,X_{t_{n}})$ for $X$ having the FOED property in terms of
distribution of $X_{0}$.

######  Theorem 4.3.

Assume that $X$ is a Markov process having the FOED property. Let
$n\in\mathbb{N}$ and $0=t_{0}<t_{1}<\ldots<t_{n}$. Assume that
$f_{1},\ldots,f_{n}$ are bounded Borel functions. Then

$\displaystyle\mathbb{E}f_{1}(X_{t_{1}})=\mathbb{E}\Xi^{(1)}_{[f_{1}]}(t_{1},X_{0}).$

Moreover, if for $n\geq 2$ and $j\in\\{1,\ldots,n-1\\}$ it holds

$f_{n-1}(\cdot)\cdots
f_{j}(\cdot)e^{\sum_{i=j}^{n-1}\int_{0}^{t_{i}-t_{i-1}}F(u,\cdot)du}\in\mathcal{K},$

then for $n>1$ we have

$\displaystyle\mathbb{E}\prod_{i=1}^{n}f_{i}(X_{t_{i}})=\mathbb{E}$
$\displaystyle\Big{[}\Xi^{(n)}_{[f_{1},\ldots,f_{n}]}(t_{1},\ldots,t_{n},X_{0})$
(36) $\displaystyle-
f_{n}(X_{0})e^{\int_{0}^{t_{n}-t_{n-1}}F(u,X_{0})du}\big{(}\Xi^{(n-1)}_{[f_{1},\ldots,f_{n-1}]}(t_{1},\ldots,t_{n-1},\cdot)\big{)}^{\bot}_{t_{n}-t_{n-1}}(X_{0})\Big{]}.$

###### Proof.

Step 1. Let $n=1$. By definition, for $t_{1}>0$ and $z\in E$

$\displaystyle\Xi^{(1)}_{[f_{1}]}(t_{1},z)=f_{1}(z)e^{\int_{0}^{t_{1}}F(u,z)du}.$

Hence

$\displaystyle\mathbb{E}\Big{[}f_{1}(X_{0})e^{\int_{0}^{t_{1}}F(u,X_{0})du}\Big{]}=\mathbb{E}f_{1}(X_{t_{1}}),$

where in that last equality we used the definition of FOED. This proves the
assertion for $n=1$.

Step 2. We recall formula (31) for $f_{0}\equiv 1$

$\displaystyle\mathbb{E}f_{1}(X_{t_{1}})$ $\displaystyle\ldots
f_{n}(X_{t_{n}})=\mathbb{E}\Big{(}f_{n}(X_{0})e^{\int_{0}^{t_{n}-t_{n-1}}F(u,X_{0})du}$
(37)
$\displaystyle\times\Lambda_{t_{n}-t_{n-1}}\Big{(}f_{n-1}(\cdot)e^{\int_{0}^{t_{n-1}-t_{n-2}}F(u,\cdot)du}\Lambda_{t_{n-1}-t_{n-2}}\Big{(}\cdots\Lambda_{t_{1}}1\Big{)}(\cdot)\Big{)}(X_{0})\Big{)}.$

Take $n\geq 2$. Set

$\displaystyle\widetilde{\Lambda}_{1}(z)$ $\displaystyle=\Lambda_{t_{1}}1(z);$
$\displaystyle\widetilde{\Lambda}_{i}(z)$
$\displaystyle=\Lambda_{t_{i}-t_{i-1}}\Big{(}f_{i-1}(\cdot)e^{\int_{0}^{t_{i-1}-t_{i-2}}F(u,\cdot)du}\Lambda_{t_{i-1}-t_{i-2}}\Big{(}\cdots\Lambda_{t_{1}}1\Big{)}(\cdot)\Big{)}(z),\quad
i\geq 2,\ z\in E,$

so, using (37), we may write

$\displaystyle\mathbb{E}f_{1}(X_{t_{1}})\ldots
f_{n}(X_{t_{n}})=\mathbb{E}\Big{(}$ $\displaystyle
f_{n}(X_{0})e^{\int_{0}^{t_{n}-t_{n-1}}F(u,X_{0})du}$
$\displaystyle\times\Lambda_{t_{n}-t_{n-1}}\Big{(}f_{n-1}(\cdot)e^{\int_{0}^{t_{n-1}-t_{n-2}}F(u,\cdot)du}\widetilde{\Lambda}_{n-1}(\cdot)\Big{)}(X_{0})\Big{)},$

which is equal to

$\displaystyle\mathbb{E}$
$\displaystyle\Big{(}f_{n}(X_{0})e^{\int_{0}^{t_{n}-t_{n-1}}F(u,X_{0})du}$
$\displaystyle\times\Lambda_{t_{n}-t_{n-1}}\Big{(}f_{n-1}(\cdot)e^{\int_{0}^{t_{n-1}-t_{n-2}}F(u,\cdot)du}\Lambda_{t_{n-1}-t_{n-2}}\Big{(}f_{n-2}(\cdot)e^{\int_{0}^{t_{n-2}-t_{n-3}}F(u,\cdot)du}\widetilde{\Lambda}_{n-2}(\cdot)\Big{)}(X_{0})\Big{)}.$

We rewrite the last expression using decomposition (32) as

$\displaystyle\mathbb{E}$
$\displaystyle\Big{(}f_{n}(X_{0})e^{\int_{0}^{t_{n}-t_{n-1}}F(u,X_{0})du}$
$\displaystyle\times\Lambda_{t_{n}-t_{n-1}}\Big{(}f_{n-1}(\cdot)e^{\int_{0}^{t_{n-1}-t_{n-2}}F(u,\cdot)du}\Big{(}f_{n-2}(\cdot)e^{\int_{0}^{t_{n-2}-t_{n-3}}F(u,\cdot)du}\widetilde{\Lambda}_{n-2}(\cdot)$
$\displaystyle\quad-\Big{(}f_{n-2}(\cdot)e^{\int_{0}^{t_{n-2}-t_{n-3}}F(u,\cdot)du}\widetilde{\Lambda}_{n-2}(\cdot)\Big{)}^{\bot}_{t_{n-1}-t_{n-2}}\Big{)}(X_{0})\Big{)}=:\mathcal{I}.$

By assumption
$f_{n-1}(\cdot)e^{\int_{0}^{t_{n-1}-t_{n-2}}F(u,\cdot)du}\in\mathcal{K}$, so
we obtain

$\displaystyle\mathcal{I}$
$\displaystyle=\mathbb{E}\Big{(}f_{n}(X_{0})e^{\int_{0}^{t_{n}-t_{n-1}}F(u,X_{0})du}$
$\displaystyle\times\Lambda_{t_{n}-t_{n-1}}\Big{(}f_{n-1}(\cdot)e^{\int_{0}^{t_{n-1}-t_{n-2}}F(u,\cdot)du}f_{n-2}(\cdot)e^{\int_{0}^{t_{n-2}-t_{n-3}}F(u,\cdot)du}\widetilde{\Lambda}_{n-2}(\cdot)\Big{)}(X_{0})\Big{)},$

and, after iterating in the analogous way, we conclude that

$\displaystyle\mathbb{E}f_{1}(X_{t_{1}})\ldots f_{n}(X_{t_{n}})$
$\displaystyle=\mathbb{E}\Big{(}f_{n}(X_{0})e^{\int_{0}^{t_{n}-t_{n-1}}F(u,X_{0})du}\Lambda_{t_{n}-t_{n-1}}\Big{(}f_{n-1}(\cdot)f_{n-2}(\cdot)\cdots
f_{1}(\cdot)e^{\sum_{i=1}^{n-1}\int_{0}^{t_{i}-t_{i-1}}F(u,\cdot)du}\widetilde{\Lambda}_{1}(\cdot)\Big{)}(X_{0})\Big{)}$
$\displaystyle=\mathbb{E}\Big{(}f_{n}(X_{0})e^{\int_{0}^{t_{n}-t_{n-1}}F(u,X_{0})du}\Lambda_{t_{n}-t_{n-1}}\Big{(}f_{n-1}(\cdot)f_{n-2}(\cdot)\cdots
f_{1}(\cdot)e^{\sum_{i=1}^{n-1}\int_{0}^{t_{i}-t_{i-1}}F(u,\cdot)du}\Lambda_{t_{1}}1(\cdot)\Big{)}(X_{0})\Big{)}$

