stochastic-calculusstochastic-differential-equationsstochastic-processes
Consider the one-dimensional stochastic differential equation:
$$dX_t = {\bf 1}_{\{X_t>0\}}\big(b(t,X_t)dt + a(t,X_t)dW_t\big),\quad \forall t>0,$$
or equivalently
$$dX_t = b(t,X_t)dt + a(t,X_t)dW_t,\quad \forall t\le \tau,\quad \mbox{with } \tau:=\inf\{t\ge 0: X_t\le 0\},$$
where $(W_t)_{t\ge 0}$ is a standard Brownian motion and $\mathbb P(X_0>0)=1$. Under which conditions on the coefficients $b, a$ does pathwise uniqueness or even uniqueness in law hold?
Let $\tau = \inf\{ t>0 : W_t = 1 \}$. The conjecture is true and the essence of the proof outlined below appears to be the following peculiar property of the hitting time $\tau$: $$ \lim_{\epsilon \searrow 0} P_{\epsilon}[\tau > \epsilon-\epsilon^{3/2}] = 1 \;. $$ (This can be computed directly using the fact that distribution of $\tau$ is inverse gamma with parameters $1/2$ and $1/2$.) In order to leverage this property, one must carefully split up $e_t:= X_t - Y_{t/\epsilon}$ as outlined below.
Theorem. It holds that: $$\lim_{\epsilon \searrow 0} E[ \sup_{0 \le t \le \epsilon} |e_t|^2 ] = 0 \;.$$
Proof. The proof shows that for any $T \in [0, \epsilon]$ $$ E_{\epsilon} [ \sup_{0 \le t \le T} |e_t|^2 ] \le C_1(\epsilon) + C_2 (\epsilon) \int_0^T E_{\epsilon} [ \sup_{0 \le r \le s} |e_r|^2 ] ds $$ where $C_1(\epsilon)$ and $C_2 (\epsilon) $ are non-negative and $\lim_{\epsilon \searrow 0} C_1(\epsilon) =0$ and $\lim_{\epsilon \searrow 0} C_2(\epsilon) \epsilon = O(1)$. By Grönwall's inequality, $$ E_{\epsilon} [ \sup_{0 \le t \le \epsilon} |e_t|^2 ] \le C_1(\epsilon) \exp(C_2 (\epsilon) \epsilon) \;, $$ and then passing to the limit gives the required result. The remaining details follow.
By Itô's formula, \begin{align*} & |e_t|^2 = \mathrm{I} + \mathrm{II} + \mathrm{III} \quad \text{where} \\ & \mathrm{I}:= \frac{2}{\epsilon} \int_0^t e_s (\sigma(X_s) - \sigma(Y_{s/\epsilon})) ds \;, \\ & \mathrm{II}:= \frac{2}{\epsilon} \int_0^t e_s \sigma(X_s) (\epsilon dW_s - ds) \;, \\ & \mathrm{III}:= \int_0^t \sigma(X_s)^2 ds \;. \end{align*}
Estimate for $\mathrm{I}$.
This term exclusively contributes to $C_2(\epsilon)$. Since $\sigma$ is $L$-Lipschitz for some $L>0$ $$ I \le \frac{2 L}{\epsilon} \int_0^t |e_s|^2 ds $$ and thus $$ \sup_{0 \le t \le \epsilon} I \le \frac{2 L}{\epsilon} \int_0^{\epsilon} |e_s|^2 ds \le \frac{2 L}{\epsilon} \int_0^{\epsilon} \sup_{0 \le r \le s} |e_r|^2 ds \;. $$ Thus, $C_2(\epsilon) = 2 L / \epsilon$.
Estimate for $\mathrm{II}$.
This term contributes to $C_1(\epsilon)$, and here is where we leverage the aforementioned peculiar property of $\tau$.
\begin{align*}
& \lim_{\epsilon \searrow 0} E_{\epsilon} [ \sup_{0 \le t \le \epsilon} \left| \mathrm{II} \right| ] =
\lim_{\epsilon \searrow 0} E_{\epsilon} [ \sup_{0 \le t \le \epsilon} \left| \mathrm{II} \right| \mathbf{1}_{ \{ \tau > \epsilon - \epsilon^{3/2} \} } ] \\
& \quad =
\lim_{\epsilon \searrow 0} E [ \sup_{0 \le t \le \epsilon} \left| 2 \int_0^t e_s \sigma(X_s) dW_s \right| ] = 0
\end{align*}
Here we took 3 steps that are explained in detail below.
In the first step, we used Cauchy-Schwarz to show that $$ \left( E_{\epsilon} \sup_{0 \le t \le \epsilon} |\mathrm{II}| \mathbf{1}_{\{\tau < \epsilon - \epsilon^{3/2} \} } \right)^2 \le \underbrace{E[\sup_{0 \le t \le \epsilon} |\mathrm{II}|^2 ]}_{\to O(1)} \, \underbrace{P_{\epsilon}[ \tau < \epsilon - \epsilon^{3/2}]}_{\to 0} $$
In the second step, we used a natural splitting and the triangle inequality to write, \begin{align*} & E_{\epsilon} \sup_{0 \le t \le \epsilon} |\mathrm{II}| \mathbf{1}_{\{\tau > \epsilon - \epsilon^{3/2} \} } \le \mathrm{II}_a + \mathrm{II}_b \quad \text{where} \\ & \mathrm{II}_a := E_{\epsilon} \sup_{0 \le t \le \epsilon} \frac{2}{\epsilon} \left| \int_0^{t \wedge \tau} e_s \sigma(X_s) (\epsilon dW_s - ds) \right| \mathbf{1}_{\{\tau > \epsilon - \epsilon^{3/2} \} }\\ & \mathrm{II}_b := E_{\epsilon} \sup_{0 \le t \le \epsilon} \frac{2}{\epsilon} \left| \int_{t \wedge \tau}^t e_s \sigma(X_s) (\epsilon dW_s - ds) \right| \mathbf{1}_{\{\tau > \epsilon - \epsilon^{3/2} \} } \;. \end{align*} To estimate these terms, there are two cases to consider:
In other words, conditioned on the event $(\tau < \epsilon)$, the law of $\epsilon W_s - s$ is equal to the law of a standard Brownian bridge.
