# Does the density of a stopped drifted Brownian motion vanish at zero?

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 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.

• Could you please provide the reference for the joint density of $(\sup_{0\le s\le t}(-s+W_s), t+W_t)$? My feeling is that the related arguments might be adapted to your case, while I am unable to find its derivation...
– user420828
Dec 1, 2021 at 20:39

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 $$0$$.

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 $$0$$.

• If you mean $W_t$ vs. $W(t)$ — done. (I am so much used to mixing both notations that I do not even notice the difference, sorry.) Dec 2, 2021 at 11:35
• I mean $\tilde W_t$ (or $\tilde W(t)$). $W(t)$ is fine as it is consistent here. Thanks so much! Dec 2, 2021 at 11:37
• Really nice idea! Dec 2, 2021 at 11:37
• Dear Mateusz, I returned to the case where $b$ is an adapted process, i.e. $b=g(t,Y_t)$ for some suitable function $g$. Could you please detail how to show the density function continuously vanishing at zero? Thank you very much Apr 4 at 15:04
• If needed, I can formulate my question in another post Apr 4 at 15:05