Since $\Lambda_{t_{1}}1=1$ we have

$\displaystyle\mathbb{E}f_{1}(X_{t_{1}})\ldots f_{n}(X_{t_{n}})$
$\displaystyle=\mathbb{E}\Big{(}f_{n}(X_{0})e^{\int_{0}^{t_{n}-t_{n-1}}F(u,X_{0})du}\Lambda_{t_{n}-t_{n-1}}\Big{(}f_{n-1}(\cdot)f_{n-2}(\cdot)\cdots
f_{1}(\cdot)e^{\sum_{i=1}^{n-1}\int_{0}^{t_{i}-t_{i-1}}F(u,\cdot)du}\Big{)}(X_{0})\Big{)}$
$\displaystyle=\mathbb{E}\Big{(}f_{n}(X_{0})e^{\int_{0}^{t_{n}-t_{n-1}}F(u,X_{0})du}\Big{(}f_{n-1}(\cdot)f_{n-2}(\cdot)\cdots
f_{1}(\cdot)e^{\sum_{i=1}^{n-1}\int_{0}^{t_{i}-t_{i-1}}F(u,\cdot)du}$
$\displaystyle\quad-\big{(}f_{n-1}(\cdot)f_{n-2}(\cdot)\cdots
f_{1}(\cdot)e^{\sum_{i=1}^{n-1}\int_{0}^{t_{i}-t_{i-1}}F(u,\cdot)du}\big{)}^{\bot}_{t_{n}-t_{n-1}}\Big{)}(X_{0})\Big{)}$
$\displaystyle=\mathbb{E}\Big{(}e^{\sum_{i=1}^{n}\int_{0}^{t_{i}-t_{i-1}}F(u,\cdot)du}\prod_{i=0}^{n}f_{i}(X_{0})$
$\displaystyle\quad-
f_{n}(X_{0})e^{\int_{0}^{t_{n}-t_{n-1}}F(u,X_{0})du}\big{(}f_{n-1}(\cdot)f_{n-2}(\cdot)\cdots
f_{1}(\cdot)e^{\sum_{i=1}^{n-1}\int_{0}^{t_{i}-t_{i-1}}F(u,\cdot)du}\big{)}^{\bot}_{t_{n}-t_{n-1}}(X_{0})\Big{)}$
$\displaystyle=\mathbb{E}\Big{[}\Xi^{(n)}_{[f_{1},\ldots,f_{n}]}(t_{1},\ldots,t_{n},X_{0})$
$\displaystyle\quad-
f_{n}(X_{0})e^{\int_{0}^{t_{n}-t_{n-1}}F(u,X_{0})du}\big{(}\Xi^{(n-1)}_{[f_{1},\ldots,f_{n-1}]}(t_{1},\ldots,t_{n-1},\cdot)\big{)}^{\bot}_{t_{n}-t_{n-1}}(X_{0})\Big{]}.$

The proof is complete. ∎

######  Remark 4.4.

Let us comment on a possible use of formula (36). Suppose that we know the
distribution $\gamma$ of $X_{0}$ and the transition density
$p_{t}(\cdot,\cdot)$ of a Markov process $X$. In the classical forward
approach, in order to compute
$\mathbb{E}\big{(}\prod_{i=1}^{n}f_{i}(X_{t_{i}})\big{)}$ we need to compute a
multiple integral involving a chain of transition densities. The alternative
backward approach is completely different and uses the notion of bridge
operator and its kernel. We have the following algorithm:

Step 1. We determine $e^{\int_{0}^{t}F(u,z)du}$ as

$\displaystyle
e^{\int_{0}^{t}F(u,z)du}=\frac{\mu_{t}(dz)}{\gamma(dz)},\quad\mu_{t}(dz)=\int_{E}p_{t}(v,dz)\gamma(dv).$

This is needed for the computation of the family of objects
$(\Xi^{(i)}_{[f_{1},\ldots,f_{i}]}(t,z))_{t>0,i\leq n}$.

Step 2. Using (33) we compute
$(\Xi^{(i)}_{[f_{1},\ldots,f_{i}]}(t_{0},\ldots,t_{i},z))_{0<t_{1}<..<t_{i}\leq
t,i\leq n}$ in straightforward recursive way.

Step 3. To determine the distribution of $(X_{0},X_{t_{1}},\ldots,X_{t_{n}})$
by applying Theorem 4.3 we need to verify that for $n\geq 2$ and
$j\in\\{1,\ldots,n-1\\}$ it holds

$f_{n-1}(\cdot)\cdots
f_{j}(\cdot)e^{\sum_{i=j}^{n-1}\int_{0}^{t_{i}-t_{i-1}}F(u,\cdot)du}\in\mathcal{K}.$

Once this is verified, we then follow the recurrence (36) using formulas (30)
and (32).

In the corollary below we see that under appropriate assumptions the FOED
property reduces the complexity of computational problems related to finite-
dimensional distributions of a Markov process. The standard approach relies on
computations based on chain rule of transition densities. The FOED backward
approach offers the new possibility of dealing with such computations.

######  Corollary 4.5.

Under assumptions of Theorem 4.3 and under the assumption that $X_{0}$ has a
density $g>0$, formula (36) can be written as

$\displaystyle\mathbb{E}\prod_{i=1}^{n}f_{i}(X_{t_{i}})=\mathbb{E}\Big{[}\frac{f_{n}(X_{0})}{g(X_{0})}\Xi^{(n-1)}_{[f_{1},\ldots,f_{n-1}]}(t_{1},\ldots,t_{n},\widehat{X}_{0})p_{t_{n}-t_{n-1}}(\widehat{X}_{0},X_{0})\Big{]},$
(38)

where $\widehat{X}_{0}$ is an independent copy of $X_{0}$.

