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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2011 (2011), No. 110, pp. 1--6.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2011 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2011/110\hfil Comparison and existence theorems]
{Comparison and existence theorems for backwards stochastic
DE's with discontinuous generators}

\author[N. Halidias, P. E. Kloeden\hfil EJDE-2011/110\hfilneg]
{Nikolaos Halidias, Peter E. Kloeden}  % in alphabetical order

\address{Nikolaos Halidias\newline
Department of Statistics and Actuarial-Financial Mathematics\newline
University of the Aegean\\
Karlovassi  83200  Samos, Greece}
 \email{nick@aegean.gr}

\address{Peter E. Kloeden \newline
Institut f\"ur Mathematik, Goethe-Universit\"at \\
D-60054 Frankfurt am Main, Germany}
\email{kloeden@math.uni-frankfurt.de}

\thanks{Submitted May 25, 2010. Published August 26, 2011.}
\thanks{This work was done while N. Halidias
visited the Institut f\"ur Mathematik, Goethe-Universit\"at,
\hfill\break\indent
partially supported by  DAAD}
\subjclass[2000]{60H10, 60H20}
\keywords{Backward stochastic differentail equation;
 comparison theorem; discontinuous generator}

\begin{abstract}
 An existence result is proved for backwards stochastic
 differential equations (BSDEs) with a generator
 $f(t,x,z)$ which is possibly discontinuous in the $x$ variable.
 For this comparison results are first established
 for BSDEs with the generator satisfying a generalized Lipschitz
 condition in its $x$ variable.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{remark}[theorem]{Remark}

\section{Introduction}

Let $W_t$ be a standard one-dimensional Wiener process defined on
the complete probability space $( \Omega, \mathcal{F}, \mathbb{P})$, let
 $\{ \mathcal{F}_t^W \}$ be the natural filtration
generated by the Wiener process and let $\{ \mathcal{F}_t \}$ be
the augmentation under $\mathbb{P}$ of this natural filtration.
In addition, let  $\mathcal{P}$ denote the $\sigma$-algebra of $\mathcal{F}_t$
progressively measurable subsets of $[0,T]\times\Omega$ and let
$H^p(\mathbb{R})$ be  the space  of $\mathcal{P}$-measurable
$X:[0,T] \times \Omega\to\mathbb{R}$ with $\| X \|^p:=\mathbb{E}
\int_0^T |X_s|^p \,ds <\infty$.

Suppose that  the mapping $f:[0,T]\times \mathbb{R} \times
\mathbb{R}\to\mathbb{R}$ is
$\mathcal{B}\otimes \mathcal{B}\otimes \mathcal{B}$-measurable and
consider the scalar backward
stochastic differential equation (BSDE) with
\begin{equation} \label{bsde}
x_t = \xi + \int_t^T f(s, x_s,z_s)\, ds - \int_t^T z_s \,dW_s.
\end{equation}


The classical existence theorem for BSDE states that if the
generator is globally Lipschitz in both variables then there
exists a strong solution. Lepeltier and San Martin
\cite{Lepeltier3} prove an existence result for the case where the
generator is only continuous in both variables. To do that they
use the classical comparison theorem and monotonicity arguments.
The comparison theorem proved in this note allows  the conditions
on the generator to be further relaxed. In particular, an
existence theorem is established in the case where
$x\to f(t,x,z)$ is left continuous. A similar result appears
in the paper by Jia \cite{Jia}, however our assumptions here
are more general because we use a comparison theorem which
holds for generators having super-linear growth  in the $x$ variable.

It is well known that BSDEs are related to PDEs, see for example
\cite{Mao}.   Benth  et
al \cite{Karlsen} showed that the nonlinearity in the semilinear
Black and Scholes equation depends discontinuously on the American
option value. Moreover, this discontinuity then appears in the
generator of the  associated BSDE, see also Karoui et al \cite{Karoui}.


