This question can be seen as a continuation of [Lipschitz continuity of $\mathbb P[\tau>t]$ with respect to $t$](https://mathoverflow.net/questions/396424/lipschitz-continuity-of-mathbb-p-taut-with-respect-to-t)


Consider the martingale given as


$$X\_t=1+\int\_0^t a(s,X\_s)dW\_s,\quad \forall t\ge 0.$$


Denote $\tau:=\inf\{t\ge 0: X\_t\le0\}$. My question is whether $t\mapsto \mathbb P[\tau>t]$ is Holder continuous, i.e. $\exists C>0, \alpha\in (0,1]$ s.t.


$$0\le \mathbb P[\tau>t]-\mathbb P[\tau>t+\Delta t]\le C\Delta t^{\alpha}.$$


Any solution, references or comments are appreciated. Here we assume that $0<\underline a \le \inf\_{(t,x)} a(t,x)\le \sup\_{(t,x)} a(t,x)\le \overline a$, $t\mapsto a(t,x)$ is continuous and $|a(t,x)-a(t,y)|\le L|x-y|$ for some $L>0$.


PS : My idea is as follows :


\begin{eqnarray}
\mathbb P[\tau>t]-\mathbb P[\tau>t+\Delta t] &=& \mathbb P\left[\inf\_{0\le s\le t}X\_s>0, X\_t+\inf\_{t\le u\le t+\Delta t}\int\_t^ua(s,X\_s)dW\_s\le 0\right] \\
&=& \mathbb P\left[\inf\_{0\le s\le t}X\_s>0, X\_t>0, X\_t+\inf\_{t\le u\le t+\Delta t}\int\_t^ua(s,X\_s)dW\_s\le 0\right] \\
&\le &\mathbb P\left[X\_t>0, X\_t+\inf\_{t\le u\le t+\Delta t}\int\_t^ua(s,X\_s)dW\_s\le 0\right] \\
&=& \int\_{(0,\infty)}\mathbb P\left[x+\inf\_{t\le u\le t+\Delta t}\int\_t^ua(s,X\_s)dW\_s\le 0\Big|X\_t=x\right]\mathbb P[X\_t\in dx].
\end{eqnarray}


Therefore, it suffices to estimate the conditional probability


$$\mathbb P\left[x+\inf\_{t\le u\le t+\Delta t}\int\_t^ua(s,X\_s)dW\_s\le 0\Big|X\_t=x\right]$$



First hitting time for non-homogeneous diffusion martingale
For $h:=\Delta t>0$, you had
$$P(\tau>t)-P(\tau>t+h)=\int\_{(0,\infty)}
P(M\le-x|X\_t=x)\,P(X\_t\in dx),$$
where
$$M:=\inf\_{t\le u\le t+h}J\_u,\quad J\_u:=\int\_t^u a(s,X\_s)\,dW\_s.$$
By [Doob's martingale inequality](https://en.wikipedia.org/wiki/Doob%27s_martingale_inequality#Statement_of_the_inequality),
$$P(M\le-x|X\_t=x)\le x^{-2}\,E(J\_{t+h}^2|X\_t=x)
\le x^{-2}\,ha\_2^2,$$
where $a\_2:=\overline a$ and $a\_1:=\underline a$.


The crucial point is that the pdf $p\_t$ of $X\_t$ for $t>0$ is [bounded](https://mathoverflow.net/a/405686/36721) so that
$$p\_t(x)\le\frac c{\sqrt t}\,e^{-bx^2/t}\le\frac c{\sqrt t}$$
for some positive real constants $c,b$ depending only on $a\_1,a\_2,L$ and for all real $x$. So,
$$
\begin{align}
P(\tau>t)-P(\tau>t+h)&\le
\int\_{(0,\infty)}\min(1,x^{-2}\,ha\_2^2)\,\frac c{\sqrt t}\,dx \\ 
&=
\sqrt h\int\_{(0,\infty)}\min(1,u^{-2}\,a\_2^2)\,\frac c{\sqrt t}\,du \\ 
&=\frac C{\sqrt t}\,\sqrt h,
\end{align}
$$
where $C>0$ is a real constant depending only on $a\_1,a\_2,L$.




---


To get now a uniform Hölder continuity, we can reason as follows:
$$P(\tau>t)-P(\tau>t+h)\le1-P(\tau>t+h)=P(\tau\le t+h)\le a\_2^2(t+h)/1^2,$$
again by Doob's martingale inequality. So,
$$P(\tau>t)-P(\tau>t+h)\le\min\Big(1,\frac C{\sqrt t}\,\sqrt h,a\_2^2(t+h)\Big)\le C\_1h^{1/3},$$
where $C\_1>0$ is a real constant depending only on $a\_1,a\_2,L$ (to verify the latter inequality, consider separately the three cases when (i) $h\ge1$, (ii) $t\le h^{1/3}<1$, or (iii) $t>h^{1/3}$).


Using here an exponential inequality (see e.g. [Theorem 3.1](https://projecteuclid.org/download/pdf_1/euclid.aop/1176988477)) instead of Doob's one, one can improve the factor $h^{1/3}$ to $h^{1/2}\ln\frac1h.$