###### Proof.

Using (32) we rewrite (36) in the form

$\displaystyle\mathbb{E}$
$\displaystyle\prod_{i=1}^{n}f_{i}(X_{t_{i}})=\mathbb{E}\Big{[}\Xi^{(n)}_{[f_{1},\ldots,f_{n}]}(t_{1},\ldots,t_{n},X_{0})-f_{n}(X_{0})e^{\int_{0}^{t_{n}-t_{n-1}}F(u,X_{0})du}$
$\displaystyle\times\Big{(}\Xi^{(n-1)}_{[f_{1},\ldots,f_{n-1}]}(t_{1},\ldots,t_{n-1},X_{0})-\Lambda_{t_{n}-t_{n-1}}(\Xi^{(n-1)}_{[f_{1},\ldots,f_{n-1}]}(t_{1},\ldots,t_{n-1},\cdot))(X_{0})\Big{)}\Big{]}$
$\displaystyle=\mathbb{E}\Big{(}f_{n}(X_{0})e^{\int_{0}^{t_{n}-t_{n-1}}F(u,X_{0})du}\Lambda_{t_{n}-t_{n-1}}(\Xi^{(n-1)}_{[f_{1},\ldots,f_{n-1}]}(t_{1},\ldots,t_{n-1},\cdot))(X_{0})\Big{)}.$

By (30) the last expression is equal to

$\displaystyle\mathbb{E}$
$\displaystyle\Big{(}\frac{f_{n}(X_{0})}{g(X_{0})}\int_{E}(\Xi^{(n-1)}_{[f_{1},\ldots,f_{n-1}]}(t_{1},\ldots,t_{n-1},v)p_{t_{n}-t_{n-1}}(v,X_{0})g(v)dv)\Big{)}$
$\displaystyle=\mathbb{E}\Big{(}\frac{f_{n}(X_{0})}{g(X_{0})}\mathbb{E}\left(\Xi^{(n-1)}_{[f_{1},\ldots,f_{n-1}]}(t_{1},\ldots,t_{n-1},\widehat{X}_{0})p_{t_{n}-t_{n-1}}(\widehat{X}_{0},X_{0})|X_{0}\right)\Big{)}$
$\displaystyle=\mathbb{E}\Big{(}\frac{f_{n}(X_{0})}{g(X_{0})}\Xi^{(n-1)}_{[f_{1},\ldots,f_{n-1}]}(t_{1},\ldots,t_{n-1},\widehat{X}_{0})p_{t_{n}-t_{n-1}}(\widehat{X}_{0},X_{0})\Big{)},$

where the first equality is a consequence of the independence lemma (see e.g.
[21], page 72, Rule 7). This finishes the proof. ∎

Using Theorem 4.3 we are able to determine the finite-dimensional
distributions of the process $e^{X}$ for various processes $X$ satisfying the
FOED property.

######  Example 4.6.

Let $X$ be a Gauss-Gauss or an Ornstein-Uhnlenbeck process. We saw in Example
2.8 that $e^{\int_{0}^{t}F(u,z)du}$ is a function of order $e^{cz^{2}}$ with
respect to $z$ for a constant $c>0$ dependent on $t$. Since, as shown in
Example 4.2, $\mathcal{K}=L^{1}(\gamma)$, then to apply Theorem 4.3 we need to
verify integrability of the functions

$f_{n-1}(\cdot)\cdots
f_{j}(\cdot)e^{\sum_{i=j}^{n-1}\int_{0}^{t_{i}-t_{i-1}}F(u,\cdot)du}$

for $j\leq n-1$. This will always hold if we choose
$f_{i}(z)=e^{-\lambda_{i}e^{z}}$ for $\lambda_{i}>0$, $i=1,..,n$. In this case
formula (38) enables us to establish a formula for Laplace transform for
vector of exponents of $X_{t_{i}}$, $i=1,..,n$, namely

$\displaystyle\mathbb{E}e^{-\sum_{i=1}^{n}\lambda_{i}e^{X_{t_{i}}}}=\mathbb{E}\Big{(}e^{-\lambda_{n}e^{X_{0}}-\sum_{i=1}^{n}\lambda_{i}e^{\widetilde{X}_{0}}}e^{\sum_{i=1}^{n-1}\int_{0}^{t_{i}-t_{i-1}}F(u,\widetilde{X}_{0})du}\frac{p_{t_{n}-t_{n-1}}(\widetilde{X}_{0},X_{0})}{g(X_{0})}\Big{)},$

where $\widetilde{X}_{0}$ is an independent copy of $X_{0}$. In terms of
computation the above equality provides a reduction of a multivariate
integration to bivariate integration.

Under additional assumptions we are able to describe the distribution of
vector $(X_{1},\ldots,X_{n})$ in terms of distribution of $X_{0}$.

######  Corollary 4.7.

Assume that $X$ is a Markov process with the FOED property,
$\mathcal{K}=L^{1}(\gamma)$ and
$e^{\sum_{i=1}^{n-1}\int_{0}^{t_{i}-t_{i-1}}F(u,\cdot)du}\in L^{1}(\gamma)$
for $0<t_{1}<\ldots<t_{n}$. Then for bounded Borel functions
$f_{1},\ldots,f_{n}$ we have

$\displaystyle\mathbb{E}\prod_{i=1}^{n}f_{i}(X_{t_{i}})=\mathbb{E}\Xi^{(n)}_{[f_{1},\ldots,f_{n}]}(t_{1},\ldots,t_{n},X_{0}).$
(39)

###### Proof.

If $\mathcal{K}=L^{1}(\gamma)$, then $\operatorname{Ker}\Lambda_{t}=\\{0\\}$
for each $t>0$. Indeed, take $\alpha\in\operatorname{Ker}\Lambda_{t}$. Since
$\alpha\in L^{1}(\gamma)=\mathcal{K}$, by definition of $\mathcal{K}$ it
follows that $\alpha^{2}\in\operatorname{Ker}\Lambda_{t}$ which easily yields
that $\alpha\equiv 0$. The assertion follows from Theorem 4.3. ∎

Then formula (39) becomes powerful. Let us see this on the following example.

######  Example 4.8.

Consider a Gauss-Gauss process $(X_{t})_{t\geq 0}=(B_{t+a})_{t\geq 0}$ for a
fixed $a>0$. Hence

$\displaystyle\gamma(dz)=\frac{1}{\sqrt{2\pi a}}e^{-\frac{z^{2}}{2a}},\qquad
e^{\int_{0}^{t}F(u,z)du}=\sqrt{\frac{a}{a+t}}e^{\frac{z^{2}t}{2a(a+t)}}.$

Let for a given natural $n$, $t_{i}=i\leq n$ and let $f_{1},\ldots,f_{n}$ be
bounded Borel functions. Then we easily find that the condition
$e^{\sum_{i=1}^{n-1}\int_{0}^{t_{i}-t_{i-1}}F(u,\cdot)du}\in L^{1}(\gamma)$
holds if $n\leq a+1$. Since $\mathcal{K}=L^{1}(\gamma)$ the formula (39) is
true in that case.

######  Example 4.9.

Let $\gamma(dx)=g(x)dx$ with $g>0$. Suppose that $X$ is a Markov process with
the transition density $p$ such that $t\mapsto
p_{t}(x,y)\in\mathcal{C}^{1}[0,\infty)$ for $x,y\in E$. Then by Proposition
2.3 the process $X$ has the FOED property and

$\displaystyle
e^{\int_{0}^{t}F(u,x)du}=\frac{\int_{E}p_{t}(z,x)g(z)dz}{g(x)}\leq\mathbb{E}k(t,X_{0}),$

where $k(t,z)=\sup_{x}\frac{p_{t}(z,x)}{g(x)},\ t\geq 0,\ z\in E.$ If,
moreover, $\mathbb{E}k(t_{i}-t_{i-1},X_{0})<\infty$ for each $i$, then

$\displaystyle
e^{\sum_{i=1}^{n}\int_{0}^{t_{i}-t_{i-1}}F(u,z)du}\leq\prod_{i=1}^{n}\mathbb{E}k(t_{i}-t_{i-1},X_{0}),$

which means that $e^{\sum_{i=1}^{n-1}\int_{0}^{t_{i}-t_{i-1}}F(u,\cdot)du}\in
L^{1}(\gamma)$. If additionally $\mathcal{K}=L^{1}(\gamma)$, assumptions of
Corrolary 4.7 are satisfied.

