Let $W$ be a one-dimensional Brownian motion. Consider the stochastic differential equation (SDE)


$$dX\_t = C(t)(1-X\_t)dW\_t,\quad \forall t\ge 0,$$


where $C$ is a continuous and bounded function. Under which condition (on $C$),


1. the above SDE has an explicit solution?
2. any solution $X$ satisfies $\lim\_{t\to\infty} X\_t=1$ almost surely?



On a martingale defined via some SDE
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](https://doi.org/10.1007/978-3-662-06400-9),* third edition (1993).