\section{Comparison theorems}

Consider the  following scalar BSDEs:
\begin{gather}\label{bsde1}
y_t  =  \xi_1 + \int_t^T f(s,y_s,z^1_s)\, ds - \int_t^T z^1_s \,
dW_s,\\ \label{bsde2}
x_t  =  \xi_2 + \int_t^T g(s,x_s,z^2_s) \, ds - \int_t^T z^2_s
\, dW_s,
\end{gather}
and suppose that each admits  a unique solution, which is denoted
by $(y_t,z^1_t)$ and $(x_t,z^2_t)$, respectively.
(Note that $y_t$ and $x_t$ have continuous modifications).
 The generator $g$  satisfies the  assumption

\begin{itemize}

\item[(A1)] There exists   a constant   $K$
 such that
\[
  |g(t,x_1,z_1)-g(t,x_2,z_2)|^2 \leq \kappa (|x_1-x_2|^2)+K|z_1-z_2|^2,
 \quad \mbox{a.s.},
\]
for all $t, x_1,x_2,z_1,z_2$, where
$\kappa:\mathbb{R}_+\to\mathbb{R}_+$ is a concave increasing
function with $\kappa(0)=0$  and $\kappa (u)>0$ for
$u>0$ such that
\[
\int_{0^+} \frac{du}{\kappa (u)} = \infty.
\]
\end{itemize}

The following comparison theorem will be used later to prove the
existence of a solution to  \eqref{bsde} with the
generator $f$ being discontinuous in its second variable.


\begin{theorem} \label{thm1}
Let $(y_s,z^1_s)$ and $(x_s,z^2_s)$ be  the unique solutions of
 \eqref{bsde1} and \eqref{bsde2}, respectively.
 Suppose {\rm (A1)}  holds and, in addition, that
 $$
 f(t,y_s,z_s^1) \leq g(t,y_s,z^1_s) \quad  \mbox{for all }
  t \in [0,T], \mbox{ a.s.} \,.
 $$
Finally, suppose that
$\xi_1, \xi_2\in L^2(\Omega,\mathcal{F}, \mathbb{P})$ satisfy
 $\xi_1\leq \xi_2$. Then,  $y_t\leq x_t$,  a.s.,
 for all $t\in [0,T]$.
\end{theorem}


\begin{proof}  Consider the  auxiliary problem
\begin{equation} \label{aux}
h_t = \xi_2 + \int_t^T g(s,\max\{h_s,y_s\},z_s^h)ds
- \int_t^T z^h_s dW_s.
\end{equation}
The generator of this BSDE  is the random function
$$
\hat{g}(s,\omega,x,z) := g(s,\max\{x,y_s(\omega)\},z).
$$
 Since the function  $x\mapsto \max\{x,y_s(\omega)\}$
satisfies the Lipschitz condition with constant one and a growth
condition $|\max\{x,y_s(\omega)\}|\leq |x| + |y_s(\omega)|$,
it follows that $\hat{g}$ is Lipschitz in  $x$ and has linear growth.
Hence by  \cite[Theorem 7.4.1]{Mao} this auxiliary BSDE \eqref{aux}
has a unique solution.

We want to compare this solution $h_t$ with $y_t$. First,  define
$$
H_t := \int_t^T [f(s,y_s,z^1_s) -
g(s,\max\{h_s,y_s\},z^1_s)] \, ds
+\int_t^T b_s \widehat{Z}_s \, ds
-\int_{t}^{T} \widehat{Z} \, dW_s,
$$
where
$$
b_s :=
\frac{g(s,\max\{h_s,y_s\},z^1_s)-g(s,
\max\{h_s,y_s\},z^2_s)}{z^1_s-z^2_s}, \quad
 \widehat{Z}_s := z^1_s-z^2_s.
$$
Note  that $b_s$ is uniformly bounded by Assumption (A1).
Then write
\[
M(t) = \exp\Big(\int_0^t b_s dW_s - \frac{1}{2} \int_0^t |b_s|^2ds\Big),
\quad t \in [0,T],
\]
and define a new probability measure $\hat{\mathbb{P}}$ by
\[
\frac{d  \widehat{\mathbb{P}}}{d  \mathbb{P}} = M(T).
\]
By  Girsanov's theorem,
 $\widehat{W}_t:=W_t - \int_0^t b_s \, ds$ is a
$\widehat{\mathbb{P}}$-Wiener process.
Hence, under $\widehat{\mathbb{P}}$, the difference $y_t-h_t$
satisfies the equation
\begin{equation}\label{diff}
y_t-h_t =  \xi^1-\xi^2+\int_t^T \left[f(s,y_s,z^1_s) -
g(s,\max\{h_s,y_s\},z^1_s)\right] \, ds - \int_t^T \widehat{Z}_s \,
d\widehat{W}_s.
\end{equation}