##  5 Conditional structures and FOED

One of the consequences of the FOED property and of the definition of bridge
operators is a convenient computation of conditional structures of Markov
processes. By a conditional structure of a Markov process we mean a function
of the form

$\displaystyle\mathbb{E}\Big{(}\Psi(X_{0},X_{t_{1}})\cdots,X_{t_{n}})\Big{|}X_{s+t_{1}+\ldots
t_{n}}=z\Big{)},\quad z\in E,$ (40)

for an integrable Borel function $\Psi:E^{n+1}\rightarrow\mathbb{R}$,
$0<t_{1}<\ldots<t_{n}$ and $s\geq 0$. A particular conditional structure takes
the form

$\displaystyle\mathbb{E}\Big{(}f_{0}(X_{0})f_{1}(X_{t_{1}})\cdots
f_{n}(X_{t_{n}})\Big{|}X_{s+t_{1}+\ldots t_{n}}=z\Big{)},\quad z\in E,$ (41)

for a sequence $(f_{i})_{i=0,\ldots,n}$ of bounded Borel functions,
$0<t_{1}<\ldots<t_{n}$ and $s\geq 0$. Such conditioning is motivated by use of
the FOED property in what follows. The conditional structures and theory of
stochastic bridges are strictly connected. For a Markov process with the FOED
property using the bridge operators $(\Lambda_{t})_{t\geq 0}$ we can give a
full characterization of objects given by (41). For $s\geq 0$ and
$0<t_{1}<\ldots<t_{n}$ let

$\displaystyle T^{s}_{k,n}=s+\sum_{i=1}^{k}t_{i},$

and let $\mu_{t}$ be the distribution of $X_{t}$.

######  Theorem 5.1.

Assume that $X$ is a Markov process having the FOED property and the initial
distribution $\gamma$. Let $n\in\mathbb{N}$, $s\geq 0$ and
$0<t_{1}<\ldots<t_{n}$. Assume that $f_{0},\ldots,f_{n}$ are bounded Borel
functions. Then a.e. with respect to the Lebesgue measure

$\displaystyle\mathbb{E}\Big{(}f_{0}(X_{0})$ $\displaystyle
f_{1}(X_{t_{1}})\cdots
f_{n}(X_{t_{n}})\Big{|}X_{T^{s}_{n,n}}=z\Big{)}=e^{-\int_{T^{s}_{n-1,n}}^{T^{s}_{n,n}}F(u,z)du}$
$\displaystyle\quad\times\Lambda_{T^{s}_{n-1,n}}\Big{(}f_{n}(\cdot)e^{\int_{0}^{t_{n}-t_{n-1}}F(u,\cdot)du}\Lambda_{t_{n}-t_{n-1}}\Big{(}\ldots\Lambda_{t_{1}}f_{0}\Big{)}(\cdot)\Big{)}(z).$
(42)

###### Proof.

Let $B\in\mathcal{E}$. By the Markov property and Theorem 4.1 we have

$\displaystyle\mathbb{E}$
$\displaystyle\Big{(}f_{0}(X_{0})f_{1}(X_{t_{1}})\cdots
f_{n}(X_{t_{n}})1_{\\{X_{T^{s}_{n,n}}\in B\\}}\Big{)}$
$\displaystyle=\mathbb{E}\Big{(}f_{0}(X_{0})f_{1}(X_{t_{1}})\cdots
f_{n}(X_{t_{n}})\mathbb{P}_{X_{t_{n}}}\big{(}X_{T^{s}_{n-1,n}}\in
B\big{)}\Big{)}$
$\displaystyle=\mathbb{E}\Big{(}f_{n}(X_{0})e^{\int_{0}^{t_{n}-t_{n-1}}F(u,X_{0})du}\mathbb{P}_{X_{0}}\big{(}X_{T^{s}_{n-1,n}}\in
B\big{)}$
$\displaystyle\qquad\times\Lambda_{t_{n}-t_{n-1}}\big{(}f_{n-1}(\cdot)e^{\int_{0}^{t_{n-1}-t_{n-2}}F(u,\cdot)du}\Lambda_{t_{n-1}-t_{n-2}}\big{(}\cdots\Lambda_{t_{1}}f_{0}\big{)}\big{)}(X_{0})\Big{)}$
$\displaystyle=\mathbb{E}\Big{(}f_{n}(X_{0})e^{\int_{0}^{t_{n}-t_{n-1}}F(u,X_{0})du}\mathbb{E}\big{(}1_{\\{X_{T^{s}_{n-1,n}}\in
B\\}}\big{|}\mathcal{F}^{X}_{0}\big{)}$
$\displaystyle\qquad\times\Lambda_{t_{n}-t_{n-1}}\big{(}f_{n-1}(\cdot)e^{\int_{0}^{t_{n-1}-t_{n-2}}F(u,\cdot)du}\Lambda_{t_{n-1}-t_{n-2}}\big{(}\cdots\Lambda_{t_{1}}f_{0}\big{)}\big{)}(X_{0})\Big{)}$
$\displaystyle=\mathbb{E}\Big{(}1_{\\{X_{T^{s}_{n-1,n}}\in
B\\}}f_{n}(X_{0})e^{\int_{0}^{t_{n}-t_{n-1}}F(u,X_{0})du}$
$\displaystyle\qquad\times\Lambda_{t_{n}-t_{n-1}}\big{(}f_{n-1}(\cdot)e^{\int_{0}^{t_{n-1}-t_{n-2}}F(u,\cdot)du}\Lambda_{t_{n-1}-t_{n-2}}\big{(}\cdots\Lambda_{t_{1}}f_{0}\big{)}\big{)}(X_{0})\Big{)}.$

Rewriting the last expression by conditioning with respect to
$X_{s+t_{1}+\ldots+t_{n-1}}$ we obtain

$\displaystyle\mathbb{E}$
$\displaystyle\Big{(}\mathbb{E}\big{(}f_{n}(X_{0})e^{\int_{0}^{t_{n}-t_{n-1}}F(u,X_{0})du}\Lambda_{t_{n}-t_{n-1}}\big{(}f_{n-1}(\cdot)e^{\int_{0}^{t_{n-1}-t_{n-2}}F(u,\cdot)du}$
$\displaystyle\quad\times\
\Lambda_{t_{n-1}-t_{n-2}}\big{(}\cdots\Lambda_{t_{1}}f_{0}\big{)}\big{)}(X_{0})\big{|}X_{T^{s}_{n-1,n}}\big{)}\
1_{\\{X_{T^{s}_{n-1,n}}\in B\\}}\Big{)}$
$\displaystyle=\int_{B}\Lambda_{T^{s}_{n-1,n}}\Big{(}f_{n}(X_{0})e^{\int_{0}^{t_{n}-t_{n-1}}F(u,X_{0})du}$
$\displaystyle\quad\times\Lambda_{t_{n}-t_{n-1}}\big{(}f_{n-1}(\cdot)e^{\int_{0}^{t_{n-1}-t_{n-2}}F(u,\cdot)du}\Lambda_{t_{n-1}-t_{n-2}}\big{(}\cdots\Lambda_{t_{1}}f_{0}\big{)}\big{)}(z)\Big{)}\mathbb{P}\big{(}X_{T^{s}_{n-1,n}}\in
dz\big{)}.$