In the third and last step, we used Doob's martingale inequality, Itô isometry, and (standard) a priori bounds on $X_t$ and $Y_{t/\epsilon}$ over $(0,\epsilon)$. Since the estimate of this term is almost identical to the estimate of $\mathrm{III}$ given below, the details are suppressed.
Estimate for $\mathrm{III}$.
This term also contributes to $C_1(\epsilon)$. Noting that $\sigma$ is $L$-Lipschitz, \begin{align*} E_{\epsilon} [ \sup_{0 \le t \le \epsilon} \mathrm{III} ] &= E_{\epsilon}[ \int_0^{\epsilon} [ \sigma(X_s)^2 ds ] \\ &\le 2 \epsilon \sigma(0)^2 + 2 L^2 E_{\epsilon}[ \int_0^{\epsilon} |X_s|^2 ds ] \\ &\le 2 \epsilon ( \sigma(0)^2 + L^2 |x_0|^2 ) + 2 L^2 \epsilon \int_0^{\epsilon} \sigma(X_s)^2 ds \\ & \quad + 2 L^2 \epsilon \int_0^{\epsilon} X_s \sigma(X_s) dW_s \end{align*} and as long as $2 L^2 \epsilon \le 1/2$, it follows that $$ E_{\epsilon} [ \sup_{0 \le t \le \epsilon} \mathrm{III} ] \le 4 \epsilon ( \sigma(0)^2 + L^2 |x_0|^2 ) + 4 L^2 \epsilon \int_0^{\epsilon} X_s \sigma(X_s) dW_s $$ The last term in this expression can be treated in a similar way as the last step in the estimate for $\mathrm{II}$, namely Doob's martingale inequality, Cauchy-Schwarz, Itô isometry, and (standard) a priori bounds on $X_t$ over $(0,\epsilon)$. In particular, \begin{align*} \left( E_{\epsilon} [ \sup_{0 \le t \le \epsilon} \left| \int_0^t X_s \sigma(X_s) d W_s \right| ] \right)^2 &\le E \sup_{0 \le t \le \epsilon} \left| \int_0^t X_s \sigma(X_s) dW_s \right|^2\\ &\le 4 E \left| \int_0^{\epsilon} X_s \sigma(X_s) dW_s \right|^2 \\ &\le 4 E \int_0^{\epsilon} X_s^2 \sigma(X_s)^2 ds \\ &\le 4 \tilde{C}_2 (1+ x_0^4) e^{\tilde{C}_1 \epsilon} \epsilon \end{align*} where in turn we used Cauchy-Schwarz, Doob's martingale inequality with $p=2$, Itô's isometry, and then an a priori bound on the second/fourth moment of $X_t$ over $(0, \epsilon)$.
$\Box$
If $X_0$ and $C$ are deterministic (or independent of $W$), then $$X_t = 1-(1-X_0)\exp\left(-\int_0^tC(s)\mathrm dW_s-\frac12\int_0^tC(s)^2\mathrm ds\right).$$ This will go to 1 if the argument of the exponential goes to $-\infty$. I claim that this is equivalent to the condition $I:=\int_0^\infty C(s)^2\mathrm ds=+\infty$.
If $I<\infty$, then the argument of the exponential converges in $\mathrm L^2$ to the same integral for $t=\infty$, which has distribution $\mathcal N(-I/2,I)$; this makes it impossible for it to converge almost surely to $-\infty$.
If $I=\infty$, then $M:t\mapsto\int_0^t C(s)\mathrm dW_s$ is a continuous martingale. By the Dambis-Dubins-Schwarz theorem [RY, Theorem V-(1.6) in my edition], there exists a Brownian motion $B$ on the same probability space such that $M_t=B_{\langle M,M\rangle_t}$ for all $t\geq0$. In these terms, the argument of the exponential is $B_{\langle M,M\rangle_t}-\langle M,M\rangle_t/2$. Since almost surely we have $\lim_{\tau\to\infty}(B_\tau-\tau/2)=-\infty$ and $\lim_{t\to\infty}\langle M,M\rangle_t=I=+\infty$, the exponential goes to zero and $X$ goes to 1 almost surely.
[RY] D. Revuz and M. Yor, Continuous Martingales and Brownian Motion, third edition (1993).
Best Answer
The solution of the SDE in question is an example of a time-inhomogeneous diffusion process that is stopped/killed/terminated when it first hits the origin. For pathwise uniqueness until time $t \wedge \tau$, Theorem 3.7 in Chapter 5 of the following classic should do the trick.
Ethier, Stewart N.; Kurtz, Thomas G., Markov processes. Characterization and convergence, Wiley Series in Probability and Mathematical Statistics. New York etc.: John Wiley & Sons. X, 534 p. \textsterling 49.10 (1986). ZBL0592.60049.
In this setting the theorem precisely states the following.
To be sure, choose the open set to be $U = \{ x > 0 \}$. The proof of this theorem is standard, and involves computing the mean-squared difference between $X_{t \wedge \tau}$ and $Y_{t \wedge \tau}$, applying Lipschitz continuity of the coefficients, and then invoking Grönwall's inequality.