Consider a particle whose position is driven by the following stochastic differential equation:
$$Y_t = y + t + W_t + C\min\big(1,(Y_t+1)^+\big)\Lambda_t,\quad \mbox{for all } 0\le t<\tau_*,$$
where $y>0$, $0<C<1$, $(W_t)_{t\ge 0}$ is a standard Brownian motion,
$$\Lambda_t:=\log\big(\mathbb P(\tau>t)\big),\quad \tau:=\inf\{t\ge 0: Y_t\le 0\}$$
and
$$\tau_*:=\inf\left\{t\ge 0: \mathbb P\big(\inf_{0\le s\le t} Y_s\le 0\big)=1\right\}.$$
We say the particle is absorbed once it hits zero. Could we show $\mathbb P(\tau=\infty)=\mathbb P(Y_t>0, \forall t\ge 0)>0$? Any answers, references or remarks are highly appreciated!
Personal thoughts : Consider an alternative SDE
$$X_t = y + t + W_t + C\min\big(1,(X_t+1)^+\big)\log(\lambda),\quad \mbox{for all } t\ge 0,$$
where $\lambda:=\mathbb P(X_t>0, \forall t\ge 0)$. Using a comparison argument, it can be shown that $Y_t\ge X_t$ for all $t\ge 0$. Therefore,
$$\mathbb P(Y_t>0, \forall t\ge 0)\ge \mathbb P(X_t>0, \forall t\ge 0)=\lambda.$$
It remains to show $\lambda>0$. As noted that, on the event $\left\{X_t>0, \forall t\ge 0\right\}$, it follows that $X$ has the same law as $(y + t + W_t + C\log(\lambda))_{t\ge 0}$ knowing that $Z_t>0, \forall t\ge 0$. A straightforward computation yields
$$ 1 -\lambda^{-2C}e^{-2y}=\mathbb P(Z_t>0, \forall t\ge 0)=\mathbb P(X_t>0, \forall t\ge 0)=\lambda\quad \quad \quad \quad \quad (\ast)$$
As shown that, under suitable conditions (e.g. when $y$ is large enough), $(\ast)$ always admits a strictly positive solution, which proves partially my desired result. However, for general $y>0$, I don't know how to proceed.
