Let $D$ be the space of functions on $\mathbb R_+$ that are right-continuous with left limits. Consider two measurable functions $a,b: \mathbb R_+\times D\to [0,1]$ that are non-anticipative, i.e.

$$a(t,f)=a(t,f^t) \quad\mbox{and}\quad b(t,f)=b(t,f^t),\quad \forall (t,f)\in \mathbb R_+\times D,$$

where $f^t\in D$ is defined as $f^t(s):=f\big(\min(t,s)\big)$. Given two independent Brownian motions $B,W$ starting at zero, consider the following two-dimensional SDE:

\begin{eqnarray} X_t &=& x + B_t + \int_0^ta(u,X)du - \alpha X_{\sigma} {\bf 1}_{\{\sigma\le t\}},\quad \forall 0\le t\le \tau \\ Y_t &=& y + W_t + \int_0^tb(u,Y)du - \beta Y_{\tau} {\bf 1}_{\{\tau\le t\}},\quad \forall 0\le t\le \sigma \\ \end{eqnarray}

where $\tau:=\inf\{t\ge 0: X_t \le 0\}$, $\sigma:=\inf\{t\ge 0: Y_t \le 0\}$, and $\alpha, \beta \in (0,1)$, $x,y>0$ are constants. Does the above SDE admit a unique solution $(X_t,Y_t)_{0\le t\le \max(\tau,\sigma)}$?

PS : The existence is straightforward. Construct two independent processes $\bar X, \bar Y$ as the solution of

\begin{eqnarray} \bar X_t &=& x + B_t + \int_0^ta(u,\bar X)du,\quad \forall t\ge 0 \\ \bar Y_t &=& y + W_t + \int_0^tb(u,\bar Y)du,\quad \forall t\ge 0. \\ \end{eqnarray}

Define by $\bar \tau, \bar\sigma$ the first hitting time at zero of $\bar X, \bar Y$. Then we define $X,Y$ by distinguishing the following two cases

  1. If $\bar \tau\le \bar\sigma$, then $X_t:=\bar X_t$, $Y_t:=\bar Y_t$ for $t<\bar \tau$ and for $t\ge \bar\tau$, $Y_t$ solves

$$Y_t=(1-\beta)Y_{\bar\tau-}+ (W_t-W_{\bar\tau})+\int_{\bar\tau}^tb(u, Y)du$$

  1. If $\bar \sigma\le \bar\tau$, then $Y_t:=\bar Y_t$, $X_t:=\bar X_t$ for $t<\bar \sigma$ and for $t\ge \bar\sigma$, $X_t$ solves

$$X_t=(1-\alpha)X_{\bar\sigma-}+ (B_t-B_{\bar\sigma})+\int_{\bar\sigma}^ta(u, X)du.$$

Is there any mistake in my construction? If not, is this the unique solution?



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