\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 $$\label{bsde} x_t = \xi + \int_t^T f(s, x_s,z_s)\, ds - \int_t^T z_s \,dW_s.$$ 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 $$\label{aux} h_t = \xi_2 + \int_t^T g(s,\max\{h_s,y_s\},z_s^h)ds - \int_t^T z^h_s dW_s.$$ 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 $$\label{diff} y_t-h_t = \xi^1-\xi^2+\int_t^T \left[f(s,y_s,z^1_s) - g(s,\max\{h_s,y_s\},z^1_s)\right] \, ds - \int_t^T \widehat{Z}_s \, d\widehat{W}_s.$$ 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 $$\label{eqnh} \begin{split} &\mathbb{I}_A (y_{t^*}-h_{t^*}) \\ & = \int_{t^*}^{\tau} \mathbb{I}_A \left[f(s,y_s,z^1_s) - g(s,\max\{h_s,y_s\},z^1_s)\right]\, ds - \int_{t^*}^{\tau} \mathbb{I}_A \widehat{Z}_s \,d\widehat{W}_s, \end{split}$$ 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 $$H_t = \int_t^T \left[f(s,y_s,z^1_s) - g(s,x_s,z^1_s)\right]\, ds - \int_t^T \widehat{Z}_s \, d \widehat{W}_s$$ 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 \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} 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} \begin{thebibliography}{0} \bibitem{Briand} P. Briand, B. Delyon and J. Memin; {Donsker-Type theorem for BSDEs}, \emph{Electronic Communications in Probability,} \textbf{6} (2001), 1--14. \bibitem{Karlsen} F. E. Benth, K. H. Karlsen and K. Reikvam; A semilinear Black and Scholes partial differential equation for valuing American options, \emph{Finance and Stochastics,} \textbf{7} (2003), 277--298. \bibitem{Jia} G. 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