It will now be shown  that $y_t\leq h_t$, a.s.,
for all $t\in [0,T]$. Suppose that this is not true.
Then, there exists some
$t^*$ such that $y_{t^*}>h_{t^*}$  on an event $A$ with
$\widehat{\mathbb{P}}(A)>0$. Note, that $A$ is
$\mathcal{F}_{t*}$ measurable, so the  indicator function
$\mathbb{I}_A$ of the event $A$ is  $\mathcal{F}_{t*}$ measurable.
Then, by \cite[Lemma 1.5.10]{Mao},
$$
\mathbb{I}_A \int_{t^*}^{\tau} \widehat{Z}_s \, d\hat{W}_s
= \int_{t^*}^{\tau} \mathbb{I}_A \widehat{Z}_s \, d\widehat{W}_s.
$$
 Define the  stopping time,
\[
\tau := \inf \{ t \in [t^*,T]  :  y_t  \leq h_t \}.
\]
Since  $y_{\tau}=h_{\tau}$ by continuity of $y_t$ and $h_t$,
it follows  from  \eqref{diff} that
$$
y_{t^*}-h_{t^*}
 = \int_{t^*}^{\tau} [f(s,y_s,z^1_s) -
g(s,\max\{h_s,y_s\},z^1_s)]\, ds - \int_{t^*}^{\tau} \widehat{Z}_s \,
d\widehat{W}_s.
$$
Multiplying this equation by $\mathbb{I}_A$ gives
\begin{equation}\label{eqnh}
\begin{split}
&\mathbb{I}_A (y_{t^*}-h_{t^*}) \\
& = \int_{t^*}^{\tau} \mathbb{I}_A
\left[f(s,y_s,z^1_s) - g(s,\max\{h_s,y_s\},z^1_s)\right]\, ds -
\int_{t^*}^{\tau} \mathbb{I}_A \widehat{Z}_s \,d\widehat{W}_s,
\end{split}
\end{equation}
Now $y_t\geq h_t$ on the stochastic interval $[t^*,\tau]$, so
$\max\{h_t,y_t \}=y_t$.
Finally,  taking the  expectation  on both sides of \eqref{eqnh}
gives
\begin{gather*}
\mathbb{E}(\mathbb{I}_A (y_{t^*}-x_{t^*})) \geq 0, \quad
\mathbb{E}\Big(\int_{t^*}^{\tau} \mathbb{I}_A \widehat{Z}_s
\,d\widehat{W}_s\Big) = 0, \\
\mathbb{E}\Big( \int_{t^*}^{\tau} \mathbb{I}_A
[f(s,y_s,z^1_s) - g(s,y_s,z^1_s)]\, ds\Big) \leq 0.
\end{gather*}
It follows that
 $\mathbb{E}\big(\mathbb{I}_A (y_{t^*}-h_{t^*})\big)=0$
and hence that $\hat{\mathbb{P}}(A)=0$, which is a contradiction.
 Thus, $h_t \geq y_t$ for all $t$, a.s. This means that $h_t=x_t$,
where $x_t$ is the unique solution of the BSDE  \eqref{bsde2}.
\end{proof}


\begin{remark} \label{rem1}\rm
Theorem \ref{thm1} is also valid for a random  generator $f$; i.e.,
for   a $\mathcal{P} \otimes \mathcal{B} \otimes
\mathcal{B}$-measurable  $f:[0,T] \times  \Omega  \times
\mathbb{R} \times \mathbb{R} \to\mathbb{R}$ with
values $f(t,\omega,x,z)$.
\end{remark}

A similar argument gives a comparison theorem that
applies to backward stochastic differential inequalities.
Backward stochastic differential inequalities are closely related
to self-financing super-strategies  in mathematical finance,
see for example  \cite[Definition 1.2]{Karoui}.

Consider the following two backward stochastic differential
inequalities:
\begin{gather}
y_t \leq \xi_1 + \int_t^T f(s,y_s,z^1_s)ds - \int_t^T z^1_s dW_s,
\\
x_t \geq \xi_2 + \int_t^T g(s,x_s,z^2_s)ds - \int_t^T z^2_s dW_s.
\end{gather}



\begin{theorem} \label{thm2}
Suppose that $x\mapsto f(t,x,z)$ is non-increasing and that
$x\mapsto g(t,x,z)$ is non-decreasing.
In addition, suppose that   $\xi_1\leq \xi_2$ with
$\xi_1$, $\xi_2\in L^2(\Omega,\mathcal{F}, \mathbb{P})$ and that
 $$
 f(t,x,z) \leq g(t,x,z) \quad \mbox{for all } t \in [0,T], \;
 (x,z) \in \mathbb{R}^2.
$$
Then $y_t\leq x_t$,  a.s., for all $t\in [0,T]$.
\end{theorem}