Denoting

$\displaystyle R_{s,t_{1},\ldots,t_{n}}(z)$
$\displaystyle=\Lambda_{T^{s}_{n-1,n}}\big{(}f_{n}(X_{0})e^{\int_{0}^{t_{n}-t_{n-1}}F(u,X_{0})du}$
$\displaystyle\times\Lambda_{t_{n}-t_{n-1}}\big{(}f_{n-1}(\cdot)e^{\int_{0}^{t_{n-1}-t_{n-2}}F(u,\cdot)du}\Lambda_{t_{n-1}-t_{n-2}}\big{(}\cdots\Lambda_{t_{1}}f_{0}\big{)}\big{)}(z),$

we conclude after comparing the first and the last term of the above sequence
of equalities that

$\displaystyle\mathbb{E}\big{(}f_{1}(X_{t_{1}})\cdots f_{n}(X_{t_{n}})$
$\displaystyle 1_{\\{X_{T^{s}_{n,n}}\in
B\\}}\big{)}=\int_{B}R_{s,t_{1},\ldots,t_{n}}(z)\mu_{T^{s}_{n-1,n}}(dz).$ (43)

On the other hand

$\displaystyle\mathbb{E}\big{(}f_{1}(X_{t_{1}})$ $\displaystyle\cdots
f_{n}(X_{t_{n}})1_{\\{X_{T^{s}_{n,n}}\in B\\}}\big{)}$
$\displaystyle=\mathbb{E}\big{(}\mathbb{E}\big{(}f_{1}(X_{t_{1}})\cdots
f_{n}(X_{t_{n}})\big{|}X_{T^{s}_{n,n}}\big{)}1_{\\{X_{T^{s}_{n,n}}\in
B\\}}\big{)},$ (44)

so denoting

$\displaystyle S_{s,t_{1},\ldots,t_{n}}(z)=\mathbb{E}\big{(}f_{1}(X_{t_{1}})$
$\displaystyle\cdots f_{n}(X_{t_{n}})\big{|}X_{T^{s}_{n,n}}=z\big{)},\quad
z\in E,$

we can write (5) in the form

$\displaystyle\mathbb{E}\Big{(}f_{1}(X_{t_{1}})\cdots f_{n}(X_{t_{n}})$
$\displaystyle 1_{\\{X_{T^{s}_{n,n}}\in
B\\}}\Big{)}=\int_{B}S_{s,t_{1},\ldots,t_{n}}(z)\mu_{T^{s}_{n,n}}(dz).$

Compare this and (43) to obtain

$\displaystyle\int_{B}R_{s,t_{1},\ldots,t_{n}}(z)\mu_{T^{s}_{n-1,n}}(dz)=\int_{B}S_{s,t_{1},\ldots,t_{n}}(z)\mu_{T^{s}_{n,n}}(dz),$

which by the FOED property is equivalent to

$\displaystyle\int_{B}R_{s,t_{1},\ldots,t_{n}}(z)\mu_{T^{s}_{n-1,n}}(dz)=\int_{B}S_{s,t_{1},\ldots,t_{n}}(z)e^{\int_{T^{s}_{n-1,n}}^{{T^{s}_{n,n}}}F(u,z)du}\mu_{T^{s}_{n-1,n}}(dz).$

The arbitrary choice of $B$ yields $z$\- a.e.

$\displaystyle
S_{s,t_{1},\ldots,t_{n}}(z)=e^{-\int_{T^{s}_{n-1,n}}^{{T^{s}_{n,n}}}F(u,z)du}R_{s,t_{1},\ldots,t_{n}}(z),$

which is exactly our assertion. ∎

In the next theorem we give a computation of the conditional structure (40).
We start with an auxiliary result.

######  Lemma 5.2.

Let $X$ be a Markov process having the FOED property. For any $s,t>0$ and
$x\in E$ we have

$\displaystyle p_{s+t}(x,w)\mathbb{P}(X_{s}\in
dw)=p_{s}(x,w)\mathbb{P}(X_{t+s}\in dw).$ (45)

###### Proof.

By the FOED property, and the time homogeneity of $X$, we have for any $s,t>0$

$\displaystyle\mathbb{P}(X_{s+t}\in dw)=\mathbb{P}(X_{s}\in
dw)e^{\int_{s}^{t+s}F(u,w)du},$

which actually means that

$\displaystyle
p_{s+t}(x,w)\gamma(dx)=p_{s}(x,w)e^{\int_{s}^{t+s}F(u,w)du}\gamma(dx).$

This yields $\gamma$\- a.s.

$\displaystyle p_{s+t}(x,w)=p_{s}(x,w)e^{\int_{s}^{t+s}F(u,w)du},$

so

$\displaystyle p_{s+t}(x,w)\mathbb{P}(X_{s}\in
dw)=p_{s}(x,w)e^{\int_{s}^{t+s}F(u,w)du}\mathbb{P}(X_{s}\in dw),$

which by the FOED property, and the time homogeneity of $X$, becomes

$\displaystyle p_{s+t}(x,w)\mathbb{P}(X_{s}\in
dw)=p_{s}(x,w)\mathbb{P}(X_{t+s}\in dw).$

The proof is complete. ∎

For a transition density $p_{t}(\cdot,\cdot)$ and a measure $\gamma$ we will
use the notation

$\displaystyle\gamma^{n+1}(d(z_{0},\ldots,z_{n}))=\gamma(dz_{0})\otimes\ldots\otimes\gamma(dz_{n}),\qquad
g_{t}(z)=\int_{E}p_{t}(v,z)\gamma(dv).$

######  Theorem 5.3.

Assume that $X$ is a Markov process having the FOED property with the initial
distribution $\gamma$ and the transition density $p_{t}(\cdot,\cdot)$. Let
$n\in\mathbb{N}$, $s\geq 0$ and $0=t_{0}<t_{1}<\ldots<t_{n}$. Let
$\Psi:E^{n+1}\rightarrow\mathbb{R}$ be a Borel function such that
$\mathbb{E}|\Psi(X_{0},\ldots,X_{t_{n}})|<\infty$. Then

$\displaystyle\mathbb{E}$
$\displaystyle\big{(}\Psi\big{(}X_{0},\ldots,X_{t_{n}}\big{)}\big{|}X_{T^{s}_{n,n}}=w\big{)}$
$\displaystyle=e^{-\int_{T^{s}_{n-1,n}}^{T^{s}_{n,n}}F(u,w)du}\int_{E^{n+1}}\Psi(z_{0},\ldots,z_{n})\prod_{i=1}^{n-1}\frac{p_{t_{i}}(z_{i-1},z_{i})}{g_{t_{i}}(z_{i-1})}\frac{p_{t_{n}}(z_{n},w)}{g_{t_{n}}(w)}$
$\displaystyle\times
e^{\sum_{i=1}^{n}\int_{0}^{t_{i}-t_{i-1}}F(u,z_{i})du}\gamma^{n+1}(d(z_{0},\ldots,z_{n})).$
(46)

###### Proof.

First we prove that formula (5.3) holds for

$\Psi(z_{0},\ldots,z_{n})=f_{0}(z_{0})\cdots f_{n}(z_{n}),$

where $f_{0},\ldots,f_{n}$ are bounded Borel functions.

