# Martingale Problem

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 Revision as of 00:35, 20 November 2012 (view source)Nestor (Talk | contribs) (Created page with "{{stub}} (Classical Martingale Problem) Given a (local) linear operator $\mathcal{L}: C^2(\mathbb{R}^n) \to C(\mathbb{R}^n)$ Martingale Problem for $\mathcal{L}$ consists in fin...")← Older edit Latest revision as of 00:48, 20 November 2012 (view source)Nestor (Talk | contribs) Line 1: Line 1: {{stub}} {{stub}} - (Classical Martingale Problem) Given a (local) linear operator $\mathcal{L}: C^2(\mathbb{R}^n) \to C(\mathbb{R}^n)$ Martingale Problem for $\mathcal{L}$ consists in finding for each $x_0 \in \mathbb{R}^d$ a probability measure $\mathbb{P}^{x_0}$ over the space of all continuous functions $X: [0,+\infty) \to \mathbb{R}^d$  such that + (Classical Martingale Problem) Given a (local) linear operator $\mathcal{L}: C^2(\mathbb{R}^n) \to C(\mathbb{R}^n)$ Martingale Problem for $\mathcal{L}$ consists in finding for each $x_0 \in \mathbb{R}^d$ a probability measure $\mathbb{P}^{x_0}$ over the space of all continuous functions $X: [0,+\infty) \to \mathbb{R}^d$  such that \begin{equation*} \begin{equation*} Line 7: Line 7: \end{equation*} \end{equation*} - and whenever $f \in C^2(\mathbb{R}^2)$ we have that + and whenever $f \in C^2(\mathbb{R}^d)$ we have that \begin{equation*} \begin{equation*} Line 13: Line 13: \end{equation*} \end{equation*} - is a [[Martingale| Local Martingale]] under $\mathbb{P}^{x_0}$ + is a [[Martingale| Local Martingale]] under $\mathbb{P}^{x_0}$. If for any $x \in \mathbb{R}^d$ there exists a unique $\mathbb{P}^x$ satisfying the above conditions we say that the Martingale Problem for $\mathcal{L}$ is well posed. + + Note that existence for the Martingale Problem implies existence of solutions to the Cauchy problem for the operator $\mathcal{L}$ with initial data in $C^2(\mathbb{R}^d)$, indeed, if $f \in C^2(\mathbb{R}^d)$ then we can define + + \begin{equation*} + u(x,t) = \mathbb{E}_{\mathbb{P}^x} [ f(X_t) ] + \end{equation*} + + Then it can be shown that $u(x,t)$ is a classical solution to the problem + + \begin{equation*} + \left \{ \begin{array}{ll} + u_t  = \mathcal{L}u & \text{ in } \mathbb{R}^d\times \mathbb{R}_+\\ + u\;\;  = f & \text{ for } t=0 + \end{array}\right. + \end{equation*} + + Thus at first glance it would seem that  well-posedness for the Martingale Problem for $\mathcal{L}$ is a stronger fact than that the well-posedness for the corresponding Cauchy problem, but often one can turn this around and show the well-posedness for the Martingale Problem after understanding the Cauchy problem for $\mathcal{L}$ well enough. + + The oldest example of a Martingale Problem is when $\mathcal{L}= \Delta$, the classical Laplacian, in this case the unique solution $\mathbb{P}^{x}$ corresponds to the standard Brownian motion or Wiener process. Using this as a base case and using basic Ito Calculus one can solve the corresponding problem for a generic second order operator + + \begin{equation*} + \mathcal{L} u = \text{Tr}(A(x)D^2u(x))+b(x)\cdot \nabla u(x)+c(x)u(x) + \end{equation*} + + where $A(x)\geq 0$, $c(x)\leq 0$ and $A,b,c$ are all Lipschitz functions of $x$. The case where the coefficients are merely continuous is much harder and it is an important result of Stroock and Varadhan that says still in this case the Martingale Problem is well posed.

## Latest revision as of 00:48, 20 November 2012

(Classical Martingale Problem) Given a (local) linear operator $\mathcal{L}: C^2(\mathbb{R}^n) \to C(\mathbb{R}^n)$ Martingale Problem for $\mathcal{L}$ consists in finding for each $x_0 \in \mathbb{R}^d$ a probability measure $\mathbb{P}^{x_0}$ over the space of all continuous functions $X: [0,+\infty) \to \mathbb{R}^d$ such that

\begin{equation*} \mathbb{P}^{x_0}\left ( X(0)=x_0 \right ) = 1 \end{equation*}

and whenever $f \in C^2(\mathbb{R}^d)$ we have that

\begin{equation*} f(X(t))-f(X(0))-\int_0^t \mathcal{L} f(X(s)) \;ds \end{equation*}

is a Local Martingale under $\mathbb{P}^{x_0}$. If for any $x \in \mathbb{R}^d$ there exists a unique $\mathbb{P}^x$ satisfying the above conditions we say that the Martingale Problem for $\mathcal{L}$ is well posed.

Note that existence for the Martingale Problem implies existence of solutions to the Cauchy problem for the operator $\mathcal{L}$ with initial data in $C^2(\mathbb{R}^d)$, indeed, if $f \in C^2(\mathbb{R}^d)$ then we can define

\begin{equation*} u(x,t) = \mathbb{E}_{\mathbb{P}^x} [ f(X_t) ] \end{equation*}

Then it can be shown that $u(x,t)$ is a classical solution to the problem

\begin{equation*} \left \{ \begin{array}{ll} u_t = \mathcal{L}u & \text{ in } \mathbb{R}^d\times \mathbb{R}_+\\ u\;\; = f & \text{ for } t=0 \end{array}\right. \end{equation*}

Thus at first glance it would seem that well-posedness for the Martingale Problem for $\mathcal{L}$ is a stronger fact than that the well-posedness for the corresponding Cauchy problem, but often one can turn this around and show the well-posedness for the Martingale Problem after understanding the Cauchy problem for $\mathcal{L}$ well enough.

The oldest example of a Martingale Problem is when $\mathcal{L}= \Delta$, the classical Laplacian, in this case the unique solution $\mathbb{P}^{x}$ corresponds to the standard Brownian motion or Wiener process. Using this as a base case and using basic Ito Calculus one can solve the corresponding problem for a generic second order operator

\begin{equation*} \mathcal{L} u = \text{Tr}(A(x)D^2u(x))+b(x)\cdot \nabla u(x)+c(x)u(x) \end{equation*}

where $A(x)\geq 0$, $c(x)\leq 0$ and $A,b,c$ are all Lipschitz functions of $x$. The case where the coefficients are merely continuous is much harder and it is an important result of Stroock and Varadhan that says still in this case the Martingale Problem is well posed.