\begin{proof}  Define $H_t$ by
$$
H_t := \int_t^T [f(s,y_s,z^1_s) - g(s,x_s,z^1_s)] \, ds
+\int_t^T b_s \widehat{Z}_s \, ds -\int_{t}^{T} \widehat{Z} \, dW_s,
$$
where
$$
b_s := \frac{g(s,x_s,z^1_s)-g(s,x_s,z^2_s)}{z^1_s-z^2_s}, \quad
\widehat{Z}_s := z^1_s-z^2_s.
$$
As before, using Girsanov's theorem,  $H_t$ can be rewritten as
\begin{equation}
H_t = \int_t^T \left[f(s,y_s,z^1_s) - g(s,x_s,z^1_s)\right]\, ds -
\int_t^T \widehat{Z}_s \, d \widehat{W}_s
\end{equation}
with respect to the new probability measure $\widehat{\mathbb{P}}$
and corresponding Wiener process $ \widehat{W}_t$. Using the
assumptions that $x\to f(t,x,z)$ is non-increasing and
that $x\to g(t,x,z)$ is non-decreasing, it follows  that
$$
H_t \leq \int_t^T \left[f(s,\min\{x_s,y_s\},z^1_s) -
g(s,\min\{x_s,y_s\},z^1_s)\right] \, ds - \int_t^T \widehat{Z}_s \,
d\widehat{W}_s.
$$


Suppose now that there exists a $t^*$ such that $H_{t^*}>0$ on an
event $A$ with $\widehat{\mathbb{P}}(A)>0$. Then, multiplying the
above inequality by $\mathbb{I}_A$ and taking the expectation leads
to  a contradiction, since
$\mathbb{E}( \int_{t^*}^T \mathbb{I}_A \widehat{Z}_s\,
\widehat{W}_s)=0$. Hence, $H_t$ $\leq 0$ for all $t\in [0,T]$ and
it  follows that
$$
y_t - x_t \leq \xi_1 - \xi_2 + H_t \leq 0
$$
for all $t \in [0,T]$, a.s.
\end{proof}

\section{An existence theorem}

The first comparison theorem, Theorem \ref{thm1},  will now be applied
to scalar BSDEs for which the generator $f$ is  not necessarily
continuous in $x$.
In particular, $f$ is now assumed to  satisfy the following
assumptions:
\begin{itemize}
\item[(A2)]   The generator $f:[0,T] \times
 \mathbb{R} \times \mathbb{R} \to\mathbb{R}$
is $\mathcal{B} \otimes \mathcal{B} \otimes
\mathcal{B}$-measurable and  satisfies
\begin{itemize}
\item[(i)] The mapping $f$ has the form
$f(t,x,z)=f_1(t,x,z)+f_2(t, x,z)$, where $f_1(t,x,z)$
is continuous in all variables and satisfies (A1),
while  $f_2$ is continuous in $t$ and $z$, is an increasing
function of $x$, and satisfies a linear growth condition for
both variables; i.e.,
$$
|f_2(t,x,z)| \leq C(|x|+|z|+1),
$$
and is possibly discontinuous in $x$, but is right or left
continuous.


\item[(ii)] There exist  a  $K$  such that
\[
|f_2(t,x,z_1)-f_2(t,x,z_2)| \leq K |z_1-z_2|
\]
for all $t,x, x_1,x_2,z,z_1,z_2$.

\end{itemize}

\item[(A3)] There exists  functions $g_1(t,x,z)$ and $g_2(t,x,z)$
satisfying (A1)   such that
$$
g_1(t,x,z) \leq f(t,x,z) \leq  g_2(t,x,z)  \quad \mbox{for all }
  t,x,z.
$$


\item[(A4)]  $\mathbb{E} (|\xi|^2 )<\infty$.

\end{itemize}


\begin{theorem} \label{thm3}
Suppose that   Assumptions {\rm (A2), (A3), (A4)} hold. Then
 \eqref{bsde}  has at least one solution.
\end{theorem}

\begin{proof}
The solution will be obtained as the limit of   an increasing or a
decreasing sequence, which is constructed as follows.