For a bounded Borel function $f$ we will determine the form of $\Lambda_{t}f$.
By (30) and Proposition 2.3 we have, for any $t>0$ and $w\in E$,

$\displaystyle\Lambda_{t}f(w)=\int_{E}f(v)\frac{p_{t}(v,w)\gamma(dv)}{\int_{E}p_{t}(r,w)\gamma(dr)}.$

Let us define a function $(t,w)\rightarrow\eta_{t,w}$, where $\eta_{t,w}$ is a
measure on $[0,\infty)$ given by

$\displaystyle\eta_{t,w}(dv)=\frac{p_{t}(v,w)\gamma(dv)}{\int_{E}p_{t}(r,w)\gamma(dr)},$
(47)

so

$\displaystyle\Lambda_{t}f(w)=\int_{E}f(v)\eta_{t,w}(dv).$ (48)

Now, we compute the succesive inner components of the RHS of (5.1). By (48) we
have

$\displaystyle\Lambda_{t_{2}-t_{1}}\big{(}f_{1}(\cdot)e^{\int_{0}^{t_{1}}F(u,\cdot)du}\Lambda_{t_{1}}f_{0}\big{)}(w)=\int_{E}f_{1}(v)e^{\int_{0}^{t_{1}}F(u,v)}\Lambda_{t_{1}}f_{0}(v)\eta_{t_{2}-t_{1},w}(dv).$
(49)

Next, for $n=3$, using formulas (49) and (48) we obtain

$\displaystyle\Lambda_{t_{3}-t_{2}}\big{(}f_{2}(\cdot)e^{\int_{0}^{t_{2}-t_{1}}F(u,\cdot)du}\Lambda_{t_{2}-t_{1}}\big{(}f_{1}(\cdot)e^{\int_{0}^{t_{1}}F(u,\cdot)du}\Lambda_{t_{1}}f_{0}\big{)}\big{)}(w)$
$\displaystyle=\Lambda_{t_{3}-t_{2}}\Big{(}f_{2}(\cdot)e^{\int_{0}^{t_{2}-t_{1}}F(u,\cdot)du}\Big{(}\int_{E}f_{1}(z_{1})e^{\int_{0}^{t_{1}}F(u,z_{1})du}\Lambda_{t_{1}}f_{0}(z_{1})\eta_{t_{2}-t_{1},\cdot}(dz_{1})\Big{)}\Big{)}(w)$
$\displaystyle=\int_{E}\Big{(}f_{2}(z_{2})e^{\int_{0}^{t_{2}-t_{1}}F(u,z_{2})du}\Big{(}\int_{E}f_{1}(z_{1})e^{\int_{0}^{t_{1}}F(u,z_{1})du}\Lambda_{t_{1}}f_{0}(z_{1})\eta_{t_{2}-t_{1},z_{2}}(dz_{1})\Big{)}\Big{)}\eta_{t_{3}-t_{2},w}(dz_{2})$
$\displaystyle=\int_{E}\Big{(}f_{2}(z_{2})e^{\int_{0}^{t_{2}-t_{1}}F(u,z_{2})du}\Big{(}\int_{E}f_{1}(z_{1})e^{\int_{0}^{t_{1}}F(u,z_{1})du}$
$\displaystyle\qquad\qquad\qquad\qquad\times\int_{E}f_{0}(z_{0})\eta_{t_{1},z_{1}}(dz_{0})\eta_{t_{2}-t_{1},z_{2}}(dz_{1})\Big{)}\Big{)}\eta_{t_{3}-t_{2},w}(dz_{2}),$

which, using Fubini theorem, we rewrite as

$\displaystyle\int_{E}\int_{E}\int_{E}f_{2}(z_{2})f_{1}(z_{1})f_{0}(z_{0})e^{\int_{0}^{t_{2}-t_{1}}F(u,z_{2})du+\int_{0}^{t_{1}}F(u,z_{1})du}\eta_{t_{1},z_{1}}(dz_{0})\eta_{t_{2}-t_{1},z_{2}}(dz_{1})\eta_{t_{3}-t_{2},w}(dz_{2}).$

By formula (47) and the form of $g_{t}$ this is equal to

$\displaystyle\int_{E}$
$\displaystyle\int_{E}\int_{E}f_{2}(z_{2})f_{1}(z_{1})f_{0}(z_{0})e^{\int_{0}^{t_{2}-t_{1}}F(u,z_{2})du+\int_{0}^{t_{1}}F(u,z_{1})du}$
$\displaystyle\times\quad\frac{p_{t_{3}-t_{2}}(z_{2},w)p_{t_{2}-t_{1}}(z_{1},z_{2})p_{t_{1}}(z_{0},z_{1})}{g_{t_{3}-t_{2}}(z_{2})g_{t_{2}-t_{1}}(z_{1})g_{t_{1}}(z_{0})}\gamma(dz_{2})\gamma(dz_{1})\gamma(dz_{0}).$

Continuing in an analogous way we get the formula

$\displaystyle\Lambda_{t_{n}-t_{n-1}}\big{(}f_{n-1}(\cdot)e^{\int_{0}^{t_{n-1}-t_{n-2}}F(u,\cdot)du}\Lambda_{t_{n-1}-t_{n-2}}\big{(}\ldots\big{)}\big{)}(w)$
$\displaystyle=\int_{E}\ldots\int_{E}f_{n-1}(z_{n-1})\ldots
f_{0}(z_{0})e^{\sum_{i=1}^{n-1}\int_{0}^{t_{i}-t_{i-1}}F(u,z_{i})du}\frac{\prod_{i=1}^{n-1}p_{t_{i}-t_{i-1}}(z_{i-1},z_{i})}{\prod_{i=1}^{n-1}g_{t_{i}-t_{i-1}}(z_{i-1})}$
$\displaystyle\quad\quad\times\frac{p_{t_{n}-t_{n-1}}(z_{n-1},w)}{g_{t_{n}-t_{n-1}}(z_{n-1})}\gamma(dz_{0})\ldots\gamma(dz_{n-1}).$
(50)

Denote the last expression by $\Theta(n)(w)$. Using (48) we compute

$\displaystyle\Lambda_{T^{s}_{n-1,n}}$
$\displaystyle\Big{(}f_{n}(\cdot)e^{\int_{0}^{t_{n}-t_{n-1}}F(u,\cdot)du}\Theta(n)(\cdot)\Big{)}(w)$
$\displaystyle=\int_{E}f_{n}(z_{n})e^{\int_{0}^{t_{n}-t_{n-1}}F(u,z_{n})du}\Theta(n)(z_{n})\frac{p_{T^{s}_{n-1,n}}(z_{n},w)}{g_{T^{s}_{n-1,n}}(w)}\gamma(dz_{n}).$

Hence, using formula (5) we conclude from (5.1) that

$\displaystyle\mathbb{E}$ $\displaystyle\big{(}f_{1}(X_{t_{1}})\cdots
f_{n}(X_{t_{n}})\big{|}X_{T^{s}_{n,n}}=w\big{)}$
$\displaystyle=e^{-\int_{T^{s}_{n-1,n}}^{T^{s}_{n,n}}F(u,w)du}\int_{E^{n+1}}f_{0}(z_{0})\cdots
f_{n}(z_{n})\frac{\prod_{i=1}^{n-1}p_{t_{i}-t_{i-1}}(z_{i-1},z_{i})}{\prod_{i=1}^{n-1}g_{t_{i}-t_{i-1}}(z_{i-1})}\frac{p_{T^{s}_{n-1,n}}(z_{n},w)}{g_{T^{s}_{n-1,n}}(w)}$
$\displaystyle\times
e^{\sum_{i=1}^{n}\int_{0}^{t_{i}-t_{i-1}}F(u,z_{i})du}\gamma^{n+1}(d(z_{0},\ldots,z_{n})).$

Using Lemma 5.2 we find that

$\displaystyle\frac{p_{t_{i}-t_{i-1}}(z_{i-1},z_{i})}{g_{t_{i}-t_{i-1}}(z_{i-1})}=\frac{p_{t_{i}}(z_{i-1},z_{i})}{g_{t_{i}}(z_{i-1})},\qquad\quad\frac{p_{T^{s}_{n-1,n}}(z_{n},w)}{g_{T^{s}_{n-1,n}}(w)}=\frac{p_{t_{n}}(z_{n},w)}{g_{t_{n}}(w)},$

so formula (5.3) holds for

$\Psi(z_{0},\ldots,z_{n})=f_{0}(z_{0})\cdots f_{n}(z_{n}),$

where $f_{0},\ldots,f_{n}$ are bounded Borel functions.

