7
$\begingroup$

Suppose that $(\Omega,\mathscr{F},P)$ is a complete probability space equipped a filtration $\{\mathscr{F}_t\}$ satisfying the usual conditions. $B_t$ is a 1-dimentional Brownian motion with respect to the filtration $\{\mathscr{F}_t\}$. Consider the 1-dimentional SDE

\begin{equation}dx(t)=f(x(t),t)dt+g(x(t),t)dB(t),\ (*)\end{equation}

with the initial value $x(t_0)=\xi\in L^2(\Omega;\mathbb{R})$, which statisfies the Lipschitz condition and the Linear growth condition.

Let \begin{equation}\Omega_0=\{\omega,\ there\ exist\ at\ least\ two\ different\ sample\ paths\ x_{\alpha}(t,\omega), x_{\beta}(t,\omega)\}\end{equation} Where $x_{\alpha}(t)$, $x_{\beta}(t)$ are both solutions of $(*)$.

By the existence and uniqueness theorem, $(*)$ has the pathwise unique solution, which means that if there are two solutions $x(t)$ and $\overline{x}(t)$, then $P\{x(t)=\overline{x}(t),\forall t\ge t_0\}=1$. However, from the proof(...$E\big(\sup_{t_0\le t\le T}|x(t)-\overline{x}(t)|^2\big)=0$) of existence and uniqueness theorem(cf. Stochastic Differential Equations and Applications(second Edition) by Xuerong Mao, page 53), we know that for any two solutions $x(t)$, $\overline{x}(t)$, there is a $P$-null set $\Omega_0(x,\overline{x})$ such that $\forall\omega\in\Omega\setminus\Omega_0(x,\overline{x})$, $x(t)=\overline{x}(t), \forall t\ge t_0$, if there are another two solutions $y(t)$, $\overline{y}(t)$, the $P$-null set $\Omega_0(y,\overline{y})$ may not equal to $\Omega_0(x,\overline{x})$.

It is straightforward to see that \begin{equation}\Omega_0=\bigcup_{x,y\ are\ solutions\ of\ (*)} \Omega_0(x,y)\end{equation}

My question is: whether $P(\Omega_0)=0$. In other words, whether we have a unique solution to the following equation for almost surely fixed $\omega$?

\begin{equation}x(t,\omega)=x(t_0,\omega)+\int_{t_0}^tf(x(s,\omega),s)ds+\int_{t_0}^tg(x(s),s)dB(s)(\omega)\end{equation}

Of course, if $P(\Omega_0)\ne0$, we can construct two new processes destroying the pathwise uniqueness, but are these prosesses $\mathscr{F}_t$-adapted?

Thank you.

$\endgroup$
1
  • 1
    $\begingroup$ In general there is a modification of the solution that is actually continuous in (x,t), and generates a flow of diffeomorphism. See Kunita's St Flour lecture notes for a discussion of this theory. $\endgroup$ – ofer zeitouni Dec 8 '13 at 17:56
3
$\begingroup$

The answer really depends on what you mean exactly by "a solution" to your SDE. If it is a solution in the classical (Itô calculus) sense then, since these can always be modified on a null set, you can never hope to have a property of the type you want, as correctly pointed out by thomas. Maybe a more interesting question then is whether there does exist a notion of solution that is sufficiently strong so that it completely determines the solution on some fixed set of full measure.

If $f$ and $g$ are sufficiently smooth ($\mathcal{C}^{2,\delta}$ for some $\delta > 0$ is enough), then this is indeed the case. In the case you consider (i.e. when $B$ is one-dimensional) this was pointed out by Doss and Sussmann in the seventies. (See Doss, Ann. IHP 13 1977 and Sussmann, AoP 6 1978.) In the case when $B$ is multidimensional, this can still be done, but it is a little bit more complicated (see Lyons, Rev. Mat. Iberoamericana 14 1998 or my recent book with Peter Friz).

$\endgroup$
1
$\begingroup$

I do not provide a proof here, but I guess the answer to the question "Is it true that $P(\Omega_0)=0$" is NO. I will give an answer to a simpler, yet similar question which is inspired by a great article of Tom Kurtz (see http://ejp.ejpecp.org/article/view/431).

It is helpful to clarify what we mean by the solution of the stochastic equation. Let $S_1,S_2$ be Polish spaces and $f:S_1 \times S_2 \to \mathbb{R}$ be measurable. Suppose that $\nu = \mathcal{L}[Y] \in \mathcal{M}_1 (S_2)$ is given, i.e. $Y$ is an $S_2$-valued random variable. We want to find a solution $X$ to the equation: $$ f(X,Y) = 0 .$$ More precisely, we want a law $\mu \in \mathcal{M}_1(S_1 \times S_2)$ such that its projection on the second coordinate is identical to the law $\nu$ of $Y$:

  • $(\pi_y)_* \mu = \nu$,

  • $ f(x,y) = 0, \qquad \mu (dx, dy) \text{-a.s.} $

The set of such solution measures $\mu$ will be called $\mathcal{S}_{f,\nu}$. This can also be formulated more abstractly (similar to the definition of SDE solutions): A solution to $f(X,Y) = 0$ is any probability space $(\Omega, \mathcal{A}, P)$ which hosts two random variables $X$ and $Y$ such that

  • $\mathcal{L}[Y] = \nu,$

  • $f(X,Y) = 0, \quad P \text{-a.s.} $

Kurtz defines in his article the notion of pointwise uniqueness (which is just pathwise uniqueness, if there really is a path): If $(X_1,Y)$ and $(X_2,Y)$ are two solutions on the same probability space, then for $\lambda_{1,2} := \mathcal{L} (X_1,X_2,Y) \in \mathcal{M}_1(S_1\times S_1 \times S_2)$ it holds that: $$ \lambda_{1,2}(X_1 \neq X_2 ) =0 .$$

Your question in the setting here: Does the pointwise uniqueness imply that for $\lambda_I \in \mathcal{L}[(X_i)_{i\in I}, Y] \in \mathcal{M}_1(S_1^I \times S_2)$ such that all projections are solutions measures $(\pi_{x_i},\pi_y)_* \lambda_I \in \mathcal{S}_{f,\nu}$ it holds that $$ \lambda_I (\exists i,j \in I : \, x_i \neq x_j ) = 0 \ ? $$

The answer in general is no as can be seen by the example which is always used when the probability of an uncountable union is shown not always to be the (uncountable) sum of the probabilities: Let $\Omega = [0,1], P$ = Lebesgue measure restricted to $\Omega$ and $f(x,y) = x-y$, $x,y \in S_1=S_2 =\mathbb{R}$. Suppose $\nu = \delta_a$ i.e. $Y=a$ and $X_i(\omega) = b 1(\omega = i) + a 1(\omega \neq i)$, $\omega \in \Omega$, $i \in I = [0,1]$ and $a\neq b \in \mathbb{R}$. Then for all $i \in I$: the pair $(X_i, Y)$ is a solution to the equation $f(X,Y) = 0$ in the previous sense. However for $\mu_I = \mathcal{L}[(X_i)_{i\in I},Y]$: $$ \mu_I(\exists i,j \in I : \, x_i \neq x_j ) \geq \mu_I (\exists j \in I:\, x_0 \neq x_j) = P( [0,1] ) = 1. $$

The answer to the general SDE question should go along similar lines. This should also show you that even if you can construct new solutions, e.g. $Z_{i,j} = b 1(\omega = i) + b 1(\omega =j) + a 1(\omega \neq i, \omega \neq j)$ you will not get away from the pointwise uniqueness.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.