Prove that $dX_t=1_{(0,\infty )}(X_t)dW_t$ has a strong solution

Question

Let $(W_t)$ a Brownian motion. How can I prove that $$dX_t=\boldsymbol 1_{(0,\infty )}(X_t)dW_t,\quad X_0=x_0$$ has a strong solution ? ($x_0$ may be not deterministic). I know that the Tanaka's equation $dX_t=sgn(X_t)dW_t$ has no strong solution, and the equation above looks the same in the sense that $\boldsymbol 1_{(0,\infty )}$ oscillate between $0$ and $1$ instead of $-1$ and $1$. So, for me, there is no strong solution. But my TA told me that there is one. Can some explain how I could prove it ?

Answer 1

In the case that $x_0 \le 0$, this has the pathwise unique trivial solution $X_t = x_0$ so assume $x_0 > 0$.

Let $W$ be a Brownian motion and define $\tau := \inf\{t \ge 0 : W_t = -x_0\}$. Then $X_t := x_0 + W_{t\wedge \tau}$ satisfies $X_0 = x_0$ and $dX_t = 1_{(0,\infty)}(X_t)dW_t$ because for $t \ge \tau$ we have $X_t = X_\tau = 0$ and $$dX_t = 0 = 1_{(0,\infty)}(X_t)dW_t.$$ Furthermore, since $X$ is adapted to the filtration generated by $W$, $X$ is a strong solution.