This question can be seen as a variant of the post [Bounded density for diffusions with diffusion coefficients bounded away from $0$](https://mathoverflow.net/questions/405624/bounded-density-for-diffusions-with-diffusion-coefficients-bounded-away-from-0) by Iosif Pinelis. Namely, consider the diffusion


$$X\_t=\int\_0^t a(s,X\_s){\bf 1}\_{\{|X\_s|<1\}}dW\_s,\quad \forall t\ge 0,$$


where $(W\_t)\_{t\ge 0}$ is a standard Brownian motion and $a$ is smooth s.t. $\inf\_{(t,x)}a(t,x)\ge c>0$. Denote by $\mu\_t$ the distribution of $X\_t$. Can we write (under suitable conditions)


$$\mu\_t(dx) = q^+\_t\delta\_{1}(dx)+q^-\_t\delta\_{-1}(dx)+ p\_t(x)dx?$$


Here $q^{\pm}\_t=\mathbb P[X\_t=\pm 1]$ and $p\_t$ (up to a normalization) is the conditional density function of $X\_t$ knowing $\{\tau>t\}$, where $\tau:=\inf\{t\ge 0: |X\_t|\ge 1\}$. So an alternative formulation is whether $X\_t$ admits a density on the event $\{\tau>t\}$.


Any solution, references or comments are appreciated.



On the marginal distributions of an absorbed diffusion
The answer is yes. Indeed, let
$$Y\_t=\int\_0^t a(s,Y\_s)\,dW\_s\quad \forall t\ge 0.$$
Then $X\_t=Y\_t$ on the event $\{\tau>t\}$. So, for any Borel set $A\subseteq(-1,1)$, we have
$$P(X\_t\in A)=P(Y\_t\in A,\tau>t)\le P(Y\_t\in A).$$
So, the distribution of $X\_t$ is absolutely continuous with respect to the distribution of $Y\_t$. By the [previous answer](https://mathoverflow.net/a/405686/36721), for $t>0$, the distribution of $Y\_t$ has a density (with respect to the Lebesgue measure). Thus, for $t>0$, the distribution $\mu\_t$ of $X\_t$ has a density $p\_t$ on the interval $(-1,1)$ as well, so that indeed
$$\mu\_t(dx) = q^+\_t\delta\_{1}(dx)+q^-\_t\delta\_{-1}(dx)+ p\_t(x)dx$$
for some nonnegative $q^\pm\_t$.


