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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.

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  • $\begingroup$ 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... $\endgroup$
    – user420828
    Dec 1, 2021 at 20:39

1 Answer 1

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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 $0$.

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  • $\begingroup$ 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.) $\endgroup$ Dec 2, 2021 at 11:35
  • $\begingroup$ I mean $\tilde W_t$ (or $\tilde W(t)$). $W(t)$ is fine as it is consistent here. Thanks so much! $\endgroup$
    – GJC20
    Dec 2, 2021 at 11:37
  • $\begingroup$ Really nice idea! $\endgroup$
    – GJC20
    Dec 2, 2021 at 11:37
  • $\begingroup$ 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 $\endgroup$
    – GJC20
    Apr 4 at 15:04
  • $\begingroup$ If needed, I can formulate my question in another post $\endgroup$
    – GJC20
    Apr 4 at 15:05

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