Firstly, note that the  BSDEs
\begin{gather*}
L_t  =  \xi + \int_t^T g_1(s,L_s,z^L_s)\, ds - \int_t^T z^L_s \,dW_s,
\\
U_t  = \xi + \int_t^T g_2(s,U_s,z^U_s)\, ds - \int_t^T z^U_s \,dW_s
\end{gather*}
admit  unique solutions. Then consider a sequence $(y^ n_t,z^ n_t)$ of
stochastic processes obtained as the solutions of the BSDEs
\[
y_t^n = \xi + \int_t^T [f_1(t,y^n,z^n) + f_2(t,y^{n-1},z^n)]\, ds -
\int_t^T z^n \, dW_s,
\]
with $y_t^0=L_t$. These solutions exist by
\cite[Theorem 7.4.1]{Mao}.
In particular, note that the Assumption   (A4) on the final
value $\xi$ ensures that $y_t^n\in H^2(\mathbb{R})$.


We will now prove that $y^1_t\geq L_t$. For this we have to compare
a BSDE with generator $g_1(t,x,z)$ and
a BSDE with random generator
$$
\hat{f}(t,\omega,x,z) := f_1(t,x,z) + f_2(t,L_t(\omega),z).
$$
It is clear that $\hat{f}$ satisfies the conditions of the comparison
theorem, Theorem  \ref{thm1}, and  that
$g_1(t,L_t,z^L_t)\leq \hat{f}(t,L_t,z^L_t)$. Hence
$L_t\leq y^1_t$. It follows by the same argument that
$y^n_t\geq y^{n-1}_t$, a.s., for each $n\in \mathbb{N}$.

It also follows similarly  that $y^1_t\leq U_t$ and
 hence that $y^{n}_t\leq U_t$ for each $n\in \mathbb{N}$.


Now it is easy to show that $y^n_t\to y^*_t$ in $H^2(\mathbb{R})$,
where $y^n_t\leq y^*_t\leq y_t$, using the Lebesgue
Dominated Convergence Theorem and the fact that $y^n$ is an
increasing and bounded sequence. To show that $z^n\to z$ in
$H^2(\mathbb{R})$
we apply the  It\^{o} formula to $|y_n-y_m|^2$ and  obtain
\begin{equation} \label{ito1}
\begin{aligned}
&\mathbb{E}|y_n-y_m|^2 + \mathbb{E}\int_t^T|z_n-z_m|^2\, ds \nonumber\\
&=  2 \mathbb{E} \int_t^T(y_n-y_m)\big(f_1(s,y_n,z_n)
- f_1(s,y_m,z_m)\big)\, ds   \\
&\quad+ 2 \mathbb{E} \int_t^T(y_n-y_m)\big(f_2(s,y_n,z_n)
- f_2(s,y_m,z_m)\big)\, ds. %\label{ito2}
\end{aligned}
\end{equation}
Using the inequality $2 |yz|\leq \frac{y^2}{\varepsilon}
+\varepsilon z^2$, the first term  on right-hand side
can be estimated by
\begin{align*}
&2 \mathbb{E} \int_t^T(y_n-y_m)
\Big(f_1(s,y_n,z_n) - f_1(s,y_m,z_m)\Big)\, ds\\
& \leq  \varepsilon \mathbb{E} \int_t^T |y_n-y_m|^2ds
  + \frac{1}{\varepsilon}\mathbb{E} \int_t^T |f_1(s,y_n,z_n)
 - f_1(s,y_m,z_m)|^2 \, ds
\\
& \leq   \varepsilon \mathbb{E} \int_t^T |y_n-y_m|^2 \, ds
+  \frac{1}{\varepsilon} \mathbb{E} \int_t^T \kappa (|y_n-y_m|^2)\, ds
+ \frac{1}{\varepsilon} \mathbb{E} \int_t^T |z_n-z_m|^2\, ds.
\end{align*}
The same arguments can be used to estimate the second term  on
right-hand side of \eqref{ito1}. Finally, choosing a suitable
$\varepsilon$,  it follows that $\{z^n\}$ is a Cauchy sequence
in $H^2(\mathbb{R})$. Hence, $(y_n,z_n)\to(y^*,z^*)$,
which is a solution of \eqref{bsde}.
\end{proof}

\begin{remark} \label{rmk2} \rm
The above  results can be applied for BSDEs with a  generator
of the form
$$
f(t,x,z) =  f_1(t,x,z)+ H(x-1)x+z;
$$
where $H(x)$ is the Heaviside function and $f_1(t,x,z)$ satisfies assumption (\textbf{A1}). Here one can take
$$
g_1(t,x,z) = f_1(t,x,z)+z, \quad g_2(t,x,z) = f_1(t,x,z)+ H(x)x+z.
$$
\end{remark}


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\end{document}