Let


$$Y\_t:=1+\int\_0^t b(s)ds + W\_t,\quad\forall t\ge 0,$$


where $b:\mathbb R\_+\to[1,2]$ is continuous and $(W\_t)\_{t\ge 0}$ is a standard Brownian motion. Denote $\tau:=\{t\ge 0: Y\_t\le 0\}$ and $X\_t:=Y\_{t\wedge \tau}$. It is known from [On the marginal distributions of an absorbed diffusion](https://mathoverflow.net/questions/407535/on-the-marginal-distributions-of-an-absorbed-diffusion) that the law of $X\_t$, denoted by $\mu\_t$, has the following decomposition:


$$\mu\_t(dx) = \alpha(t)\delta\_0(dx) + p\_t(x)dx,\quad \forall t>0.$$


Can we show (under suitable conditions) $p\_t(0+):=\lim\_{x\to 0+}p\_t(x)=0$ for every $t>0$?


PS : When $b$ is constant, e.g. $b\equiv 1$, we have


$$\int\_x^{\infty}p\_t(y)dy = \mathbb P[X\_t>x] = \mathbb P[\inf\_{0\le s\le t}Y\_s>0, Y\_t>x]= \mathbb P[\sup\_{0\le s\le t}(-s+W\_s)<1, -t+W\_t<1-x],\quad \forall t,x>0.$$


Using the joint density of the drifted Brownian Motion and its running maximum, one has


$$\mathbb P[\sup\_{0\le s\le t}(-s+W\_s)<1, -t+B\_t<1-x]=\int\_0^1 dm \int\_{-\infty}^{1-x}{\bf 1}\_{\{y\le m\}} e^{-t/2-y}\frac{2(2m-y)}{\sqrt{2\pi t^3}}e^{-(2m-y)^2/2t}dy,$$


which yields by differentiating with respect to $x$


$$p\_t(0+)=-\lim\_{x\to 0+} \frac{\partial \mathbb P[X\_t>x]}{\partial x}=0.$$


Can we extend to the general function $b$? The key is to show the existence of the joint density of $(Y\_t, \inf\_{0\le s\le t}Y\_s)$ but I do not know how prove it.



Does the density of a stopped drifted Brownian motion vanish at zero?
We can re-write the problem in terms of $W(t)$ alone, or, even better, in terms of the drifted Brownian motion $\tilde W(t) = W(t) - M t$, where $M$ is the supremum of $|b(s)|$.


Define
$$ B(t) = -1 - \int\_0^t b(s) ds - M t ,$$
so that $X(t) = \tilde W(t) - B(t)$ up to time
$$ \tau = \inf \{ t > 0 : \tilde W(t) \leqslant B(t) \} . $$
Fix $t\_0 > 0$ and define
$$ \sigma = \inf \{ t \in (0, t\_0] : \tilde W(t) \leqslant B(t\_0) \} . $$
Since $B$ is a non-increasing function, we clearly have $\sigma \geqslant \tau$, and hence the measure
$$\mu(dx) = \mathbb P(t\_0 < \tau, \tilde W(t\_0) - B(t\_0) \in dx)$$
is dominated by the measure
$$\nu(dx) = \mathbb P(t\_0 < \sigma, \tilde W(t\_0) - B(t\_0) \in dx) .$$
The latter is, however, just the distribution at time $t\_0$ of the drifted Brownian motion $\tilde W(t) - B(t\_0)$, killed upon hitting $0$. As you write in the statement of the problem, this is known to have a density function continuously vanishing at zero, and hence $\mu(dx)$ also has a density function continuously vanishing at zero.


It remains to note that $\mu$ is precisely the distribution of $X(t\_0)$, up to an extra atom at $0$.




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*Remark:* A more general approach to the problem, which seems to work also when $b(s)$ is an (adapted) stochastic process rather than a deterministic function, would involve showing first that the distribution of $Y(t)$ — and thus also that of $X(t)$ — has a bounded density function (save for an atom at $0$), and then using Chapman–Kolmogorov equation and a comparison argument similar to the one given above to conclude that the density function of the distribution of $X(t)$ goes to zero at $0$.