The general case, that is (5.3) for a Borel functions
$\Psi:E^{n+1}\rightarrow\mathbb{R}$ such that
$\mathbb{E}|\Psi(X_{0},\ldots,X_{t_{n}})|<\infty$ follows by using a monotone
convergence theorem.

The proof is complete. ∎

##  6 FOED and Kolmogorov distance: example

In our considerations in this section we use the FOED property to solve the
following problem: Consider a Brownian motion $B$ with the initial
distribution $\gamma$ on $\mathbb{R}$. Suppose that $X$ is a Markov process
having the FOED property. Our task is to measure the similarity between
$B_{t}$ and $X_{t}$, for fixed $t$, if $X_{0}$ has a distribution $\gamma$. In
statistics, the Kolmogorov distance is a convenient way of measuring
similarity of distributions (see [31] for an overview). For a couple of nice
recent results measuring the Kolmogorov distance between the general Poisson
random variables and Gaussian distribution see for example [5]. Recall that
the Kolmogorov distance between real valued random variables $X$ and $Y$ is
defined as

$\displaystyle d_{Kol}(X,Y)=\sup_{x\in\mathbb{R}}\big{|}\mathbb{P}(X\leq
x)-\mathbb{P}(Y\leq x)\big{|}.$

We will show that for computation Kolmogorov distance $d_{Kol}(X_{t},B_{t})$
it suffices to consider the supremum over a special set that is related to the
FOED function and the distribution $\gamma$. We use below the common notation:
by $\Phi$ we denote the cumulative distribution function of a standard
Gaussian random variable $G$ and by $\phi$ the p.d.f of $G$.

######  Theorem 6.1.

Suppose that $X$ is a Markov process having the FOED property with the
function of evolution of distribution $F$, $X_{0}$ has a distribution $\gamma$
and a differentiable density $g$. Let $B$ be a Brownian motion with the
initial distribution $\gamma$. Fix $t>0$. Let $\mathcal{X}_{0}$ be the set of
solutions of the following equation with respect to variable $x$

$\displaystyle g(x)e^{\int_{0}^{t}F(u,x)du}=\mathbb{E}1_{\\{G\leq
0\\}}\big{(}g(x-\sqrt{t}G)+g(x+\sqrt{t}G)\big{)}.$ (51)

Then

$\displaystyle d_{Kol}$ $\displaystyle(X_{t},B_{t})$ (52)
$\displaystyle=\sup_{x_{0}\in\mathcal{X}_{0}}\Big{|}\int_{-\infty}^{x_{0}}\Big{(}e^{\int_{0}^{t}F(u,z)du}-\Phi\Big{(}\frac{x_{0}-z}{\sqrt{t}}\Big{)}\Big{)}g(z)dz-\int_{x_{0}}^{\infty}\Phi\Big{(}\frac{x_{0}-z}{\sqrt{t}}\Big{)}g(z)dz\Big{|}.$

###### Proof.

For any $x\in\mathbb{R}$ we have

$\displaystyle\mathbb{P}(B_{t}\leq
x)=\int^{\infty}_{-\infty}\Big{(}\int_{-\infty}^{x}\frac{1}{\sqrt{2\pi
t}}e^{-\frac{(z-y)^{2}}{2t}}dy\Big{)}g(z)dz=\int^{\infty}_{-\infty}\Phi\Big{(}\frac{x-z}{\sqrt{t}}\Big{)}g(z)dz.$

Hence and from the fact that $X$ has the FOED property we have

$\displaystyle\psi(x)$ $\displaystyle:=\mathbb{P}(X_{t}\leq
x)-\mathbb{P}(B_{t}\leq
x)=\int_{-\infty}^{x}e^{\int_{0}^{t}F(u,z)du}g(z)dz-\int^{\infty}_{-\infty}\Phi\Big{(}\frac{x-z}{\sqrt{t}}\Big{)}g(z)dz,$
$\displaystyle=\int_{-\infty}^{\infty}\Big{(}1_{(-\infty,\
x]}(z)e^{\int_{0}^{t}F(u,z)du}-\Phi\Big{(}\frac{x-z}{\sqrt{t}}\Big{)}\Big{)}g(z)dz$
$\displaystyle=\int_{-\infty}^{x}\Big{(}e^{\int_{0}^{t}F(u,z)du}-\Phi\Big{(}\frac{x-z}{\sqrt{t}}\Big{)}\Big{)}g(z)dz-\int_{x}^{\infty}\Phi\Big{(}\frac{x-z}{\sqrt{t}}\Big{)}g(z)dz.$
(53)

Substituting in the last integral $w=2x-z$ yields

$\displaystyle\int_{x}^{\infty}\Phi\Big{(}\frac{x-z}{\sqrt{t}}\Big{)}g(z)dz=\int_{-\infty}^{x}\Phi\Big{(}\frac{w-x}{\sqrt{t}}\Big{)}g(2x-w)dw.$

Hence

$\displaystyle\psi(x)$
$\displaystyle=\int_{-\infty}^{x}\Big{[}\Big{(}e^{\int_{0}^{t}F(u,z)du}-\Phi\Big{(}\frac{x-z}{\sqrt{t}}\Big{)}\Big{)}g(z)-\Phi\Big{(}\frac{z-x}{\sqrt{t}}\Big{)}g(2x-z)\Big{]}dz$
$\displaystyle=\int_{-\infty}^{x}\Big{[}\Big{(}e^{\int_{0}^{t}F(u,z)du}-\Phi\Big{(}\frac{x-z}{\sqrt{t}}\Big{)}\Big{)}g(z)-\Big{(}1-\Phi\Big{(}\frac{x-z}{\sqrt{t}}\Big{)}\Big{)}g(2x-z)\Big{]}dz$
$\displaystyle=\int_{-\infty}^{x}\Big{[}e^{\int_{0}^{t}F(u,z)du}g(z)-\Phi\Big{(}\frac{x-z}{\sqrt{t}}\Big{)}\Big{(}g(z)-g(2x-z)\Big{)}-g(2x-z)\Big{]}dz.$

The last representation of $\psi$ clearly shows that $\psi$ is differentiable.
We have

$\displaystyle\psi^{\prime}(x)=e^{\int_{0}^{t}F(u,x)du}g(x)-g(x)+\int_{-\infty}^{x}\Big{[}$
$\displaystyle-\frac{1}{\sqrt{t}}\phi\Big{(}\frac{x-z}{\sqrt{t}}\Big{)}\Big{(}g(z)-g(2x-z)\Big{)}$
$\displaystyle+2\Phi\Big{(}\frac{x-z}{\sqrt{t}}\Big{)}g^{\prime}(2x-z)-2g^{\prime}(2x-z)\Big{]}dz.$

So if $\psi^{\prime}(x_{0})=0$, then

$\displaystyle
g(x_{0})\Big{(}1-e^{\int_{0}^{t}F(u,x_{0})du}\Big{)}=\int_{-\infty}^{x_{0}}\Big{[}$
$\displaystyle\frac{1}{\sqrt{t}}\phi\Big{(}\frac{x_{0}-z}{\sqrt{t}}\Big{)}\Big{(}g(2x_{0}-z)-g(z)\Big{)}$
(54)
$\displaystyle+2g^{\prime}(2x_{0}-z)\Big{(}\Phi\Big{(}\frac{x_{0}-z}{\sqrt{t}}\Big{)}-1\Big{)}\Big{]}dz=:I+II.$

We compute $I$ substituting $w=z-x_{0}$

$\displaystyle I$ $\displaystyle=\int^{0}_{-\infty}\frac{1}{\sqrt{2\pi
t}}e^{-\frac{w^{2}}{2t}}(g(x_{0}-w)-g(x_{0}+w))dw$
$\displaystyle=\mathbb{E}1_{\\{G\leq
0\\}}\big{(}g(x_{0}-\sqrt{t}G)-g(x_{0}+\sqrt{t}G)\big{)}.$

To compute $II$ we integrate by parts and substitute $w=z-x_{0}$

$\displaystyle II$
$\displaystyle=\Big{[}-2g(2x_{0}-z)\Big{(}\Phi\Big{(}\frac{x_{0}-z}{\sqrt{t}}\Big{)}-1\Big{)}\Big{]}^{x_{0}}_{-\infty}-2\int_{-\infty}^{x_{0}}g(2x_{0}-z)\frac{1}{\sqrt{t}}\phi\Big{(}\frac{x_{0}-z}{\sqrt{t}}\Big{)}dz$
$\displaystyle=g(x_{0})-2\int_{-\infty}^{0}\frac{1}{\sqrt{2\pi
t}}e^{-\frac{w^{2}}{2t}}g(x_{0}-w)dw=g(x_{0})-2\mathbb{E}1_{\\{G\leq
0\\}}g(x_{0}-\sqrt{t}G).$

Putting together just computed $I$, $II$ and comparing their sum with the LHS
of (54) yields (51). From this and from the fact that
$\lim_{x\rightarrow-\infty}\psi(x)=\lim_{x\rightarrow\infty}\psi(x)=0$, we
obtain

$\displaystyle\sup_{x\in\mathbb{R}}|\psi(x)|=\Big{|}\int_{-\infty}^{x_{0}}\Big{(}e^{\int_{0}^{t}F(u,z)du}-\Phi\Big{(}\frac{x_{0}-z}{\sqrt{t}}\Big{)}\Big{)}g(z)dz-\int_{x_{0}}^{\infty}\Phi\Big{(}\frac{x_{0}-z}{\sqrt{t}}\Big{)}g(z)dz\Big{|},$

for some $x_{0}$ solving equation (51). This proves (52), which finishes the
proof. ∎

## References

  * [1] Barrieu P., Rouault A., Yor M. A study of the Hartman-Watson distribution motivated by numerical problems related to the pricing of Asian options, J. Appl. Probab. 41, 1049-1058 (2004).
  * [2] Berestycki, H., J. Busca, and I. Florent _Computing the implied volatility in stochastic volatility models_ , Communications on Pure and Applied Mathematics 57(10), 1352–1373 (2004).
  * [3] Borodin A., Salminen P., Handbook of Brownian Motion - Facts and Formulae, Birkhäuser (2nd ed.), 2002.
  * [4] Carmona R., Durrleman V. Pricing and hedging spread options, SIAM Rev. 45, 627-685 (2003).
  * [5] Dobler Ch., Peccati G. The fourth moment theorem on the Poisson space, Ann. Probab. 46 (4) 1878-1916 (2018).
  * [6] Eqworld - The World of Mathematical equations,
http://eqworld.ipmnet.ru/index.htm

  * [7] Fouque Jean-Pierre, George Papanicolaou, and K Ronnie Sircar, _Derivatives in financial markets with stochastic volatility_ , Cambridge University Press, (2000).
  * [8] Fouque, J.-P., G. Papanicolaou, R. Sircar, and K. Solna _Multiscale stochastic volatility for equity, interest rate, and credit derivatives_ , Cambridge: Cambridge University Press, (2011).
  * [9] Gulisashvili, A. Analytically Tractable Stochastic Stock Price Models, Springer Finance, 2012.
  * [10] Gerhold, S. The Hartman-Watson distribution revisited: asymptotics for pricing Asian options, J. Appl. Probab. Volume 48, Number 3, 892-899 (2011).
  * [11] Hagan P., Kumar D., Lesniewski A., Woodward D. Managing smile risk. Wilmott Magazine, September, 84-108 (2002).
  * [12] Hull, J., White, A. The pricing of options on assets with stochastic volatilities. J. Finance 42, 281-300 (1987).
  * [13] Ikeda N., Watanabe S. Stochastic Differential Equations and Diffusion Processes. North-Holland Kodansha 1981.
  * [14] Jakubowski J., Wisniewolski M., _Another Look at the Hartman-Watson Distributions_ , Potential Anal 53, 1269–1297 (2020).
  * [15] Jakubowski J., Wisniewolski M., _A convolution formula for the local time of an Itô diffusion reflecting at $0$ and a generalized Stroock-Williams equation_, Bernoulli 27(3), 2021, 1870–1898.
  * [16] Jakubowski J., Wiśniewolski M. Probabilistic representations of the density function of the asset price and of vanilla options in linear stochastic volatility models. (2009) arXiv:0909.4765.
  * [17] Jakubowski J., Wiśniewolski M.  Revisiting linear and lognormal stochastic volatility models. Banach Center Publications 122, 169-185 (2020).
  * [18] Jourdain B. Loss of martingality in asset price model with log-normal stochastic volatility. ENPC-CERMICS, Working paper (2004).
  * [19] Lacour C. _Nonparametric estimation of the stationary density and the transition density of a Markov chain_ , Stochastic Processes and their Applications, Volume 118, Issue 2, 2008, Pages 232-260,
  * [20] Lebedev, N.N. Special Functions and their Applications, Dover,(1972). New York.
  * [21] Mikosch, T. Elementary Stochastic Calculus with Finance in View, World-Scientific (1998).
  * [22] Musiela M., Rutkowski M. _Martingale methods in financial modelling_ , Vol. 36. Springer Science and Business Media, (2006).
  * [23] Matsumoto H., Yor M. Exponential functionals of Brownian motion, I, Probability laws at fixed time. Probab. Surveys 2, 312-347 (2005).
  * [24] Rebonato R.  Volatility and Correlation. The Perfect Hedger and the Fox. Wiley (2nd ed.) 2004.
  * [25] Revuz D., Yor M. Continous Martingales and Brownian Motion. Springer-Verlag (3rd ed.). 2005.
  * [26] Rudin W., _Functional Analysis_ , McGraw-Hill (2nd ed.), 1991.
  * [27] Sart M. _Estimation of the transition density of a Markov chain_ , Ann. Inst. H. Poincaré Probab. Statist. 50 (3) 1028 - 1068, August 2014.
  * [28] Sin C. Complications with stochastic volatility models. Adv. in Appl. Probab. 30, 256-268, (1998).
  * [29] Sheu, S.-J. _Some Estimates of the Transition Density of a Nondegenerate Diffusion Markov Process_ , The Annals of Probability, 19(2), 538–561, (1991)
  * [30] Stanislaw, J. S., Werner, E. (1999). _A nonsymmetric correlation inequality for Gaussian measure_ , Y. Multivariate Anal., 68, 193-211.
  * [31] Wilcox R. _Introduction to Robust Estimation and Hypothesis Testing_ , Academic Press (Fourth Edition), 2017.